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

location  Ann Arbor  
age  34  
visits  member for  5 years, 2 months 
seen  1 hour ago  
stats  profile views  31,429 
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 21stcentury 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 $ABBA$ (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 
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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 
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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 
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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 
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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. 