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bio website math.lsa.umich.edu/~speyer
location Ann Arbor
age 34
visits member for 5 years, 6 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.


2d
awarded  Nice Answer
Apr
22
comment “Diagonalizing” an associative algebra
Regarding the infinite dimensional case: $k[x]$ does not have a basis of orthogonal idempotents: The only idempotent elements are $0$ and $1$.
Apr
22
revised A Polynomial With Positive Prime Density
added 18 characters in body
Apr
22
awarded  Nice Answer
Apr
21
revised A Polynomial With Positive Prime Density
added 4 characters in body
Apr
21
comment A Polynomial With Positive Prime Density
Oh! I didn't realize that the square brackets mean round down!
Apr
21
answered A Polynomial With Positive Prime Density
Apr
20
answered Factoring constant rank maps into a submersion and an immersion
Apr
18
awarded  Altruist
Apr
18
awarded  Nice Question
Apr
18
awarded  Nice Answer
Apr
17
comment Factoring constant rank maps into a submersion and an immersion
Thanks! That does it, then. My only thought for a hypothesis to add that might save the day is to ask that $\phi: X \to Z$ be proper.
Apr
17
accepted Factoring constant rank maps into a submersion and an immersion
Apr
17
comment Factoring constant rank maps into a submersion and an immersion
So Exercise 5-13 is true, but the map $X \to Y$ may not be continuous with respect to the topology for $Y \to Z$ is an immersion. (Again, think about the line mapping to a nodal cubic.)
Apr
17
comment Factoring constant rank maps into a submersion and an immersion
@IgorRivin I'm looking though Lee's book at webmath2.unito.it/paginepersonali/sergio.console/lee.pdf . I didn't find the Theorem you reference, but I did find Exercise 5-13, which makes this claim. At first I thought this had to be wrong, (think about the line mapping to a nodal cubic) but then I checked his definition of immersed submanifold (p. 119). His definition is very generous: It just says that $\phi(X)$ can be given some topology such that $\phi(X) \to Z$ is an injective immersion. (continued)
Apr
17
comment Ordering subsets of the cyclic group to give distinct partial sums
I'm only saying that the low degree terms are $0 \bmod 2$. So $P \equiv 0 \bmod \langle 2, x_1^6, x_2^6,\ \ldots,\ x_6^6 \rangle$. Doesn't seem obviously wrong to me.
Apr
16
comment Factoring constant rank maps into a submersion and an immersion
Presumably, the way you'd want to do this is to define an equivalence relation $\sim$ on $X$ by $x_1 \sim x_2$ if $\phi(x_1) = \phi(x_2)$ and $x_1$ and $x_2$ are in the same connected component of $\phi^{-1}(\phi(x_1))$. Then show that $X/\sim$ is a smooth manifold. But this seemed hard to me, which was when I decided to ask.
Apr
16
comment Factoring constant rank maps into a submersion and an immersion
There is something called the global constant rank theorem, but it only says that, for each $z \in Z$, the fiber $\pi^{-1}(z)$ is a submanifold of $X$. It doesn't say that the connected components of the $\phi^{-1}(z)$'s can be organized into a manifold.
Apr
16
comment Factoring constant rank maps into a submersion and an immersion
@IgorRivin All the sources I found only said that this is locally true: I.e., for any $x \in X$, there is an open $U \ni x$ where I can make such a factorization. It doesn't seem obvious to me how to glue all of these into a global $Y$.
Apr
16
comment Ordering subsets of the cyclic group to give distinct partial sums
math.lsa.umich.edu/~speyer/foo6.txt Interestingly, all the coefficients are even.