User robert bryant - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T19:35:07Z http://mathoverflow.net/feeds/user/13972 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/130959/reference-request-affine-transforms-circle-inversion/131005#131005 Answer by Robert Bryant for Reference request: affine transforms + circle inversion? Robert Bryant 2013-05-17T21:50:14Z 2013-05-19T11:47:47Z <p>$\mathcal{T}$ is not a Lie group when $n>1$. </p> <p>Actually, the OP did not say whether he wanted $\mathcal{T}$ to be all possible sequences of compositions of these generating sets, but, if he did, then it is clear that $\mathcal{T}$ is not a Lie group, in the sense that it is not defined as the set of solutions of some system of PDE for transformations of $\mathbb{R}^n$. For one thing, the group that they generate would properly contain the conformal group $\mathrm{O}(n{+}1,1)$ acting on $S^n$, which is known to be a maximal Lie group, i.e., there is no group (in Lie's sense) between the conformal group and the full diffeomorphism group. (NB: The group of analytic diffeomorphisms of $S^n$ is not a subgroup of the full diffeomorphims in Lie's sense because it is not defined as the set of solutions of some system of PDE.)</p> <p>In particular, no group $G$ that contains $\mathcal{T}$ can preserve any geometric structures of the kind the OP mentions because this would define a PDE that $G$ satisfies.</p> <p>(By the way, note that $\mathcal{T}$, as the OP defined it, does not consist of smooth transformations of $S^n$ only when $n>1$, since the non-conformal affine transformations do not extend smoothly to $\infty$ except when $n=1$.)</p> http://mathoverflow.net/questions/130438/invariants-of-a-gl3-mathbbr-action/130497#130497 Answer by Robert Bryant for Invariants of a $GL(3,\mathbb{R})$ action Robert Bryant 2013-05-13T15:52:02Z 2013-05-14T23:04:49Z <p>There's a 'quasi-normal form' on a dense open set that can be described without too much trouble. Here's one way to do it.</p> <p>First, recognize that we are looking for a normal form for elements $Q\in W=\bigl[S^2(\Lambda^2(V^\ast))\otimes V^\ast \bigr]_0$ under the action of $G=\mathrm{GL}(V)$, where $V$ is a vector space of dimension $3$ and $W$ is the kernel of the $G$-module mapping $$S^2(\Lambda^2(V^\ast))\otimes V^\ast\hookrightarrow \Lambda^2(V^\ast)\otimes \Lambda^2(V^\ast)\otimes V^\ast\to \Lambda^2(V^\ast)\otimes\Lambda^3(V^\ast)$$ defined to be the natural inclusion followed by fully skewsymmetrizing in the last two factors. This $W$ is an irreducible $G$-module of dimension $15$. </p> <p>Now, it makes things a little easier to follow if one recognizes that there is a canonical isomorphism $\Lambda^2(V^\ast) = V\otimes \Lambda^3(V^\ast)$ (given by the obvious contraction), so that $W$ can also be understood as a subspace of $S^2(V)\otimes V^\ast\otimes \bigl(\Lambda^3(V^\ast)\bigr)^2$. In fact $W = W_0\otimes \bigl(\Lambda^3(V^\ast)\bigr)^2$, where $W_0\subset S^2(V)\otimes V^\ast$ is the $15$-dimensional subspace that is the kernel of the trace mapping $S^2(V)\otimes V^\ast\to V$. So I am going to think of $Q$ as an element of $W_0\otimes \bigl(\Lambda^3(V^\ast)\bigr)^2$. </p> <p>The vector space $U = V^\ast\otimes \bigl(\Lambda^3(V^\ast)\bigr)^2$ has dimension $3$, and by pairing $Q$ with elements in $U^\ast$, one can generate a subspace $\delta Q\subset S^2(V)$ of dimension at most $3$; say that $Q$ is of <em>full quadratic rank</em> if $\delta Q$ has dimension $3$, i.e., $\delta Q$ lies in $\mathrm{Gr}_3\bigl(S^2(V)\bigr)$. The set of $3$-dimensional subspaces of a $6$-dimensional vector space has dimension $9$, and there is another way to generate a $9$-parameter family $3$-dimensional subspaces of $S^2(V)$, namely, if $C\in S^3(V)$ is a nondegenerate cubic, thought of as a cubic polynomial function on $V^\ast$, then we can let $\partial C\subset S^2(V)$ denote the $3$-dimensional space of its partial derivatives. (In fact, 'nondegenerate' in this context means exactly that $\partial C$ has dimension $3$, the maximum dimension possible.) </p> <p>Now, these two methods of generating a $3$-dimensional subspace of $S^2(V)$ are related: For <em>any</em> $3$-dimensional subspace $P\subset S^2(V)$, there is always a cubic $C$ such that $\partial C\subset P$, and, most of the time, this cubic is unique up to multiples and satisfies $\partial C = P$. One can see this as follows: There is a natural exact sequence $$0\to S^3(V)\to S^2(V)\otimes V\to V\otimes \Lambda^2(V)\to \Lambda^3(V)\to 0.$$ In particular, the image of the map $S^2(V)\otimes V\to V\otimes \Lambda^2(V)$ has dimension $8$. If $P\subset S^2(V)$ has dimension $3$, then the restriction of this map to $P\otimes V\to V\otimes \Lambda^2(V)$ has rank at most $8$, so there must be at least a $1$-dimensional kernel, i.e., there must be a nonzero $C\in S^3(V)$ whose image under the above mapping lies in $P\otimes V$. Say that a $3$-dimensional subspace $P\subset S^2(V)$ is <em>uniquely partial</em> if the kernel of $P\otimes V\to V\otimes \Lambda^2(V)$ has dimension $1$. The set of uniquely partial $P$ is a dense open set in $\mathrm{Gr}_3(S^2(V))$. In particular, for a dense open set of $Q\in W$, the subspace $\delta Q$ will be of dimension $3$ and uniquely partial. Let us say that $Q$ is <em>nonsingular</em> if, in addition, the cubic $C\in S^2(V)$ (unique up to multiples) such that $\partial C = \delta Q$ is nonsingular. The set of nonsingular $Q$ is a dense open set in $W$. </p> <p>Finally, put the nonsingular $C$ associated (uniquely up to multiples) to a given nonsingular $Q$ in normal form, i.e., take a basis $e_i\in V$ such that $$C = {e_1}^3+{e_2}^3+{e_3}^3 + 6\sigma\ e_1e_2e_3\ ,$$ where $\sigma\not=-\tfrac12$ is a real number. (In fact, one needs to disallow a few more values of $\sigma$ in order to make sure that $\partial C$ is uniquely partial, but I'll leave that to the reader. For example, $\sigma=0$ is not allowed.) The basis $e_i$ is uniquely determined by $C$ up to permutations. (Of course, $C$ is only determined up to a multiple, so the basis $e_i$ is only defined by $Q$ up to permutation and simultaneous scaling by a real number. This $S_3\times \mathbb{R}^*$-ambiguity is the reason that the normal form is only 'quasi-normal', as we will see.) </p> <p>Let $e^i$ be the dual basis. Now, by definition, $Q$ lies in $\partial C\otimes V^\ast\otimes S^2(\Lambda^3(V^\ast))$, so there are numbers such that $$Q = \tfrac13 b_{ij} \frac{\partial C}{\partial e_i}\otimes e^j \otimes (e^1\wedge e^2\wedge e^3)^{\otimes 2}$$ The $9$ entries of $b = (b_{ij})$ are subject to three linear equations caused by the trace relation (i.e., the second Bianchi identity), and these relations are found to be $b_{ii}+\sigma(b_{jk}+b_{kj})=0$, where $(i,j,k)$ is any even permutation of $(1,2,3)$. They must also satisfy $\det b\not=0$, since, otherwise, $Q$ would be degenerate. </p> <p>Obviously, these relations are invariant under the symmetric group and scaling. One could normalize the scaling away (up to a $\pm1$) by requiring that the sum of the squares of the $b_{ij}$ be equal to $1$, and this would leave a finite group $S_3\times \lbrace\pm1\rbrace$ that preserves this quasi-normal form. This brings the number of free parameters down to $6$, namely $\sigma$ and the $5$-sphere of the normalized $b_{ij}$ (which lie in a canonical $6$-dimensional subspace that depends on $\sigma$). One could then use the finite group to normalize things further or use the invariant theory of this finite group to find invariant combinations of these quantities that will yield invariants of the original $\mathrm{GL}(V)$-action on the dense open set consisting of the nonsingular $Q$s.</p> <p>With a little more work, one could actually get quasi-normal forms for all of the orbits in $W$, but that can get complicated.</p> http://mathoverflow.net/questions/130080/functional-equations/130097#130097 Answer by Robert Bryant for Functional equations Robert Bryant 2013-05-08T16:44:04Z 2013-05-10T01:30:30Z <p>You don't need to assume any differentiability (or even continuity). In fact, the domain of $f$ need have no structure other than being a set of the form $D\times D$ for some other set $D$. </p> <p>I'll assume that the range of $f$ in each case is $\mathbb{R}$, even though the OP didn't specify that. I suppose that, more generally, for the first equation, $f$ could take values in any field $\mathbb{F}$ (or even a division ring), and the answer might differ a little bit in that case, depending on the nature of $\mathbb{F}$. For the second equation, perhaps one could allow $f$ to take values in a group $\mathbb{G}$ to get a more general problem.</p> <p>For the first equation, setting $x=y=z$, one sees that $f(x,x)=\epsilon(x)/\sqrt{2}$ for some function $\epsilon$ that satisfies $\epsilon(x)=\pm 1$. However, now, setting $z=x$, one has $f(x,y)+f(y,x)=\sqrt{2}\epsilon(x)=\sqrt{2}\epsilon(y)$ for all $x$ and $y$, so $\epsilon$ is constant. Replacing $f$ by $-f$ if necessary, one can assume $\epsilon(x)\equiv1$, so that $f(x,y)+f(y,x)\equiv\sqrt{2}$. Thus, set $f(x,y) = 1/\sqrt{2}+a(x,y)$ where $a(x,y)=-a(y,x)$. Substituting this back into the equation with $x=y$ yields $\sqrt{2}+a(x,z)=1/\bigl(1/\sqrt{2}+a(x,z)\bigr)$, which gives $a(x,z)\bigl(a(x,z)+3/\sqrt{2}\bigr)=0$. Thus, either $a(x,z)=0$ or $a(x,z)=-3/\sqrt{2}$. However, the latter is not possible since then $a(z,x)=3/\sqrt{2}$, which is not allowed. Thus, $a(x,z)=0$ for all $x$ and $z$. Thus, $f(x,y)\equiv1/\sqrt{2}$ and $f(x,y)\equiv -1/\sqrt{2}$ are the only solutions.</p> <p>Similarly, for the second equation, one has $f(x,x)^3=1$, so $f(x,x)=1$. Setting $z=x$, yields $f(x,y)f(y,x)=1$, so $f(x,y)=1/f(y,x)$ for all $x$ and $y$. Suppose that there is a pair $(x,y)$ such that $f(x,y) &lt; 0$. Then for any $z$, one has $f(x,z)f(y,z) &lt; 0$, so either $f(x,z) &lt; 0$ or $f(y,z) &lt; 0$ (and not both). Let $Y$ be the set of $z$ such that $f(x,z) &lt; 0$ and let $X$ be the set of $z$ such that $f(y,z) &lt; 0$. Then $X$ and $Y$ are disjoint and nonempty and their union is everything. If $z$ and $w$ belong to $Y$, then the equation $f(x,z)f(z,w)f(x,w)=1$ implies $f(z,w)>0$. Similarly, if $z$ and $w$ belong to $X$, then $f(z,w)>0$. </p> <p>Letting $\epsilon(z,w)=1$ when $f(z,w)>0$ and $\epsilon(z,w)=-1$ when $f(z,w)&lt;0$, the function $g$ defined by $g(z,w)=f(z,w)\epsilon(z,w)=|f(z,w)|$ will satisfy the same functional equation as $f$ but will be positive everywhere. Writing $h(z,w)=\log\bigl(g(z,w)\bigr)$, one sees that $h$ satisfies the functional equation $$h(u,v) + h(v,w) + h(u,w)=0$$ But this says that $h(v,w)=-h(u,v)-h(u,w)=h(w,v)$, while one already knows that $h(v,w)=-h(w,v)$. Thus $h(v,w)\equiv0$, so $g(v,w)\equiv1$.</p> <p>Thus, the solutions of the second functional equation are obtained as follows: Write the domain $D$ as a disjoint union of two sets $X$ and $Y$ and set $f(z,w) = +1$ when $z$ and $w$ both belong to either $X$ or $Y$ and $f(z,w)=-1$ otherwise. </p> http://mathoverflow.net/questions/129538/laplacian-on-coset-spaces/129747#129747 Answer by Robert Bryant for Laplacian on coset spaces Robert Bryant 2013-05-05T18:02:16Z 2013-05-05T18:02:16Z <p>I'll give an answer in a few parts:</p> <p>First, for the $n$-sphere $S^n\subset\mathbb{R}^{n+1}$: The obvious thing to do is to consider the vector fields $$X_{ij} = x_i\frac{\partial }{\partial x_j} - x_j\frac{\partial }{\partial x_i}\ , \qquad 0\le i &lt; j\le n.$$ Then one easily computes that the operator $$L = \sum_{0\le i &lt; j\le n} {X_{ij}}^2$$ is equal to the Laplacian for the induced metric on $S^n$. </p> <p>Second, notice that one doesn't always have uniqueness of this representation. For example, when $n=3$, consider the vector fields $$Y_1^\pm = X_{01}\pm X_{23},\quad Y_2^\pm = X_{02}\pm X_{31},\quad Y_3^\pm = X_{03}\pm X_{12}.$$ Then, for functions on $S^3$, one has that the Laplacian equals $$L = (Y_1^+)^2 + (Y_2^+)^2 + (Y_3^+)^2 = (Y_1^-)^2 + (Y_2^-)^2 + (Y_3^-)^2,$$</p> <p>Third, in the general case of a Riemannian homogeneous space $M=G/K$, where $G$ acts effectively on $M$ and fixes a metric $g$ on $M$, I gather that the question is whether there always exists a quadratic polynomial $\lambda\in S^2({\frak{g}})$ (where $\frak{g}$ is the Lie algebra of $G$, thought of as $g$-Killing vector fields on $M$) such that $\lambda$, when regarded as a self-adjoint second-order differential operator on $M$, is equal to the Laplacian of the metric $g$. </p> <p>This certainly is true in a large number of cases. For example, if $G$ is semi-simple and $G/K$ is a Riemannian symmetric space, this is true. Note, though, that $\lambda$ may not be 'positive definite' in the sense that it may not be the sum of squares of a basis of $\frak{g}$. For example, see the $S^3$ case above. As another example, if $G=\mathrm{SO}(2,1)$ and $K = \mathrm{SO}(2)$, then $G/K$ is the Poincaré disk, and $\lambda$ in this case turns out to be of the form ${X_1}^2 + {X_2}^2 - {X_3}^2$ for an appropriate basis $X_i$ of ${\frak{so}}(2,1)$.</p> <p>Whether it is true in all cases (and how unique the representation is) is not clear to me at first glance, but I'll think about it if I get the chance. It seems that it probably is true when $M$ is compact, or when $M$ is isotropy irreducible, but I should check to be sure before attempting a definitive answer.