User peter michor - MathOverflow

Answer by Peter Michor for Special coordinates for periodic metrics

2013-05-20T07:11:58Z

Maybe this can help, but it might not be what you want: Add reflections at affine planes $\{x: x_i = \mathbb Z +\frac12\}$, then the $\mathbb Z^n$ action is extended to an action which is generated by reflections. Make your $\mathbb Z^n$-invariant metric also invariant under the extended reflection group (you have to average over the $n$ generating reflections only) and try to use methods from the following two papers (on the arXiv or on my homepage):

Dmitri Alekseevky, Andreas Kriegl, Mark Losik, Peter W. Michor: Reflection groups on Riemannian manifolds. Annali di Matematica 186 (2006), 25-58
Dmitri V. Alekseevsky, Peter W. Michor, Yurii A. Neretin: Rolling of Coxeter polyhedra along mirrors. 19 pages. To appear in: Geometric Methods in Physics: XXXI Workshop, Bialowieza, Poland, June 24–30, 2012. Trends in Mathematics, Birkhauser, Due: June 30, 2013. arXiv:0907.3502.

Answer by Peter Michor for Under exactly what (extra) conditions (if any) is a connected Hausdorff manifold with a Riemannian metric a metric space?

2013-05-19T17:54:54Z

There are infinite dimensional weak Riemannian manifolds with vanishing geodesic distance (in the sense as defined in the question). These are modeled on nuclear Frechet spaces, but the results extend to Sobolev completions of high enough order ($>\dim/2 +2$). They are still weak Riemannian manifolds (i.e., the Riemann metric does not generate the topology on the tangent spaces).

The first example was the $L^2$ metric on $\text{Emb}(S^1,\mathbb R^2)/\text{Diff}(S^1)$, as shown in the first paper below. Then it turned out that the right invariant $L^2$-metric on each full diffeomorphism group also has this property, also Sobolev metrics for Sobolev order $<1/2$ ($\le 1/2$ on $\text{Diff}(S^1)$). In particular, Burgers' equation and KdV are nonlinear PDE's corresponding to geodesic equations for metrics with vanishing geodesic distance.

All the papers are in the arXiv or on my homepage.

Peter W. Michor; David Mumford: Riemannian geometries on spaces of plane curves. J. Eur. Math. Soc. (JEMS) 8 (2006), 1-48.
Peter W. Michor; David Mumford. Vanishing geodesic distance on spaces of submanifolds and diffeomorphisms. Documenta Math. 10 (2005), 217--245 (written later)
Martin Bauer, Martins Bruveris, Philipp Harms, Peter W. Michor: Vanishing geodesic distance for the Riemannian metric with geodesic equation the KdV-equation. Ann. Glob. Anal. Geom. 41, 4 (2012) 461-472.
Martin Bauer, Martins Bruveris, Philipp Harms, Peter W. Michor: Geodesic distance for right invariant Sobolev metrics of fractional order on the diffeomorphism group. Ann. Glob. Anal. Geom. 44, 1 (2013), 5-21.
Martin Bauer, Martins Bruveris, Peter W. Michor: Geodesic distance for right invariant Sobolev metrics of fractional order on the diffeomorphism group. II. 7 pages. To appear in: Ann. Glob. Anal. Geom.

Answer by Peter Michor for Geodesics on a Grassmannian

2013-05-19T17:27:02Z

Have a look at the paper:

Y.~A. Neretin: On Jordan angles and the triangle inequality in Grassmann manifold}, Geometriae Dedicata, 86 (2001).

There are explicit formulas for geodesics and even for the geodesic distance on real Grassmannians. This ties in with Greg Kuperberg's answer.

Answer by Peter Michor for Reference request : dimensions of real representations of Lie groups

2013-05-17T15:37:21Z

MR2041548 Onishchik, Arkady L. Lectures on real semisimple Lie algebras and their representations. ESI Lectures in Mathematics and Physics. European Mathematical Society (EMS), Zürich, 2004. x+86 pp.

Answer by Peter Michor for Reference request: Riemannian manifold of linear isometries from $\mathbb{C}^n$ into $\mathbb{C}^m$

2013-05-17T08:00:26Z

They are called Stiefel manifolds, and are principal $U(n)$-bundles over Grassmann manifolds. They are homogeneous Riemannian manifolds, whereas the Grassmannian are symmetric spaces.

