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1d
comment Topological idea of orientability of manifold
@Mike: You know more about topological manifolds than I do!
2d
comment Topological idea of orientability of manifold
At least for a smooth manifold, this idea is equivalent to the usual notion of orientability. The idea of the proof uses normal bundles of embedded closed curves.
2d
comment How to show $[\omega]=0$ implies $[\omega^n]=0$?
Thanks! And sorry to resurrect such an old question!
Aug
28
comment How to show $[\omega]=0$ implies $[\omega^n]=0$?
Perhaps I misunderstood the question or your answer, but I don't think the point is to show that $d(\eta^n) = 0$ (it must be in order for $[\eta^n]$ to make sense) but rather that $[\eta^n]=0$, meaning $\eta^n = d\alpha$ for some $n-1$ form $\alpha$.
Aug
28
comment How to show $\omega^n$ is a volume form in a symplectic manifold $(M, \omega)$?
For those wondering why I edited this old post, apparently a typo at the end caused confusion to at least one person (math.stackexchange.com/questions/1411971/…)
Aug
28
comment Symplectic form and wedge sum
Sorry to have made that mistake at the end of my answer! (and thanks to Ted for suggesting I correct it)
Aug
28
comment How to show $\omega^n$ is a volume form in a symplectic manifold $(M, \omega)$?
@Ted: Done. To whomever I caused confusion - sorry!
Aug
28
revised How to show $\omega^n$ is a volume form in a symplectic manifold $(M, \omega)$?
Fixed the type in penultimate line, as it's caused some confusion.
Aug
25
comment Groups of order $8n$ have at least five distinct conjugacy classes
I'm sorry, but I'm still not sure why the maximal Sylow $p$-group is normal. What about a simple group, like $A_n$ ($n\geq 5$)? Such a group has order $n!/2$ which is of the form $8k$ once $n \geq 6$.
Aug
25
comment Groups of order $8n$ have at least five distinct conjugacy classes
I am probably missing something, but how do you use Sylow's first theorem to prove $H$ is a normal subgroup?
Aug
25
revised Topological spaces admitting an averaging function
added 103 characters in body
Aug
25
comment Topological spaces admitting an averaging function
@Qiaochu: You're right, of course. Thanks for catching that! (Fixed now)
Aug
25
revised Topological spaces admitting an averaging function
added 646 characters in body
Aug
24
awarded  Nice Answer
Aug
24
comment Topological spaces admitting an averaging function
@Eric: Very nice.
Aug
24
comment Topological spaces admitting an averaging function
So, having ruled out many manifolds from admitting averaging functions, the natural question is: Does anything other than Euclidean space support an averaging function? Note that if a compact manifold admits an averaging function, then every $\pi_k$ with $k\geq 2$ is a finite group of odd order. Can this actually occur?
Aug
24
answered Topological spaces admitting an averaging function
Aug
21
comment Algebraic Topology Challenge: Homology of an Infinite Wedge of Spheres
Very nice! (and more characters)
Aug
21
comment Vector valued 2-forms which satisfy Jacobi Identity
Well, perhaps that is what I mean :-). On a related note, I don't think there are any simple Lie algebras in dimension 4,5,6, 7, 9, 11, ..., so there should be plenty of examples of 1) in these dimensions. On the other hand, every orientable three manifold is parallelizable, so these are all examples of 2).
Aug
20
comment Vector valued 2-forms which satisfy Jacobi Identity
My understanding is that there are, up to isomorphism, precisely two 2-dim Lie algebras: the abelian one and one with $[x,y] = x$. But neither is simple. So it seems as though any surface gives an example of 1). That said, you mention in the linked MO question that every parallelizable manifold has the "tensorial property", so perhaps I'm missing something...