User macbeth - MathOverflow

"Mathai-Quillen-type" form on $M\times M$?

2013-02-27T15:02:37Z

Let $(M,g)$ be a compact, oriented, $(2n)$-dimensional Riemannian manifold. I'm wondering whether there is a "canonical" construction of a $(2n)$-form $\eta_g$ on $M\times M$, such that

$\eta_g$ is a de Rham representative of the Poincare dual of the diagonal;
The pullback under the diagonal inclusion $\iota:M\to M\times M$ is proportional to the curvature Pfaffian: $\iota^*\eta_g=\tfrac{1}{(2\pi)^n}\text{Pf}(\text{Rm})$.

There are plenty of "non-canonical" such forms, I think. For instance, pick a neighbourhood of the diagonal which is diffeomorphic to $TM$, and transfer the Mathai-Quillen Thom form on $TM$ to this neighbourhood using a suitable diffeomorphism. (By the way, the Mathai-Quillen Thom form on the total space of a bundle-with-metric-and-connection is the model I have in mind here for what "canonical" should mean -- a form whose value at each point depends only on local invariants.)

Motivation: Such a form would yield a proof of the Chern-Gauss-Bonnet theorem which is both quick and natural. Namely,

$\chi(M)=PD(\Delta)\cup PD(\Delta)=\int_\Delta \eta_g =\tfrac{1}{(2\pi)^n}\int_M \text{Pf}(\text{Rm})$.

Dolbeault cohomology of Hopf manifolds

2010-05-23T21:51:40Z

This should be straightforward; I'm sorry if it's too much so. Can someone point me to a reference which computes the Dolbeault cohomology of the Hopf manifolds?

Motivation: I'd like to work through a concrete example of the Hodge decomposition theorem failing for non-Kähler manifolds. The textbook I have handy (Griffiths & Harris) doesn't treat this, and the obvious Google search was unhelpful.

Answer by macbeth for Constant scalar curvature metrics in a conformal class

2012-06-22T14:36:04Z

Plenty is known! For instance,

(Schoen 1989) For any $N$, there exists a product of round spheres whose conformal class' set of such constants has size at least $N$.
(Brendle-Marques 2009, based on earlier work of Brendle) In any dimension $n\geq 25$, there exist conformal classes on $S^n$ for which the set of such constants is infinite. Specifically, there exists a sequence of them tending upwards to the Yamabe constant of the round sphere.
(Khuri-Marques-Schoen 2009) For any spin manifold $M$ of dimension $\leq 24$, for any conformal class and any real $c$, the set of metrics with constant scalar curvature $c$ is compact in the $\mathcal{C}^2$ topology.

For a recent survey see this article of Brendle-Marques.

Can a PDE constrain the degree of a $C^\infty$ map germ?

2012-03-14T00:26:42Z

Let $\pi:E\to M$ be a smooth vector bundle over a smooth manifold, with $\text{rank}(E)=\text{dim}(M)$. For a section $\sigma$ of $E$ with a zero at $p\in M$, define the degree of the zero at $p$ to be the topological degree of the induced map from a small sphere in $T_pM$ to a small sphere in $E_p$.

One motivation for studying degrees of zeros is that they contain information about the topology of $E$. I think the following is true, although I couldn't find a good reference:

Theorem 1 (Hopf index theorem). Suppose the zeroes of $\sigma$ are the isolated points $p_1, \ldots p_k$, with degrees $d_1,\ldots d_k$ respectively. Then the Euler class of $E$ is $\chi(E)=\sum_{i=1}^kd_i$.

With this as motivation, my first question, the one stated in the title, is roughly (see the Example for an idea of what I'm getting at, and feel free to suggest a sharper version):

Are there conditions on [say, the symbol of] a linear differential operator $D:E\to F$, such that [some constraint] is satisfied by degree of any zero $p\in M$ of any local solution $\sigma\in\Gamma(E)$ to the PDE $D\sigma=0$?

Example. If $M$ is a Riemann surface and $E$ a holomorphic line bundle over it, the kernel of the delbar operator $\overline{\partial}:E\to T^{0,1}M\otimes E$ is precisely the holomorphic sections of $E$. By complex analysis, zeroes of holomorphic functions have positive degree.

Theorem 1 then yields the standard result that if a line bundle admits a global holomorphic section then its Euler class (aka first Chern class) is nonnegative.

Here's an idea I had for trying to prove a theorem of the sort I ask for in Question 1. Recall the definition of the local ring of a zero $p\in M$ of a section of E:

Write $\mathcal{O}_p$ for the ring of germs of smooth functions about $p$.