Riemannian geometry on homogeneous manifolds is governed by the Nomizu operator. See page 364-367 of here.

Explicit geodesics for Grassmannians, even a formula for geodesic distance, in the form of horizobtal geodesics of the Stiefel manifold, are described in the following paper:

Y.~A. Neretin: On Jordan angles and the triangle inequality in Grassmann manifold}, Geometriae Dedicata, 86 (2001).

Geodesics on an infinite dimensional Stiefel manifold of isometries of $\mathbb R^2$ into a Hilbert space were used for shape space ananlysis in the paper

Laurent Younes, Peter W. Michor, Jayant Shah, David Mumford: A Metric on Shape Space with Explicit Geodesics. Rend. Lincei Mat. Appl. 9 (2008) 25-57. arXiv:0706.4299 (pdf)

Answer by Peter Michor for How to characterize this particular kind of bundle?

2013-05-15T19:35:19Z

If $G$ is compact, then $P\to M_4$ is a locally trivial fiber bundle by a theorem of Ehresmann (see 17.2 in here). Take a principal $S^1$-connection of the the $S^1$-bundle $M_5\to M_4$, and a principal $G$-connection of $P\to M_5$. Then smooth curves in $M_4$ have horizontal lifts first to $M_5$, unique by choosing an initial point, and one can lift this further horizontally to $P$ by choosing another initial point. So you have a complete Ehresmann connection for the fiber bundle $P\to M_4$. Now we have to compute the holonomy Lie algebra of this Ehresmann connection; if it is finite dimensional, then we have a principal bundle by 17.11 loc.cit, whose structure group has the holonomy Lie algebra as Lie algebra. It seems to me that the holonomy group is an extension of $G$ with kernel $S^1$, but I did not check the details.

The whole construction should also work without the assumption that $G$ is compact.

Answer by Peter Michor for Special kind of operators

2013-05-15T19:17:39Z

See page 228ff of

Albrecht Pietsch: Operator ideals, Elsevier 1980. (pdf here)

Maybe, your operators are the $(\infty, p, \infty)$-summing operators there.

Answer by Peter Michor for Constructing a special infinite-dimensional vector bundle

2013-05-01T16:11:46Z

In "Convenient Setting" the construction that you want is done in 42.17.

There is also the old book, where 10.10 is relevant and your construction (for the special case $E=TN$) is done in (too many) details in 10.12:

Peter W. Michor: Manifolds of differentiable mappings. Shiva Mathematics Series 3, Shiva Publ., Orpington, (1980), iv+158 pp., MR 83g:58009, ZM 433.58001, (pdf)

If $M$ is not compact you have to use a very fine topology where the set maps which differ from a given map only on a compact set are open. Otherwise the space might be not locally contractible and you cannot find open charts.

Then you just take $C^\infty(M,E)$ as total space with $C^\infty(M,p)=p_*:C^\infty(M,E)\to C^\infty(M,N)$ as projection, or an open subspace therein inf $M$ is not compact. The fiber over $\phi:M\to N$ are the maps $M\to E$ which project to $\phi$, i.e., the sections (with compact support if $M$ is not compact) of $\phi^*E$.

Edit:

Okay, here is a proof of local triviality ($U$ will always denote a neighborhood of the zero section): We need:

A Riemann metric on $N$ and its Riemannian exponential mapping $\exp^N: TN\supset U \to N$ such that $(\pi_N,\exp^N):TN\supset U \to N\times N$ is a diffeomorphism onto a neighborhood of the diagonal. If $f\in C^\infty(M,N)$ and if $g$ is near enough to $f$ (in a neighborhood $U_f$ of $f$ in $C^\infty(M,N)$), then $$u_f(g)(x) = (\pi_N,\exp^N)^{-1}(f(x),g(x))$$ defines a chart which maps $g\in U_f$ to $u_f(g)\in\Gamma(f^*TN)$. Note that $$ t\mapsto c_{f,g,x}(t) := u_f^{-1}(t.u_f(g))(x) = \exp^N_{f(x)}(t.u_f(g)(x)) $$ is a smooth curve from $f(x)$ to $g(x)$ in $N$ which depends smoothly on $x$ and on $g$.
A linear connection on $E$ with its parallel transport $Pt(c,t):E_{c(0)}\to E_{c(t)}$ along each smooth curve $c$ in $N$. The parallel transport is smooth in $c$ (easy to proof with convenient calculus) and in $t$. In detail, the following mapping is smooth by convenient calculus and usual results about linear ODEs in finite dimension:

$$Pt: C^\infty(\mathbb R,N)\times_{ev_0,p} E \times \mathbb R \to E,\quad (c,v_{c(0)},t)\mapsto Pt(c,t)v_{c(0)}$$

