Timeline for Description of determinantal varieties in $\mathbb{P}^n$ that are linear sections of determinantal varieties in $\mathbb{P}^{n+1}$
Current License: CC BY-SA 4.0
9 events
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| Apr 23, 2021 at 15:12 | history | edited | Sergey Guminov | CC BY-SA 4.0 |
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| Apr 23, 2021 at 14:50 | comment | added | skeptic | @Serge you are absolutely right .. in the "complexity theory" situation, they are working with affine forms .... so in particular in the affine situation we can express any hypersurface as the determinant hypersurface | |
| Apr 23, 2021 at 14:46 | comment | added | Sergey Guminov | @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. | |
| Apr 23, 2021 at 14:43 | history | edited | Sergey Guminov | CC BY-SA 4.0 |
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| Apr 23, 2021 at 14:40 | comment | added | Sergey Guminov | @WillSawin You are absolutely correct, I stated the question poorly. Gonna fix now. | |
| Apr 23, 2021 at 14:38 | comment | added | skeptic | for eaxmple here cs.utexas.edu/~benleevo/pdf/det-complexity.pdf | |
| Apr 23, 2021 at 14:38 | comment | added | Will Sawin | 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 $\mathbb P^{n^2-1}$. | |
| Apr 23, 2021 at 14:34 | comment | added | skeptic | 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 $f$ | |
| Apr 23, 2021 at 13:52 | history | asked | Sergey Guminov | CC BY-SA 4.0 |