78,708 reputation
5175392
bio website math.lsa.umich.edu/~speyer
location Ann Arbor
age 35
visits member for 5 years, 10 months
seen 39 mins ago

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.


2d
answered May integration spoil real-analyticity?
2d
awarded  Revival
Aug
28
revised degeneration of reductive group
deleted 1 character in body
Aug
28
answered degeneration of reductive group
Aug
27
comment A Linear Order from AP Calculus
Related mathoverflow.net/questions/29624/…
Aug
25
answered Examples of common false beliefs in mathematics
Aug
25
awarded  galois-theory
Aug
24
awarded  Nice Answer
Aug
24
revised Does $\mathbb{K}[G]\simeq\mathbb{K}[H]$ for some field $\mathbb{K}$ of characteristic $p$, imply $\mathbb{F}_p[G]\simeq\mathbb{F}_p[H]$?
added 2527 characters in body
Aug
24
comment Does $\mathbb{K}[G]\simeq\mathbb{K}[H]$ for some field $\mathbb{K}$ of characteristic $p$, imply $\mathbb{F}_p[G]\simeq\mathbb{F}_p[H]$?
Section 1.3 of igt.uni-stuttgart.de/LstDiffgeo/Hertweck/preprints/… shows how to canonically turn a group algebra $\mathbb{F}_p[G]$ into a $p$-Lie algebra $Jen(G)$. I am reasonably confident that I have constructed an example where $Jen(G_1) \not \cong Jen(G_2)$ but $\mathbb{F}_{p^2} \otimes Jen(G_1) \cong \mathbb{F}_{p^2} \otimes Jen(G_2)$; the remaining question is whether I can lift that last fact back to $\mathbb{F}_{p^2}[G]$.
Aug
24
comment Does $\mathbb{K}[G]\simeq\mathbb{K}[H]$ for some field $\mathbb{K}$ of characteristic $p$, imply $\mathbb{F}_p[G]\simeq\mathbb{F}_p[H]$?
Hmmm, so either I am wrong, or it is worth putting in the time to actually think through the deformation theory. Maybe you can help: My intuition is that group algebras of two step $p$-torsion nilpotent Lie groups are very similar to enveloping algebras of two-step nilpotent $p$-Lie algebras where the $p$-th power map is zero. This really is a counterexample in the $p$-Lie algebra world. Do you know theorems making this analogy precise?
Aug
24
answered Does the Galois group of a Pisot polynomial contain the alternating group?
Aug
24
answered Does $\mathbb{K}[G]\simeq\mathbb{K}[H]$ for some field $\mathbb{K}$ of characteristic $p$, imply $\mathbb{F}_p[G]\simeq\mathbb{F}_p[H]$?
Aug
23
answered Why does $d^n \exp(-x-x^{-1})/(dx)^n$ only have $n$ positive real zeroes?
Aug
22
comment Is there a version of the Titchmarsh Convolution theorem to find singular support?
@TerryTao But that gives you a much weaker statement, unless I am confused. Looking at the case of $f \ast f$, the Minkoski sum of two copies of $\partial H$ is the entire interior of $2H$, not the $1$-dimensional thing we want.
Aug
22
answered Is there a version of the Titchmarsh Convolution theorem to find singular support?
Aug
21
comment Why does $d^n \exp(-x-x^{-1})/(dx)^n$ only have $n$ positive real zeroes?
@TerryTao That sounds like a really cool approach, which I'll think about when I get some time (unless someone else does it first). Just to check that I understand your strategy, that third order ODE has coefficients which depend on $x$, right? So we don't get a fixed vector field in $\mathbb{RP}^2$, but one that varies according to the "time" $x$. Or did I miss something?
Aug
21
comment Three involutions on the set of 6-box Young diagrams
Quick proof that $(4)$ and $(4,2)$ are fixed by $t$: One of them is in the kernel of the unique nontrivial map $S_6 \to \mathbb{Z}/2$, and the other isn't. (The unique map is taking the sign of the permutation.)
Aug
21
awarded  Good Question
Aug
21
comment Algebraically independent matrix invariants
A modification which I don't feel like editing into the original: For $n$ prime, the invariants $Tr(A^i B^j)$ with $0 \leq i,j \leq n-1$ and $(i,j) \neq (0,0)$, together with $Tr(A^n)$ and $Tr(B^n)$ provide $n^2+1$ algebraically independent functions. Proof: Copy the above and, in addition, consider differentiation with respect to $B_{n1}$. You get $n$ blocks of size $n \times n$ as before, and a $1 \times 1$ block in position $(Tr(B^n), B_{n1})$.