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

location  Ann Arbor  
age  35  
visits  member for  5 years, 10 months 
seen  39 mins ago  
stats  profile views  34,877 
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 realanalyticity? 
2d

awarded  Revival 
Aug
28 
revised 
degeneration of reductive group
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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  galoistheory 
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.unistuttgart.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 twostep 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 $d^n \exp(xx^{1})/(dx)^n$ 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 $d^n \exp(xx^{1})/(dx)^n$ 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 6box 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 $0 \leq i,j \leq n1$ 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 $(Tr(B^n), B_{n1})$. 