16,189 reputation
33172
bio website rybu.org
location Victoria, Canada
age 41
visits member for 4 years, 5 months
seen Jan 22 at 14:27

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


Jan
12
comment Barycentric subdivision
Also asked here, simultaneously: mathoverflow.net/questions/193744/barycentric-subdivision
Jan
11
comment When should I be doing cohomology?
I think learning a subject like obstruction theory is very informative on this kind of question that you have. Obstruction theory in a sense was the reason people took cohomology seriously -- that it wasn't just some formal trinket. It has to do with the problem of assembling types of "global objects" from things like local data -- or any kind of intermediate data -- stuff that's smaller than the entire space.
Dec
30
comment Actual Classification re Nielsen-Thurston Theorem (how to)?
The question in your title does not agree with the question in the body of your post. There is of course oodles of uses of the Thurston classification. For example, you can describe when two mapping classes commute in terms of it. You can describe when a bundle over the circle is a hyperbolic 3-manifold in terms of it. And so on.
Dec
16
comment Is every self-homeomorphism homotopic to a diffeomorphism?
That theorem in Hirsch's text is much stronger than what you quote. Hirsch proves a $C^1$ diffeo is isotopic to a $C^\infty$ diffeo, not just homotopic (a strange relation on diffeomorphisms). Are you sure this is the question you really want to ask?
Dec
15
awarded  Caucus
Dec
15
awarded  Enlightened
Dec
15
awarded  Nice Answer
Dec
12
comment The missing vector derivative operation
The linear transformation of the plane $(x,y) \longmapsto (y,-x)$ is rotation by 90 degrees, or in the complex plane, multiplication by $i$. That's what I meant.
Dec
12
comment The missing vector derivative operation
This is the Hodge star operator applied to the gradient. In my calculus courses I just call the operation "rotation by 90 degrees". In ODE courses sometimes this is called a planar Hamiltonian system. I think that's the bulk of the verbiage surrounding this construction.
Dec
11
revised Limit of solution of linear system of ODEs as $t\to \infty$
edited body
Dec
11
revised Limit of solution of linear system of ODEs as $t\to \infty$
added 1018 characters in body
Dec
10
answered Limit of solution of linear system of ODEs as $t\to \infty$
Nov
25
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.
Nov
24
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.
Nov
24
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).