I have a determinantal hypersurface defined by $\det(A)=0$, with $a_{ij}$ homogeneous polynomial of fixed degree $d$ in $n$ variables. $A$ is not diagonal. How can I find out whether the hypersurface is irreducible or not?
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$\begingroup$ Do you have specific polynomials $a_{ij}$? Or are you asking for general criteria on the $a_{ij}$ to determine (ir)reducibility? $\endgroup$– Lazzaro CampeottiCommented Jun 28, 2016 at 12:44
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$\begingroup$ I have some hypotheses on them, but I have no clue how to tackle the problem.. $\endgroup$– user46071Commented Jun 28, 2016 at 12:47
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4$\begingroup$ You can try to prove that the singular locus of the hypersurface has codimension at least $3$ inside the ambient manifold. For a "generic" determinantal variety, the singular locus has codimension $4$, and it equals the locus where the matrix has rank $\leq n-2$. So first you could try to confirm that the rank $\leq n-2$ locus of your matrix has codimension $\geq 3$. $\endgroup$– Jason StarrCommented Jun 28, 2016 at 13:09
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2 Answers
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You can try to estimate the codimension of the singular locus. If it is greater than 2, then the hypersurface is irreducible.
Note that there are two sorts of singular points on this hypersurface. One sort are the points where the corank of the matrix is 2 or more. The other is when the corank is 1, but all the derivatives of the matrix vanish on the left kernel - right kernel pair of vectors.
EDIT: Let $x_k$ be the coordinates. Assume $\det(A(p)) = 0$ and corank of $A(p)$ is 1. Then both the left and the right kernels of $A(p)$ are 1-dimensional. Let $(l_1,\dots,l_n)$ and $(r_1,\dots,r_n)$ are vectors generating the kernels. Then the singularity condition is that for all $k$ $$ \sum_{i,j = 1}^n (\partial a_{ij}/\partial x_k)(p)l_ir_j = 0. $$
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$\begingroup$ Your comment is very useful. Could you explain me better the last sentence? $\endgroup$ Commented Jun 28, 2016 at 13:22
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$\begingroup$ Could you give me a reference for that? $\endgroup$ Commented Jul 4, 2016 at 12:08
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$\begingroup$ The singular locus of determinantal hypersurface. $\endgroup$ Commented Jul 4, 2016 at 13:47
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$\begingroup$ It comes directly from the computation of tangent spaces. This is a good exercise. $\endgroup$– SashaCommented Jul 4, 2016 at 14:23
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You have to check if $det(A)$ is an irreducible polynomial. This is equivalent to the hypersurface being irreducible. You could compute this determinant by first applying the Gauss algorithm.