To describe the chart structure on $C^\infty(M,E)$ we need also a Riemannian metric on $E$. The best one now is a fiber metric on the vector bundle $E$ which is respected by the connection on $E$, and then we pull the metric from $N$ to the horizontal bundle in $TE$ and lift the fiber metric to the vertical bundle in $TE$ and declare them to be orthogonal. I will not use this explicitly now.

For $s\in C^\infty(M,E)$ with $p\circ s$ near $f$ consider the mapping $$ \Phi(s)(x) := Pt(c_{f,p\circ s,x},1)^{-1}(s(x))\in E_{f(x)}. $$
Then $$ s\mapsto (p\circ s, \Phi(s))\in U_f\times \Gamma(f^*E) $$ is a smooth local trivialization of $p_\star:C^\infty(M,E)\to C^\infty(M,N)$.

You can play with the topology on $C^\infty(M,E)$ by requiring compact support in $C^\infty(M,N)$ but using the compact $C^\infty$-topology along the fibers of $E$; or also compact support there. If you need more details here, tell me.

Answer by Peter Michor for Extensions with trivial induced outer action

2013-05-02T18:35:10Z

Edit: Diagram added, more details added.

See 15.21 in pages 177-190 of here, where I collected the results on extensions of groups and Lie groups that I could find. 15.24 summarizes your situation quite clearly: We have $\text{Inn}(N)= N/Z(N)$ where $Z(N)$ is the center of $N$. Then you have a mapping of extensions $$ \begin{array}{ccccc} Z(N) & \xrightarrow{i|_{Z(N)}} & E & \xrightarrow{\theta} & G\times\text{Inn}(N) \newline \downarrow & & \downarrow & & \downarrow \newline N & \xrightarrow{i} & E & \xrightarrow{p} & G \end{array} $$ where the down arrows are inclusion, identity, and first projection, and where $\theta(x)=(p(x),\text{Conj}_x|_N)$. The first line is a central extension since $G\times \text{Inn}(N)$ acts trivially on $Z(N)$.

Answer by Peter Michor for de Rham cohomology and flat vector bundles

2013-04-30T19:23:30Z

A view not yet mentioned is the following: The universal cover $\tilde M$ is a principal fiber bundle over $M$ with structure group $\pi=\pi_1(M)$, and a flat connection on $E$ identifies $E$ as the assiciated bundle to $\tilde M$ for the holonomy representation of $\pi$ in $V:=E_{x_0}$, the fiber over the base point. Then $\Omega(M,E)\cong \Omega(\tilde M,V)^\pi = (\Omega(\tilde M)\otimes V)^\pi$ and $H^k_{dR}(M,\nabla) = H^k((\Omega(\tilde M)\otimes V)^\pi) = (H^k_{dR}(\tilde M)\otimes V)^\pi$.

Answer by Peter Michor for Matrices whose kernel escapes a sub-vector space

2013-04-30T19:00:51Z

Write $\mathbb C^n = V \oplus W$ and $A=A|_V + A|_W$. Then $A\in X$ iff

$A|_W$ has kernel $\ne 0$ or $Im(A|_V)\cap Im(A|_W)\ne 0$.

Answer by Peter Michor for Algebraic Stratifications of $G$-varieties

2013-04-30T16:48:08Z

See:

MR0342523 (49 #7269) Luna, Domingo Slices étales. (French) Sur les groupes algébriques, pp. 81–105. Bull. Soc. Math. France, Paris, Memoire 33 Soc. Math. France, Paris, 1973.

