bio  website  math.berkeley.edu/~cgerig 

location  UC Berkeley  
age  25  
visits  member for  3 years, 10 months 
seen  42 mins ago  
stats  profile views  6,100 
After doing my BS in engineering physics, I started my PhD in experimental atomic physics. But I quit to do math, and am now a student of Michael Hutchings!
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comment 
Inverted pair of complex analytic families
I don't quite understand the dictionary between the referenced paper and the problem here, can you explain further? 
Nov 23 
revised 
Inverted pair of complex analytic families
added 416 characters in body 
Nov 22 
reviewed  Approve suggested edit on combinatorialgametheory tag wiki 
Nov 19 
revised 
Inverted pair of complex analytic families
added 362 characters in body 
Nov 19 
revised 
Embedded Contact Homology and Manifold Decompositions
added 257 characters in body 
Nov 19 
answered  Embedded Contact Homology and Manifold Decompositions 
Nov 17 
asked  Inverted pair of complex analytic families 
Nov 2 
awarded  Popular Question 
Oct 23 
comment 
Why the DoldThom theorem?
I don't, but I also tend not to gain intuition by making things more abstract (that's a statement about me). Jesse's answer, along with the comments under the question, does seem to provide an intuition about the two things directly; although I will admit that I prefer something similar in spirit to Ryan's suggestion. 
Oct 23 
accepted  Why the DoldThom theorem? 
Oct 23 
comment 
Why the DoldThom theorem?
OK, if I buy this and $S_f$'s existence, then to get an element of $X$'s homology, I take the composition $S_f\hookrightarrow S^i\times X\twoheadrightarrow X$ and pushforward the image of the fundamental class $[S_f]$? Why would this be the "correct" element? 
Oct 23 
revised 
Why the DoldThom theorem?
Consolidating the important comments 
Oct 23 
comment 
Why the DoldThom theorem?
So I think this response shifts my question onto another isomorphism, and my question is then: "Why is $\pi_*(F(X))\cong \widetilde{kO}_*(X)$ true, intuitively?" I see a proof of the isomorphism, but I think it skirts my question. 
Oct 22 
comment 
Why the DoldThom theorem?
I'm sorry I cant' wrap my head around this yet, even for $i=1$. In the first paragraph, $k$ can't vary across the cells of $S^i$, right? And I don't see how the "subspace" of $S^i\times X$ come together to form a CWcomplex. I also didn't understand the last sentence; how does the map $S_f\to S^i$ relate to "lifted maps to $X^k$"? And how do I ultimately get an element of homology of $X$? 
Oct 21 
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Teaching the fundamental group via everyday examples
Should this be Community Wiki? If applications of Brouwer fixedpoint theorem are allowed, you can use the one about placing a map on a table, or others from this similar thread: mathoverflow.net/questions/19272/… 
Oct 19 
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Is there any relationship between the Euler class and the Vandermonde determinant?
Here is a weak relationship I know of, which of course doesn't relate to Wiki's claim: You can compute the total Chern class of the flag manifold, written as a product of 2nd degree cohomology classes. Then you use Vandermonde determinants to see that its Todd genus equals 1. (Chapter III.14 of Hirzebruch's Topoloical Methods in Algebraic Topology) 
Oct 18 
reviewed  Approve suggested edit on Estimating the number of clusters 
Oct 15 
reviewed  Reject suggested edit on Refereeing a Paper 
Oct 13 
awarded  Popular Question 
Oct 13 
comment 
Why the DoldThom theorem?
I am confused. Could you elaborate in your answer on forming that new CWcomplex? I don't see how to put together the union of copies of kcells with varying k  what are the attaching maps? I would think this affects whether the resulting "complex" could even have a fundamental class. 