Definition. Let $\sigma\in\Gamma(E)$ be a smooth section which vanishes at $p$. The local ring of the germ $[\sigma]_p$, denoted $Q([\sigma]_p)$, is the quotient $\mathcal{O}_p/([\sigma]_p)$, where $([\sigma]_p)$ is the ideal of $\mathcal{O}_p$ generated by "components of $\sigma$": $([\sigma]_p)=\ <\{[v(\sigma)]_p:v\text{ a nonvanishing section of }E^*\}> \ \subseteq \mathcal{O}_p$ .

Theorem 2 (Eisenbud-Levine-Khimshiashvili). Suppose $p$ is a zero of $\sigma$, and the local ring $Q([\sigma]_p)$ is a finite-dimensional algebra over $\mathbb{R}$. Then there is a canonical quadratic form on $Q([\sigma]_p)$, such that the degree of the zero of $\sigma$ at $p$ can be calculated as this quadratic form's signature.

Because a system of PDE is precisely a constraint on the local behaviour of a section, it seems plausible that local rings of zeros of solutions of a PDE might have interesting properties.

Are there conditions on [say, the symbol of] a linear differential operator $D:E\to F$, such that [some constraint] is satisfied by the signature of the local ring $Q([\sigma]_p)$ of any zero $p\in M$ of any local solution $\sigma\in\Gamma(E)$ to the PDE $D\sigma=0$?

Example. As in the previous example, let $E$ be a holomorphic line bundle over a Riemann surface $M$. By manipulating the Cauchy-Riemann equations, one can (I think!) classify the possible local rings of zeroes of a holomorphic section, and show that all of them have positive signature.

Theorem 2 then yields an alternative proof of the quoted result that zeroes of holomorphic functions have positive degree.

Classification of symplectic surfaces

2009-12-26T23:14:15Z

Is there a classification of symplectic surfaces, i.e. of surfaces equipped with an area form? Symplectic topology references like McDuff-Salamon seem to start their discussion of open questions with dimension four.

A surface admits a symplectic form iff it is orientable.
The Moser trick seems to show that on a compact orientable surface $M$, the unique invariant of a symplectic form is its total area? So the set of symplectic forms on $M$ (up to symplectomorphism) is parametrized by $\mathbb{R}^+$?
And, for non-compact orientable surfaces, ...?

Seek "typical examples" for the structure of spaces with two-sided Ricci bounds

2012-02-07T17:14:39Z

By a 1990 paper of Michael Anderson, the following is true:

Theorem. Let the metric space $(X,d,p)$ be a pointed Gromov-Hausdorff limit of a sequence of complete pointed Riemannian manifolds $(M_i,g_i, p_i)$ satisfying the uniform bounds

[two-sided Ricci] $|Ric(g_i)|\leq \lambda$
[$L^{n/2}$ norm of curvature] $\int\limits_M |Rm(g_i)|^{n/2} \leq \Lambda$
[non-collapsing] for some $r_0$, $\inf\limits_{q\in M_i, r\leq r_0} \ vol(B(q,r))/r^n\geq v$.

Then $(X,d,p)$ is (topologically) a smooth orbifold, whose singularities are of the form $\mathbb{R}^n/\Gamma$ for some finite $\Gamma\subseteq SO(n)$.

I'm trying to get a sense of how sharp this statement is. Here's a possible converse.

Question. Suppose I have a spherical space form $S^{n-1}/\Gamma$, and a $n$-manifold $M$ with boundary $\partial M = S^{n-1}/\Gamma$. Can I construct a sequence of complete Riemannian metrics on $M$'s interior, all satisfying the bounds above, such that some pointed Gromov-Hausdorff limit is homeomorphic to $\mathbb{R}^n/\Gamma$?

A few remarks on what I know: I am aware of some standard constructions for particular $\Gamma$, such as the ALE hyperkähler metrics for $\Gamma\subseteq SU(2)$. However, I don't know any good general ways to construct a family of metrics satisfying two-sided Ricci bounds.

Length spaces with continuous length functional: is this set Gromov-Hausdorff closed?

2011-12-20T17:02:34Z

As far as I can tell, a major motivation for the study of length spaces is that they arise as Gromov-Hausdorff limits of Riemannian manifolds. Specifically,

A complete connected Riemannian manifold is a complete length space.
A Gromov-Hausdorff limit of complete length spaces is a complete length space.

