bio  website  math.berkeley.edu/~qchu 

location  Berkeley, CA  
age  24  
visits  member for  5 years 
seen  3 hours ago  
stats  profile views  53,028 
I'm a thirdyear graduate student in pure mathematics at UC Berkeley. You can contact me at qchu[at]math[dot]berkeley[dot]edu.
1h

awarded  Nice Question 
3h

answered  Is there any relationship between the Euler class and the Vandermonde determinant? 
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comment 
How to extend index theorem to infinite dimensional manifolds?
I don't really know what you have in mind on the analysis side. The indextheoretic description of the Euler characteristic involves constructing, one way or another, the Euler class of a compact oriented manifold, which lives in top cohomology. Where could the Euler class possibly live on an infinitedimensional manifold? 
1d

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How to extend index theorem to infinite dimensional manifolds?
The Euler characteristic of a contractible space is (perfectly welldefined and equal to) $1$, not $0$. Roughly speaking the problem here is a failure of monotone convergence (thinking of the Euler characteristic as a kind of measure): the spheres only have homology in two places and when passing to the limit above, the homology in the topdimensional place has escaped off to infinity. That's nothing too unexpected. 
2d

answered  Take contraction wrt a vector field twice and define kernel mod image. Does that give anything interesting? 
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Quotient of the hyperbolic plane with respect to commutator group of $\pi_1(\Sigma_g)$
I would run that argument the other way: $\mathbb{H}/[G, G]$ is an infinite covering space of $\Sigma_g$, hence it is noncompact. As a noncompact surface, it's homotopy equivalent to a $1$dimensional CW complex, hence its fundamental group is free. 
2d

awarded  Nice Answer 
Oct 17 
awarded  Excavator 
Oct 17 
revised 
Wonderful applications of the Vandermonde determinant
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Oct 17 
awarded  Popular Question 
Oct 15 
asked  Is there any relationship between the Euler class and the Vandermonde determinant? 
Oct 15 
comment 
“extended TQFT” versus “TQFT with defects”
I think you want the cobordism hypothesis with singularities (Theorem 4.3.11 in arxiv.org/abs/0905.0465). 
Oct 13 
awarded  Yearling 
Oct 8 
revised 
Why the DoldThom theorem?
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Oct 8 
revised 
Why the DoldThom theorem?
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Oct 8 
revised 
Why the DoldThom theorem?
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Oct 8 
answered  Why the DoldThom theorem? 
Oct 7 
awarded  Necromancer 
Oct 6 
comment 
Classifying space for fibrations with EilenbergMacLane space as fibers
What's true with no hypotheses on the base is that $[B, K(\pi, n+1)]$ classifies principal $K(\pi, n)$fibrations. 
Oct 6 
comment 
From the perspective of bordism categories, where does the ring structure on Thom spectra come from?
@Thomas: yes, but my comment above is not about the cartesian product of manifolds. It is about the cartesian product in $\text{Bord}$. 