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bio website math.lsa.umich.edu/~speyer
location Ann Arbor
age 34
visits member for 5 years, 2 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.


7h
comment Flatness of normalization
@user26857 The original question specified that the extension of function fields should be separable, which it isn't if $\mathrm{char}(k)=2$.
1d
revised Question about a family of semistable curves
added 12 characters in body
Dec
23
awarded  Enlightened
Dec
23
awarded  Nice Answer
Dec
20
awarded  Guru
Dec
20
comment Character table does not determine group Vs Tannaka duality
In the positive direction, I recently encountered a result of Hoehnke and Johnson, arxiv.org/abs/math/9210219 : A group is determined by the collection of functions $G^k \to \mathbb{C}$ for $k=1$, $2$ and $3$ given by $g \mapsto \chi(g)$, $(g,h) \mapsto \chi(g)\chi(h)- \chi(gh)$ and $(f,g,h) \mapsto \chi(f) \chi(g) \chi(h) - \chi(fg) \chi(h) - \chi(gh) \chi(f) - \chi(h) \chi(fg) + \chi(fgh) + \chi(fhg)$ for all characters $\chi$.
Dec
19
comment textbooks on modern algebraic geometry for 21st-century starters
@bananastack If I understand what you are looking for, <i>Methods of Homological Algebra</i> by Gelfand and Manin fits the bill. Beware the many minor typos.
Dec
17
revised Cap product à la Poincaré
edited body
Dec
17
answered Cap product à la Poincaré
Dec
15
comment when can I say that $UV^T$ is a permutation matrix?
@math2014 So, are you saying that what you want is $U^T V$ a permutation as I suggested? In that case, why not use $|AB-BA|$ (for whichever matrix norm you like best)?
Dec
14
comment when can I say that $UV^T$ is a permutation matrix?
If $U^T V$ is a permutation $P$, then $AB=BA$. (Proof: $AB = U \Lambda P \Sigma V^T = U \Lambda \Sigma' P V^T = U \Lambda \Sigma' U^T$, where $\Sigma'= P \Sigma P^{-1}$, and $BA = V \Sigma V^T U \Lambda U^T = V \Sigma P^{-1} \Lambda U^T = V P^{-1} \Sigma' \Lambda U^T = U \Sigma' \Lambda U^T$. Since $P$ is permutation, $\Sigma'$ is diagonal, and $\Lambda$ and $\Sigma'$ commute.) Conversely, if $A$ and $B$ commute, then we can choose $U$ and $V$ so that $U^T V$ is permutation.
Dec
14
comment when can I say that $UV^T$ is a permutation matrix?
Are you sure you don't want $U^T V$? If we permute the order of the entries in $\Lambda$ and $\Sigma$, this multiplies $U^T V$ by permutation matrices, so this condition is independent of how you order the entries.
Dec
13
comment Equations for points to lie on a rational normal curve
That does sound helpful. Thanks!
Dec
13
accepted Equations for points to lie on a rational normal curve
Dec
12
comment Equations for points to lie on a rational normal curve
Whoa, that's great! Probably you need to intersect that over all choices of $k+1$ points, if you want to exclude other spurious components. But that is much cleaner than anything I had thought of.
Dec
12
answered Equations for points to lie on a rational normal curve
Dec
12
asked Equations for points to lie on a rational normal curve
Dec
10
comment Whitehead for maps
Here mathoverflow.net/questions/87830 is a reference for modules over an algebra; it should work in any abelian category. Also, some good discussion at mathoverflow.net/q/8974 from the algebra side.
Dec
10
comment Whitehead for maps
Thanks, this is really useful! I think you mean $\mathrm{Ext}^i$ nonzero for $i \geq 2$, yes? If only $\mathrm{Ext}^1$ and $\mathrm{Ext}^0$ is nonzero, then I believe every complex is derived equivalent to its homology.
Dec
10
comment Whitehead for maps
@EricWofsey True, though I asked the question about finite CW complexes. If I understand the terminology correctly, that is essentially small.