But of course there are stronger metric properties of Riemannian manifolds that one might hope would carry over to their limits. One that I have been wondering about is the continuity (say in the ~~compact-open topology~~ EDIT (see below): some other topology) of the length functional. After a couple of days' thought I've decided I have absolutely no intuition for this. So, I'd be very glad to hear:

Is the length functional of a complete connected Riemannian manifold indeed continuous? (Proof in some special cases: if $\Gamma:[0,1]\times (-\epsilon,\epsilon)\to M$ is continuously differentiable, then $$ \lim_{t\to 0}\int_0^1|\frac{\partial\Gamma}{\partial s}(s,t)|ds = \int_0^1|\frac{\partial\Gamma}{\partial s}(s,0)|ds $$ by limit-swapping.)
Is a Gromov-Hausdorff limit of complete-length-spaces-with-continuous-length-functional also a complete-length-space-with-continuous-length-functional?

EDIT: It was quickly pointed out by Anton Petrunin, Pietro Majer and Vitali Kapovitch that for the compact-open topology, the answer to the first question is no (and that the second question is vacuous). Is it possible that there is some finer topology on (perhaps some subspace of) the space of curves in a length space, for which the answer to these questions is yes?

For instance, consider the following property that a length space $X$ (with length functional $\mathcal{L}$) might possess:

For any Lipschitz map $\Gamma:[0,1]\times(-\epsilon,\epsilon)\to X$, $$ \lim_{t\to 0}\ \mathcal{L}(\Gamma(\cdot,t))=\mathcal{L}(\Gamma(\cdot,0)). $$

It seems plausible to me that this would be true of complete connected Riemannian manifolds and that it would not be true of arbitrary length spaces. Is this so? And if so, is the set of length spaces which do have this property Gromov-Hausdorff closed?

Answer by macbeth for On a compact manifold, what kind of function can be the Jacobian of a diffeomorphism?

2011-12-03T23:06:53Z

Here is an "answer-version" of my comment:

Yes, this is true in general. The reference I know is Moser's 1965 paper "On the volume elements on a manifold" (http://www.jstor.org/stable/1994022).

Specifically, let $M$ be a compact connected orientable manifold, and let $\sigma$ and $\tau$ be smooth volume forms on $M$ both with integral 1. Then there exists a diffeomorphism $\varphi:M\to M$ such that $\varphi^*\tau=\sigma$.

The orientability hypothesis isn't really necessary (just use densities rather than volume forms; see Moser's footnote (2)).

Answer by macbeth for Vocabulary of 19th Century analytic projective geometry: What are "order" and "dimension"?

2011-11-22T23:51:06Z

As Francesco Polizzi remarks, the idea is to identify $\text{Sym}^2(\mathbb{P}^2)$ with a cubic hypersurface in $\mathbb{P}^5$. Frege's lecture (see link in comments) actually goes on to explain how this is done. The cubic hypersurface is (in the six variables $s_1,s_2,s_3,t_1,t_2,t_3$),

$\text{det} \begin{pmatrix} s_1 & t_3 & t_2 \cr t_3 & s_2 & t_1 \cr t_2 & t_1 & s_3 \end{pmatrix}=0$

and the identification of $\text{Sym}^2(\mathbb{P}^2)$ with this hypersurface is

$\{[x_1:x_2:x_3],[y_1:y_2:y_3]\}\mapsto [x_1y_1:x_2y_2:x_3y_3:\frac{1}{2}(x_2y_3+x_3y_2):\frac{1}{2}(x_3y_1+x_1y_3):\frac{1}{2}(x_1y_2+x_2y_1)]$ .

Quite possibly, someone more knowledgeable than me can motivate this!

Answer by macbeth for Cayley-Menger Curvature and "Flatness" for inscribed simplices in hypersurface

2011-11-17T19:25:03Z

The first question is cool, but unfortunately the answer's boring: The limit in general will not exist.

Counterexample: Consider the conic hypersurface $z=\lambda x^2+\mu y^2$. There are two symmetrical "1-parameter families of quadruples-of-points" for which the limit of the circumradius as $\epsilon\to 0$ is easy to calculate:

$\{(0,0,0), (\epsilon,0,\lambda\epsilon^2),(-\epsilon,0,\lambda\epsilon^2),(0,\epsilon,\mu\epsilon^2)\}$ and $\{(0,0,0), (\epsilon,0,\lambda\epsilon^2),(0,\epsilon,\mu\epsilon^2),(0,-\epsilon,\mu\epsilon^2)\}$ .

For the first, the limit of the circumradius is $\frac{1}{2\lambda}$; for the second, it is $\frac{1}{2\mu}$.

However, here's a weaker statement that is true:

Fact: Let $\Sigma$ be a convex hypersurface of $\mathbb{R}^n$. Then $\frac{1}{\limsup}$ and $\frac{1}{\liminf}$ of the circumradius of an $(n+1)$-tuple of points, are the minimal and maximal [absolute-values-of-] principal curvatures respectively.

