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Consider a general $4\times 4$ matrix: $$ X:=\left( \begin{array}{cccc} X_0 & X_1 & X_2 & X_3 \\ X_4 & X_5 & X_6 & X_7 \\ X_8 & X_9 & X_{10} & X_{11} \\ X_{12} & X_{13} & X_{14} & X_{15} \end{array} \right) $$ and let $Y_k\subset\mathbb{P}^{15}$ the variety of matrices of rank equal to $k$. What is the Picard group of $Y_k$?

For instance, for $k = 1$ we have that $Y_1\cong\mathbb{P}^3\times\mathbb{P}^3$ so that $\text{Pic}(Y_1)\cong\mathbb{Z}\times \mathbb{Z}$. The variety $Y_3$ is the hypersurface given by $\det(X) = 0$ from which we remove the locus of matrices of rank less than or equal to $2$.

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There is a natural morphism $$ Y_k \to Gr(k,4) \times Gr(k,4) $$ associating to a matrix its image and coimage. Moreover, this morphism factors as the composition $$ Y_k \hookrightarrow \mathbb{P}_{Gr(k,4) \times Gr(k,4)}(U_1^\vee \otimes U_2) \to Gr(k,4) \times Gr(k,4), $$ where $U_i$ are the tautological vector bundles of rank $k$ on the Grassmannians, the first morphism is an open embedding, and the comlement image is the hypersurface $$ Z_k \subset \mathbb{P}_{Gr(k,4) \times Gr(k,4)}(U_1^\vee \otimes U_2) $$ of degenerate matrices. If $H_1,H_2$ are the generators of the Picard groups of the Grassmannians and $H$ is the hyperplane class of the projective bundle, the class of $Z_k$ is equal to $$ [Z_k] = kH + H_1 - H_2. $$ Therefore, $$ \mathrm{Pic}(Y_k) = (\mathbb{Z}H \oplus \mathbb{Z}H_1 \oplus \mathbb{Z}H_2)/\mathbb{Z}(kH + H_1 - H_2) \cong \mathbb{Z}^2. $$

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  • $\begingroup$ Thank you. Your answer makes sense to me but then I am confused by the two following things. Since the closure of $Y_3\subset\mathbb{P}^{15}$ is a hypersurface arguing as in the first answer here mathoverflow.net/questions/378902/… shouldn't we get that $\text{Pic}(Y_3)\cong \mathbb{Z}$ generated by the hyperplane section? Also, why your argument does not work if we consider just symmetric matrices? If $S_k$ is the variety of $4\times 4$ symmetric matrices of rank $k$ then we still have a morphism $S_k\rightarrow Gr(k,4)\times Gr(k,4)$. $\endgroup$
    – user124234
    Jan 17, 2021 at 20:04
  • $\begingroup$ For symmetric matrices the image and coimage coincide, so the map to the product of two Grassmannians factors through the diagonal. Consequently, the Picard group is $\mathbb{Z}^{\oplus 2}/\mathbb{Z} \cong \mathbb{Z}$. Also, note that the Picard group of $Y_3$ is not equal to the Picard group of its closure, but rather to its class group of Weil divisors, so there is no contradiction with the argument you cite. $\endgroup$
    – Sasha
    Jan 18, 2021 at 4:44
  • $\begingroup$ You are right. However, both the answers here mathoverflow.net/questions/378902/… say that the Picard groups of the codimension $1$ orbit and of its closure are the same in the symmetric case. Is this due to the nature of the singularities that the orbits have along smaller dimension orbits in the symmetric and in the general case? $\endgroup$
    – user124234
    Jan 18, 2021 at 9:54
  • $\begingroup$ For $S_{k,n}$ (the locus of rank $k$ symmetric matrices of size $n$) the class of the divisor $Z_k$ defined as above is $kH + 2H_1$, which is not primitive if $k$ is even. Therefore, the Picard group of $S_{k,n}$ is $\mathbb{Z}$ for odd $k$ and $\mathbb{Z} \oplus \mathbb{Z}/2\mathbb{Z}$ for even $k$. In particular, for $S_{2,3}$ in the answer you cite the answer is $\mathbb{Z} \oplus \mathbb{Z}/2\mathbb{Z}$ (I will correct my answer there). $\endgroup$
    – Sasha
    Jan 18, 2021 at 11:11
  • $\begingroup$ OK. Now, everything is clear. Thank you very much. $\endgroup$
    – user124234
    Jan 18, 2021 at 11:22

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