Answer by Peter Michor for Vector fields on $(4n+1)$-spheres

2013-04-30T05:33:32Z

The Radon-Hurwitz number $k(n)$ is the largest $k$ such that there exists an orthogonal multiplication $\mathbb R^k\times \mathbb R^n\to \mathbb R^n$; so for an ONB $x_1,\dots, x_k$ of $\mathbb R^k$ and a unit vector $y\in \mathbb R^n$ the vectors $y, x_1.y, x_2.y,\dots x_k.y$ are orthogonal in $\mathbb R^n$. This describes vector fields on $S^{n-1}$. The orthogonal multiplications were constructed by Radon [Lineare Scharen orthogonaler Matrizen, Abh. Math. Sem. Univ. Hamburg 1, 1-14, 1921] who extended the construction of Hurwitz for $k=n=1,2,4,8$. They extend to representations of Clifford algebras $C(\mathbb R^k, -\langle\quad,\quad\rangle)$ which explains the "periodicity" in $k(n)$ with respect to 8 and 2.

Radon writes: "For real matrices the solution offered itself to me by a particular reduction method using matrices whose elements are complex numbers or quaternions." (My translation) Maybe, there is an inkling of the fiber method you are looking at.

Adams showed that there are not more linearly independent vector fields on the sphere $S^{n-1}$.

Answer by Peter Michor for Tangent space in Algebraic geometry and Differential geometry

2013-04-28T08:36:53Z

$T_xX$ as the dual of $\mathfrak m_x/\mathfrak m_x^2$ corresponds to tangent vectors acting as derivations over the point evaluation at $x$ on the function algebra. $T_xX$ as equivence classes of curves through $x$ correspond to velocity vectors at $x$, so they help kinematic visualization. On (at least $C^1$-) finite dimensional manifolds (including smooth points of varieties) both notions coincide. In singular points there are many more "operational" than "kinematic" tangent vectors.

But even on an infinite dimensional Hilbert space the same is true: see 28.3 and 28.4 of here.

Answer by Peter Michor for What is visualization of gradient flow of a functional?

2013-04-27T19:31:39Z

A functional is just a function on a space of functions, so it is defined on an infinite dimensional space. The visualization is the same as in finite dimensions, but there are many more things that can go wrong. The functional must be suitably differentiable. You need an inner product which might be continuous in the function space topology, biut it might not generate the topology, so the gradient a priory lies in the Hilbert space completion of the functions space -- so it might point outside of the function space. Beyond Banach spaces ODE's behave differently, yuou might loose existence or uniqueness of solution curves. But often it works.

Answer by Peter Michor for naming for the map $T = x \mapsto a x b$

2013-04-25T07:37:06Z

For $a,b,z\in Mat(n\times n)$ you have $T_{a,b}(z) = azb^\top$, $T_{a,b}\in Mat(n^2\times n^2)$. The set $H$ of all these operators is a monoid (a semigroup with unit) under multiplication, and it contains the dense group $H^o$ of all invertible ones. Note that $H$ is a quadratic cone, and through each point there are two affine subspaces (fix $b$ or fix $a$) of $\dim n^2$ inside of $H$. So this looks like a cone over a hyperboloid.

In fact the group is $GL(n)\times GL(n)/\lbrace-1,1\rbrace$ under the representation $(g,h)\mapsto M_g. M^{h^\top}= M^{h^\top}.M_g$ where $M(a,b) = a.b = M_a(b) = M^b(a)$. The set $H^o$ is the cone (without 0) through the hyperboloid $SL(n)\times SL(n) \cong \lbrace T_{a,b}: a,b \in SL(n)\rbrace$. $H$ is its closure. Can we have a nice compact equation for this hyperboloid?

Aside: We may use the inner product $Tr(x.y^\top)$ on $Mat(n\times n)$. The transpose of $T_{a,b}$ with respect to this inner product is $T_{a,b}^\top = T_{a^\top, b^\top}$. Then we can work out $Tr(T_{a,b}.T_{c,d}^\top)$ and try to express the equation for the hyperboloid in terms of that.

Edit: The mapping $T$ could be seen as a polarized version of an affine quatratic Veronese map.