Answer by macbeth for Riemann surfaces with bounded curvature

2011-11-16T14:01:06Z

Whatever notion of limit you're using, you need a few more things in your "limit set." Consider the sequence of flat tori $\mathbb{R}^2/\Lambda_n$, where $\Lambda_n$ is the lattice generated by $(0,n)$ and $(1/n,0)$. We have uniform bounds 0 on curvature and 1 on area. However,

The pointed Gromov-Hausdorff limit is a line.
Pulling back by the diffeomorphisms $(x,y)\mapsto (nx, 1/n y)$, we get a sequence of metrics $n^2dx^2+1/n^2dy^2$ all on the "same" torus $\mathbb{R}^2/\Lambda_1$, satisfying the above and also your extra assumption of having the same area element. These metric tensors have no ($\mathcal{W}^{k,p}$, say) limit.

$\mathbb{Z}/2$ is to $\mathbb{Z}/3$ as K3 is to what?

2011-11-15T19:50:45Z

I'd like to know "what" (say, in the classification of complex surfaces) the following complex manifold $X$ is:

Construction: Let $\Lambda$ be the hexagonal lattice in $\mathbb{C}$; that is, the lattice generated by $(1, \omega)$ where $\omega=e^{2i\pi/3}$ is a third root of unity.

Observe that the lattice $\Lambda^2\subseteq \mathbb{C}^2$ is invariant under the $\mathbb{Z}/3\subseteq SU(2)$ action $\begin{pmatrix} \omega & 0 \cr 0 & \omega^2 \end{pmatrix}$.

I'm curious about the complex manifold $X$ obtained by quotienting $\mathbb{C}^2/\Lambda^2$ by $\mathbb{Z}/3$, and then blowing up at the $9=3^2$ singular points.

Remarks:

Doing this with $\mathbb{Z}/4\subseteq SU(2)$ instead of $\mathbb{Z}/3$, and the square (i.e., generated by $(1, i)$) lattice rather than the hexagonal lattice, and the resulting $4=2^2$ singular points, is something I'm equally curious about and equally unable to answer.
Doing this with $\mathbb{Z}/2$ instead of $\mathbb{Z}/3$, and any lattice in $\mathbb{C}^2$ at all (since all are invariant under this action), and the resulting $16=4^2$ singular points, gives a K3 surface; this was my motivation for the question.
The exceptional divisor at each of the 9 blowups is a pair of $\mathbb{P}^1$'s intersecting at a point.

A "Cauchy integral formula" for the Poisson kernel?

2011-04-10T22:07:01Z

The inspiration for this question is in a certain breakdown of the analogy between holomorphic and harmonic functions.

First recall the Cauchy integral formula: Let $U$ be an open subset of $\mathbb{C}$, let $p$ be a point in $U$, let $f$ be a holomorphic function on $U$. For closed curves $\gamma$ in $U\setminus\{p\}$, define

$F(\gamma)=\frac{1}{2\pi i}\displaystyle\int_\gamma \frac{f(z)}{z-p}dz$.

Then if two curves $\gamma_1$ and $\gamma_2$ are homotopic in $U\setminus\{p\}$, we have $F(\gamma_1)=F(\gamma_2)$.

Let's try a naive generalization of this to harmonic functions. Let $U$ be an open subset of $\mathbb{R}^n$, let $p$ be a point in $U$, let $f$ be a harmonic function on $U$, and let $\mathcal{A}$ be the "space of spheres in $U$ containing $p$ in their interior": $\mathcal{A}=\{(x, R)\in U \times \mathbb{R}^+ : S_R(x) \subseteq U, \ p \in B_R(x) \}$.

The Poisson kernel gives a function $F:\mathcal{A}\to \mathbb{R}$, mapping a sphere $S_R(x)$ to the value at $p$ of the harmonic function on $B_R(x)$ whose boundary values are the same as $f$'s:

$F(x,R) = \frac{R^2-|p-x|^2}{n\omega_n R} \displaystyle\int_{S_R(x)} \frac{f(y)}{|y-p|^n}$.

For spheres $S_R(x)$ such that all of the ball $B_R(x)$ is in $U$, the harmonic function on $B_R(x)$ whose boundary values are the same as $f$'s is just the function $f$. So $F(x,r)= f(p)$. Thus $F$ is constant on $\{(x, R) : S_R(x) \subseteq U, \ p \in B_R(x), \ B_R(x)\subseteq U \}\subseteq\mathcal{A}$, the "homotopy equivalence class" of spheres whose interiors are completely in $U$ -- just like in the Cauchy integral formula!

