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Complex, contact, Riemannian, pseudo-Riemannian and Finsler geometry, relativity, gauge theory, global analysis.

9 votes

Geodesics on a Grassmannian

This is a variation on the 1st answer, but I find it more straightforward and have explained it to EE students. Consider the map $U(n) \to Gr(k,n)$ from the unitary group to the Grassmannian by $g …
Richard Montgomery's user avatar
2 votes

connections on AN(3) Lie group

The space of connections forms an affine space, so you can take convex combinations and average. Thus you can take the average of your connections also.
Richard Montgomery's user avatar
1 vote

Variational characterization of curvature?

Take a circle of radius $r$ about a point $p$ (metric concept). Compute its circumference (metric concept). Compare $C(r)$ to $2 \pi r$ in the limit as $r \to 0$ to get the curvature $K(p)$ at $p$. …
Richard Montgomery's user avatar
3 votes

Morse-Bott homology for infinite-dimensional manifolds

See Atiyah and Bott's papers on Yang-Mills for Riemann surfaces. The essence of it is that the Yang-Mills functional acts like a perfect Morse functional after quotienting by gauge transformations. So …
Richard Montgomery's user avatar
3 votes

Walking around Santa Cruz, track around the soccer field

Allow self-intersections. Take the symmetric figure eight. That does the trick. And it has the bonus that each half is convex. This is what Kloeckner might have been trying to say. ( As a kid, we …
Richard Montgomery's user avatar
2 votes

Deformation of Lagrangian manifolds

In general, deformations of a submanifold L of an ambient space M are identified with sections of L's normal bundle: $TM|_{L}/TL$. For your case, the normal bundle is canonically isomorphic to $T^*L …
Richard Montgomery's user avatar
0 votes

A strongly non-integrable distribution

I believe the standard example $dz - ydx = 0$ does the trick. This is the normal form for what many people call a ``quasi-contact distribution'-the even dimensional analogue of a contact form. Suppo …
Richard Montgomery's user avatar
4 votes

cotangent bundle symplectic reduction and fibre bundles

This is more a comment towards Gourishankar than an answer to the original question. It was part of my thesis, (UCB, about 1986), so, apologies, I chime in. For simplicity, I take the case $G$ Abeli …
Richard Montgomery's user avatar
0 votes

Solving the geodesic equation for a singularity crossing curve

Perhaps you mean to say that your metric behaves rather like those arising in Snell's law from geometric optics where to model the refraction of light as it passes from one medium to another you t …
Richard Montgomery's user avatar
1 vote

Integrable compatible complex structures

Yes! It can happen and is quite important. The Teichmüller space of a Riemann surface provides a counterexample to your discreteness assertion. See, eg. Tromba's book, Teichmüller Theory in Riemannian …
Richard Montgomery's user avatar
16 votes
Accepted

Why is proving $C^{\infty}$ regularity of sub Riemannian geodesics so hard?

The reason the problem is hard is that we do not have a good handle on what abnormal (=singular) geodesics can look like. See the chapter of my book that describes abnormal geodesics. Progress is …
Richard Montgomery's user avatar
16 votes

Frobenius Theorem for subbundle of low regularity?

You can even get a Frobenius theorem for Lipshitz vector fields, which need not span everywhere (the rank can drop along closed subsets). The state of the art in this domain is done by the control th …
Richard Montgomery's user avatar
4 votes

parallel transport along $W^{1,2}$-curves

You will find a proof in Appendix D (Theorem D.1) of my book `A Tour of SubRiemannian Geometry'. It may not look like what you want at first glance, since that theorem is stated in a more general c …
Richard Montgomery's user avatar
5 votes

G-bundles in classical mechanics

Take the planar three-body problem. Or, said a bit differently, take that 'cat' to consist of three point masses moving about in the plane -- a triangle! Fix the center of the mass at the origin by …
4 votes

How to motivate and interpret the geometric solutions of Hamilton-Jacobi equation?

The famous KAM tori arose out of HJ considerations. They are Lagrangian torii. They were found by attempting to solve the HJ equation generally, and then finding one can only solve it when certain …
Richard Montgomery's user avatar

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