Answer by Peter Michor for Volume of a convex set

2013-04-25T06:10:14Z

Have a look at: [JFM 34.0649.01 Minkowski, H. Volumen und Oberfläche. (German) Math. Ann. 57, 447-495 (1903)] and [MR0478079 (57 #17572)
Pogorelov, Aleksey Vasilʹyevich. The Minkowski multidimensional problem. Translated from the Russian by Vladimir Oliker. Introduction by Louis Nirenberg. Scripta Series in Mathematics. V. H. Winston & Sons, Washington, D.C.; Halsted Press [John Wiley & Sons], New York-Toronto-London, 1978. 106 pp.]

Representing immersions from a surface into 3-space

2013-04-23T06:40:04Z

Let $\mathbb T^2=(S^1)^2$ be the 2-torus, for convenience. $\def\Imm{\operatorname{Imm}}$ Consider the Frechet manifold of immersion $\Imm(\mathbb T^2, \mathbb R^3)$ and the smooth mapping $R:\Imm(\mathbb T^2,\mathbb R^3)\to C^\infty(\mathbb T^2,\mathbb R^3)$ which is given by $$ R(f) := \partial_x f\times \partial_y f = f_x\times f_y, $$ where $\times$ denotes the vector product. Note that $R(f)$ is the unit normal to $f(\mathbb T^2)$ times the the density function of the area measure, parameterized.

Question: Roughly, given $g\in C^\infty(\mathbb T^2,\mathbb R^3)$, can one solve $R(f)=g$ for $f$? Can one reconstruct $f$ from $R(f)=g$ up to translations? Are there local solutions in $\mathbb R^2$? Of course one expects that the image $R$ is at best a codimension 6 submanifold of $C^\infty(\mathbb T^2,\mathbb R^3)$ because of the two periodicity conditions.

The tangent mapping of $R$ is $T_fR.h = f_x\times h_y + h_x \times f_y = f_x\times h_y - f_y\times h_x$. This is not elliptic in $h$. The symbol of $T_fR$ is $$ \sigma(T_fR)(\xi,\eta) = \begin{pmatrix} 0 & \gamma & -\beta \\ -\gamma & 0 & \alpha \\ \beta & -\alpha & 0 \end{pmatrix} \quad\text{ where }\quad \begin{matrix} \alpha = f^1_y.\xi - f^1_x.\eta \\ \beta = f^2_y.\xi - f^2_x.\eta \\ \gamma = f^3_y.\xi - f^3_x.\eta \end{matrix}. $$

Easier question: Is $T_fR$ injective? Is it surjective onto a codimension 6 linear subspace?

The interest in this problem comes from the fact that variants of $R$ like $f\mapsto |R(f)|^{-1/2}.R(f)$ pull back the flat $L^2$-metric on $C^\infty(\mathbb T^2,\mathbb R^3)$ to a weak Riemannian metric on $\Imm(\mathbb T^2,\mathbb R^3)$ which is invarint under the action of the diffeomorphism group of $\mathbb T^2$.

The related mapping for closed curves has been used successfully here. For the diffeomorphism group of $\mathbb R$ it has been used here.

The related problem for the unit normal for convex surfaces is classically called the Minkowski problem and there are solutions available, see [JFM 34.0649.01 Minkowski, H. Volumen und Oberfläche. (German) Math. Ann. 57, 447-495 (1903)] and [MR0478079 (57 #17572)
Pogorelov, Aleksey Vasilʹyevich. The Minkowski multidimensional problem. Translated from the Russian by Vladimir Oliker. Introduction by Louis Nirenberg. Scripta Series in Mathematics. V. H. Winston & Sons, Washington, D.C.; Halsted Press [John Wiley & Sons], New York-Toronto-London, 1978. 106 pp.]

Answer by Peter Michor for Higher derivatives than Jacobi fields.

2013-04-24T06:16:48Z

You might be interested in the Jacobi flow on $TTM$ whose flow lines project to geodesics, velocity fields of geodesics, and Jacobi fields. You can continue to higher order.

Peter W. Michor: The Jacobi Flow. Rend. Sem. Mat. Univ. Pol. Torino 54, 4 (1996), 365-372 (pdf)

Answer by Peter Michor for Lagrangian submanifolds

2013-04-23T07:32:01Z

Take the projection $\mathbb C^n\to P$ with kernel $iP$ and restrict it to $Q$. Its inverse, composed with the projection $\mathbb C^n\to iP$ with kernel $P$ is the linear map $\Phi$ you look for. See 31.7 of here for more information.