But the analogy fails once we try to extend this for spheres $S_R(x)$ whose interior isn't completely in $U$. For example, if $U$ is $\mathbb{R}^n\setminus\{0\}$, and the harmonic function is the Green's function $\Gamma$, then on the "homotopy equivalence class" of spheres containing both $p$ and $0$, the function $F$ can take many values (certainly all values less than $\Gamma(p)$).

I'd like to know whether there are some conditions I can impose on my harmonic function $f$ to make the analogy true. One possibility that I can think of is to demand that

$\displaystyle\int_{S_R(x)}\nabla f \cdot \nu = 0$

for any sphere $S_R(x)$ in $\mathcal{A}$ -- the idea being that I think in two dimensions this is the condition for a harmonic function to admit a harmonic conjugate.

Answer by macbeth for The conformal group of $S^n$.

2011-04-05T15:14:38Z

Try Lecture One of Eastwood, "Notes on Conformal Differential Geometry" (http://dml.cz/dmlcz/701576).

Connected components of space of maps between two manifolds

2010-05-01T20:27:10Z

Question: What are the connected components of the familiar spaces of functions between two (let's say compact and smooth, for simplicity) manifolds $M$ and $N$?

Specifically, I'm thinking of the Hölder spaces $\mathcal{C}^{k,\alpha}(M, N)$ and the Sobolev spaces $\mathcal{W}^{k,p}(M, N)$.

Some comments:

For a smooth function $f:M\to N$, it seems clear that, at least, all continuous functions homotopic to $f$ will be connected to it.
This question is inspired by the discussion of $\mathcal{W}^{k,p}(M, N)$ in McDuff-Salamon's book on $J$-holomorphic curves. There it's stated as an offhand remark that the connected components of $\mathcal{W}^{k,p}(M, N)$ (in the case of $M$ oriented & two-dimensional; I'm not sure if this makes a difference) are the completions of the sets {$f:M\to N \text{ smooth}: f_*[M]=A$}, for $A\in H_{\dim M}(N)$.
If the McD-S factoid is true, there should exist sequences of smooth not-all-mutually-homotopic functions which converge in $\mathcal{W}^{k,p}(M, N)$. (This isn't too counterintuitive, since $\mathcal{W}^{k,p}(M, N)$ presumably contains functions which aren't continuous, & so don't themselves have a homotopy class). Can someone give me an example of this phenomenon?

Please feel free to re-tag -- I can't think of anything really appropriate.

Answer by macbeth for Defining Quotient Bundles

2010-06-08T03:11:59Z

I agree this isn't completely obvious. Here's a slightly different take on it. Our intended vector bundle is

$E/E' :=\coprod E_x/E'_x$,

the disjoint union of the quotient vector spaces of fibres. We just need to specify the topology on it. We do this by describing a family of maps which we intend to be continuous local trivializations for the bundle.

So, take a point $p\in B$, and a nhd $U$ of $p$ on which we have a frame $(e_1, \cdots e_{n+k})$ for $E$. Choose $1\leq i_1<\ldots< i_k\leq n+k$ such that on the fibre $p$,

$E_p=E_p' + span(e_{i_1},\ldots e_{i_k})$.

By the continuity of the determinant function, in fact there's a neighbourhood of $p$ on which this is true; that is, there's a (perhaps smaller) nhd $V$ of $p$ such that for all $x\in V$,

$E_x=E_x' + span(e_{i_1},\ldots e_{i_k})$.

So at each $x\in V$, we have a basis

$(e_{i_1}+E_x',\ \ldots \ e_{i_k}+E_x')$

for $E_x/E'_x$. We demand that this collection of bases give a (continuous) frame for $E/E'$ over $V$. It's an easy check that the transition functions between two thus-constructed local trivializations are continuous, as required.

Maslov index of a pullback bundle

2010-05-01T11:28:02Z

This question has bugged me as I read McDuff-Salamon's book on pseudoholomorphic curves. I'll use their terminology.

Let $\Sigma$ be a compact surface possibly with boundary, $M$ an almost-complex manifold, and $L$ a totally real submanifold of $M$. A map

$u:(\Sigma, \partial\Sigma)\to(M, L)$

gives rise to a bundle pair over $\Sigma$: a complex vector bundle $u^*TM$ over $\Sigma$, together with a totally real sub-bundle $u^*TL$ over $\partial\Sigma$.

Question: Is there a nice description for the Maslov index of this bundle pair, in terms of a topological invariant of $u$? For instance, in terms of the homology class $u_*[\Sigma]\in H_2(M, L)$, or in terms of the homotopy equivalence class of $u$?