</p> http://mathoverflow.net/questions/129365/normal-forms-for-homogeneous-cubic-polynomials-in-mathbbrx-1-x-2-x-3/129404#129404 Answer by Robert Bryant for Normal forms for homogeneous cubic polynomials in $\mathbb{R}[x_1, x_2, x_3]$ Robert Bryant 2013-05-02T11:20:36Z 2013-05-03T08:54:30Z <p>Hi, Jeanne! It may help to have some geometric explanation of the normal forms over $\mathbb{R}$. The standard account is this:</p> <p>If the projective cubic curve $F(x_1,x_2,x_3)=0$ is nonsingular (the 'generic' case), it has exactly three real flexes, they are distinct and lie on a line. One can make a linear change to make them lie on the line $x_1{+}x_2{+}x_3=0$ and have this line intersect the curve at the three points where some $x_i=0$ (so that these are the three flexes). From this, one sees that one can make a real linear change of variables so that $$F = {x_1}^3 + {x_2}^3 + {x_3}^3 + 6\sigma\ x_1x_2x_3\ ,$$<br> where $\sigma\not=-\tfrac12$ is a real number. (When $\sigma=-\tfrac12$, the above cubic factors as a line $x_1{+}x_2{+}x_3=0$ and an irreducible quadratic form.)</p> <p>Enumerating the singular cases over the reals gets a little messy, but the final result is that when the curve is irreducible and singular there is exactly one singular point, which is necessarily real, and it is either a hyperbolic node, elliptic node, or a cusp. These have the normal forms $$F = {x_2}^2x_3 - \epsilon\ {x_1}^2 x_3 - {x_1}^3$$ where $\epsilon$ is $1$, $-1$, or $0$, respectively.</p> <p>If the curve is the union of a line and a smooth quadric, i.e., $F = LQ$, where $L$ is linear and $Q$ is nonsingular (possibly without real points), you can put the quadric in normal form, $Q = {x_1}^2+{x_2}^2\pm{x_3}^2$ and then use the stabilizer group of the quadric to normalize $L$. In the $+$ case, you can always rotate, using $\mathrm{SO}(3)$, to make $L=x_1{+}x_2{+}x_3$ (say). In the $-$ case, there are three cases, and you can rotate, using $\mathrm{SO}(2,1)$ to make $L$ be one of $x_1$, $x_1{+}x_3$, or $x_3$.</p> <p>Finally, if the curve is the union of three (complex) lines, it depends on whether the lines are all distinct or not and whether they are concurrent or not. In the distinct case, you get that $F$ lies on the orbit of either $F=x_1x_2x_3$ (all real, not concurrent), $x_1x_2(x_1{+}x_2)$ (all real, concurrent), $x_1({x_2}^2{+}{x_3}^2)$ (two complex conjugate, not concurrent), or $x_1({x_1}^2{+}{x_2}^2)$ (two complex, concurrent). If they are not all distinct, you get either $x_1{x_2}^2$ (two distinct) or ${x_1}^3$ (all the same).</p> <p>These are the normal forms for all the nonzero cubics in $3$ variables.</p> <p><em>Remark:</em> You also asked about the 'orbit space' of $\mathrm{GL}(3,\mathbb{R})$ acting on $\mathrm{Sym}^3(\mathbb{R}^3)$. The naïve quotient, endowed with the quotient topology, is not very nice because it is highly non-Hausdorf, as every open neighborhood of the fixed point $0$ meets every orbit. Even when you remove $0$, so that the remainder can be thought of as $\mathrm{PGL}(3,\mathbb{R})\simeq \mathrm{SL}(3,\mathbb{R})$ acting on $\mathbb{P}\bigl(\mathrm{Sym}^3(\mathbb{R}^3)\bigr)\simeq \mathbb{RP}^9$, the orbit space with the quotient topology is <em>still</em> non-Hausdorf. The whole point of developing Geometric Invariant Theory (aka GIT) is to figure out the right way to remove the bad points so that the remaining 'good' points (aka, the 'semi-stable' points) will make a nice quotient space when you divide out by the action of the group. I don't know how good this theory is in the real case, though. In any case, the nonsingular cubics are all 'good' points for the moduli space.</p> http://mathoverflow.net/questions/129150/derham-cohomology-of-sn-without-mayer-vietoris/129153#129153 Answer by Robert Bryant for deRham cohomology of $S^n$ without Mayer-Vietoris Robert Bryant 2013-04-29T21:45:03Z 2013-04-29T21:45:03Z <p>Have you done any integration theory? (I assume you have, otherwise you wouldn't necessarily know what the deRham cohomology does for you.) The fastest proof I know is: </p> <ol> <li><p>Take a closed $k$-form $\omega$ on $S^n$, note that $g^\ast\omega$ is cohomologous to $\omega$ for all $g\in \mathrm{SO}(n{+}1)$ (since $\mathrm{SO}(n{+}1)$ is connected. </p></li> <li><p>Conclude that $\omega$ is cohomologous to $\bar\omega$, the average over $\mathrm{SO}(n{+}1)$ of $g^\ast\omega$ as $g$ varies over $\mathrm{SO}(n{+}1)$. </p></li> <li><p>But $\bar\omega$ is invariant under the action of $\mathrm{SO}(n{+}1)$, so its value at $x\in S^n$ must be invariant under the subgroup (isomorphic to $\mathrm{SO}(n)$) that stabilizes $x$. </p></li> <li><p>However, $\mathrm{SO}(n)$ acting on $\mathbb{R}^n$ only fixes nonzero forms in degree $0$ and $n$. </p></li> <li><p>Thus, if $\bar\omega$ is not zero, it must be either a constant function ($k=0$) or a multiple of the volume form ($k=n$).</p></li> </ol> http://mathoverflow.net/questions/128980/linearization-of-vector-fields/128998#128998 Answer by Robert Bryant for Linearization of vector fields Robert Bryant 2013-04-28T13:01:54Z 2013-04-28T13:01:54Z <p>Since you don't say what your 'some other conditions are', it is hard to know what result you are claiming. However, it is not difficult to show that, for $\mathbb{R}^4$ endowed with the symplectic form $\omega = dp\wedge dx + dq\wedge dy$, the vector field $$Z = x\ \partial_x + (2y{+}x^2)\ \partial_y - (p{+}2xq)\ \partial_p - 2q\ \partial_q$$ is Hamiltonian and yet it cannot be smoothly (let alone real-analytically) linearized near the origin. The reason is the presence of $C^1$ integral curves of $Z$ passing through the origin that are not $C^2$, let alone smooth. For example, the curve $(x,y,p,q)=(x,x^2\ln|x|,0,0)$ is an integral curve of $Z$. (For the linearized vector field (which is also Hamiltonian) $$Z_0 = x\ \partial_x + 2y\ \partial_y - p\ \partial_p - 2q\ \partial_q\ ,$$ every integral curve passing through the origin is smooth.)</p> http://mathoverflow.net/questions/128698/exact-or-numerical-solutions-of-a-system-of-differential-equatios/128781#128781 Answer by Robert Bryant for Exact or Numerical solutions of a system of differential equatios Robert Bryant 2013-04-26T00:54:40Z 2013-04-27T10:58:25Z <p>This isn't much of an answer, but I thought that I'd put down a few observations here that you may find helpful. </p> <p>First of all, if $b=0$, then you have a first integral, in that the ratio $x^{3c}/y^{3a}$ is constant, so you can use this to eliminate, say, $y$ from the first equation, and then the result becomes an autonomous first order equation for $x$ as a function of $t$, which can, in theory, be integrated by quadrature, so let's set this case aside. </p> <p>When $b\not=0$, you can scale $t$ (the independent variable) so as to make $b=1$, which I'll assume from now on. Also, I'll replace your $3a$ and $3c$ by $a$ and $c$, respectively, because I don't see what good the $3$s do; they just clutter the formulae.</p> <p>If you have been trying to do numerics, you may have run into a bit of trouble because the equations as you have written them aren't differentiable, which leads to instabilities in the numerics and nonuniqueness in the solutions. One way to deal with this is to 'resolve' the singularities by 'unfolding' the domain. Your equations require that $x{+}y$ and $xy$ both be nonnegative, which can only happen when $x$ and $y$ are nonnegative. The solutions that have either $x$ or $y$ vanishing identically are easy to find, so set those aside. To see what happens to the other solutions, make the substitution $(x,y) = \bigl(4u^2v^2,(u^2-v^2)^2\bigr)$. The differential equations then become polynomial: \begin{align} 4u' &amp;= \phantom{-}v -av(2uv) -cu(u^2{-}v^2),\\ 4v' &amp;= -u -au(2uv) +cv(u^2{-}v^2). \end{align} \tag1 (A word of caution: In this parameterization, $\sqrt{xy}=2uv(u^2-v^2)$, which changes sign as one crosses the lines $u=0$, $v=0$, $u=v$ and $u=-v$, so you have to interpret $\sqrt{xy}$ as a signed quantity in your original equations. Of course, $\sqrt{x{+}y}=u^2+v^2$, so it does not change sign.)</p> <p>The linear terms on the right hand side of $(1)$ describe the rotation vector field in the $uv$-plane, so the solutions near the origin are either a center or a spiral sink or source. Hence, the corresponding curves in the $xy$-plane 'bounce' between $x=0$ and $y=0$ an infinite number of times.) </p> <p>In fact, $$\frac{2(u^2+v^2)'}{(u^2{+}v^2)^2} = -c-(a{-}c)\frac{4u^2v^2}{(u^2{+}v^2)^2},$$ and the right hand side of this expression varies between $-c$ and $-a$, so, unless $a$ and $c$ have opposite signs, a nonzero solution curve always spirals towards (when $a,c>0$) or away from (when $a,c&lt;0$) the origin.</p> <p>It's also useful to look at things in polar coordinates. Set $$x=\tfrac12 r^2 (1+\cos\phi)\qquad\text{and}\qquad y = \tfrac12 r^2 (1-\cos\phi).$$ Then, interpreting $\sqrt{xy}$ as $\tfrac12 r^2\sin\phi$ and $\sqrt{x{+}y}$ as $r$, one computes that \begin{align} r' &amp;= -\tfrac14\ r^2\bigl( (a{+}c)+(a{-}c)\cos\phi\bigr),\\ \phi' &amp;= b + \tfrac12(a{-}c)\ r\sin\phi. \end{align} \tag2 When $a$ and $c$ are positive, this shows that $r$ is strictly decreasing. In fact, when $r_0>0$ is small, one has $\phi\approx bt$, so integrating in the first equation gives $$r \approx \left(\frac1{r_0} + \frac{(a{+}c)}4\ t + \frac{(a{-}c)}{4b}\ \sin bt\right)^{-1}.$$ This gives a reasonably good approximation to the solutions when $r_0 = \sqrt{x_0{+}y_0}$ is small.</p> <p>Meanwhile, if $a &lt; 0 &lt; c$ and $\sqrt{xy}$ is interpreted to be positive (respectively, when $c &lt; 0 &lt; a$ and $\sqrt{xy}$ is interpreted to be negative), then there is a fixed point in the first quadrant of the $xy$-plane at $$(x,y) = \left(\frac1{a(a{-}c)},\frac1{c(c{-}a)}\right),$$ and this gives rise to fixed points in the $uv$-plane, somewhat symmetrically arranged around the origin. Stability analysis of these fixed points will probably tell you something useful. They appear to be saddles (I haven't checked this for sure), but I have no idea where the separatrices go.</p> http://mathoverflow.net/questions/128441/representing-immersions-from-a-surface-into-3-space/128552#128552 Answer by Robert Bryant for Representing immersions from a surface into 3-space Robert Bryant 2013-04-23T22:52:16Z 2013-04-24T18:03:12Z <p>I may have to enter this as a sketch and fill in details later, but I thought that I'd go ahead and get the main ideas out there.</p> <p>The first thing to notice is that the given problem is equivalent to the problem of solving $\rho(f)=g$ where $\rho:\mathrm{Imm}(T^2,\mathbb{R}^3)\to\Omega^2(T^2,\mathbb{R}^3)$ is the differential operator $$\rho(f) = f_x\times f_y\ \ dx{\wedge}dy$$ taking values in (nonvanishing) $\mathbb{R}^3$-valued $2$-forms on $T^2$. The advantage of considering $\rho$ instead of $R$ is that $\rho$ is well-defined, independent of coordinates on $T^2$.</p> <p>The second thing to notice is that the inner product on $\mathbb{R}^3$ is something of a red herring. The cross product is really just the wedge product followed by the orientation-and-metric induced canonical isomorphism between the second exterior power of $\mathbb{R}^3$ and $\mathbb{R}^3$ itself. </p> <p>Let's undo all this, and consider, instead, a $3$-dimensional vector space $V$ and a surface $\Sigma$ and define an operator $\rho:\mathrm{Imm}(\Sigma,V)\to\Omega^2\bigl(T^2,\Lambda^2(V)\bigr)$ that, in local coordinates $(x,y)$ on $\Sigma$, has the expression $$\rho(f) = f_x{\wedge}f_y\ \ dx{\wedge}dy.$$</p> <p>Now we want to ask the question "Given a nonvanishing $\Lambda^2(V)$-valued $2$-form $g$ on a surface $\Sigma$, when can it be written in the form $g = \rho(f)$ for some immersion $f:\Sigma\to V$ and in how many ways?" This question is interesting both locally and globally. Locally, it is three (nonlinear) first-order equations for three unknowns, but, as we shall see, it is highly degenerate, even when $g$ is 'generic', so that, for example, there is no hope of writing it in Cauchy form.</p> <p>One distinct advantage of formulating the equation in this way is that its full equivariance is manifest. Not only does $\rho$ not depend on a choice of local coordinates on the surface, if $L(v) = Av + b$ is any invertible affine transformation of $V$, then one has $\rho(L\circ f) = \Lambda^2(A)\bigl(\rho(f)\bigr)$. In particular, replacing $g$ by $\Lambda^2(A)(g)$ does not change the problem, which shows that one can use the geometry of linear transformations of $V$ to uncover invariants of $g$ that will determine properties of the solution.</p> <p>For example, in the (very degenerate) case in which $g$ is a $2$-form that takes values in a line in $\Lambda^2(V)$, one can write $g = e_1{\wedge}e_2\ \Phi$ where $e_1$ and $e_2$ are linearly independent and $\Phi$ is (nonvanishing) $2$-form on $\Sigma$. Then the only possible solutions $f$ to $\rho(f)=g$ are to have $f$ immerse $\Sigma$ into a $2$-plane $P$ parallel to the subspace spanned by $e_1$ and $e_2$ in such a way that it pulls back the volume form on $P$ dual to $e_1{\wedge}e_2$ to be $\Phi$. This is now essentially one equation for two unknowns and is always locally solvable, though there may well be no global solution. This is a trivial special case, though, and doesn't give you any sense of the general case.</p> <p>On this first pass, I will avoid discussing the singular and intermediate cases and go directly to the 'generic' case, the case in which the projectivization of $g$, i.e., $[g]:\Sigma\to \mathbb{P}\bigl(\Lambda^2(V)\bigr)\simeq\mathbb{RP}^2$ is an immersion. In this case, the problem reduces to either an elliptic or a hyperbolic problem, depending on the geometry of the mapping $g$. </p> <p><em>Example:</em> Before considering the general case, it's worth pausing to consider a simple example to illustrate where the analysis will take us: Let $\Sigma$ be the $xy$-plane and consider the two maps $f_\pm:\Sigma\to\mathbb{R}^3$ defined by $$f_\pm = \bigl(x,y,\tfrac12(x^2\pm y^2)\bigr).