Let $f:E\to F$ be a morphism of vector bundles on an irreducible algebraic variety $X$. Does anybody know any results about the irreducibility or smoothness of the degeneracy locus of $f$? I know only the Connectedness Theorem due to Fulton.
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2 Answers
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The degeneracy locus can be reducible, and even non-reduced. For instance, take $X= \mathbb{P}^2$ and consider a morphism
$\mathcal{O}(-1)^2 \stackrel{f} \to \mathcal{O}^2$.
$f$ is given by a $2 \times 2$ matrix of linear forms, so its degeneracy locus is a conic. For a general choice of $f$ this conic will be smooth, but for special choice of the matrix it can become singular or even a double line.
The best result I am aware of can be found in Ottaviani's book "Varieta' proiettive di codimensione piccola" [projective varieties of small codimension, unfortunately I do not think an english translation is available].
Set
$D_k(f):=\{x \in X \; | \; \textrm{rank}(f_x) \leq k \}$
Then we have the following
Theorem (of Bertini's type)
Set $\textrm{rank}(E)=m$, $\textrm{rank}(F)=n$. Assume that $E^{*} \otimes F$ is globally generated. Then for the generic morphism $f \colon E \to F$, the locus $D_k(f)$ is either empty or it has the expected codimension $(m-k)(n-k)$, and the singular locus of $D_k(f)$ is contained in $D_{k-1}(f)$.
In particular, if
$\dim X < (m-k+1)(n-k+1)$
then $D_k(f)$ is smooth for a general choice of $f$.
answered Nov 4, 2010 at 15:05
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$\begingroup$ If a have a specific $f$, how can I know whether it is generic in this sense? $\endgroup$ Commented Nov 4, 2010 at 15:35
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$\begingroup$ In that case, I'm afraid you must use the geometric information that you have in order to calculate\describe explicitly the degeneracy locus of $f$. Bertini's type results do not say anything about specific objects. $\endgroup$ Commented Nov 4, 2010 at 15:50
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3$\begingroup$ Another result is that, when $D_{k}(f)$ has the expected dimension, it is Cohen-Macaulay. This is, of course, far from being non-singular, but limits how bad the singularities can be (e.g., no embedded points and all irreducible components have the same dimension). $\endgroup$– jlkCommented Nov 4, 2010 at 17:30
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2$\begingroup$ I think jlk's comment should become an answer, if possible with a reference added. $\endgroup$– BarbaraCommented Nov 5, 2010 at 9:26
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$\begingroup$ @Barbara: Done. I just noticed that "ginevra86" is only assuming that $X$ is an irreducible variety. The statement I made ("$D_{k}(f)$ is determinantal") is only true if $X$ is Cohen-Macaulay. $\endgroup$– jlkCommented Nov 10, 2010 at 0:32
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At Barbara's request, I am posting this as an answer (with a correction).
Assume that $X$ is Cohen-Macaulay. If $D_{k}(f)$ has the expected dimension, then it is Cohen-Macaulay. This is, of course, far from being non-singular, but limits how bad the singularities can be (e.g., no embedded points and all irreducible components have the same dimension).
I learned about this result from Chapter 4 of Geometry of algebraic curves. I do not have the book handy, but I think this result was originally proven in [Hochster-Eagon, "A class of perfect determinantal ideals"].
Chapter 4 has a general discussion of determinantal varieties that might be helpful. I seem to remember that they prove, if $D_{k}(f)$ has expected dimension, then the singular locus of $D_{k}(f)$ is contained in $D_{k-1}(f)$ (part of the theorem cited by Francesco Polizzi, but without genericity), but I could be misremembering.
I was misremembering.