Motivating special case: if $\partial\Sigma=\emptyset$, then the Maslov index of the bundle pair $(u^*TM, \emptyset)$ is $2\langle c_1(TM), u_*[\Sigma]\rangle$.

Answer by macbeth for Parallel translation in Lie groups

2010-04-29T23:07:13Z

For a general $Y$, it will be matrix-exponential in $t$ with initial conditions determined by $Y$. Here's an explicit computation. Pick a left-invariant global frame $(E_1, \ldots E_n)$ for the group, and define structure constants

$[E_i, E_j]=\sum c_{ij}{}^kE_k$.

The covariant derivative of $E_i$ along the geodesic $\exp(tX)$ from 0 is the constant

$\frac{1}{2}[X, E_i]=\frac{1}{2}\sum X^jc_{ji}{}^kE_k$

(see eg Lee "Riemannian Manifolds" problem 5-11). Therefore a vector field

$t\mapsto \sum f^i(t)E_i$

along this geodesic is parallel if it is a solution to

$0=D_t\left(\sum f^i(t)E_i\right)=\sum_k\left[(f^k)'(t)+\frac{1}{2}\sum_{i,j} f^i(t)X^jc_{ji}{}^k\right]E_k$,

i.e., to $\forall k \ 0=(f^k)'(t)+\frac{1}{2}\sum_{i,j} f^i(t)X^jc_{ji}{}^k$.

The parallel transport of $Y$ will be the solution to this linear system with initial value $Y$. Perhaps there's a nice basis-invariant way of expressing this? I can't think offhand.

Analogy of Liouville conformal mapping theorem with Mostow rigidity?

2010-03-26T07:08:06Z

I often hear mention of two theorems, Mostow's rigidity theorem and Liouville's theorem on conformal mappings, which superficially sound similar: they say that a set of geometric structures is, if nonempty, big in dimension 2, but small in dimension greater than 2.

(For Mostow's theorem, the set of structures in question is the set of hyperbolic metrics on a manifold; for Liouville's, it's the set of germs of flat metrics in a conformal equivalence class.)

I know that hyperbolic and conformal geometry are closely connected, at least in dimension 2. I'm curious as to whether this analogy is hinting at one such connection. Is there a "good reason" for this analogy?

Answer by macbeth for Non-affine, projective vector field on $R^n$

2010-02-22T14:06:50Z

Recall two connections on a manifold are said to be projectively equivalent, if they have the same geodesics. You want to know what local diffeos $\mathbb{R}^n\mapsto\mathbb{R}^n$ preserve geodesics; that is, what flat metrics on $\mathbb{R}^n$ are projectively equivalent to the Euclidean one. I think your conjecture, that the only such metrics are those obtained by affine-linear transformations of the Euclidean metric, is correct.

Let's discuss this first locally, then globally.

Local question: What flat connections on a neighbourhood of $\mathbb{R}^n$ are projectively equivalent to the Euclidean one?

Answer: If $n=1$, all. If $n\geq 2$, exactly those obtained by what depending on your terminology you might call a a perspective transformation or a projective linear transformation or something else.

Comment: Sketch proof below. The case distinction (which explains your colleague's observation) comes from a factor of $1/(n-1)$ in the formula for the appropriate Schouten curvature tensor. This is analogous to the case distinction $n\leq 2$ vs $n\geq 3$ in conformal geometry, see http://en.wikipedia.org/wiki/Liouville's_theorem_(conformal_mappings)

Now we can deal with

Global question: What flat connections on $\mathbb{R}^n$ are projectively equivalent to the Euclidean one?

Answer: Just the standard one. The others "blow up"/"go off to infinity"/involve-division-somewhere-by-zero if you try to extend them to all $\mathbb{R}^n$.

Sketch proof of local version: Use the notation and some formulae from, eg., http://www.maths.adelaide.edu.au/michael.eastwood/projective.pdf

Let $\nabla$ be the standard connection on $\mathbb{R}^n$. Consider a projectively equivalent connection defined by a 1-form $\Upsilon_i$ (so the new connection acts on a 1-form $\omega_i$ by, $\nabla_i\omega_j - \Upsilon_i\omega_j -\Upsilon_j\omega_i$.) The projective Schouten curvature tensor of $\nabla$ is zero (since it's flat). If the new connection's also flat, then, applying appropriate transformation laws, we have that $\nabla_i\Upsilon_j = \Upsilon_i\Upsilon_j$, and that $\Upsilon_i$ is closed, so locally exact.

