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bio website math.berkeley.edu/~qchu
location Berkeley, CA
age 24
visits member for 4 years
seen 7 mins ago

I'm a second-year graduate student in pure mathematics at UC Berkeley. You can contact me at qchu[at]math[dot]berkeley[dot]edu.


23h
revised What does $GL_n(R)$ look like?
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23h
comment Spherical geometry as an example of non euclidean geometry
All of the difficulties are conceptual; the hard part is recognizing that spherical geometry deserves to be called geometry at all. Obviously arcs aren't straight lines, in the same way that obviously negative numbers aren't numbers (e.g. if you think that numbers measure lengths, obviously negative lengths are meaningless).
23h
comment What are the continuous automorphisms of $\Bbb T$?
Yes, that's correct. (Pontryagin duality tells you this as well.)
1d
revised What does $GL_n(R)$ look like?
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1d
comment What does $GL_n(R)$ look like?
@user: if $\lambda$ means the diagonal matrix with diagonal values $\lambda$, then this is not true. You need to pick a splitting of the exact sequence $\text{SL}_n \to \text{GL}_n \to \text{GL}_1$ and there are many such splittings.
1d
answered What does $GL_n(R)$ look like?
1d
comment In $\mathbb{Z}/(n)$, does $(a) = (b)$ imply that $a$ and $b$ are associates?
It should more precisely be "if you want something, try to make it happen, then verify that it actually happened." The universal counterexample is a counterexample iff any counterexample is a counterexample, so without more work (which you do in the parentheses), exhibiting the universal counterexample doesn't by itself settle the question.
1d
comment homotopy class of maps in terms of homotopy groups of spectra
The naive guess is much too naive; if it were true then there would be essentially no difference between spectra and sequences (not even chain complexes) of abelian groups. In particular the naive guess would imply that there are no interesting maps between Eilenberg-MacLane spectra, so in particular no interesting stable cohomology operations. The generating hypothesis fixes things by restricting to finite spectra and keeping track of a natural source of extra structure on the homotopy groups.
2d
awarded  Notable Question
Aug
14
awarded  Good Answer
Aug
13
awarded  Nice Answer
Aug
13
comment Doubts about the fundamental group of a mapping torus
@Daniel: I certainly need $f$ to be a homotopy equivalence; sorry for omitting that condition. I'm not sure how important it is that $f$ is a homeomorphism.
Aug
12
comment Can $R \times R$ be isomorphic to $R$ as rings?
@blue: a direct sum of unital rings need not be unital, though.
Aug
12
comment Doubts about the fundamental group of a mapping torus
You mean $f : X \to X$. It is possible to compute the fundamental group of a generic mapping torus: I claim it is always the semidirect product of $\pi_1(X)$ by $\mathbb{Z}$, with $1 \in \mathbb{Z}$ acting by $\pi_1(f) : \pi_1(X) \to \pi_1(X)$.
Aug
12
answered Why do we care about simple groups rather than indecomposable groups?
Aug
12
comment Is this a fruitful enrichment of $R[X]$?
See the paragraph above "a quick remark" at qchu.wordpress.com/2013/06/09/operations-and-lawvere-theories.
Aug
11
comment Finite topologies — what are they good for?
mathoverflow.net/questions/177461/…
Aug
11
comment Is this a compact group?
That's not specific enough. For example, if the underlying Hilbert space is allowed to be infinite-dimensional, then it's unclear what you mean by "the Euclidean topology." There are various topologies you can put on the space of operators on an infinite-dimensional Hilbert space for various reasons.
Aug
11
revised Does loga/logb = log(a^(1/logb))?
edited tags
Aug
11
revised How to solve “So Who's Counting” problem using Markov Decision Process?
edited tags