bio | website | math.stanford.edu/~church |
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location | Stanford | |
age | ||
visits | member for | 5 years |
seen | 2 days ago | |
stats | profile views | 3,719 |
Oct 12 |
awarded | Yearling |
Aug 30 |
comment |
Corvallis 1979 proceedings
The AMS pages for these books are now ams.org/books/pspum/033.1 and ams.org/books/pspum/033.2; however they no longer seem to be freely available. |
Jun 26 |
reviewed | Reject suggested edit on nontrivial theorems with trivial proofs |
Jun 17 |
reviewed | Reject suggested edit on nontrivial theorems with trivial proofs |
Jun 15 |
revised |
$G_\mathbb{Z}$-homotopy type of rational Tits building $\Delta_{G, \mathbb{Q}}$
clarify result for orbits of chains, add remark on PID |
Jun 15 |
revised |
$G_\mathbb{Z}$-homotopy type of rational Tits building $\Delta_{G, \mathbb{Q}}$
correct ell = |K_0| to ell = |K_0| - 1 |
Jun 15 |
answered | $G_\mathbb{Z}$-homotopy type of rational Tits building $\Delta_{G, \mathbb{Q}}$ |
Jun 8 |
comment |
Integral of sin(x)/sqrt(x) from 0 to \pi
@NoamD.Elkies I agree that the downvotes may have been hasty; I didn't downvote, but I easily could have. However questions like this about evaluating definite integrals seem to be more at home at math.SE (even quite difficult integrals, which often receive extremely impressive answers there!). So the best outcome may be for this question to be moved to math.SE. |
Jun 4 |
awarded | Custodian |
Jun 4 |
reviewed | Reject suggested edit on nontrivial theorems with trivial proofs |
Jun 4 |
reviewed | Reject suggested edit on nontrivial theorems with trivial proofs |
May 14 |
comment |
Some question about polynomial representations of $GL(V)$
This is just about the difference between "rational" and "polynomial". Let's consider $\text{GL}(1)$. The irreducible rational representations of $\text{GL}(1)=\mathbb{C}^\times$ (or $\mathbb{G}_m$) are indexed by integers $k\in \mathbb{Z}$, with $k\in \mathbb{Z}$ corresponding to the 1-dimensional representation on which $z\in \mathbb{C}^\times$ acts by $z^k$. This is a rational function of $z$, matching the terminology. But $z^k$ is only a polynomial when $k\geq 0$, so the irreducible polynomial representations are indexed by $k\geq 0\in \mathbb{Z}$. |
May 12 |
comment |
How many proofs of the Polya's recurrent theorem are there?
For those who have access, this paper can be accessed online on the journal's website here: jstor.org/stable/10.4169/000298910X480072 For those who don't, it can also be accessed freely on the author's website here: pages.uoregon.edu/dlevin/polya.pdf |
Apr 3 |
awarded | Custodian |
Jan 25 |
comment |
Background to understand Gromov's green book
@Misha: I hadn't heard that this exercise had been solved. Can you point me in the right direction? I spent some weeks as a grad student feeling quite dumb for being unable to solve it. ;) |
Dec 21 |
comment |
When taking the fixed points commutes with taking the orbits
Dear Will, Instead of "not residually finite", do you perhaps mean "not having any finite quotients"? (e.g. H infinite simple) |
Dec 7 |
comment |
Find a simple closed curve in $S$ which represents a commutator in $\pi_1 S$
You're completely right -- thanks, Andy! |
Dec 3 |
comment |
Find a simple closed curve in $S$ which represents a commutator in $\pi_1 S$
Just to clarify, the term "conjugate" here means not that they are conjugate as elements of the fundamental group, but that they are equivalent under some automorphism of the fundamental group. The same argument shows for general $g$ that every simple closed curve in the commutator subgroup is equivalent to $[a_1,b_1]\cdots[a_k,b_k]$ for some $1\leq k<g$. |
Nov 21 |
comment |
Action of $\pi_1(S)$ on its commutator subgroup
Alex is correct (according to the questioner's comment on Al Tai's answer). However, since $G_1/G_2$ surjects to $G_1/[G,G_1]\cong \textstyle{\bigwedge}^2 \mathbb{Z}^{2g}/\mathbb{Z}$, Andy's argument does still prove that there are infinitely many orbits. |
Nov 1 |
awarded | Nice Answer |