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bio website rybu.org
location Victoria, Canada
age 41
visits member for 4 years, 3 months
seen 10 hours ago

I am a professor of mathematics at the University of Victoria, in Canada.


1d
comment Are compact complete geodesics closed?
Technically we don't need to know that, as we've already cornered ourselves into a contradiction using the E&U theorem.
1d
comment Are compact complete geodesics closed?
If $\vec v(t)$ is a solution, $\vec v(n)$ for $n$ a natural number is a sequence. So it has a convergent subsequence, being in a compact space. I call this the "endpoint" of the curve. Now apply the existence and uniqueness theorem to that endpoint.
1d
comment Are compact complete geodesics closed?
There's a simpler proof of the theorem. If the geodesic is compact, a little playing around with limits will tell you it has a well-defined endpoint, i.e. the limit as $t \to \infty$ exists. Check to see if the vector field is non-zero at this "end point" and apply the existence and uniqueness theorem to complete the proof.
Nov
12
comment Orientability of integrable plane fields
Orientability isn't enough in general. For 2-manifolds it is, but not for 3-manifolds. For example, $S^1 \times \mathbb R^2$ has an integrable plane field where the foliation consists of a Moebius band and a 1-parameter family of cylinders. You should think of this as a regular neighbourhood of a Moebius band.
Nov
11
comment Orientability of integrable plane fields
The tangent bundle of the Klein bottle splits as a direct sum of a trivial bundle and a non-trivial bundle. It's that non-trivial line bundle you're interested in, so your problem is equivalent to finding a non-zero vector field on the Klein bottle -- just take the complementary line bundle once you've found that. The vectors field comes from a fixed point free action of $S^1$ on the Klein bottle.
Nov
10
comment Orientability of integrable plane fields
Nope. It's the same argument in dimension $1$, if you replace plane field by line field.
Oct
24
comment Does Stokes' Theorem hold on spaces with singular points?
I suppose it depends on how you would like to state it, but I suspect it would naturally be stateable pretty much exactly like the normal Stokes theorem. One would only have to worry about special properties of residues when integrating singular objects, distributions and such.
Oct
23
comment the 0-th homology of a simplical complex
Well, it's not always nontrivial. If your notion of simplicial complex allows for the empty complex, then the 0th homology can be trivial. But provided your complex is non-empty yes, you can prove it is always non-trivial. Look up the definition of reduced homology and the proof that the reduced homology has rank one less than the normal homology (in dimension zero).
Oct
23
comment Does Stokes' Theorem hold on spaces with singular points?
Stokes' theorem is one that likes to be true in almost any context you can imagine. Manifolds are the "deluxe" place where the proofs are particularly easy. But there are versions of Stokes' theorem on all kinds of relatively degenerate spaces. Stratified spaces, orbifolds and such. I suspect the only reason why there's no proof in "full generality" is it's not clear how general a theorem people would like. So people just prove versions of it as they need it. Off the top of my head I don't see any problems proving it for orbifolds, but I have to run off to teach now...
Oct
23
comment Show every $f_t$ is Morse for $t$ is sufficiently small
You are trying to argue that Morse functions form an open set in the space of functions (with the $C^k$-topology $k \geq 1$). This is a basic exercise. One way to think about it is a function is Morse if the derivative (thought of a certain way) is transverse to the 0-section of the cotangent bundle. Transversality is an open condition, so that would finish the job. Alternatively you could express it locally in terms of the non-degeneracy of the Hessian.
Oct
22
comment Programmatically recognizing symmetries of a polyhedron
It sounds like maybe you haven't given us enough information. Is your polyhedron 2-dimensional? Is it sitting in a euclidean space, or is it abstract? I suspect you're dealing with a polyhedron in 3-space. Determining the symmetry type amounts to computing on each face things like: fixed vertices, fixed points on edges, fixed points on the 2-dimensional facets. These ultimately boil down to understanding the permutations on the vertices and thinking through all the possibilities.
Oct
22
comment Fundamental group of an orientable surface of infinite genus.
I was describing homotopy-type, not homeomorphism type.
Oct
19
awarded  Custodian
Oct
19
reviewed Looks OK How to find equations of tangent lines to the graph passing through a line
Oct
19
reviewed Looks OK approximations of discrete distributions
Oct
19
reviewed Looks OK Prove $2n+3 \le 2^n$ for all integers $n \ge 4$.
Oct
19
reviewed Looks OK Evaluating the integral of $1+z+1/\tan z$ over a circle
Oct
19
reviewed Close How to use quotient rule to differentiate $f(t)=\frac{\cos t}{t^3}$?
Oct
19
reviewed Close Give the parity check matrix H of a binary Hamming code of length 15.
Oct
19
comment Can a region always be parametrized by a single function?
Without specifying what you mean, answering your question is impossible. The answer is yes or no depending on what you mean by "nice" and what you mean by parametrized.