Answer by Peter Michor for Homogeneous Spaces and Equivariant Hodge Maps

2013-04-21T18:39:16Z

A $G$-invariant metric $g$ on $G/H$ is uniquely determined by its $H$-invariant value $g_o$ at $T_o(G/H)$ for the base point $o\in G/H$. The Riemannian volume form $vol(g)$ is $G$-invariant, and $\star$ is given by $\phi^k\wedge \psi^{n-k} = (\Lambda^{n-k}g^{-1})(\star\phi^k,\psi^{n-k}).vol(g)$ where $\Lambda^{n-k}g^{-1}$ is the induced inner product on $\Lambda^{n-k}T^*(G/H)$. So $\star$ is $G$-equivariant.

See 25.11 and 28.2, 28.3 of here.

Edit: I was a little too fast. As Robert remarked, this is okay if $G$ also preserves the orientation. If $G/H$ is not orientable then one may go to the orientable double cover of $G/H$ where the the Hodge map exists and exchanges "formes pairs" (in the sense of the De Rham) which are invariant under the covering map, and "formes impaired", the eignespace of eigenvalue -1 under the covering map.

If $G/H$ is orientable but $G$ does not respect the orientation, then $G$ respects the volume density, and there is a homomorphism $s:G\to \lbrace-1,1\rbrace$ such that $g^*vol(g)= s(g).vol(g)$ and one can use that.

@Mihail: You are right. If you view $\Lambda^{k}g^{-1}:\Lambda^{k}T^\star M\to (\Lambda^{k}T^\star M)^\star =\Lambda^kTM$ then $\star \phi^k = i(\Lambda^{k}g^{-1}(\phi^k)) vol(g)$ -- in the orientable case.

Answer by Peter Michor for analytic structure on lie groups

2013-04-21T16:46:00Z

$\def\ad{\text{ad}}$ This follows immediately from the Baker-Campbell-Hausdoff formula: For complex $z$ near $1$ we consider the function $$f(z):= \frac{\log(z)}{z-1} = \sum_{n\geq0}\frac{(-1)^n}{n+1}(z-1)^n$$ Then for $X$, $Y$ near $0$ in $\mathfrak g$ we have $\exp X.\exp Y= \exp C(X,Y)$, where $$ C(X,Y) = Y + \int_0^1 f(e^{t. \ad X}.e^{ \ad Y}).X\,dt $$ $$ = X + Y + \sum_{n\geq1}\frac{(-1)^n}{n+1}\int_0^1\biggl( \sum_{{k,\ell\geq0, k+\ell\geq1}} \frac{t^k}{k!\,\ell!}\; ( \ad X)^k( \ad Y)^\ell\biggr)^nX\;dt $$ $$ = X + Y + \sum_{n\geq1}\frac{(-1)^n}{n+1} \sum_{{k_1,\dots,k_n\geq0, \ell_1,\dots,\ell_n\geq0,k_i+\ell_i\geq1}} \frac{(\ad X)^{k_1}(\ad Y)^{\ell_1}\dots (\ad X)^{k_n}(\ad Y)^{\ell_n}} {(k_1+\dots+k_n+1)k_1!\dots k_n!\ell_1!\dots\ell_n!}X $$ $$ = X + Y + \tfrac12[X,Y] +\tfrac1{12}([X,[X,Y]]-[Y,[Y,X]]) + \cdots . $$ For a short proof of this formula see 4.29 of here. This series has radius of convergence $\pi$ in each norm on the Lie algebra in which the bracket (as bilinear operator) is bounded by 1. This even works for $C^2$-Lie groups, since $C^2$ suffices to get the Lie bracket at the tangent space of the identity.

Answer by Peter Michor for who invented projective space $\mathbb{P}^n$?

2013-04-19T20:02:44Z

One should also mention Karl Georg Christian von Staudt (1798 – 1867), a German mathematician.

His book "Geometrie der Lage (1847)" was a landmark in projective geometry.

Staudt was the first to adopt a fully rigorous approach. Without exception his predecessors still spoke of distances, perpendiculars, angles and other entities that play no role in projective geometry.

Answer by Peter Michor for Cross section for closed Lie subgroup in a Lie group

2013-04-19T18:01:59Z

See page 65 of here.