Write $\Upsilon = df$. Then in standard Euclidean co-ordinates we have the system of PDE $\partial_i\partial_j f = \partial_i f \ \partial_j f$ for the function $f$, which we can solve to get $f(x^1, ... x^n) =-\log (a_1 x^1+ \cdots + a_nx^n + c)$ for some fixed constants $a_i$ and $c$. Hence $\Upsilon_i = \frac{-a_i}{a_1 x^1+ \cdots + a_nx^n + c}$. It should probably turn out that the family of connections this gives are all indeed flat, and correspond to the projective-linear-transformations of the Euclidean metric on $\mathbb{R}^n$.

Divisors and vector bundles in various categories

2010-02-09T09:35:30Z

I'm taking a first course on complex manifolds, and am trying to square what I hear with what I know of (real) differential geometry. Please forgive me if this question is misguided!

Here are two examples of ways of making vector bundles from codimension-one submanifolds.

(The tensor powers of its associated line bundle) In a complex manifold $M$ with a codimension-1 complex submanifold $D$, take as an atlas a system $(U_\alpha)$ of slice-coordinate charts for $V$, together with other charts $(V_\beta)$ covering $M\setminus D$. For each $n\in \mathbb{Z}$, define a line bundle via the following transition functions: $\phi_{\beta_1\beta_2}:V_{\beta_1}\cap V_{\beta_2}\to GL_1(\mathbb{C})$ is uniformly =1; $\phi_{\alpha_1\beta_2}:U_{\alpha_1}\cap V_{\beta_2}\to GL_1(\mathbb{C})$ is $z_1^n$, and $\phi_{\alpha_1\alpha_2}:U_{\alpha_1}\cap U_{\alpha_2}\to GL_1(\mathbb{C})$ is $z_1^n/w_1^n$, where $z_1$ and $w_1$ are the coordinates whose vanishing determines $D$ on $U_{\alpha_1}$ and $U_{\alpha_2}$ respectively.

Comment: This also seems to work fine if we replace "complex" by "smooth (real)" throughout. However, the family of line bundles isn't so interesting: the even ones are all trivial; the odd ones are mutually isomorphic.

(Vector bundles on spheres) For each homotopy class of maps $S^{n-1}\to GL_k(\mathbb{R})$, we can construct a vector bundle of rank $k$ on $S^n$, by using a representative of this class to define a transition function on the intersection of the "north" and "south" stereographic projection charts (which has $S^{n-1}$ as a retract).

I'd like to know: are these indeed analogous? Are they special cases of, say, a general method for constructing a smooth (respectively, complex) rank-$k$ vector bundle on a smooth (resp., complex) manifold out of a map from a codimension-one submanifold into $GL_k(\mathbb{R})$ (resp., $GL_k(\mathbb{C})$?

Answer by macbeth for What is an immersed submanifold?

2010-01-25T00:51:39Z

I think the answer to your final question is no, and more generally: countable unions of embedded submanifolds are precisely the images of (not-necessarily-injective) immersions.

Sketch proof: A countable union of manifolds is a manifold, so a countable union of embeddings is an immersion. Conversely, by the Inverse Function Theorem, an immersion $f: M\to N$ is locally-in-$M$ an embedding; we thus obtain a "cover of the immersion by embeddings", and since manifolds are Lindelöf there's a countable subcover.

To answer your second-last question, we then need to analyse whether there are images-of-immersions that aren't immersed submanifolds (= images-of-injective-immersions).

(edit: fixed typo)

Answer by macbeth for When does local invertibility imply invertibility?

2009-12-30T17:03:59Z

A local isometry between complete connected Riemannian manifolds must be a covering map. So a local isometry between complete connected Riemannian manifolds, with simply-connected range, should be a global isometry?

Answer by macbeth for What are some results in mathematics that have snappy proofs using model theory?

2009-12-27T12:11:26Z

I like what I've heard about this 2005 GAFA paper on topological dynamics by Kechris-Pestov-Todorcevic: http://www.springerlink.com/content/hp2h78112hg31733/

It uses the Fraïssé limit construction to produce a family of topological groups with interesting universal minimal flows.

Comment by macbeth

2012-10-18T03:20:28Z

(2.) Yes, we're interested in the diffeomorphisms satisfying $\varphi \circ f = \varphi$. Such a diffeo $f$ has the property that, on each level set $\{x:|x|^2=r\}$, $f$ restricts to a diffeo of the level set. Moreover the possible $f$ are basically characterized by this property. Think of $f$ as a (germ of a) 1-parameter family of diffeomorphisms of the $(n-1)$-sphere, modulo some boundary conditions (tending to the identity near 0) to ensure smoothness at $p$.

Comment by macbeth

2012-10-18T03:19:00Z

Hi Will (and Kofi). (1.) It seems to me that one can work with the set of germs of charts near p, and the group of germs of diffeomorphisms fixing p. Then the action is well-defined and free and transitive.