$$ These have $g_\pm = \rho(f_\pm) = (1,0,x){\wedge}(0,1,\pm y)\ dx{\wedge}dy$. We want to find the immersions $f:\Sigma\to\mathbb{R}^3$ that satisfy $\rho(f)=g_\pm$. Well, such an $f$ will have to satisfy $$df = (1,0,x)\ \xi + (0,1,\pm y)\ \eta = (\xi,\eta,\ x\xi{\pm}y\eta\ )$$ for some $1$-forms $\xi$ and $\eta$ that satisfy $\xi\wedge\eta=dx\wedge dy$ and are such that $d(df)=0$. Now, this latter equation implies $d\xi=d\eta=d(x\xi{\pm}y\eta)=0$, and so, assuming that a local solution is defined on a simply-connected open set $U\subset\Sigma$, one can then write $\xi = dp$ and $\eta = \pm dq$ for some functions $p$ and $q$ on $U$. The remaining equations on $p$ and $q$ then become $0=d(x\xi{\pm}y\eta)=dx\wedge dp + dy\wedge dq$ and $dp\wedge dq = \pm dx\wedge dy$. This is two PDE on $p$ and $q$ that can be written as a single Monge-Ampère equation by setting $p = u_x$ and $q = u_y$ (which works because $d(p\ dx + q\ dy) = 0$) and then the remaining equation becomes $u_{xx}u_{yy}-u_{xy}^2 = \pm 1$, the classic Monge-Ampère equation. This is, of course, elliptic for $g_+$ and hyperbolic for $g_-$. (We will see how this is reflected in the general analysis below.) The general solution to $\rho(f)=g_\pm$ is then $$f = (u_x,\ \pm u_y,\ xu_x+yu_y-u )$$ where $u$ satisfies $u_{xx}u_{yy}-u_{xy}^2 = \pm 1$.</p> <p>Now, I'm going to do the general analysis in terms of the moving frame and exterior differential systems, simply because that's the way I understand it and am most comfortable computing. Of course, one can avoid this, but I'm too lazy (and short on time) to do that now.</p> <p>So fix a nondegenerate $g\in \Omega^2\bigl(\Sigma,\Lambda^2(V)\bigr)$. I'm going to define a bundle $P_g\to\Sigma$ whose elements will consist of the quadruples $(x;e_1,e_2,e_3)\in\Sigma\times V\times V\times V$ such that the $e_i$ are a basis of $V$ and, moreover, $e_1\wedge g(x) = e_2\wedge g(x) = 0$. On $P_g$, there exists a nonvanishing $2$-form $\Phi$ such that $g = e_1{\wedge}e_2\ \Phi$. This $\Phi$ is semi-basic for the projection to $\Sigma$ and hence is a multiple of any nonvanishing $2$-form on $\Sigma$ pulled up to $P_g$. Note that $P_g$ is a right $G_1$-bundle over $\Sigma$, where $G_1\subset\mathrm{GL}(3,\mathbb{R})$ is the subgroup that consists of matrices $A\in \mathrm{GL}(3,\mathbb{R})$ that are $(2,1)$-block upper triangular, i.e., $A^3_1=A^3_2=0$.</p> <p>Now, one has the usual structure equations $de_a = e_b\ \eta^b_a$, which satisfy $d\eta^a_b = - \eta^a_c\wedge\eta^c_b$. (The summation convention always in force. I'll use the index ranges $1\le a,b,c \le 3$ and $1\le i,j,k \le 2$.) Since $g$ is a vector-valued $2$-form on a surface, one has $dg=0$, and expanding this using the structure equations yields $$0 = dg = e_1{\wedge}e_2\ \bigl(d\Phi + (\eta^1_1+\eta^2_2)\wedge\Phi\bigr) + e_3{\wedge}e_2\ \eta^3_1\wedge\Phi + e_1{\wedge}e_3\ \eta^3_2\wedge\Phi,$$ so $\eta^3_1\wedge\Phi=\eta^3_2\wedge\Phi=0$. Now, the hypothesis that $[g]:\Sigma\to\mathbb{RP}^2$ be an immersion is equivalent to $\eta^3_1{\wedge}\eta^3_2$ being nonvanishing, Consequently, $\Phi = \lambda\ \eta^3_1{\wedge}\eta^3_2$ for some nonvanishing function $\lambda$ on $P_g$. One easily computes that, for $A\in G_1$, $$R_A^*\lambda = \left(\frac{A^3_3}{A^1_1A^2_2-A^2_1A^1_2}\right)^2\lambda,$$ so the sign of lambda is constant on the fibers of $P_g$ and the equation $\lambda=\pm1$ defines a subset $P_g'\subset P_g$ that will be a right $G_2$-bundle over $\Sigma$, where $G_2\subset G_1$ is the subgroup consisting of those $A\in G_1$ that satisfy $A^3_3 = \pm(A^1_1A^2_2-A^2_1A^1_2)$. </p> <p>I'll say that $g$ is of <em>elliptic</em> (resp. <em>hyperbolic</em>) type if $\lambda = +1$ (resp. $\lambda=-1$) on $P_g'$. (The reasons for these designations will become apparent below.)</p> <p>Since $\Phi = \pm \eta^3_1{\wedge}\eta^3_2$ on $P_g'$, the structure equations plus the identity $0=d\Phi + (\eta^1_1+\eta^2_2)\wedge\Phi$ now show that $(\eta^3_3-\eta^1_1-\eta^2_2)\wedge\eta^3_1{\wedge}\eta^3_2=0$, so $\eta^3_3=\eta^1_1+\eta^2_2+b^1\eta^3_1+b^2\eta^3_2$ for some functions $b^1$ and $b^2$ on $P_g'$. The structure equations now show that the equations $b^1=b^2=0$ define a $G_3$-subbundle $P_g''\subset P_g'$, where $G_3\subset G_2$ is the subgroup consisting of those $A\in G_2$ with $A^1_3=A^2_3=0$. (This is the last reduction I will need to do to complete the calculation.)</p> <p>Our goal now is to find $1$-forms $\eta^1$ and $\eta^2$ on $P_g''$ satisfying the equations $$d(e_1\ \eta^1 + e_2\ \eta^2) = 0\qquad\text{and}\qquad e_1{\wedge}e_2\ \eta^1{\wedge}\eta^2 = g = \pm e_1{\wedge}e_2\ \eta^3_1{\wedge}\eta^3_2\ .$$<br> If we can do this, then, at least locally, $e_1\ \eta^1 + e_2\ \eta^2 = df$ for some $V$-valued function $f:\Sigma\to V$ such that $\rho(f) = g$, and this will solve our problem.</p> <p>Of course, the second equation is just $\eta^1\wedge\eta^2 = \pm\ \eta^3_1{\wedge}\eta^3_2$, while using the structure equations to expand the first equation yields $$0 = e_1 (d\eta^1 + \eta^1_1{\wedge}\eta^1+ \eta^1_2{\wedge}\eta^2) +e_2 (d\eta^2 + \eta^2_1{\wedge}\eta^1+ \eta^2_2{\wedge}\eta^2) +e_3 (\eta^3_1{\wedge}\eta^1 + \eta^3_2{\wedge}\eta^2)$$ Looking at the $e_3$-coefficient in this equation and applying Cartan's Lemma, we see that there must exist $h^{ij}=h^{ji}$ such that $\eta^i = h^{ij}\ \eta^3_j$, which, substituted into the second equation, says that $\det(h) = \pm 1$. </p> <p>Now, let $S_\pm$ denote the surface in the $3$-dimensional space of symmetric $2$-by-$2$ matrices that consists of the matrices with determinant $\pm 1$. ($S_+$ is a hyperboloid of 2 sheets, while $S_-$ is a hyperboloid of 1 sheet.) Let $h:P_g''\times S_\pm\to S_\pm$ be the projection onto the second factor. Set $\eta^i = h^{ij}\eta^3_j$ as $1$-forms on $P_g''\times S_\pm$ and consider the pair of $2$-forms $$\Upsilon^i = d\eta^i + \eta^i_1{\wedge}\eta^1+ \eta^i_2{\wedge}\eta^2.$$ The formulae derived so far show that $\Upsilon^i = \pi^{ij}\wedge\eta^3_j$, where $\pi^{ij}=\pi^{ji}$ are $1$-forms defined by the relations $$\begin{pmatrix}\pi^{11}&amp;\pi^{12}\\ \pi^{21}&amp;\pi^{22}\end{pmatrix} = dh + \begin{pmatrix}\eta^1_1&amp;\eta^1_2\\ \eta^2_1&amp;\eta^2_2\end{pmatrix} h + h\begin{pmatrix}\eta^1_1-\eta^3_3&amp;\eta^2_1\\ \eta^1_2&amp;\eta^2_2-\eta^3_3\end{pmatrix}.$$ Moreover, the equation $\det(h)=\pm1$ and the normalizations made so far imply that $\mathrm{tr}(h^{-1}\pi)=0$. </p> <p>These equations show that, on the quotient manifold $Q = (P_g''\times S_\pm)/G_3$, which is a bundle over $\Sigma$ with $2$-dimensional fibers, there is an ideal $\mathcal{I}$ that is locally generated by a pair of $2$-forms that pull back up to $P_g''\times S_\pm$ to be independent linear combinations of the $\Upsilon^i$, and that the integral surfaces of this ideal that are transverse to the fibers of $Q\to\Sigma$ represent solutions to our problem. When $\det(h)=1$, the above relations show that this is an elliptic ideal, while, when $\det(h)=-1$, this ideal is hyperbolic. In either case, the ideal is involutive, with Cartan characters $s_1=2$, $s_2=0$.</p> <p>The Cartan-Kähler Theorem now applies, at least when $g$ is real-analytic, to show that there are local integral manifolds transverse to the fibers of $Q\to\Sigma$. In the hyperbolic case, a smooth version of the Cartan-Kähler theorem guarantees local solvability while, in the elliptic case, there are existence theorems (basically coming from the theory of pseudo-holomorphic curves) that guarantee existence of local solutions. Thus, the equations are locally solvable if $g$ is smooth.</p> <p>Anyway, that's a sketch of an analysis of the nondegenerate case. The point is that the original determined system of three equations for three unknowns is not formally integrable, but, in the generic case, it can be prolonged to either an elliptic or a hyperbolic system that is involutive and hence formally integrable.</p> <p>One should still do the case when the rank of the differential of $g$ is $1$, and one should look for a way to reinterpret the system $\mathcal{I}$ as a single scalar equation of Monge-Ampere type, to which more standard PDE methods would apply. There may even be a way to linearize this equation as either an elliptic or hyperbolic single scalar equation, but that would require more thought than I have had time to put into it so far.</p> http://mathoverflow.net/questions/104573/open-problems-in-sub-riemannian-geometry/128013#128013 Answer by Robert Bryant for Open problems in sub-Riemannian geometry Robert Bryant 2013-04-18T21:08:57Z 2013-04-18T21:08:57Z <p>Andrei Agrachev recently posted an article on the arXiv: <a href="http://arxiv.org/abs/1304.2590" rel="nofollow">http://arxiv.org/abs/1304.2590</a> entitled "Some open problems" that you might find interesting. He discusses several open problems in control theory and sub-Riemannian geometry that are of considerable interest.</p> http://mathoverflow.net/questions/127319/intuition-for-levi-civita-connection-via-hamiltonian-flows/127363#127363 Answer by Robert Bryant for Intuition for Levi-Civita connection via Hamiltonian flows Robert Bryant 2013-04-12T13:33:27Z 2013-04-15T17:41:52Z <p>I think that you are asking for a symplectic interpretation of the splitting of the tangent bundle of $T^\ast M$ that is induced by the Levi-Civita connection on a Riemannian $n$-manifold $(M,g)$. I.e., to know how one can see, from the symplectic geometry of the cotangent bundle and the information of the Hamiltonian, the splitting $T(T^\ast M) = D\oplus V$, where $D$ is the $n$-plane field that is transverse to $V$, the tangent bundle to the fibers of $\pi:T^\ast M\to M$, that has the property that parallel transport of covectors with respect to the Levi-Civita connection gives curves in $T^\ast M$ that are tangent to $D$.</p> <p>There is a simple way to get $D$ in this case: Set $\gamma = \pi^\ast g$, and let $\dot \gamma$ be the Lie derivative of $\gamma$ with respect to the Hamiltonian vector field associated to the Hamiltonian $H$ computed from $g$. Then $\dot\gamma$ is a nondegenerate quadratic form on $T^\ast M$ of split type $(n,n)$, and one can easily show that $D\subset T(T^\ast M)$ is the unique $n$-plane field that is Lagrangian, transverse to $V$, and $\dot\gamma$-null.</p> <p><em>Historical Background:</em> This sort of construction was discussed thoroughly by Patrick Foulon in his (mid-1980s) thesis, and, yes, something like it works more generally for pseudo-Riemannian and Finsler geometries (see below). The key extra ingredient, besides knowing the symplectic form $\omega$ and the Hamiltonian $H$, is that one also needs to know the Lagrangian foliation $\mathcal{F}$ that is defined by the fibers of $\pi:T^\ast M\to M$. With just these, one can construct the complementary $n$-plane field $D$ (which, of course, is Lagrangian but not integrable) in a canonical manner. Thus, one knows what it means to have a 'parallel transport' in these more general cases. </p> <p>É. Cartan did something along these lines in his book on the geometry of Finsler spaces, but it's not written in a modern manner and so it's a bit hard to read for most geometers nowadays.</p> <p>Here is one way to describe the construction: The starting data are $(X,\omega,\mathcal{F},H)$, where $X$ is a manifold of dimension $2n$, $\omega$ is a symplectic form on $X$, $\mathcal{F}$ is an $\omega$-Lagrangian foliation, and $H$ is a function on $X$. The goal is to construct an $n$-plane field $D\subset TX$ that is $\omega$-Lagrangian and transverse to $\mathcal{F}$ in a canonical way. This will require making a nondegeneracy assumption on $H$; for example, this clearly cannot be done if $H$ is constant.</p> <p>Now, the data $(X,\omega,\mathcal{F})$ has no local geometry, in the sense that all Lagrangian foliations of a symplectic manifold are locally equivalent. This means that, locally, one can always choose canonical coordinates $(q,p)=(q^i,p_i)$ in which the symplectic form $\omega$ is $dp_i\wedge dq^i$ and the leaves of $\mathcal{F}$ are given by $dq^i=0$. Let's consider the derivatives of $H$ in these coordinates: Define $H^i$ and $H^{ij}=H^{ji}$ by the conditions $$dH \equiv H^i\ dp_i\ \text{mod}\ dq\qquad\text{and}\qquad dH^i \equiv H^{ij}\ dp_j\ \text{mod}\ dq$$ It is easy to check that, if one were to make a different choice $\bar q = F(q)$ for the $\mathcal{F}$-null coordinates and complete to canonical $\omega$-coordinates with a corresponding new $\bar p$, then the Jacobian change of variables formula yields $$\bar H^{ij} d\bar p_i\circ d\bar p_j \equiv H^{ij}\ dp_i\circ dp_j\ \text{mod}\ dq$$ so that the quadratic form $\eta_H = H^{ij}\ dp_i\circ dp_j$ is well-defined on the leaves of $\mathcal{F}$. (What is going on here is that, for any Lagrangian foliation $\mathcal{F}$ on a symplectic manifold $M$, there is a canonical, torsion-free flat connection on each leaf of $\mathcal{F}$. The quadratic form $\eta_H$ is merely the leafwise Hessian of $H$ with respect to this canonical connection.)