Answer by Peter Michor for Density of smooth functions in Sobolev spaces on manifolds

2013-04-19T12:52:12Z

According to pages 14 and 15 of:

MR2343536 Reviewed Eichhorn, Jürgen Global analysis on open manifolds. Nova Science Publishers, Inc., New York, 2007. x+644 pp. ISBN: 978-1-60021-563-6; 1-60021-563-7 (Reviewer: Yuri A. Kordyukov)

The poof is given in:

MR1066741 Reviewed Eichhorn, Jürgen Elliptic differential operators on noncompact manifolds. Seminar Analysis of the Karl-Weierstrass-Institute of Mathematics, 1986/87 (Berlin, 1986/87), 4–169, Teubner-Texte Math., 106, Teubner, Leipzig, 1988. (Reviewer: Steven Rosenberg)

The results are:

The closure of smooth functions with compact support, the closure of smooth function in the Sobolev space, and the Sobolev space are all different in general on open Riemannian manifolds. If the manifold is of bounded geometry (of order $k$) then all these spaces coincide up to Sobolev order $k+2$. This holds even for sections of vector bundles. Thus on compact manifolds all these spaces coincide also.

Answer by Peter Michor for geometric meaning of Ricci-flatness

2013-04-19T12:34:14Z

You find in Wikipedia:

"Indeed, if $\xi$ is a vector of unit length on a Riemannian n-manifold, then $Ric(\xi,\xi)$ is precisely (n−1) times the average value of the sectional curvature, taken over all the 2-planes containing $\xi$."
In Riemann normal coordinates, the Taylor expansion of the Riemannian volume has vanishing first order term, and the second order term is $1/6$ times the Ricci curvature.

Answer by Peter Michor for Contractibility of a configuration space

2013-04-17T19:04:31Z

It seems that $GVect(f)$ is a manifold without boundary. Building on my answer to your last question (did you prove all of it?) let us argue as follows: A vector field $X$ is in $GVect(f)$ if:

$X(p)=0$ for each critical point $p$ of $f$. This describes a closed linear subspace $G_1$ of the Frechet space $\mathfrak X(M)$.
Near each critical point $p$ the function $df(X)$ is Morse with a maximum at $p$. This is a $C^2$-open condition in $G_1$: The differential must be transversal to the zero section, at $p$.
Off the critical points we have $df(X)<0$. This would be a $C^1$-open condition in $G_1$ if it holds on a closed subset of $M$. We can take the closed subset as the complement of the union of small open neighborhoods of the critical points of $f$. But these neighborhoods depend on $X$. So we have to look at all of them and take the union. Since the union of open sets is open, we are done.

So $GVect(f)$ is open in the Frechet space $G_1$.

The rest seems to be done by the comments to your question, by Geoffroy.

Answer by Peter Michor for geometric interpretation of Lie bracket

2013-04-17T05:36:52Z

It is correct "that $\mathcal{L}_{X}Y=[X,Y]$ calculates changes of $Y$ along integral curve of $X$".

Edit: Namely, if $Fl^X_t$ is the flow of $X$, then $\mathcal L_XY = [X,Y] = \frac{d}{dt}|_{t=0} (Fl^X_t)^*Y$, the derivative of a smooth curve in the space of vector fields on the manifold (or on open subsets if $X$ does not have a global flow).

Spivak's description is another view. A general version of this view that "infinitesimal versions of group commutators are Lie brackets" is here:

Markus Mauhart, Peter W. Michor: Commutators of flows and fields. Archivum Mathematicum (Brno) 28,3-4 (1992), 228--236, arXiv:math.DG/9204221 (pdf).

Answer by Peter Michor for Connected groupoids and action groupoids

2013-04-17T04:57:52Z

In the $C^\infty$ case, the Lie groupoid is the action of an identity neighborhood of the Lie group. You can enlarge the action by enlarging $X$, but you may loose Hausdorff. This goes back to Palais. See (and references therein):

Franz W. Kamber, Peter W. Michor: Completing Lie algebra actions to Lie group actions. Electron. Res. Announc. Amer. Math. Soc. 10 (2004) 1-10.(pdf)

Answer by Peter Michor for A basis of the symmetric power consisting of powers

2013-04-17T09:04:40Z

Can you work with polarization?
$2 x_1x_2 = (x_1-x_2)^2 -x_1^2-x_2^2$.
If $k=2$ then $(x_i+x_j)^2$ for $i\le j$ is already a basis.
In general, $(x_{i_1}+\dots +x_{i_k})^k$ for $1\le i_1\le\dots\le i_k\le n$ might do the job.

