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bio website math.lsa.umich.edu/~speyer
location Ann Arbor
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visits member for 4 years, 11 months
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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
comment Connected components of the complement of a degree-d 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
comment 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
comment 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 NP-hard, but nonetheless there are smarter things to do than pure brute force.
Sep
22
comment 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
comment when are the Kahler quotient and the GIT quotient the same?
The basic result here is the Kirwan-Ness 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
comment 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 $\mathbb{A}^{k(n-k)}$ 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
comment Conceptual algebraic proof that Grassmannian is closed in Plucker embedding
@AndyB But $PGL$ isn't projective, so why is the image closed?
Sep
18
comment different proofs of the fact that compact riemann surface has a non-trivial 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
comment 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
comment 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
comment 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
comment 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
comment 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
comment 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
comment 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 $(n-r+1) \times (n-r+1)$ 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
comment 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?