bio  website  math.lsa.umich.edu/~speyer 

location  Ann Arbor  
age  34  
visits  member for  4 years, 11 months 
seen  2 hours ago  
stats  profile views  30,137 
Associate Professor of Mathematics at the University of Michigan. My research interests are in combinatorial algebraic geometry, particularly Schubert calculus, matroids and cluster algebras. I also enjoy thinking about number theory and computational mathematics.
6h

awarded  Revival 
17h

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Connected components of the complement of a degreed affine hypersurface
If I were trying to show that there were a collection of hyper surfaces of fixed degree and growing number of variables, I would use Viro's patchworking method arxiv.org/abs/math/0611382 . I don't have time to try it, but I might as well make sure you know about this method. 
1d

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complexity of proof of p(n) grows greater with n if for all x P(x) is unprovable?
It looks to me like the first example shows that the length of proofs of $P(n)$ can grow quite slowly (something like polynomial in $\log n$, I think) even while $\forall x : P(x)$ is independent. That's something I didn't know, so I appreciate your answer for this reason. 
2d

awarded  Nice Question 
Sep 24 
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List of integers without any arithmetic progression of n terms
For those who want to code this in a less brute force way, this is an example of a set cover problem en.wikipedia.org/wiki/Set_cover_problem: Let $X$ be the set of all $n$ term arithmetic progressions contained in $[n]$ and let $Y_i \subset X$ be the set of progressions containing $i$. Then we want to find a minimal way of writing $X = Y_{i_1} \cup Y_{i_2} \cup \cdots \cup Y_{i_r}$. If I understand the Wikipedia article correctly, the problem is NPhard, but nonetheless there are smarter things to do than pure brute force. 
Sep 22 
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when are the Kahler quotient and the GIT quotient the same?
@DanielBarter The way I learned this, one starts with the equivariant line bundle $L$ on the complex manifold $(M,J)$ and defines $\omega$ to be the curvature $(1,1)$ form of $L$. 
Sep 22 
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when are the Kahler quotient and the GIT quotient the same?
The basic result here is the KirwanNess theorem. See, for example, Theorem 5.2.1 in arxiv.org/abs/0912.1132 or Theorem 1 in arxiv.org/abs/math.LA/9911088 . I'm not sure if you are asking for something beyond this; could you spell out what more you need? 
Sep 19 
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Conceptual algebraic proof that Grassmannian is closed in Plucker embedding
I was thinking about just "image is closed" because I don't have abstract varieties available. So I can't put a variety structure on $G(k,n)$ without putting it somewhere. My thinking was, this term, to define $G(k,n)$ in the Plucker embedding, prove that each of the open affine charts is an embedding of $\mathbb{A}^{k(nk)}$ and say "we would really like to abstractly define $G(k,n)$ by gluing these $\binom{n}{k}$ many affine spaces, but we won't have the vocabulary for that until next term". So Qiaochu guessed my intended meaning, but certainly both are important to talk about. 
Sep 18 
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Conceptual algebraic proof that Grassmannian is closed in Plucker embedding
@AndyB But $PGL$ isn't projective, so why is the image closed? 
Sep 18 
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different proofs of the fact that compact riemann surface has a nontrivial meromorphic function
Related mathoverflow.net/questions/19649/… 
Sep 18 
awarded  Nice Question 
Sep 18 
answered  Conceptual algebraic proof that Grassmannian is closed in Plucker embedding 
Sep 18 
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Conceptual algebraic proof that Grassmannian is closed in Plucker embedding
Thanks! I had figured out a lot of this and was coming back to write it up, but you have found more of it. 
Sep 18 
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Conceptual algebraic proof that Grassmannian is closed in Plucker embedding
I'm not familiar with the term "dominant mapping theorem", but I suspect you mean $\phi: X \to Y$ dominant implies $\phi(X)$ contains a dense open, usually proved as a lemma on the way to Chevalley's theorem. I can rearrange things so that gets done first. At first I was worried about things like $\mathbb{A}^2$ decomposed into the image of $(x,y) \mapsto (x, xy)$ and its complement, where the lowest dimensional constructible set is not closed, but the observation that $G$invariant sets contain orbits means that $\bar{X} \setminus X$ contains an orbit. 
Sep 18 
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Conceptual algebraic proof that Grassmannian is closed in Plucker embedding
Here is how I'd rewrite this. Step 1: If $G$ acts on a projective variety $X$, there is a closed orbit. Step 2: Consider $GL(V)$ acting on $\mathbb{P} \bigwedge^k V$; let $X$ be the closed orbit. Step 3: Consider $T$ acting on $X$. Let $O$ be the closed orbit. Step 4: $O$ is closed in $\mathbb{P}\bigwedge^k V$. But explicit computation shows that the only closed $T$ orbits are the obvious fixed points, and the $GL_k$ orbit through one of them is the Grassmannian. I like it. I need to think about whether Step 1 is as straightforward as it seems, but all the rest is elementary. 
Sep 18 
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Conceptual algebraic proof that Grassmannian is closed in Plucker embedding
I love combinatorics and determinants, but this seemed a bit much (especially when I can give easier brute force proofs that omit the explicit Plucker relations.) 
Sep 18 
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Conceptual algebraic proof that Grassmannian is closed in Plucker embedding
@QiaochuYuan The proof which I was imagining is (1) write down some quadratic polynomials. There are a lot of sign issues, even if you use good notation with exterior algebras. (2) Prove that a rank $1$ wedge obeys these relations, which involves trickery with determinants. (3) Let $p_I$ obey these relations and WLOG $p_{[k]} \neq 0$. Define a $k \times n$ matrix whose first $k$ columns are the identity and whose other entries are $\pm p_{[k] \cup \{ i \} \setminus \{ j \}}/p_{[k]}$. Show that the minors of this matrix are the $p_I$. 
Sep 18 
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Conceptual algebraic proof that Grassmannian is closed in Plucker embedding
I like this, because it is analogous to the Shavarevich (and many other sources) proof that projective maps are closed, expressing the concept that $f_1$, $f_2$, ..., $f_r$ have a common zero as a certain map between graded pieces of a symmetric algebra having low rank. I can analogize that this is a similar argument in the exterior algebra. 
Sep 18 
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Conceptual algebraic proof that Grassmannian is closed in Plucker embedding
@darijgrinberg Ah, thanks for the write up. It looks to me like this proof doesn't use the Plucker relations, but rather uses the $(nr+1) \times (nr+1)$ minors of the map $\phi_x: V \to \bigwedge^{r+1} V$. (I always understood the Plucker relations to be quadratic polynomials.) As a result, this proof probably doesn't give a generators for the saturated ideal of the Grassmannian, but it does give generators for some other ideal representing the Grassmannian in a very slick way. 
Sep 18 
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Conceptual algebraic proof that Grassmannian is closed in Plucker embedding
@user52824 I'm trying to unravel your meaning. I know that Grothendieck's dual convention is that $G(k,n)(R)$ parametrizes surjections $R^n \to R^k$. What actual morphism are you saying I should consider here? 