Comment by Peter Michor

2013-05-17T13:51:26Z

Sorry, my mistake.

Comment by Peter Michor

2013-05-15T05:29:22Z

@Tom: Your last example is a sum of matrix products, not one product. Note that the rank of $AXB$ is $\le$ the minimum of the ranks, since it is the dimension of the image.

Comment by Peter Michor

2013-05-14T12:13:20Z

If (after evaluating $x_1,\dots,x_r$ at some points in $\mathbb Z$ or $\mathbb R$) the rank of $X$ is smaller than the rank of $Y$ there can be no solution. So there is a gap in your proof.

Comment by Peter Michor

2013-05-03T07:05:10Z

It is not so simple, since it also involves the central extension $Z(N)\to N\to \text{Inn}(N)$. I did not work it out.

Comment by Peter Michor

2013-04-24T06:35:17Z

@Robert: Many thanks, this is beautiful. It also seems to work in the more general case of an orientable $m$-manifold $M$ and an immersion $f:M‚Üí‚Ñù^n$ by using $$\rho(f) = f_{x_1}\wedge\dots\wedge f_{x_m}\otimes dx_1\wedge\dots\wedge dx_m \in\Omega^m(M,\Lambda^m \mathbb R^n).$$ One can use the Hodge star or not. But note that it takes values in the subset of decomposable mutlivectors in $\lambda^m\mathbb R^n$, so the Pluecker relations hold. (see (4) in mat.univie.ac.at/~michor/plue-lon.pdf why it plays no role in the the $\Lambda^2 \mathbb R^3$ case).

Comment by Peter Michor

2013-04-23T14:32:34Z

@Robert: I look forward to your description of the local solvability, many thanks.

Comment by Peter Michor

2013-04-17T18:19:21Z

@katz: Your statement "If f2 has a zero of infinite order at p, then its square root f does, as well, and hence f is smooth at p" is wrong. See section 2 of mat.univie.ac.at/~michor/roots.pdf. (But 2.5 is wrong - corrected in mat.univie.ac.at/~michor/roots2.pdf)

Comment by Peter Michor

2013-04-17T11:53:53Z

Is not $(x_1+x_1+x_2)^3$ as good as $(x_1+x_2)^3$?

Comment by Peter Michor

2013-04-17T09:25:56Z

How do you deform the group action? the members of your flat family again orbit spaces?

Comment by Peter Michor

2013-04-14T12:41:22Z

The algebra of Kaehler differentials of $A\circledS M$ generalizes the exterior algebra from vector spaces to modules over a commutative algebra, or even to bimodules over a non-commutative algebra.

Comment by Peter Michor

2013-04-09T13:05:57Z

replaced _ by _ in the last displayed equation.

Comment by Peter Michor

2013-03-25T20:56:05Z

$A$ is positive definite (since invertible). Viewing $A$ as an inner product on $\mathbb R^n$, your formula describes the natural induced inner product on the space of symmetric bilinear forms. This is the background for my description of the natural invariance group above.

Comment by Peter Michor

2013-03-25T08:32:25Z

A is positive, thus symmetric. The problem is invariant under the $GL(n)$-action $X\mapsto g^T.X g$ for $g\in GL(n)$. Any $g$ is of the form $g=n.a.k$ where $n$ is lower triangular with 1's on the diagonal, $a$ is diagonal with positive entries, and $k$ is orthogonal (Gram-Schmidt or Iwasawa). Choose $a=I$ so to not mess up too much your assumptions. There is an $n$ such that $n^T.A.n$ is diagonal with positive entries. Compute this entries. Check $n^t.B.n$. I hope this helps.

Comment by Peter Michor

2013-03-19T18:58:20Z

@Piotr: That is true for continuous functions. Spaces of smooth functions are not normable or Banach spaces.

Comment by Peter Michor

2013-03-18T18:32:58Z

Many thanks. That settles it.