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Fix an algebraically closed field $k$ of characteristic 0. Consider an $n$-tuple $(A_1,\ldots, A_n)$ of $n\times n$ matrices over $k$ and assign to it the determinantal surface in $\mathbb{P}_k^{n-1}$ cut out by the polynomial $\det(x_1A_1+\ldots x_nA_n).$ Now consider a fixed non-zero matrix $G$ of rank $r<n$.

Now let $x$ be the vector with components $x_i$ and $G_{ij}$ be the components of $G$. We may then consider the variety cut out by the polynomial $\det(\sum\limits_{j=1}^nx_jA_iG_{ij}).$ With $y=Gx$, this is then a plane section of the variety cut out by $\det(y_1A_1+\ldots y_nA_n)$ with the $\mathbb{P}_k^{r-1}$ corresponding to the image of $G$, and also a determinantal hypersurface of degree $n$ itself in said $\mathbb{P}_k^{r-1}$.

Now fix a 'generic' $n$-tuple $(A_1,\ldots,A_n)$. 'Generic' means I don't exactly know yet in which sense, but let's say that at least all the matrices are invertible. My question is then whether a generic determinantal hypersurface of degree $n$ in $\mathbb{P}_k^{r-1}$ is a linear section of the determinantal hypersurface of degree $n$ in $\mathbb{P}_k^{n-1}$ corresponding to $(A_1,\ldots,A_n)$. Actually, even knowing whether some fixed determinantal hypersurface has infinitely many nonisomorphic plane sections would be huge for me. This does not seem like a strong statement, but it's non-obvious for me.

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  • $\begingroup$ It is well known in the "complexity. theory circles" that given a polynomial $f \in k[x_1,\ldots,x_n]$, then one can come up with a matrix of linear forms (in variables $x_i$) so that the corresponding matrix has determinant $f$ .. I wonder if won't this imply we can always express a any hypersurface as a determinant hypersurface? perhaps I am not thinking right .. the smallest dimension matrix you need for $f$ is called detrminant complexity of $f$ $\endgroup$
    – skeptic
    Commented Apr 23, 2021 at 14:34
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    $\begingroup$ Wouldn't a determinantal hypersurface be determined by a tuple of matrices $A_1,\dots, A_r$, and can't you just extend it arbitrarily by additional matrices $A_{r+1},\dots, A_n$, and then set the new variables to $0$, to represent it as a linear section? In fact it seems to me determinantal hypersurfaces are, by definition, linear sections of the determinant hypersurface in $\mathbb P^{n^2-1}$. $\endgroup$
    – Will Sawin
    Commented Apr 23, 2021 at 14:38
  • $\begingroup$ for eaxmple here cs.utexas.edu/~benleevo/pdf/det-complexity.pdf $\endgroup$
    – skeptic
    Commented Apr 23, 2021 at 14:38
  • $\begingroup$ @WillSawin You are absolutely correct, I stated the question poorly. Gonna fix now. $\endgroup$
    – Sergey Guminov
    Commented Apr 23, 2021 at 14:40
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    $\begingroup$ @skeptic I think (still new to algebraic geometry, may be very wrong) that determinantal varieties in $\mathbb{P}_k^n$ are all singular for $n>3$, so not every hypersurface is a determinantal hypersurface. $\endgroup$
    – Sergey Guminov
    Commented Apr 23, 2021 at 14:46

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