Comment by macbeth

2012-09-23T04:31:43Z

Very interesting and relevant. Thanks!

Comment by macbeth

2012-08-28T18:59:36Z

Regarding Walker's comment on Hawley's paper: The Bochner paper which is cited by Hawley is "Curvature in Hermitian metric" (1947). In this paper Bochner proves the local version of the result: that the metric of constant holomorphic bisectional curvature $b$ is unique up to local isometry. Maybe Walker felt that passing to the global version (as done by Hawley/Igusa) was straighforward.

Comment by macbeth

2012-05-09T18:25:25Z

$\mathbb{Z}$ does act freely on $S^{2k+1}$ in the same way. This doesn't contradict Wolf's theorem -- the point is that this action is not properly discontinuous, so the quotient by this action is not a manifold.

Comment by macbeth

2012-05-09T16:02:38Z

In your new answer, you're missing one group in the even-$n$ case: the group $\mathbb{Z}/2$, generated by the antipodal map $x\mapsto -x$. (Its eigenvalues are all $-1$, which is real.)

Comment by macbeth

2012-05-09T15:56:24Z

This problem was solved by Joseph Wolf in his book "Spaces of Constant Curvature" (1967). (That is, if I understand the problem correctly -- so it's equivalent to the classification of complete manifolds of constant positive curvature.) I don't remember the answer -- its statement is quite involved

Comment by macbeth

2012-05-09T15:45:57Z

It is not true that if $n>1$ then rotations fix points. There are counterexamples in all odd dimensions. For instance, the element of $SO(2n)$ whose nonzero entries are $n$ $2\times 2$ blocks along the diagonal, each some nonzero rotation $\begin{pmatrix} \cos t & -\sin t \cr \sin t & \cos t\end{pmatrix}$.

Comment by macbeth

2012-04-17T22:30:08Z

Maybe you know this already: if $(Q, g)$ is not flat, then the sectional curvatures of the Sasaki metric on $TQ$ are both unbounded below and unbounded above. (See eg Propositions 7.6-7.8 in Gudmundsson-Kappos "On the geometry of tangent bundles" ams.org/mathscinet-getitem?mr=1888866 .) To me, the latter makes it seem unlikely that $\tilde\rho>0$.

Comment by macbeth

2012-04-16T13:41:26Z

Is this the same question as mathoverflow.net/questions/31920/… ? (Joseph O'Rourke and Will Jagy, I'm second-guessing myself because you both also commented on that question!)

Comment by macbeth

2012-04-02T19:32:07Z

Regarding "Is there a more standard name for a geometry consisting of a G-structure and a connection?": Perhaps the concept you want is a "Cartan geometry of type $(G, H)$." (Here $H\subseteq G$ are arbitrary Lie groups.) Standard references are Sharpe (www.ams.org/mathscinet-getitem?mr=1453120) or Čap-Slovák (www.ams.org/mathscinet-getitem?mr=2532439). The definition of a Cartan geometry appears on p71 of Čap-Slovák, which is visible in the AMS preview pdf at www.ams.org/bookstore-getitem/item=surv-154.

Comment by macbeth

2012-03-25T20:22:16Z

I see. I don't know your terminology, then -- would you mind defining the "radial part" of an $O(n)$-invariant operator on [functions on] this space? Is it the induced operator on [functions on] its space of $O(n)$-orbits?

Comment by macbeth

2012-03-25T12:47:59Z

Ah -- maybe I've misread your question? My interpretation was: operator is Monge-Ampère operator $\det(\frac{\partial^2}{\partial x_i\partial x_j})$, on the space $\mathcal{C}^\infty(\mathbb{R}^n)$ (say). Is your operator actually $\det(\frac{\partial}{\partial x_{ij}})$, on the space $\mathcal{C}^\infty(Mat_n(\mathbb{R}))$?

Comment by macbeth

2012-03-25T02:02:16Z

By the way, I don't know the answer to your general question, but by my calculation the radial part of $\det(\partial_{ij})$ (aka the Monge-Ampère operator) is given by: the operator sends $f(|\cdot|^2)$ to $[2f'(|\cdot|^2)]^{n-1}[4f''(|\cdot|^2)|\cdot|^2+2f'(|\cdot|^2)]$. Proof: The endomorphism induced from the Hessian is $4f''(|\cdot|^2)|\cdot|^2\text{Proj}_{\cdot}+2f'(|\cdot|^2)]I$.

Comment by macbeth

2012-03-24T21:22:32Z

In (a), you mean your differential operator to be $O(n)$-invariant, right?