</p> <p>Let us say that $H$ is $\mathcal{F}$-<em>nondegenerate</em> if $\eta_H$ is a nondegenerate quadratic form (i.e., pseudo-Riemannian metric) on each $\mathcal{F}$-leaf. </p> <p>From now on, assume that $H$ is $\mathcal{F}$-nondegenerate. Let $G = (G_{ij})$ be the inverse matrix of $(H^{ij})$. The change of variables formula then shows that the quadratic form $$\gamma_H = G_{ij}\ dq^i\circ dq^j$$ is well-defined on $X$, independent of the choice of coordinates. (What is going on here is that, for each $x\in X$, with $V_x\subset T_xX$ being the tangent to the $\mathcal{F}$-leaf through $x$, one has a canonical isomorphism ${V_x}^\ast \simeq T_xX/V_x$, and $\gamma_H$ at $x$ is simply the canonical quadratic form on $T_x/V_x$ induced by this isomorphism from the canonical dual quadratic form $\eta_H^\ast$ (on ${V_x}^\ast$) of the quadratic form $\eta_H$ on $V_x$.)</p> <p>Let $\dot\gamma_H$ denote the Lie derivative of $\gamma_H$ with respect to the Hamiltonian vector field $X_H$. One can now easily verify that $\dot\gamma_H$ is a nondegenerate quadratic form of type $(n,n)$ on $X$. Moreover, there is a unique $n$-plane field $D\subset TX$ on $X$ that is transverse to the leaves of $\mathcal{F}$, Lagrangian with respect to $\omega$, and null with respect to $\dot\gamma_H$. This $D$ is the desired splitting.</p> <p><strong>Example:</strong> As stated at the beginning, in the case that $g$ is a pseudo-Riemannian metric on $M^n$, when one defines the associated Hamiltonian $H$ on $X=T^\ast M$, then one finds that $\gamma_H$ is simply ${\pi^\ast}g$, and then $D$ is indeed the plane field on $T^\ast M$ induced by the Levi-Civita connection. </p> <p><em>Remark:</em> Note that, in the pseudo-Riemannian case, $\gamma_H$ is expressed in terms of the original metric using no derivatives, so that $D$, which is defined algebraically from $\dot\gamma_H$, uses only one derivative of the metric, which is as it should be, since the Levi-Civita connection depends on only one derivative of the metric. However, in the general case outlined above, $\gamma_H$ depends on two derivatives of $H$ and then $\dot\gamma_H$ depends on <em>three</em> derivatives. Thus, for example, in the general case, the so-called 'nonlinear connection' in Finsler geometry depends on three derivatives of the Finsler structure, and 'curvature' depends on <em>four</em> derivatives. This is one reason that the general Finsler case is more difficult to study than the Riemannian case.</p> http://mathoverflow.net/questions/127530/surfaces-filled-densely-by-a-geodesic/127535#127535 Answer by Robert Bryant for Surfaces filled densely by a geodesic Robert Bryant 2013-04-14T13:39:43Z 2013-04-15T15:55:17Z <ol> <li><p>Any surface of revolution in $3$-space with poles will have this property. The reason is that, in this case, any geodesic either goes through a pole (i.e., a point where the axis of revolution meets the surface) and is a profile curve that lies in a plane or else, because of the Clairaut integral, it avoids that pole by some positive distance. Thus, no geodesic on the surface is dense in the surface.</p></li> <li><p>You mention ellipsoids, which furnish examples of these special surfaces. These are examples of so-called 'Liouville surfaces', i.e., Riemannian surfaces $(S,g)$ for which there exist two independent first integrals of the geodesic flow on $T^\ast S$ that are quadratic functions on the fibers of $T^\ast S\to S$, one of which is the co-metric associated to $g$ and the other of which is an independent first integral. As you probably know, surfaces of revolution are surfaces for which there exist a first integral of the geodesic flow that is <em>linear</em> on the fibers of $T^\ast S\to S$, namely the Clairaut integral. It has been known for some time that there are metrics on the $2$-sphere that don't possess any 'extra' first integrals that are linear or quadratic functions on the fibers of $T^\ast S\to S$, but do possess first integrals that are cubic or quartic functions on the fibers of $T^\ast S\to S$. These are due to Goryachev-Chaplygin (early 20th century) and Dullin-Matveev (2004). These are also examples for which no geodesic winds densely over the surface. All of these work because there are 'conservation laws' for the geodesic flow of a particular kind, and they properly generalize the Liouville surfaces (which includes the famous case of ellipsoids). </p></li> </ol> http://mathoverflow.net/questions/127306/differential-form-with-empty-zero-locus/127533#127533 Answer by Robert Bryant for differential form with empty zero locus Robert Bryant 2013-04-14T13:33:22Z 2013-04-14T13:33:22Z <p>There are many compact examples that are not Kähler. For example, by a theorem of Borel, every complex semisimple Lie group $G$ contains a discrete, cocompact subgroup $\Gamma\subset G$. If you let $X = G/\Gamma$, then $X$ will be a holomorphically parallelizable compact complex manifold. The ring of right-invariant holomorphic differential forms on $G$, say, $R$, descends to be a ring (still called $R$) of holomorphic forms on $X$. This ring is closed under exterior derivative, of course, and every holomorphic differential form on $X$ belongs to $R$. </p> <p>Now, taking different examples of $G$ and $\Gamma$, one can construct many examples of holomorphic forms on compact manifolds that are not closed, are closed but not exact, etc. Each nonzero form in $R$ has empty vanishing locus. </p> <p>Nontrivial Kähler examples can be constructed this way: Let $Z$ be a compact Kähler manifold of complex dimension $2n$ that supports a holomorphic symplectic $2$-form $\Omega$, and let $X\subset Z$ be a smooth subvariety of dimension $d>n$. Then the pullback of $\Omega$ to $X$ cannot vanish anywhere on $X$ for dimension reasons, but there is no reason to believe (except when $d=2n{-}1$) that the rank of this pullback needs to be constant. $X$ will be Kähler, but, if the rank of the pullback is not constant, then you won't be able to write $X$ as a product of Kähler manifolds, some of which support the holomorphic $2$-form as a symplectic form.</p> http://mathoverflow.net/questions/126989/sun-invariant-subring-of-lambda-mathbbr2n/127001#127001 Answer by Robert Bryant for $SU(n)$-invariant subring of $\Lambda^{*}\mathbb{R}^{2n}$ Robert Bryant 2013-04-09T17:21:58Z 2013-04-09T17:21:58Z <p>Actually, as you have written it, the ring $R$ is generated by $\omega_0$ and the real and imaginary parts of $\Omega_0$, since $R$ consists of real-valued $1$-forms. You should probably consider, instead the ring $R^\mathbb{C} = \mathbb{C}\otimes R$ of complex-valued exterior forms on $\mathbb{R}^{2n}=\mathbb{C}^n$. Then the proof goes in two stages:</p> <p>First, consider the forms that are invariant under the diagonal maximal torus in $\mathrm{SU}(n)$, which acts as $g\cdot dz^i = e^{\lambda_i} dz^i$ and $g\cdot d\bar z^i = e^{-\lambda_i} d\bar z^i$ where $\lambda_i$ are purely imaginary and satisfy $\lambda_1+\cdots+\lambda_n=0$. It is easy to see that the only complex-valued forms that are invariant under this torus action are the ones generated by $\omega_i = dz^i\wedge d\bar z^i$ and $\Omega_0$ and $\overline{\Omega_0}$. In particular, this shows that $R$ lies in the ring generated by the $(1,1)$-forms and $\Omega_0$ and $\overline{\Omega_0}$. Note that $\Omega_0$ (and $\overline{\Omega_0}$, too) wedged with any $(1,1)$-form must vanish, so it's clear that, outside of $\Omega_0$ and $\overline{\Omega_0}$, the ring $R$ must lie in the ring that is the sum of the $(p,p)$-forms. </p> <p>Now, each $(p,p)$-form $\Psi$ can be written uniquely in the form $$\Psi = \Psi_p + \omega_0\wedge\Psi_{p-1} + {\omega_0}^2\wedge\Psi_{p-2} + \cdots + {\omega_0}^p\wedge\Psi_0$$ where each $\Psi_j$ is an $\omega_0$-<em>primitive</em> $(j,j)$-form, and these primitive subspaces are $\mathrm{SU}(n)$-irreducible. (This follows from a root space representation.) In particular, if $\Psi$ is invariant under $\mathrm{SU}(n)$, then $\Psi = {\omega_0}^p\wedge \Psi_0$, so $\Psi$ is a complex multiple of ${\omega_0}^p$.</p> http://mathoverflow.net/questions/126275/invariance-of-the-l-h-s-of-euler-lagrange-equation/126298#126298 Answer by Robert Bryant for Invariance of the l.h.s. of Euler-Lagrange equation Robert Bryant 2013-04-02T17:19:53Z 2013-04-04T12:09:15Z <p>There is a coordinate-free description using only natural objects on $TM$. Here is one way to do it.</p> <p>First, consider the basepoint submersion $\pi:TM\to M$. For each $v\in TM$, the linear map $\pi'(v):T_v(TM)\to T_{\pi(v)}M$ is surjective, and the $\pi$-fiber through $v$ is equal to $T_{\pi(v)}M$, a vector space. It follows that the kernel of $\pi'(v)$ is naturally isomorphic to $T_{\pi(v)}M$. Call this isomorphism $\iota_v: T_{\pi(v)}M\to \mathrm{ker}\bigl(\pi'(v)\bigr)\subset T_v(TM)$, and let $\nu_v : T_v(TM) \to T_v(TM)$ be the nilpotent endomorphism $\nu_v = \iota_v\circ \pi'(v)$. </p> <p>Next, consider a Lagrangian $L:TM\to \mathbb{R}$, which I will assume to be smoothly differentiable. Define a $1$-form $\omega_L$ on $TM$ by $$\omega_L(w) = dL\bigl(\nu_v(w)\bigr)$$ for all $w\in T_v(TM)$. Let $R$ be the vector field on $TM$ that is tangent to the fibers of $\pi$ and that is the natural radial vector field on each such (vector space) fiber, and set $E_L = dL(R) - L$. (If $L$ is quadratic homogeneous on each $\pi$-fiber, then, by Euler's relation, $E_L = L$.)</p> <p>Finally, if $\gamma:[a,b]\to M$ is a twice differentiable curve, with $\gamma':[a,b]\to TM$ its velocity vector and $\gamma'':[a,b]\to T(TM)$ the velocity vector of $\gamma'$, then, for each $t\in[a,b]$, consider the co-vector $\beta(t)\in T^\ast_{\gamma'(t)}TM$ defined by the rule $$\beta(t)(w) = d\omega_L(\gamma''(t),w)+ dE_L(w)$$ for $w\in T_{\gamma'(t)}TM$. Then $\beta(t)(w)=0$ for $w\in \mathrm{ker}\bigl(\pi'(\gamma'(t))\bigr)$, so $\beta(t) = \pi'(\gamma'(t))^\ast(\delta\gamma(t))$ for a unique co-vector $\delta\gamma(t)\in T^\ast_{\gamma(t)}M$. </p> <p>This assignment $t\mapsto \delta\gamma(t)$ is the canonical 'variation $1$-form' of the Lagrangian $L$ along $\gamma$. It vanishes identically if and only if $\gamma$ satisfies the Euler-Lagrange equation for $L$.</p> http://mathoverflow.net/questions/126193/darboux-like-theorem-for-non-degenerate-3-forms-in-6-manifolds/126197#126197 Answer by Robert Bryant for Darboux like theorem for non-degenerate 3-forms in 6-manifolds Robert Bryant 2013-04-01T17:25:56Z 2013-04-02T04:33:12Z <p>This depends on what you mean by 'Darboux-like'. It is certainly not true that a closed nondegenerate 3-form on a 6-manifold is necessarily locally equivalent to one of the 'flat' models, so there is no direct analog of the Darboux' theorem in this case. </p> <p>As I remark in my article "Remarks on the geometry of almost complex $6$-manifolds" (Asian Journal of Mathematics <strong>10</strong> (2006), 561–606, also available on the arXiv as arXiv:math/0508428), the closed $3$-forms of elliptic type in dimension $6$ essentially depend on 4 arbitrary functions of 6 variables (modulo diffeomorphism). </p> <p>However, there is an analog if you are willing to consider something stronger: If $\phi\in\mathcal{A}^3_+(M^6)$ is a $3$-form of elliptic type on an <em>oriented</em> $6$-manifold $M$, then there is a unique $J(\phi)\in\mathcal{A}^3_+(M^6)$ with the property that the complex $3$-form $\Upsilon = \phi+i \ J(\phi)$ is decomposable and hence of type $(3,0)$ with respect to an almost complex structure $J_\phi$ on $M$ that induces the given orientation of $M$. All this is algebra. However, now, if one adds the hypothesis that $d\Upsilon = 0$, which is, of course, the same as $d\phi = d\bigl(J(\phi)\bigr)=0$, then one has that there always exist local complex functions $z^1,z^2,z^3$ such that $\Upsilon = dz^1\wedge dz^2\wedge dz^3$. In particular, $\phi = \mathrm{Re}(dz^1\wedge dz^2\wedge dz^3)$, so this is a sort of Darboux-like theorem; it's just that you need more hypothesis than the closure of the original form.</p> <p>There is a similar result for nondegenerate $3$-forms of hyperbolic type. This is the case in which the form $\phi$ can be written locally as $\phi = \phi_+ + \phi_-$ where each of $\phi_\pm$ is decomposable while $\phi_+\wedge\phi_-\not=0$. (These two summands are unique up to permutation.) In this case, the 'Darboux-like' theorem is that $\phi$ can be put in normal form if and only if $d\phi_+=d\phi_-=0$, which is stronger than $d\phi=0$ (which is not sufficient by itself).</p> <p>Finally, there is the case of nondegenerate $3$-forms of what might be called 'nilpotent type' (which is not a stable type in Hitchin's sense, but is nondegenerate in the sense described by the OP). A $3$-form $\phi$ on $M^6$ is nondegenerate of nilpotent type if and only if each point lies in some open set $U$ on which there exists a coframing $\omega^1,\ldots,\omega^6$ for which $$\phi = \omega^4\wedge\omega^2\wedge\omega^3 +\omega^5\wedge\omega^3\wedge\omega^1 +\omega^6\wedge\omega^1\wedge\omega^2.$$ For such a $\phi$, the conditions that it can locally be put in this form with $\omega^i = dx^i$ for some coordinates $x = (x^1,\ldots,x^6)$ consist of two things: First, the condition $d\phi=0$, which is clearly necessary; second, the condition that $3$-plane field $D\subset TM$ defined by $\omega^1=\omega^2=\omega^3=0$ (which is well-defined by $\phi$) should be Frobenius. It is not hard to show that these necessary conditions are also sufficient, so this is the 'Darboux-like' normal form theorem in this case.</p> <p>Note the interesting fact that, in each of these three cases, the 'Darboux-like' conditions are all first order equations on $\phi$. This does not continue in higher dimensions. In dimension $7$, the two stable types of $3$-forms each have examples that are flat to first order but not flat to second order, so the 'Darboux-like' theorems in this case turn out to involve a mixture of first and second order conditions.