bio | website | vmm.math.uci.edu |
---|---|---|
location | Univ. of California at Irvine | |
age | 83 | |
visits | member for | 4 years, 7 months |
seen | Jan 21 at 16:50 | |
stats | profile views | 6,876 |
I'm a Professor at UC Irvine, but spent most of my career at Brandeis. My research areas: Differential Topology, Transformation Groups, and Global Analysis. Recently I developed a math visualization program, 3D-XplorMath, freely available at http://3D-XplorMath.org) and a companion website, the Virtual Math Museum at http://VirtualMathMuseum.org. Last year I co-authored a differential equations text with my son Bob, most of which is downloadable from http://ode-math.com .
Jan 13 |
answered | Lie group action with no slice |
Jan 10 |
awarded | Popular Question |
Dec 28 |
awarded | Good Answer |
Dec 28 |
awarded | Nice Question |
Dec 9 |
awarded | Enlightened |
Dec 7 |
awarded | Nice Answer |
Dec 7 |
comment |
Notes for Bott's 1963 lectures on Morse theory
I did some more searching with Google and found that Amazon has a reference to: "Lectures on Morse theory: [revised and expanded version of notes of lectures delivered at Professor R. Bott's topology seminar at Harvard in February and March of 1963 [Unknown Binding] Richard S Palais (Author)" with Brandeis Univ, listed as the publisher. I suspect this must have been an early preprint version of "Morse Theory on Hilbert Manifolds" |
Dec 7 |
answered | Notes for Bott's 1963 lectures on Morse theory |
Nov 7 |
answered | Any good books on numerical methods for ordinary differential equations? |
Oct 16 |
comment |
Proof synopsis collection
@mathahada Where do you see a geometric series in my proof? (One of the points I had in publishing the above proof was to show that the geometric series argument (used since Banach's original proof) is really not necessary. |
Oct 14 |
comment |
Can the level set of a critical value be a regular submanifold?
You probably should say amend the statement of the theorem to say that a non-empty level set of a regular value of a smooth function f:M→ℝ on a smooth manifold is a regular submanifold of codimension one. (That takes care of the problem with f identically zero.) |
Oct 11 |
answered | Proofs for doubly ruled surfaces |
Sep 21 |
revised |
Definition of Sobolev spaces as a space of sections of certain type
Added a link to a copy of referenced work; deleted 1 characters in body; added 1 characters in body |
Sep 20 |
answered | Definition of Sobolev spaces as a space of sections of certain type |
Sep 19 |
comment |
Collapsing of Riemannian manifolds with a group action
"...Consider the fixed point set F, it is of course a submanifold of M by the slice theorem". Note that it is really simpler than that; in geodesic coordinates at a point p of F, the fixed point set is locally the linear subspace left fixed by the linearized action at p. |
Sep 15 |
comment |
First known proof of $\sqrt 2$ is irrational with prime factorization?
Your right Franz, it doesn't. It's just that there seems to be a belief that you NEED unique prime factorization to prove the irrationality of non-square integers, and when I first saw this (much more elementary) proof I found it an eye-opening experience. |
Sep 14 |
answered | First known proof of $\sqrt 2$ is irrational with prime factorization? |
Sep 3 |
comment |
Area of union of random circles in a plane
You will probably get a more "natural" answer if you choose a "torus", i.e., identify opposite edges of a square, to eliminate edge effects. |
Aug 31 |
awarded | Good Answer |
Aug 29 |
awarded | Nice Answer |