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bio website math.berkeley.edu/~cgerig
location UC Berkeley
age 25
visits member for 3 years, 10 months
seen 42 mins ago

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!


2d
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 combinatorial-game-theory 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 Dold-Thom 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 Dold-Thom theorem?
Oct
23
comment Why the Dold-Thom 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 push-forward the image of the fundamental class $[S_f]$? Why would this be the "correct" element?
Oct
23
revised Why the Dold-Thom theorem?
Consolidating the important comments
Oct
23
comment Why the Dold-Thom 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 Dold-Thom 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 CW-complex. 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
comment Teaching the fundamental group via everyday examples
Should this be Community Wiki? If applications of Brouwer fixed-point 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
comment 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 Dold-Thom theorem?
I am confused. Could you elaborate in your answer on forming that new CW-complex? I don't see how to put together the union of copies of k-cells with varying k -- what are the attaching maps? I would think this affects whether the resulting "complex" could even have a fundamental class.