</p> http://mathoverflow.net/questions/126203/conservation-law-and-generalized-symplectic-monge-ampere-equation-arising-from-3/126204#126204 Answer by Robert Bryant for conservation law and generalized Symplectic Monge-Ampere equation arising from 3-variables Robert Bryant 2013-04-01T20:03:37Z 2013-04-01T20:03:37Z <p>I think that you should be careful to define your terms, but let me guess: A <em>Jacobi PDE system</em> for three unknown functions $h_1,h_2,h_3$ of three independent variables $x_1,x_2,x_3$ is a set of three partial differential equations, each of which can be written as a linear combination of the minors (of any rank, including $0$ and $3$) of the Jacobian matrix $J_x(h) = \frac{\partial h}{\partial x}$ with coefficients that are explicit functions of the $6$ variables $x_1,x_2,x_3,h_1,h_2,h_3$. In other words, on $M=\mathbb{R}^6$ (or some open subset) with coordinates $x_1,x_2,x_3,h_1,h_2,h_3$, you have specified three $3$-forms $\Upsilon_i$ ($i=1,2,3$), and a graph $$\Gamma_u = \bigl\lbrace\bigl(x,u(x)\bigr)\ \mid\ x\in D\subseteq\mathbb{R}^3\bigr\rbrace$$ solves your system with $h=u(x)$ for some function $u:D\to\mathbb{R}^3$ if and only if the pullback of each of the $\Upsilon_i$ to $\Gamma_u$ vanishes. To get a decent theory, you'll also need to impose some kind of nondegeneracy condition, such as the condition that, for a 'generic' pair of tangent vectors $v_1,v_2\in T_pM$, the $1$-forms $\theta_i(v) = \Upsilon_i(v_1,v_2,v)$ are linearly independent at $p$.</p> <p>Now, it turns out that such a system is <em>never</em> equivalent to a so-called <em>symplectic Monge-Ampère system</em>, which is locally described on a $6$-manifold $M$ by a choice of a symplectic $2$-form $\Omega$ on $M$ and a nonzero $3$-form $\Upsilon$ that is $\Omega$-primitive, i.e., such that $\Omega\wedge\Upsilon=0$. (In this case, the $3$-manifolds you want to study are the $\Omega$-Lagrangian submanifolds $L\subset M$ to which $\Upsilon$ pulls back to be the zero $3$-form. </p> <p>The reason is that for Jacobi PDE systems as above (that satisfy the nondegeneracy condition), the general solution depends locally on <em>three</em> arbitrary functions of $2$-variables while, for the symplectic Monge-Ampère systems as defined above, the general solution depends locally on <em>two</em> arbitrary functions of $2$-variables.</p> <p>On the other hand, you could consider a generalization of 'symplectic Monge-Ampère system' in which you have a differential ideal $\mathcal{I}$ on a $6$-manifold that is generated by a nondegenerate $2$-form $\Omega$ and a nonvanishing $\Omega$-primitive $3$-form $\Upsilon$. The condition that they generate a <em>differential</em> ideal is just that there exist a $1$-form $\alpha$ and a function $a$ such that $$d\Omega = \alpha\wedge\Omega + a\ \Upsilon.$$ (If $a\equiv0$, then one finds that $d\alpha=0$, so that, at least locally, one can scale $\Omega$ to make it be closed, and you are back in the symplectic Monge-Ampère case. However, if $a\not=0$, you cannot do this.) In this generalized situation, the general solution will depend on two arbitrary functions of $2$ variables, just as in the symplectic Monge-Ampère case. It would now make sense to think of conservation laws as closed $3$-forms in $\mathcal{I}$, and there are interesting question to ask there.</p> http://mathoverflow.net/questions/125388/positive-sectional-curvature-of-submanifold-in-rn/126064#126064 Answer by Robert Bryant for positive sectional curvature of submanifold in $R^n$? Robert Bryant 2013-03-31T01:44:52Z 2013-03-31T01:44:52Z <p>If the extrinsic diameter of $N$ in $\mathbb{R}^n$ is $D = 2/\rho$ where $\rho>0$, then for any two points $p$ and $q$ in $N$ that are $D$ units apart, all the sectional curvatures at $p$ and $q$ are are at least $\rho^2$, so the Ricci curvature (in any tangent direction) at those two points will be at least $(n{-}1)\rho^2$.</p> http://mathoverflow.net/questions/126003/does-partial-overline-partialf0-imply-f-equiv-c-for-particular-kind-of-f/126011#126011 Answer by Robert Bryant for Does $\partial\overline{\partial}f=0$ imply $f\equiv c$ for particular kind of $f$? Robert Bryant 2013-03-30T13:33:06Z 2013-03-30T20:36:13Z <p>Yes, this is true. The point is that $\partial\bar\partial f = 0$ on the complement $C$ of a ball in $\mathbb{C}^n$ ($n\ge 2$) implies that $f = h_+ + \overline{h_-}$ for some functions $h_\pm$ that are holomorphic on $C$. (They are unique up to adding a constant to one and subtracting it from the other.) Now Hartogs' extension theorem implies that $h_\pm$ extend to be holomorphic on all of $\mathbb{C}^n$. Your boundedness assumption now says that the real and imaginary parts of $f$, which are harmonic, are bounded, which implies, by Liouville's Theorem, that they are constant.</p> <p><em>Requested explanation:</em> Since $d(\partial f) = (\partial + \bar\partial)(\partial f) = -\partial\bar\partial f =0$, it follows that $\partial f$ is a closed holomorphic $(1,0)$-form on $C$ and hence is of the form $\partial f =\partial h_+ = dh_+$ for some holomorphic function $h_+$ on $C$, unique up to an additive constant. (NB: $C$ is simply connected, since $n\ge 2$, so closed $1$-forms on $C$ are exact.) Since $\partial (f-h_+) = 0$, it follows that the function $f-h_+$ must be antiholomorphic, so $f = h_+ + \overline{h_-}$ for some holomorphic function $h_-$.</p> http://mathoverflow.net/questions/125304/variant-of-the-riemann-mapping-theorem-for-conf-mathbb-h2/125457#125457 Answer by Robert Bryant for Variant of the Riemann Mapping Theorem for $Conf(\mathbb H^2)$? Robert Bryant 2013-03-24T14:17:27Z 2013-03-24T14:17:27Z <p><em>A cultural remark to begin:</em> The comments asking for clarification of your question may have sounded a bit rough, but please understand that they weren't meant personally. In the culture of mathematics, definitions are <em>very</em> important because we have found, time and again, that questions are often based on some misunderstanding, and clarifying the question is usually the first step in seriously concentrating on finding a solution. Training students to ask clear questions is a major part of training young mathematicians, and, for most of us who do this for a living, it becomes second nature to begin by asking the questioner to define the terms of discussion more carefully and/or pointing out that there's some confusion going on. It doesn't necessarily stop happening to you when you finish graduate school either. When I was a postdoc at the Institute for Advanced Study, my first position after graduate school, I thought I had the best opportunity in the world to learn more about Lie groups because Armand Borel was there. I found, though, that whenever I went to ask him a question, he would invariably respond "What do you mean??" in what I then thought was a stern, almost indignant, voice, as though I had just revealed how foolish and ignorant I really was. It took me some time to realize that this was almost <em>always</em> his response, even if he thought it was an excellent question. After I did realize that, though, we got along fine and, indeed, I learned an enormous amount from him.</p> <p>Anyway, to get to your question: It seems that, by $Conf(\mathbb{H}^2)$, you mean the (real) vector fields on $\mathbb{H}^2$ whose (local) flows are holomorphic. Such a vector field $X$ is the real part of a unique <em>holomorphic</em> vector field $Z$ of the form $$Z = h(z)\ \frac{\partial\ }{\partial z} = h(x+iy)\ \frac12\left(\frac{\partial\ }{\partial x}-i\frac{\partial\ }{\partial y}\right)$$ where $h$ is holomorphic in the upper half plane. People frequently write $Z$ when they mean $X = Z +\bar Z$, which is why some folks were confused by your expression for an element of $Conf(\mathbb{H}^2)$. </p> <p>The problem with thinking of this infinite dimensional vector space as the Lie algebra of a Lie group is that most of these vector fields only define local flows on $\mathbb{H}^2$, not global ones, so they don't really generate automorphisms of $\mathbb{H}^2$. In fact, it's a theorem that the only ones that do are the ones for which<br> $$h(z) = a + bz + cz^2$$ where $a$, $b$, and $c$ are real numbers, and this is a Lie algebra isomorphic to ${\frak{sl}}(2,\mathbb{R})$. The flows that are generated in this way generate the Lie group of linear fractional transformations that carry $\mathbb{H}^2$ into itself, and this happens to be the group of automorphisms of $\mathbb{H}^2$ as a complex manifold.</p> <p>Given this, it seems that, instead, you want $Conf(\mathbb{H}^2)$ to mean something else, namely the Lie algebra of vector fields of the above form in which $h$ is holomorphic on the entire complex plane $\mathbb{C}$ and real-valued on $\mathbb{R}$. In other words, $h$ should have its power series in $z$ have all real coefficients and have infinite radius of convergence. Then, indeed, $Conf(\mathbb{H}^2)$ injects into (but not onto) the Lie algebra $\frak{X}(\mathbb{R})$ of real analytic vector fields on $\mathbb{R}$. </p> <p>By the way, if $B\subset\mathbb{C}$ is a connected and simply connected open subset, then the set of all holomorphic vector fields on $B$ will still be of the above form for some set of $h\in\mathcal{O}(B)$ (the holomorphic functions on $B$), but picking out the $3$-dimensional subalgebra whose flows generate the automorphisms of $B$ (which, by the Riemann Mapping Theorem, is a $3$-dimensional Lie subalgebra of this space) is, generally, a very difficult thing to do.</p> http://mathoverflow.net/questions/124877/symmetry-of-generationg-function-of-pde/125014#125014 Answer by Robert Bryant for symmetry of generationg function of PDE Robert Bryant 2013-03-19T20:09:44Z 2013-03-20T01:52:30Z <p>There are several methods, but let me describe (what is perhaps the simplest) one: On $\mathbb{R}^4$ with coordinates $(t,x,u,p)$, consider the pair of $2$-forms \begin{aligned} \Upsilon_0 &amp;= (du-p\ dx)\wedge dt,\\ \Upsilon_1 &amp;= du\wedge dx + d(u^{4/3}p)\wedge dt + \lambda v\ dx\wedge dt. \end{aligned} It is easy to see that a graphical surface $\bigl(t,x,u(t,x), p(t,x)\bigr)$ in $\mathbb{R}^4$ is an integral manifold of $\Upsilon_0$ and $\Upsilon_1$ if and only if $p(t,x)=u_x(t,x)$ and the function $v=u(t,x)$ satisfies the given nonlinear PDE. Thus, the symmetries of the given PDE correspond to the self-diffeomorphisms of $\mathbb{R}^4$ that preserve the ideal $\mathcal{I}$ generated by $\Upsilon_0$ and $\Upsilon_1$. </p> <p>Now, the flow of the vector field $$Z = T\ \frac{\partial\ }{\partial t} + X\ \frac{\partial\ }{\partial x} +U\ \frac{\partial\ }{\partial u}+ P\ \frac{\partial\ }{\partial p}$$ preserves $\mathcal{I}$ if and only if $\mathrm{Lie}_Z(\Upsilon_i)\in \mathcal{I}$ holds for $i=0,1$. When written out, this condition is $8$ linear PDE for the $4$ unknown functions $T$, $X$, $U$, and $P$, and this overdetermined system is easily solved by inspection, showing that the vector space of solutions has dimension $4$, with a basis given by \begin{aligned} Z_1 &amp;= \frac{\partial\ }{\partial t}\\ Z_2 &amp;= \frac{\partial\ }{\partial x}\\ Z_3 &amp;= 2x\ \frac{\partial\ }{\partial x} +3u\ \frac{\partial\ }{\partial u}+ p\ \frac{\partial\ }{\partial p}\\ Z_4 &amp;= e^{-4/3\lambda\ t}\ \frac{\partial\ }{\partial t} +\lambda e^{-4/3\lambda\ t} u\ \frac{\partial\ }{\partial u} + \lambda e^{-4/3\lambda\ t} p\ \frac{\partial\ }{\partial p}\\ \end{aligned}</p> <p>Thus, there is a $4$-dimensional group of symmetries of the original equation, generated by the flows of the vector fields $Z_i$. Note that these four vector fields are linearly independent on the locus where $\lambda up\not=0$. In particular, when $\lambda\not=0$, the group acts with open orbits away from the locus $up=0$. (By contrast, when $\lambda=0$, one has $Z_4=Z_1$, so there really is only a $3$-dimensional group of symmetries in this case, and the action is not transitive.)</p> <p>If you want to think of this in terms of the full contact manifold and generating functions, introduce a fifth coordinate $q$, let $\theta = du - p\ dx - q\ dt$ be the contact form on $\mathbb{R}^5$, and then, to each $Z_i$ you have an associated 'generating function' $f_i = \theta(Z_i)$ (in the classical language).</p> <p>Finally, if you want to think of the solutions of the 'reduced' equation, i.e., the integral manifolds that are invariant under the flow of a symmetry vector field $Z$, then one just needs to consider the two $1$-forms $\theta_i = \iota(Z)(\Upsilon_i)$. These are, for most choices of $Z$, two linearly independent $1$-forms that define a Frobenius system $\mathcal{I}_Z$ on $\mathbb{R}^4$, and the integral surfaces are integral surfaces of $\mathcal{I}$ that are foliated by the integral curves of $Z$. These are the 'graphs' of solutions of the PDE that are invariant under the flow of $Z$.</p> <p>In fact, integrating these vector fields is rather easy because they form a Lie algebra that is linearly independent, so that the orbits are open. Thus, we are really finding Lie subgroups of a $4$-dimensional (nonabelian) Lie group, and that is easy to do.</p> http://mathoverflow.net/questions/124943/qustions-on-r-bryants-papaer-calibrated-embeddings-in-the-special-lagrangian-an/124952#124952 Answer by Robert Bryant for Qustions on R.Bryant's papaer "Calibrated embeddings in the special Lagrangian and coassociative cases" Robert Bryant 2013-03-19T12:25:05Z 2013-03-19T12:25:05Z <p>I'm afraid that that article does not do a lot of details in the introductory Section 0, just because more complete explanations were already available in earlier articles of mine. Here are some brief answers. Right now, I don't have the time to write out the explanations in greater detail.