bio | website | vmm.math.uci.edu |
---|---|---|
location | Univ. of California at Irvine | |
age | 84 | |
visits | member for | 5 years, 2 months |
seen | May 14 at 20:21 | |
stats | profile views | 7,228 |
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 .
May
11 |
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What sort of models did Bolyai and Lobachevsky use to demonstrate the consistency of their models of non-Euclidean Geometry?
>... but neither Bolyai nor Lobachevsky proved that their geometry was internally consistent... That's what I was beginning to suspect, Igor, but is that something that is "As far as I know...", or do you have some strong reason to believe this---say perhaps a well-researched historical reference. |
Sep
30 |
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About Palais' remark that an isometry of Riemannian manifolds does not induce an isometry of the Hilbert manifolds of curves
The problem lies in the line above defining $g_\sigma(λ,μ)$. This is the metric for $H_0$, NOT the metric for $H_1$. The correct definition of the metric for $H_1$ uses $⟨λ'(t),μ'(t)⟩$ rather than $⟨λ(t),μ(t)⟩$ (see page 222 of my article where both metrics are defined), and with this change it is fairly obvious why my remark in the cited paper is in fact correct. |
Aug
23 |
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Algorithm for finding inverse images of a local diffeomorphism
Yes! Newton's Method is clearly the way to go. For $k=1$, the algorithm I wrote in fact did use Newton's Method, (rather than bisection) since it is much faster. But I guess I forgot that Newton's Method works in higher dimensions too. Thanks Ryan. |
Aug
22 |
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Algorithm for finding inverse images of a local diffeomorphism
@Nanda For the application I have in mind, F will be given by a formula. |
Jul
11 |
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What is a good method to find random points on the n-sphere when n is large?
@Mark Meckes Many thanks, Mark. In retrospect, if I had been more adept with my Googling I might have discovered that the Gaussian RV approach was a "well-known" answer to my question, and would not have needed to ask it on MO. But as usual, I have come away very impressed with the utility of this great website and learned a lot from the answers I received. |
May
30 |
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A Canonical Form Theorem for $n$-forms?
@Francesco YES ! I don't think you are missing anything---that is essentially the same proof as mine. I'll have to get Kobayashi's book from the library and have a look. Many thanks for this clue. |
May
30 |
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A Canonical Form Theorem for $n$-forms?
@Mariano. Right, that is the way I think of it, and it was Darboux Theorem that made me think it might be true. |
May
22 |
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Why don't more mathematicians improve Wikipedia articles?
@Mark M YES! That is a very good approximation of what I was asking for. Thanks for telling me about it. It is now one of my "pinned" tabs and I will try to help with the project. |
May
17 |
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Are there proofs that you feel you did not “understand” for a long time?
How about the proof that Einstein is said to have (re)discovered as a teen-ager? It is clear that the areas of similar right triangles is proportional to the squares of corresponding sides---in particular to the squares of their hypotenuses. Dropping a perpendicular from the vertex of the right triangle onto the hypotenuse c divides it into two similar triangles with hypotenuses a and b, hence: kc^2 = ka^2 + k*b^2 so c^2 = a^2 + b^2 |
Apr
20 |
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Which hard mathematical problems do you have to solve to earn bitcoins ?
Is it known whether there is an algorithm using a quantum computer (i.e., qubits instead of bits) for simplifying the Bitcoin mining? |
Apr
11 |
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Fixed point theorems
...has a unique solution, provided ε is sufficiently small. |
Jan
13 |
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Lie group action with no slice
@Yves Cormulier I don't think that's so Yves. If it were then even a compact group action wouldn't have a slice at an isolated fixed point p, whereas in fact an invariant neighborhood of p is a slice at p in that case. |
Jan
13 |
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Lie group action with no slice
After thinking about my answer a bit I realized that an even simpler example is the action of the group Z of integers on the circle generated by rotation through an irrational angle. |
Dec
7 |
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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" |
Oct
16 |
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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 |
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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.) |
Sep
19 |
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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 |
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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
3 |
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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
25 |
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Square root of a positive $C^\infty$ function.
Yes, and functions of this type are discussed in section 2 of the reference I gave in my answer. |