Timeline for Picard group of $\mathrm{GL}(n)$-orbits
Current License: CC BY-SA 4.0
24 events
| when toggle format | what | by | license | comment | |
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| Jan 17, 2021 at 14:22 | vote | accept | CommunityBot | ||
| Dec 20, 2020 at 19:07 | comment | added | user168611 | Corollary 3.7 works for $Sec_2(V)$ but by removing $V$ from $Sec(V)$ we are not changing anything at the level of the Picard groups since $V$ has codimension $2$ in $Sec_2(V)$. | |
| Dec 20, 2020 at 19:05 | comment | added | user168611 | Thank you. The only explanation for this is that $G/F$ is not $Sec_2(V)\setminus V$. Otherwise, by Corollary 3.7 here arxiv.org/pdf/math/0511279.pdf the Picard group must be $\mathbb{Z}$. On the other hand, it seems to me that via this $GL(3)$-action from $I_2$ we can reach all the symmetric matrices of rank $2$. | |
| Dec 20, 2020 at 18:55 | comment | added | Mikhail Borovoi | @Fra: I have added an explicit construction of an element of order 2 in the Picard group of $G/F$. Please recheck your calculations and write what you get. | |
| Dec 20, 2020 at 18:53 | comment | added | Mikhail Borovoi | @Fra: In the case $k=2$ and $n=3$ the group $F$ is not connected. | |
| Dec 20, 2020 at 18:51 | history | edited | Mikhail Borovoi | CC BY-SA 4.0 |
Explicit calculation for $k=2$ added.
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| Dec 19, 2020 at 20:08 | comment | added | Mikhail Borovoi | My calculations for $k=2$ and $n=3$ give ${\Bbb Z}\oplus{\Bbb Z}/2{\Bbb Z}$. I will type them tomorrow or on Monday. | |
| Dec 19, 2020 at 16:33 | comment | added | user168611 | @ Mikhail Borovoi. Thank you very much for your detailed answer. Anyway, there is still something that confuses me. Take for instance $k = 2$ and $n = 3$. Then the orbit $X_2$ is $Sec_2(V)\setminus V$ where $V$ is the Veronese surface in $\mathbb{P}^5$ and $Sec_2(V)$ is its secant variety. Since $Sec_2(V)$ is a hypersurface in $\mathbb{P}^5$ its Picard group is $\mathbb{Z}$ generated by the hyperplane section. Furthermore, since $V$ has codimension two in $Sec_2(V)$ we have that the Picard group of $Sec_2(V)\setminus V$ is equal to that of $Sec_2(V)$, and hence it is again $\mathbb{Z}$. | |
| Dec 18, 2020 at 20:32 | comment | added | LSpice | Ah, I see; I forgot that $\det_0 = 0$, and reasoned incorrectly that $2{\det_k} = k\sigma$ implied that $\det_k = (k/2)\sigma$ for $k$ even. Note that you need not separate out $k = 1$ from $2 \le k \le n$ in your answer; $X$ is also generated by $\det_1$ and $c = \sigma$ subject to $2{\det_1} = k\sigma$, and, just as for the other odd values of $k$, this implies that $\det_k - \lfloor k/2\rfloor\sigma = \det_1$ is a generator of $X$. | |
| Dec 18, 2020 at 20:26 | history | edited | Mikhail Borovoi | CC BY-SA 4.0 |
Mistake corrected
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| Dec 18, 2020 at 20:19 | history | edited | Mikhail Borovoi | CC BY-SA 4.0 |
Mistake corrected
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| Dec 18, 2020 at 20:06 | comment | added | Mikhail Borovoi | I agree with your answers, see my edited answer. Note that $F$ is not connected when $k<n$ is even. | |
| Dec 18, 2020 at 20:04 | history | edited | Mikhail Borovoi | CC BY-SA 4.0 |
A new answer written.
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| Dec 18, 2020 at 16:15 | comment | added | user168611 | For $k = n$ if I consider $SL(n)$ instead of $GL(n)$ and $K = F\cap SL(n)$ instead of $F$ I get that $GL(n)/F\cong SL(n)/K$, and the abelianization of $K$ is $\mathbb{Z}/n\mathbb{Z}$ independently from the parity of $n$. | |
| Dec 18, 2020 at 9:42 | comment | added | user168611 | For $k = n$ the matrix $I_n$ is the identity matrix. Its orbit $U$ is the complement in $\mathbb{P}^N$ of the hypersurface defined by the determinant of a general $n\times n$ symmetric matrix. This is an irreducible hypersurface $D$ of degree $n$. We have an exact sequence $$\mathbb{Z}\rightarrow \operatorname{Pic}(\mathbb{P}^N)\cong\mathbb{Z}\rightarrow\operatorname{Pic}(U)\rightarrow 0$$ where the first map maps $1$ to $D$. So $\operatorname{Pic}(U)\cong\mathbb{Z}/n\mathbb{Z}$. | |
| Dec 17, 2020 at 22:08 | comment | added | LSpice | In particular, $\DeclareMathOperator\Pic{Pic}$I get $\Pic(Y) = \mathbb Z$ (spanned by $\det_1 = \det_{n - 1}$) when $k = 1$, but also $\Pic(Y) = \mathbb Z$ (spanned by $\sigma$) whenever $k$ is even, which doesn't seem to match your expectation for $k = n$ when $n$ is even. | |
| Dec 17, 2020 at 21:35 | comment | added | LSpice | The reductive quotient of $F$ (which is all the character group sees) is $\operatorname{GO}_k \times \operatorname{GL}_{n - k}$, with derived group $\operatorname{SO}_k \times \operatorname{SL}_{n - k}$. The character group of the second factor is free on $\det$. The character group of the first factor is spanned by $\det$ and the conformal character $\sigma$, where $\sigma(A) = c$ when $A A^{\mathsf T} = c\mathrm I_k$, subject to $2{\det} = k\sigma$. The map $\mathsf X^*(G) \to \mathsf X^*(F)$ just sends $\det_n$ to $(\det_k, \det_{n - k})$. | |
| Dec 17, 2020 at 21:21 | history | edited | LSpice | CC BY-SA 4.0 |
Name of paper; \rm -> \operatorname
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| Dec 17, 2020 at 20:00 | comment | added | user168611 | Very kind of you. Thank you very much. | |
| Dec 17, 2020 at 19:51 | comment | added | Mikhail Borovoi | I will compute ${\sf X}^*(F)$ for you tomorrow. | |
| Dec 17, 2020 at 19:38 | comment | added | user168611 | Thank you for the answer. I was trying to use an exact sequence similar to the one you wrote. Yes, $F$ is connected. We have $X^{*}(G) = \mathbb{Z}$. The problem is that I do not know how to compute $X^{*}(F)$. | |
| Dec 17, 2020 at 19:28 | history | edited | Mikhail Borovoi | CC BY-SA 4.0 |
Page in the link corrected
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| Dec 17, 2020 at 19:22 | comment | added | Mikhail Borovoi | Do I understand correctly that $F$ is connected in your case? If not, maybe I can find the above exact sequence for nonconnected $F$ in one of my papers... | |
| Dec 17, 2020 at 19:17 | history | answered | Mikhail Borovoi | CC BY-SA 4.0 |