</p> <ol> <li><p>This is a consequence of the claim, proved in the reference [Br$_2$], that the condition of being torsion-free is $np$ first order PDE for a section $\sigma:M\to S=F/G$ that defines a $G$-structure (where $n$ is the dimension of the manifold and $p$ is the codimension of $G$ in $\mathrm{SO}(n)$). The hypothesis of strong admissibility implies (indeed, it is equivalent to the condition) that all of these equations are captured by the condition of closure of the differential forms associated to the $G$-structure, which is exactly the condition that an $n$-plane $E$ be an integral element of $\mathcal{I}$.</p></li> <li><p>If you unwind the definitions, you will see that the spaces ${\frak{h}}_k$ were <em>defined</em> so as to make this equation true. This computation is best carried out up on $F$, where one can write out the definition of $\hat \alpha$ in a coframing of $F$ given by the structure equations and compute $d\hat\alpha$ explicitly. (When I have more time, maybe I can put in a brief description of this computation.)</p></li> <li><p>I didn't claim to <em>give</em> the proof that $\mathrm{SU}(n)\subset\mathrm{SO}(2n)$ is regularly presented for all $n$, I just said that it can be proved. You can get an indication of how the proof goes by looking a little further along in the paper where, on pages 13–14, I do (briefly) give the argument for $n=3$, the case of most interest in the article. </p></li> </ol> http://mathoverflow.net/questions/124655/visualizing-bianchi-type-homogenous-spaces/124751#124751 Answer by Robert Bryant for Visualizing Bianchi type/homogenous spaces Robert Bryant 2013-03-17T05:00:15Z 2013-03-17T05:00:15Z <p>For a different viewpoint from the excellent treatments by Scott and Thurston of 3-dimensional geometries, if you are trying to get a feel for the homogeneous Riemannian $3$-manifolds (which, as noted, were first classified by Bianchi), you might want to try looking at them from the point of view of their most basic invariants, their curvatures. This provides a natural classification and made it easy for me to organize the information when I was learning the subject. From this point of view, it doesn't quite divide the way that Bianchi did it, but you shouldn't have any trouble making the comparison. In outline, it goes like this:</p> <p>Let $(M,g)$ be a connected Riemannian $3$-manifold that is homogeneous, i.e., the group $G$ of isometries of $M$ acts transitively on $M$. Consider the Ricci curvature $\mathrm{Ric}(g)$. This is also a quadratic form on $M$ and, as such, relative to $g$, it has $3$ real eigenvalues $\lambda_1$, $\lambda_2$, and $\lambda_3$, which are constant functions on $M$ because of the homogeneity hypothesis.</p> <p>Case 1: $\lambda_1=\lambda_2=\lambda_3=\lambda$. In this case, the sectional curvature of $g$ is constant, and so $M$ is a space form, of elliptic, parabolic, or hyperbolic type, depending on whether $\lambda$ is positive, negative, or zero. These are easily visualized, as models for them are straightforward to construct.</p> <p>Case 2: $\lambda_1\not=\lambda_2=\lambda_3$. In this case, there is a well-defined $G$-invariant line field $L$ on $M$ which is the eigendirection of multiplicity $1$ for $\mathrm{Ric}(g)$. After passing to a double cover, if necessary, there is a unit vector field $Y$ tangent to $L$ and one can compute the Lie derivative of $g$ in the direction $Y$. There are two subcases: </p> <p>Case 2a: $L_Yg=0$ (i.e., $Y$ is a Killing field). In this case, the isometry group of $g$ has dimension $4$, the integral curves of $Y$ are geodesics and the set of these geodesics is a surface $S$. The projection $M\to S$ with fibers tangent to $L$ is a Riemannian submersion, and the induced metric on $S$ has constant Gauss curvature. Then $M$ has the structure of a principal bundle over $S$ (with group action generated by $Y$) with a connection whose curvature $2$-form is a constant multiple of the area form on $S$. There is a $2$-parameter family of such metrics.</p> <p>Case 2b: $L_Yg$ is nonzero, in which case, it turns out that there is a coframing $\omega_1$, $\omega_2$, and $\omega_3$ defined up to sign on $M$ so that $g = {\omega_1}^2+{\omega_2}^2+{\omega_3}^2$, while $\mathrm{Ric}(g) = \lambda_1\ {\omega_1}^2$ (i.e., $\lambda_2=0$) and $L_Yg = \mu\ ({\omega_2}^2-{\omega_3}^2)$ for some constant $\mu>0$. This coframe field is invariant under $G$ (at least, up to sign), so $G$ has dimension $3$ and, essentially, the metric $g$ is a left-invariant metric on $G$, which is a covering of $M$. Thus, this subcase, which consists of a $2$-parameter family of metrics (parametrized by $\lambda_1\not=0$ and $\mu>0$) can be covered as a limit of Case 3, which consists of left invariant metrics on $3$-dimensional Lie groups.</p> <p>Case 3: The $\lambda_i$ are all distinct, in which case, after passing to a cover and restricting $G$ to its identity component if necessary, one sees that $G$ leaves fixed a homogeneous coframing on $M$, so $g$ is essentially a left-invariant metric on $G$. By the usual algebra that classifies the $3$-dimensional Lie groups, one knows that there will be a $g$-orthogonal $G$-invariant coframing $\omega_1$, $\omega_2$, $\omega_3$ so that $g = {\omega_1}^2+{\omega_2}^2+{\omega_3}^2$ and so that \begin{aligned} d\omega_1 &amp;= a_1\ \omega_2\wedge\omega_3\\ d\omega_2 &amp;= a_2\ \omega_3\wedge\omega_1 + b\ \omega_1\wedge\omega_2\\ d\omega_3 &amp;= a_3\ \omega_1\wedge\omega_2 - b\ \omega_3\wedge\omega_1 \end{aligned} where $a_1$, $a_2$, $a_3$, and $b$ are constants satisfying either $a_1=0$ or $b=0$ (these latter alternatives are necessary in order for the Jacobi identity to be satisfied). If $b=0$, then the condition that the metric $g$ so defined on the corresponding Lie group $G$ have distinct eigenvalues of its Ricci curvature is just that the $a_i$ be distinct <em>and</em> that $a_1{+}a_2{+}a_3\not=2a_i$ for any $i$. If $a_1=0$, then the condition that the metric so defined have all Ricci eigenvalues be distinct is that $a_2\not=a_3$ <em>and</em> $b^2+a_2a_3\not=0$. Thus, there are two $3$-parameter families of these metrics in Case 3. Meanwhile, when one removes some of the inequalities, one gets limiting cases that cover Case 2b.</p> <p>Thus, in all the homogeneous cases, the curvature (together with, sometimes, derivatives of curvature eigendirections) actually produces a natural coframing and/or foliation of the metric $g$, and, because these invariants are so natural, they are a good place to start to study the local geometry of the homogeneous Riemannian $3$-manifolds.</p> http://mathoverflow.net/questions/124613/induced-riemannian-metric-on-jet-manifold/124620#124620 Answer by Robert Bryant for Induced Riemannian metric on Jet-Manifold Robert Bryant 2013-03-15T15:23:30Z 2013-03-15T19:28:24Z <p>There are, of course, several different functorially induced metrics on $J^r(M,N)$ when $M$ and $N$ are endowed with given Riemannian metrics. </p> <p>For example, $J^0(M,N)=M\times N$ and one can just take the product metric.</p> <p>Meanwhile $J^1(M,N)$ can be regarded as a vector bundle over $M\times N$ with fiber $T^\ast_xM\otimes T_yN$ over $(x,y)\in M\times N$. The metrics on $M$ and $N$ induce an inner product on the bundle $T^\ast M\otimes TN$ (since you have innter products on each factor bundle separately), and the Levi-Civita connections on the two factors can be used to define a connection on this bundle. This information is enough to specify a metric on $J^1(M,N)$. </p> <p><em>Added at the request of the OP:</em> Given two bundles $E$ and $F$ over a base that are endowed with inner products and connections, the bundle $E\otimes F$ inherits both an inner product and a connection in the usual way: The inner product is the one for which the square length of $e\otimes f$ is just $\langle e,e\rangle_E\ \langle f,f\rangle_F$, and the connection is the tensor product connection. (You should look this up if you don't know it; I won't spell it out here.) Once you have a bundle $E$ with inner product $\langle,\rangle$ and connection $\nabla$ over a Riemannian manifold $M$, there is a unique metric on the total space $E$ that makes the horizontal space provided by the connection perpendicular to the fibers of $E\to M$, makes the projection to $M$ a Riemannian submersion, and restricts to each fiber to be the given inner product. </p> <p>One can continue on this this way or just use an induction based on a certain natural inclusion of $J^{r+1}(M,N)$ into $J^1\bigl(M,J^r(M,N)\bigr)$ to finish the construction.</p> <p>This construction is one choice among several possible ones, which I won't spell out here.</p> http://mathoverflow.net/questions/124235/automorphism-group-of-a-compact-kahler-manifold/124238#124238 Answer by Robert Bryant for Automorphism group of a compact Kahler manifold Robert Bryant 2013-03-11T15:40:25Z 2013-03-11T15:40:25Z <p>The answer to $(1)$ is 'yes'. It is a Lie group. This is true for any compact complex manifold. Basically, this is because the equations for a holomorphic vector field on a compact manifold always have a finite dimensional space of solutions.</p> <p>Related to this is the question $(4)$: Is there any way to 'produce' automorphisms of $X$? If by this you mean some 'effective' way, you'll need to tell me how you are effectively describing the complex manifold $X$. Whether you can 'effectively' construct the biholomorphism group of $X$ depend on how much you know about $X$. For example, is it described by algebraic equations in some projective space?</p> http://mathoverflow.net/questions/124000/are-all-null-curves-of-a-lorentzian-metric-extrema/124080#124080 Answer by Robert Bryant for Are all null curves of a Lorentzian metric extrema? Robert Bryant 2013-03-09T14:49:12Z 2013-03-09T15:42:52Z <p>Actually, your notation is causing some confusion. In one very real sense (probably not your intended one) the answer to your question is <em>yes</em>, not <em>no</em> which is probably the answer to the question that you intended to ask. </p> <p>It depends on what you mean by $\|\dot\gamma(s)\|_g$. If you borrow this notation directly from the standard notation in Riemannian geometry, where $$\|\dot\gamma(s)\|_g = \sqrt{\langle \dot\gamma(s),\dot\gamma(s)\rangle_g}\ ,$$ what this often means in Lorentzian geometry is that you only consider curves $\gamma$ as <em>admissable</em> if this makes sense, i.e., only if $\langle \dot\gamma(s),\dot\gamma(s)\rangle_g\ge0$. Unfortunately, there is also a different convention in which $$\|\dot\gamma(s)\|_g = \sqrt{-\langle \dot\gamma(s),\dot\gamma(s)\rangle_g}\ ,$$ and one only considers curves $\gamma$ as <em>admissable</em> if <em>this</em> makes sense, i.e., only if $\langle \dot\gamma(s),\dot\gamma(s)\rangle_g\le0$. (It depends on the sign conventions one chooses, and, unfortunately, there seem to be as many sign conventions in use as there are choices of sign, each defended by its practitioners as the only sane convention, but that's another story.)</p> <p>At any rate, if you adopt one of the above conventions <em>and</em> you only consider admissable curves (which is standard in the Calculus of Variations), then, <em>yes</em>, all null curves are extrema, almost by definition; it's not a question of geodesics, which is a different matter.</p> <p>Sad to say, there is a <em>third</em> convention in pseudo-Riemannian geometry. Some books and papers define $$\|\dot\gamma(s)\|_g = \langle \dot\gamma(s),\dot\gamma(s)\rangle_g\ .$$ (They don't like the square that you would expect to see based on common mathematical convention because $\|\dot\gamma(s)\|_g^2$ suggests that the quantity is nonnegative, so they break the conventional distinction between 'norm' and 'inner product' instead. This often leads to dreadful confusion when one tries to read the papers and frequent mistakes involving the so-called Cauchy inequality, but that's another story, too.)</p> <p>If that is your convention, then the answer to your question is <em>no</em>, because, in this case, <em>all</em> sufficiently smooth curves are admissable, and all the extrema are geodesics. However, in dimensions higher than $2$ (in particular, in dimension $4$), not all null curves are geodesics, as the above commenters have noted.</p> <p>Oh, one more (pedantic, I'm afraid) comment: While it's true that all extrema are geodesics in this latter convention, I'm sure you know that not all geodesics are extrema (they <em>are</em> always critical points of the functional, but 'extrema' is a stronger condition). Moreover, in pseudo-Riemannian geometry, null geodesics (and, <em>a fortiori</em>, null curves) are <em>never</em> extrema, because you can always perturb the curve slightly to make the functional either positive or negative.</p> http://mathoverflow.net/questions/123633/characterizing-hessians-among-symmetric-bilinear-tensors/123644#123644 Answer by Robert Bryant for Characterizing Hessians among symmetric bilinear tensors Robert Bryant 2013-03-05T18:02:35Z 2013-03-07T16:10:49Z <p>There <em>are</em> local conditions, but they typically involve the curvature tensor of the underlying metric. For example, if the metric is flat, so that one can choose orthonormal coordinates $x_i$ in which the covariant derivatives are the ordinary derivatives, then the condition that a quadratic form $H = h_{ij}\ dx_idx_j$ be a Hessian is just that $$\frac{\partial h_{ij}}{\partial x_k} = \frac{\partial h_{ik}}{\partial x_j} \ .