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


7h
awarded  Nice Answer
23h
awarded  ring-theory
1d
comment Why only two binary operations?
@isomorphismes: well, who knows? We could spend all day arguing about precisely what is and is not natural. What I'm willing to say is that categories have been a very fruitful point of view historically and I expect they'll continue to be in the future.
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revised Why only two binary operations?
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1d
answered Why only two binary operations?
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comment Compactness of Lie groups
@Brenin: I mean two things by this, which is why I didn't want to be precise. First, $\text{GL}_n(\mathbb{C})$ deformation retracts onto $\text{U}(n)$, so the two are homotopy equivalent. But second, it turns out that the two have essentially the same representation theory, provided that you're careful to restrict your attention to algebraic representations of $\text{GL}_n(\mathbb{C})$.
1d
answered automorphism of the projective space $\mathbb{P}_A^n$
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comment Calculating $\pi_2$ of a certain free loop space
Okay, strictly speaking that first comment only shows that we get some semidirect product decomposition, not that we get the one associated to the usual action of $\pi_1$ on $\pi_2$. But this probably comes from thinking a bit harder about how the fibration $\Omega X \to LX \to X$ works (which I haven't, yet).
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comment Calculating $\pi_2$ of a certain free loop space
Incidentally, the simplest example of a space I know with nontrivial action of $\pi_1$ on $\pi_2$ is $X = \mathbb{RP}^2$, where $\pi_1 = \mathbb{Z}_2$ acts on $\pi_2 = \mathbb{Z}$ by reversing signs. So there's your interesting semidirect product.
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comment Calculating $\pi_2$ of a certain free loop space
Okay, I know how to prove that semidirect product decomposition now. First, with all the other $\pi_k$ computed, you can show that two important boundary maps in the long exact sequence in homotopy associated to the fibration $\Omega X \to LX \to X$ vanish (assume $X$ path-connected here or else I have to specify which basepoint I take based loops at), giving a short exact sequence $\pi_2(X) \to \pi_1(LX) \to \pi_1(X)$. Second, the inclusion $X \to LX$ of constant loops into all loops is a splitting of this sequence.
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revised Calculating $\pi_2$ of a certain free loop space
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answered Calculating $\pi_2$ of a certain free loop space
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comment When is the free loop space simply connected?
@Jason: Yes, maybe I should've mentioned that it's possible to be very explicit about what $\pi_0(LX)$ is. I am less sure about $\pi_1(LX)$ though. Perhaps it's the semidirect product of $\pi_1(X)$ and $\pi_2(X)$ in general?
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comment Could the equivalence classes in the construction of quotient group be the orbits of some group action?
Yes. It's $S$ acting on $G$ by right multiplication.
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answered Compactness of Lie groups
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revised When is the free loop space simply connected?
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comment When is the free loop space simply connected?
@Mike: ah, you're right of course. The special fact about compact Lie groups I had in mind is a corresponding fact about Lie algebra cohomology.
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comment When is the free loop space simply connected?
At least in the last paragraph you are conflating the free and based loop spaces.
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comment When is the free loop space simply connected?
Trying to contract a loop of loops a priori involves $\pi_2(X)$, not just $\pi_1(X)$. I explain this in more detail below.
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answered When is the free loop space simply connected?