$$ This is an overdetermined system, and, if it is satisfied, there are solutions $f$ of $$\frac{\partial^2 f}{\partial x_i\partial x_j} = h_{ij}$$ and they are uniquely determined (locally) up to the addition of a function linear in the $x_i$.</p> <p>In the more general case, one has local conditions in terms of the curvature of $g$, so the answer is more complicated (and more interesting). I calculate using moving frames (see below at the end for the 'Nomizu-style' interpretation), so I would explain it this way: </p> <p>Let $g = {\omega_1}^2 +\cdots + {\omega_n}^2$ be the expression of $g$ in a local orthonormal coframing $\omega$. There will exist unique $1$-forms $\theta_{ij}=-\theta_{ji}$ so that (using the summation convention) $d\omega_i = -\theta_{ij}\wedge\omega_j$. (These are the first structure equations.)</p> <p>If $f$ is a function defined in the domain of the coframing, one will have $$df = f_i\ \omega_i \qquad\text{and}\qquad df_i = -\theta_{ij}\ f_j + f_{ij}\ \omega_j$$ and, by definition, <code>$\mathrm{Hess}_g(f) = f_{ij}\ \omega_i{\circ}\omega_j$</code>. </p> <p>To determine whether a given symmetric quadratic form $H = h_{ij}\ \omega_i{\circ}\omega_j$ can be written as $H = \mathrm{Hess}_g(f)$ for some $f$, one computes the exterior derivative of the equations $df_i = -\theta_{ij}\ f_j + h_{ij}\ \omega_j$ and finds that one must have $$h_{ikl}-h_{ilk} = -R_{ijkl}\ f_j\ , \tag{1}$$ where $dh_{ij} = -\theta_{ik}\ h_{kj} + \theta_{kj}\ h_{ik} + h_{ijk}\ \omega_k$ and the $R_{ijkl}=-R_{ijlk}$ are the components of the Riemann curvature tensor in this coframing, as defined by the second structure equations $$d\theta_{ij} = -\theta_{ik}\wedge\theta_{kj} + \tfrac12\ R_{ijkl}\ \omega_k\wedge\omega_l\ .$$</p> <p>The system (1) gives necessary conditions for $f$ to exist, since, in most cases, it completely determines the only possible functions $f_i$ that could be its derivatives in this coframing. </p> <p>It may be worth pausing to interpret (1) as a global equation. In Nomizu-style notation, one has $\mathrm{Hess}_g(f) = (\nabla\nabla f)^{\flat\flat}$, but I am going to ignore the distinction between $T=TM$ and $T^\ast=T^*M$ since we have a metric $g$ that is $\nabla$-parallel and just write $\mathrm{Hess}_g(f) = \nabla\nabla f$. Applying the Bianchi identities, one has $$\sigma\bigl(\nabla\bigl(\mathrm{Hess}_g(f)\bigr)\bigr) = \mathsf{R}(\nabla f),$$ where $\sigma:\mathsf{S}^2(T^\ast)\otimes T^\ast\to T^\ast\otimes\Lambda^2(T^\ast)$ is the canonical skew-symmetrization operator, and $\mathsf{R}:T\to T^\ast\otimes\Lambda^2(T^\ast)$ is the usual mapping induced by the curvature operator. In particular, if $H = \mathrm{Hess}_g(f)$, then one must have $$\sigma\bigl(\nabla H\bigr) = \mathsf{R}(\nabla f)\tag{1'}$$ which, in my moving frames notation, is the system (1). </p> <p>Now, the image of $\sigma$ is, as usual, the kernel of the further skew-symmetrization map $\sigma: T^\ast\otimes\Lambda^2(T^\ast)\to \Lambda^3(T^\ast)$ and, fortunately, the first Bianchi identity implies that $\mathsf{R}$ also takes values in this kernel, which has dimension $n\cdot \tfrac12 n(n{-}1) - \tfrac16 n(n{-}1)(n{-}2)= \tfrac13 n(n^2{-}1)$. </p> <p>We already know what happens when $\mathsf{R}\equiv0$, namely, the first order system of equations $\sigma\bigl(\nabla H\bigr)=0$ are necessary and sufficient that $H$ be locally expressible as a Hessian. </p> <p>However, the 'generic' situation is that $\mathsf{R}$ is injective, so let's consider that case. (I'll say a bit more about what happens in the intermediate cases at the end.) When $\mathsf{R}$ is injective, let $\mathsf{R}(T)\subset T^\ast\otimes\Lambda^2(T^\ast)$ be the image subbundle of rank $n$. The equations $(1')$ then say that $H$ must satisfy the system of $\tfrac13 n(n^2{-}1) - n = \tfrac13 n(n^2{-}4)$ first order equations $$\sigma(\nabla H) \equiv 0\ \text{modulo}\ \mathsf{R}(T).\tag{1''}$$ Assuming that these equations do hold, then there is a unique vector field $F(H)$ on $M$ such that $$\sigma(\nabla H) = \mathsf{R}\bigl(F(H)\bigr),$$ and this $F(H)$, which is a linear, first order differential expression in $H$ (whose coefficients involve the curvature of $g$), is the only possible candidate for $\nabla f$. However, in order for this to solve our problem, it must satisfy $$\nabla \bigl(F(H)\bigr) - H = 0,\tag{2}$$ and this is a system of $n^2$ first-order equations on $F(H)$, which is, of course, a system of $n^2$ second-order equations on $H$. If $H$ does satisfy this system, then, because $\nabla\bigl(F(H)\bigr)=H$ is symmetric, it will follow that $F(H)$ is (locally) a gradient vector field of some function $f$, uniquely determined up to an additive constant.</p> <p>For example, when $n=2$ and the Gauss curvature of $g$ is nowhere zero, the equations $(1'')$ are trivial since $\mathsf{R}(T) = T^*\otimes\Lambda^2(T^\ast)$. Thus, the result of this analysis is that there exists a linear, second order differential operator $\mathsf{S}_g$ from symmetric quadratic forms to quadratic forms, namely $\mathsf{S}_g(H) = \nabla\bigl(F(H)\bigr)-H$, such that a symmetric quadratic form $H$ is (locally) a Hessian with respect to $g$ if and only if $\mathsf{S}_g(H)=0$. Moreover, when the first deRham cohomology group of $M$ vanishes and $\mathsf{S}_g(H)=0$, there will exist a global function $f$ such that $\mathrm{Hess}_g(f)=H$, and this $f$ is unique up to an additive constant.</p> <p>When $n>2$, the equations $(1'')$ are never trivial, so the condition that $H$ be a Hessian with respect to $g$ involves first order conditions and second order conditions. What is interesting is that, when $n$ is sufficiently large and $\mathsf{R}$ is sufficiently 'generic', it appears (I haven't checked for sure) that $(1'')$ can actually imply $(2)$, so that the conditions become first order again. (This is not true when $n=3$, though.)</p> <p>Finally, if $\mathsf{R}$ is not injective, one may have to go to higher order derivatives of $H$ to determine whether it is a Hessian. This is an interesting case, but I don't have time to go into it right now. </p> <p><em>Remark:</em> Of course, the metric $g$ is not really essential in this story, since it really depends more on $\nabla$, than on $g$. The equation $\nabla(df) = H$ makes sense without any metric and is a reasonable equation, as long as $\nabla$ is a torsion-free connection on $M$. Thus, one could ask for characterizations of those symmetric quadratic differentials $H$ that can be written in the form $H = \nabla(df)$ where $\nabla$ is a given torsion-free connection on $M$. The answer in this case would have essentially the same form as the answer above since, if one carries out the computations carefully, one sees that the metric $g$ plays essentially no rôle in the problem.</p> http://mathoverflow.net/questions/123765/symplectic-block-diagonalization-of-a-real-symmetric-hamiltonian-matrix/123784#123784 Answer by Robert Bryant for Symplectic block-diagonalization of a real symmetric Hamiltonian matrix Robert Bryant 2013-03-06T16:43:39Z 2013-03-06T16:43:39Z <p>Yes, you can always do this. The right way to understand this is to look at the complex symmetric matrix $X = x + i y$. You want to act on this by a complex matrix $C = c + is$ where $C$ is unitary, by the rule, $X\mapsto C^{T}XC$, and you want to know whether you can arrange that $C^{T}XC$ is a real, diagonal matrix. </p> <p>You can do this, in two stages: First, look at the Hermitian symmetric matrix $H = {\bar X} X = {\bar H}^T$. When you act by $C$, the matrix $H$ will transform as $H\mapsto {\bar C}^{T}HC$. Thus, one can choose $C$ to that $H$ is diagonal and hence real. In particular, one is reduced to the case in which $X=x+iy$ satisfies $xy-yx=0$. Thus, $x$ and $y$ are commuting symmetric matrices and hence they can be simultaneously diagonalized by a real orthogonal matrix. Thus, you can now assume that $x$ and $y$ are diagonal.</p> <p>Second, let $C$ be diagonal and unitary and let it act on a diagonal $X$ by $X\mapsto C^{T}XC$. Since the diagonal elements of $C$ are unit complex numbers, we can clearly choose them so that the entries of $C^{T}XC$ are real and nonnegative, which is what you wanted.</p> http://mathoverflow.net/questions/122736/kahler-metric-fundamental-forms-and-cohomology-ring-generators/123628#123628 Answer by Robert Bryant for Kahler Metric Fundamental Forms and Cohomology Ring Generators Robert Bryant 2013-03-05T16:15:57Z 2013-03-05T16:15:57Z <p>If you look at the second question, the OP is asking when the cohomology ring can be generated by some <em>set</em> of Kähler forms, not by a <em>single</em> Kähler form. Thus, while the variety $F$ of full flags in $\mathbb{C}^3$ has two generators, say $\omega_1$ and $\omega_2$ that are <em>not</em> Kähler forms, the forms $\eta_1 = 2\ \omega_1+\omega_2$ and $\eta_2=\omega_1+2\ \omega_2$ are both positive and hence Kähler forms. They generate the cohomology ring over the rationals.</p> <p>In general, if the space is of the form $X = G/P$ where $G$ is complex semi-simple and $P$ is a minimal parabolic, then $X = U/T$ where $U$ is a compact form of $G$ and $T = U\cap P$ is a maximal torus in $U$. The cohomology ring of $X$ will be generated in degree $2$ by a vector space $V$ of $U$-invariant $(1,1)$-forms, and one can find, in the positive cone in this vector space, a basis of generators of the cohomology ring of $X$ (over the rationals), and these generators will all be Kähler forms.</p> <p>Thus, for example, when $F$ is the full flag variety of $\mathbb{C}^n$, $G=\mathrm{SL}(n,\mathbb{C})$, $P$ is the upper triangular elements in $G$, $U=\mathrm{SU}(n)$, and $K=\mathbb{T}^{n-1}$ is the maximal torus of diagonal elements in $G$. The vector space $V$ has dimension $n{-}1$ and there is a positive 'orthant' in $V$ that consists of Kähler forms that are $\mathrm{SU}(n)$-invariant. These suffice to generate (rationally) the cohomology of $F$ (with, of course, some relations).</p> <p>Obviously, the case of minimal parabolics is not the only case in which this happens, but I'd have to think about this more to 'remember' the criterion, which I think I did know, at one time.</p> http://mathoverflow.net/questions/37651/riemannian-surfaces-with-an-explicit-distance-function/61846#61846 Answer by Robert Bryant for Riemannian surfaces with an explicit distance function? Robert Bryant 2011-04-15T17:48:32Z 2013-03-03T13:35:20Z <p><strong>NB:</strong> <em>For some time, I have been meaning to revise this answer to make it more complete (and, to be frank, more accurate), but I never found the time to do it. The main point is that my original answer did not take into account the difference between the cut locus and the conjugate locus, and, of course, this affects the formula for the distance between points.</em></p> <p>I'm aware of a few metrics with non-constant curvature for which one can write the distance function explicitly in terms of the coordinates. The simplest such metric I know is the (incomplete) metric $ds^2 = y\ (dx^2+dy^2)$ on the upper half plane $y>0$. The Gauss curvature of this metric is $K = 1/(2y^3)>0$, so it's not constant. </p> <p>Every geodesic of this metric in the upper half plane can be parametrized in the form $$x = a + b\ t\qquad\qquad y = b^2 + \frac{t^2}{4}$$ for some constants $a$ and $b$, and, for such a geodesic, the arclength function along the curve is $$s = c + b^2\ t + \frac{t^3}{12}\ .$$ for some constant $c$. </p> <p>Using these formulae, one finds that two points $(x_1,y_1)$ and $(x_2,y_2)$ are joinable by a geodesic segment if and only if $4y_1y_2 \ge (x_1{-}x_2)^2$. In the case of strict inequality, there are two geodesic segments joining the two points, and the length of the shorter segment is $$L_1\bigl((x_1,y_1),(x_2,y_2)\bigr) = {1\over3}\sqrt{3(x_1{-}x_2)^2(y_1{+}y_2)+4(y_1^3{+}y_2^3) - (4y_1y_2-(x_1{-}x_2)^2)^{3/2}}\ .$$ Note that, in a sense, this is better than the constant curvature case. Here, the distance function is algebraic in suitable coordinates, whereas, in the constant nonzero curvature cases, the distance function is not.</p> <p><em>However,</em> the function $L_1$ does not necessarily give the actual distance between the two points (i.e., the infinimum of the lengths of curves joining the two points), and it's not only because not every pair of points can be joined by a geodesic. To see this, one should complete the upper half plane by adding a point that represents the 'boundary' $y=0$. The Riemannian metric does not extend smoothly across this 'point', of course (after all, the Gauss curvature blows up at you approach this point), but it does extend as a metric space. The vertical lines, which are geodesics, can then be used to join $(x_1,y_1)$ to $(x_2,y_2)$ by going through the singular point, and the total length of this geodesic is $$L_2\bigl((x_1,y_1),(x_2,y_2)\bigr) = \frac{2}{3}\bigl({y_1}^{3/2}+{y_2}^{3/2}\bigr).$$ (Also, note that $L_2$ is defined for <em>any</em> pair of points in the upper half-plane.) If one doesn't like this path that goes through the singular point, one can easily perturb it slightly to avoid the singular point and not increase the length by much, so it's clear that the infimum of lengths of curves lying strictly in the upper half plane and joining the two points is no more than $L_2$. </p> <p>This suggests that the true distance function $L$ should be the minimum of $L_1$ and $L_2$ where they are both defined, i.e., where $4y_1y_2 \ge (x_1{-}x_2)^2$, and $L_2$ on the set where $4y_1y_2 &lt; (x_1{-}x_2)^2$. </p> <p>To get a sense of how these two formulae interact, one can use the fact that $x$-translation preserves the metric while the scalings $(x,y)\mapsto (ax,ay)$ for $a>0$ preserve the metric up to a homothety (and hence preserve the geodesics and scale the distances). These two actions generate a transitive group on the upper half plane, so, it suffices to see how these two functions interact when $(x_1,y_1) = (0,1)$, i.e., to see the conjugate locus and cut locus of this point.</p> <p>The conjugate locus is easy: It's just $y-x^2/4=0$, which is the boundary of the region $y-x^2/4\ge0$ consisting of the points that can be joined to $(0,1)$ by a geodesic segment. Meanwhile, the cut locus is given by points $(x,y)$ that satisfy $y-x^2/4\ge0$ and for which $L_1\bigl((0,1),(x,y)\bigr) = L_2\bigl((0,1),(x,y)\bigr)$. In fact, one has $L_1\bigl((0,1),(x,y)\bigr) &lt; L_2\bigl((0,1),(x,y)\bigr)$ only when $y > f(x)$, where $f$ is a certain even algebraic function of $x$ that satisfies $f(x) \ge x^2/4$ (with equality only when $x=0$). Moreover, for $|x|$ small, one has $$f(x) = \left({\frac{{\sqrt{3}}}{4}}x\right)^{4/3} + O(x^2)$$ while, for $|x|$ large, one has $$f(x) = \left({\frac{\sqrt{3}}{4}}x\right)^{4} + o(x^4).$$</p> <p>Thus, all of the geodesics leaving $(x,y)=(0,1)$, other than the vertical ones, meet the cut locus <em>before</em> they reach the conjugate locus (and they all do meet the conjugate locus).</p> <p>Thus, the actual distance function for this metric is explicit (it's essentially the minimum of $L_1$ and $L_2$), but it is only semi-algebraic.</p> <p><em>Remark:</em> The thing that makes this work is that, while the metric has only a 1-parameter family of symmetries, it has a 2-parameter family of homotheties (as described above), and this extra symmetry of the geodesics is critical for making this work. Of course, there are other such metrics, all the ones of the form $ds^2 = y^{a}\ (dx^2+dy^2)$ ($a$ is a constant) have this property and don't have constant curvature unless $a = 0$ or $a = -2$. You don't get algebraic answers for all values of $a$, of course, but there is a way to get $D$ implicitly defined in terms of a special function (depending on the value of $a$).</p> <p>More generally, the metrics whose geodesics admit more symmetries than the metric itself does tend to have such formulae. I'm not aware of any other cases in which one can get $D$ so explicitly. </p> http://mathoverflow.net/questions/130959/reference-request-affine-transforms-circle-inversion/131005#131005 Comment by Robert Bryant Robert Bryant 2013-05-19T11:40:45Z 2013-05-19T11:40:45Z (cont) Let $\frak{l}$ be the union of the ${\frak{l}}_i$. Then any $m$-jet of a vector field on $S^n$ is the $m$-jet of some vector field in $\frak{l}$. Since, under the conformal group, any $k$ points are equivalent to $k$ points that lie within some $\epsilon$-ball of a given point, it may be possible to use this 'jet density' to prove that the orbit under the non-Lie group of any $k$-tuple of distinct points in $S^n$ is open in $(S^n)^k$. Since the compliment of the 'diagonals' in $(S^n)^k$ is connected (assuming that $n&gt;1$), it would follow that there is only one such orbit. http://mathoverflow.net/questions/130959/reference-request-affine-transforms-circle-inversion/131005#131005 Comment by Robert Bryant Robert Bryant 2013-05-19T11:30:12Z 2013-05-19T11:30:12Z @Misha: I don't know the answer to your question above about $k$-transitivity for this latter (non-Lie) group, but, like you, I suspect that it is $k$-transitive for all $k$. Aside from the obvious observation that this holds for $k\le3$, I don't have a proof in hand, but, perhaps a proof could be devised along the following lines: Let ${\frak{l}}_0={\frak{so}}(n{+}1,1)+{\frak{sl}}(n{+}1)$ be the (finite-dimensional) space of vector fields on $S^n$ belonging to the above two Lie transformation groups and, for $i\ge0$, define ${\frak{l}}_{i+1}={\frak{l}}_i+[{\frak{l}}_i,{\frak{l}}_i]$. (cont) http://mathoverflow.net/questions/130959/reference-request-affine-transforms-circle-inversion/131005#131005 Comment by Robert Bryant Robert Bryant 2013-05-18T13:30:17Z 2013-05-18T13:30:17Z @Misha: It's not clear to me which group you mean by &quot;conformal+projective&quot;. The group $\mathcal{T}$, as defined by the OP, is not smooth on $S^n=\mathbb{R}^n\cup\lbrace\infty\rbrace$ when $n&gt;1$ because the non-conformal affine transformations don't extend smoothly to $\infty$. Are you asking instead about the (non-Lie) group generated by the two (maximal) proper Lie group extensions of $\mathrm{SO}(n{+}1)$ acting on $$S^n=\bigl(\mathbb{R}^{n+1}\setminus\lbrace0\rbrace\bigr) /\mathbb{R}^+,$$ i.e., $\mathrm{SO}(n{+}1,1)$ and $\mathrm{SL}(n{+}1)$? http://mathoverflow.net/questions/130904/can-one-block-diagonalize-the-curvature-matrix-of-2-forms-on-a-riemannian-manif/130908#130908 Comment by Robert Bryant Robert Bryant 2013-05-17T22:24:09Z 2013-05-17T22:24:09Z @Anton: Yes, but more than just having rank at most $\lfloor \frac{n}{2}\rfloor$, the curvature must take values in an <i>abelian</i> subspace of the endomorphism algebra. By the way, just because the curvature 'looks like' that of a product, that doesn't mean that the metric <i>is</i> a product. For example, the space of Riemannian metrics in dimension $3$ whose curvature operators have rank at most $1$ everywhere depends (up to diffeomorphism) on $3$ functions of $2$ variables, while the products of surfaces with lines depend (up to diffeomorphism) only on $1$ function of $2$ variables. http://mathoverflow.net/questions/130904/can-one-block-diagonalize-the-curvature-matrix-of-2-forms-on-a-riemannian-manif/130913#130913 Comment by Robert Bryant Robert Bryant 2013-05-17T12:28:22Z 2013-05-17T12:28:22Z @Deane: I think the OP is asking a different question, which is whether the curvature operator at each point must take values in a maximal torus in the Lie algebra of skew-symmetric endomorphisms of the tangent space. This is not true generally even in dimension $3$, where the condition would be that the curvature operator has rank $1$ at each point. In fact, for the generic metric at the generic point, the curvature operator is surjective onto the skew-symmetric endomorphisms of the tangent space, so this doesn't work for $n&gt;2$. http://mathoverflow.net/questions/130438/invariants-of-a-gl3-mathbbr-action/130497#130497 Comment by Robert Bryant Robert Bryant 2013-05-15T13:26:33Z 2013-05-15T13:26:33Z @Jeanne: I haven't thought about computational aspects, but I do believe that you are right that actually finding the normal form of a nonsingular $Q$ is highly nontrivial. It is known that finding the symmetric normal form of a nonsingular cubic in $3$ variables is computationally hard, and that is the hardest step in this normalization process as well. I'm not sure what reference to give for this, but it is a well-studied problem in the theory of elliptic curves, so if there were a known algorithm that dramatically improved over brute force, it would be well-known. Sorry I can't help more. http://mathoverflow.net/questions/130590/laplacian-type-operator-on-compact-lie-group Comment by Robert Bryant Robert Bryant 2013-05-14T19:35:02Z 2013-05-14T19:35:02Z Since $X$ commutes with the Laplacian on $\mathrm{SO}(n{+}1)$, which is $-S$, it must preserve the eigenspaces of $-S$ on $L^2$, which are given by the matrix coefficients of the irreducible representations of $\mathrm{SO}(n{+}1)$, by the Peter-Weyl Theorem. You just need to know the (imaginary) eigenvalues of $X$ in each of these irreducible representations (and the eigenvalue of $-S$ on that representation), and this will answer your question. This is a routine computation. Look in any book on the representation theory of Lie groups, such as Knapp's &quot;Lie groups: Beyond an introduction&quot;. http://mathoverflow.net/questions/130438/invariants-of-a-gl3-mathbbr-action/130497#130497 Comment by Robert Bryant Robert Bryant 2013-05-13T23:40:20Z 2013-05-13T23:40:20Z You're welcome. By the way, is MathJax working for you on MO? I just realized that it hasn't worked for me since this morning when I input this answer and, moreover, it doesn't seem to be working on my laptop (also a Mac) either, under any browser I have installed. Hmmm. http://mathoverflow.net/questions/37651/riemannian-surfaces-with-an-explicit-distance-function/123876#123876 Comment by Robert Bryant Robert Bryant 2013-05-10T23:58:51Z 2013-05-10T23:58:51Z Also: The metrics of the form $ds^2 = r^{2\beta}(dx^2+dy^2)$ are essentially all the same, since they are locally flat. When $\beta=-1$, one can write $x{+}iy = e^w$, where $w=u{+}iv$ is a complex coordinate, and one has $ds^2 = du^2+dv^2$. When $\beta\not=-1$, one can write $$x+iy = \bigl((1{+}\beta)(u{+}iv)\bigr)^{1/(1{+}\beta)}$$ and, again, one has $ds^2 = du^2+dv^2$. Thus, it is not surprising that one can describe the geodesics (and distances) completely in these cases, for they are (quotients of open subsets of) the plane in disguise. http://mathoverflow.net/questions/37651/riemannian-surfaces-with-an-explicit-distance-function/123876#123876 Comment by Robert Bryant Robert Bryant 2013-05-10T14:48:28Z 2013-05-10T14:48:28Z Remember that the request was for an explicit formula for the <i>distance function</i>, which is considerably harder than finding explicit integrals of the geodesic equations. There are many known metrics with explicit geodesics; the surfaces of revolution and, in particular, the metrics you list (which are characterized by having a homothetic symmetry group of dimension at least 2) are special cases of a much wider class of such, which includes the Liouville metrics (which often have no symmetries). For nearly all of these, including yours, the distance function cannot be written down explicitly. http://mathoverflow.net/questions/130037/as3-closest-point-to-cubic-bezier Comment by Robert Bryant Robert Bryant 2013-05-08T17:12:52Z 2013-05-08T17:12:52Z @unknown (google): The reason it's off-topic is that there is no mathematics research involved. Finding the closest point on a parametrized plane curve to a given point in the plane reduces immediately to an algebra problem (where is the curve tangent perpendicular to the vector pointing to the given point?), so there's no research problem involved, which is what MO is for. By the way, when the curve is a rational cubic (as in your case), the algebra problem to be solved is (usually) a 5th order polynomial equation that has no solution in radicals. (Numerical algorithms exist, though.) http://mathoverflow.net/questions/129365/normal-forms-for-homogeneous-cubic-polynomials-in-mathbbrx-1-x-2-x-3/129404#129404 Comment by Robert Bryant Robert Bryant 2013-05-06T10:59:23Z 2013-05-06T10:59:23Z You're welcome Jeanne. Probably, I should have mentioned that the reason for choosing the symmetric normal form for nonsingular cubics as I did above (rather than the more common Weierstrass normal form) is that this normal form displays the fact that the projective symmetry group of a (real) nonsingular planar cubic curve is the symmetric group on $3$ letters. This fact is not so obvious in the Weierstrass normal form. http://mathoverflow.net/questions/129282/about-kaehlerian-immersions Comment by Robert Bryant Robert Bryant 2013-05-06T00:06:53Z 2013-05-06T00:06:53Z Have you taken into account that Cortes's 'K&#228;hler' manifolds are actually pseudo-K&#228;hler? The point is that the ambient metric is not positive definite; in fact, it is of split type in his case, so there is no problem finding complex submanifolds on which the symplectic form vanishes (i.e., that are Lagrangian). Also, 'K&#228;hlerian immersion' is not quite the same as 'totally complex', because of the requirement that the metric pull back to be nondegenerate. If the ambient metric <i>were</i> positive definite, then, of course, one could not have submanifolds that are both complex and Lagrangian. http://mathoverflow.net/questions/129538/laplacian-on-coset-spaces Comment by Robert Bryant Robert Bryant 2013-05-04T11:53:38Z 2013-05-04T11:53:38Z @whatever: as J. Martel says, 'rotation about the $x_i$-axis' is not well-defined except in dimension $3$. Presumably, in the general case, you want to consider the vector fields $X_{ij}=-X_{ji}$ (which generates counterclockwise rotation in the oriented $2$-plane with basis $e_i$ and $e_j$ and has period $2\pi$) and fixes the orthogonal $(n{-}2)$-plane. (When $n=3$, $X_{12}$ can be interpreted as rotation about the $x_3$-axis, etc.) The $X_{ij}$ for $i&lt;j$ give a basis for the Lie algebra of $\mathrm{SO}(n{+}1)$. http://mathoverflow.net/questions/129150/derham-cohomology-of-sn-without-mayer-vietoris/129153#129153 Comment by Robert Bryant Robert Bryant 2013-04-30T09:01:31Z 2013-04-30T09:01:31Z @Claudio: This argument is actually a standard one, just adapted to this special case. I think it may have been originally due to Cartan. The general fact is that if $G$ is a connected, compact Lie group acting on a compact manifold $M$, then the ring of $G$-invariant differential forms on $M$ (which is closed under exterior derivative) has its $d$-cohomology, which is isomorphic to the deRham cohomology of the full ring of differential forms on $M$. If $G$ acts transitively on $M=G/H$, then this invariant subring is finite dimensional and can be computed algebraically. See Spivak for details.