# Picard group of $\mathrm{GL}(n)$-orbits

$$\DeclareMathOperator\GL{GL}\DeclareMathOperator\Mat{Mat}$$Consider the general linear group $$\GL(n) = \left\lbrace \left(\begin{array}{cc} A & C \\ M & B \end{array}\right) \text{ with } A\in \Mat(k,k),\: B\in \Mat(n-k,n-k),\: M\in \Mat(n-k,k),\: C\in \Mat(k,n-k) \right\rbrace$$ of $$n\times n$$ invertible matrices, and the following $$n\times n$$ matrix $$J = \left(\begin{array}{cc} I_{k} & 0 \\ 0 & 0 \end{array}\right)$$ where $$I_k$$ is the $$k\times k$$ identity matrix. Consider the action of $$\GL(n)$$ on the projective space $$\mathbb{P}^N$$ of $$n\times n$$ symmetric matrices modulo scalar given by $$(P,S)\mapsto PSP^T$$. Then $$PJP^T = \left(\begin{array}{cc} AA^T & AM^T \\ MA^T & MM^t \end{array}\right)$$ Now, consider the subgroup $$F\subset G$$ defined by imposing $$M = 0$$ and $$AA^T = cI_k$$ for some $$c\neq 0$$ in the base field (which we can assume algebraically closed and of characteristic zero). Let $$X_k = G/F$$ be the orbit of $$I_k$$. Then $$X_k$$ has dimension $$\frac{2nk-k^2+k-2}{2}$$.

I would like to ask if anyone knows a method to compute the Picard group of $$X_k$$. This should be $$\mathbb{Z}$$ for $$k = 1$$ and $$\mathbb{Z}/n\mathbb{Z}$$ for $$k = n$$. Thank you.

• It seems that you denote by the same letter $M\in{\rm Mat}(n-k,k)$ and also $M\in{\rm GL}(n)$. Dec 17, 2020 at 18:30
• Your description of $\operatorname{GL}(n)$ is not correct: not all the matrices you describe are invertible, and not all invertible matrices are of that form. Dec 17, 2020 at 19:14
• With this notation I just mean: pick an invertible matrix and divide it into four blocks.
– user168611
Dec 17, 2020 at 19:41
• Since you mention ${\rm GL}(n)$-orbits in the title, it would be good if you describe the action, with respect to which you consider the orbits. Dec 17, 2020 at 20:05
• @YCor, thanks for the edits! Just for reference, you can do \operatorname{GL}(n) in the title, or even $\DeclareMathOperator\GL{GL}\GL(n)$. Dec 18, 2020 at 0:00

$$\DeclareMathOperator\Pic{Pic}\DeclareMathOperator\GL{GL}\DeclareMathOperator\GO{GO}$$ Let $$G$$ be a connected linear algebraic group over an algebraically closed field $$K$$ of characteristic 0. Let $$F\subseteq G$$ be an algebraic $$K$$-subgroup, not necessarily connected, and set $$Y=G/F$$. Then there is a canonical isomorphism $$\operatorname{coker}\left[{\sf X}^*(G)\to {\sf X}^*(F)\right]\overset{\lambda}{\longrightarrow}\Pic Y,$$ where $${\sf X}^*(G)$$ denotes the character group of $$G$$, and the map $${\sf X}^*(G)\to {\sf X}^*(F)$$ is the restriction homomorphism. This follows, for instance, from Theorem 2 in the introduction of M. Borovoi and J. van Hamel, Extended equivariant Picard complexes and homogeneous spaces, Transform. Groups 17 (2012), 51-86, arXiv:1010.3414. The map $$\lambda$$ sends a character $$\chi\in{\sf X}^*(F)$$ to the class of the $${\Bbb G}_m$$-torsor $$Y_\chi\to Y$$, where $$Y_\chi=(G\times {\Bbb G}_m)/\chi_*(F)$$ and $$\chi_*\colon F\to G\times {\Bbb G}_m,\quad f\mapsto(f,\chi(f))\ \text{ for }f\in F.$$

In our case $$F=\left\{ \begin{pmatrix} A&B\\0&D \end{pmatrix} \ \ \Big |\ \ A\in\GO_k,\ B\in{\rm Mat}_{k,\,n-k}, D\in \GL_{n-k} \right\}.$$ Clearly, $${\sf X}^*(F)={\sf X}^*(\GO_k)\oplus {\sf X}^*(\GL_{n-k}).$$ If $$k, we have $${\sf X}^*(\GL_{n-k})\cong {\Bbb Z}$$ with generator $$\det_{n-k}$$. (If $$k=n$$, then of course $${\sf X}^*(\GL_{n-k})=0$$.)

We write $$X:={\sf X}^*(\GO_k)$$. For $$k=1$$ we have $$X\simeq {\Bbb Z}$$. For $$2\le k\le n$$, the group $$X$$ is generated by $$d=\det_k$$ and $$c$$ with one relation $$2d-kc=0$$. In other words, $$X\cong{\Bbb Z}^2/\langle (2,-k)\rangle.$$

If $$k$$ is odd, $$k=2p+1$$, then the element $$(2, -k)\in{\Bbb Z}^2$$ is primitive (indivisible), and hence the group $$X$$ is cyclic. Namely, we consider the following basis of $${\Bbb Z}^2$$: $$e_1=2d-(2p+1)c,\quad e_2=d-pc;$$ then $$X\cong {\Bbb Z}^2/\langle e_1\rangle \simeq {\Bbb Z}$$ with a generator of infinite order $$[e_2]$$.

If $$k$$ is even, $$k=2p$$, then the element $$(2, -k)=2(1, -p)\in{\Bbb Z}^2$$ is divisible by 2, and hence the group $$X$$ is isomorphic to $${\Bbb Z}/2{\Bbb Z}\oplus {\Bbb Z}$$. Namely, we consider the following basis of $${\Bbb Z}^2$$: $$e_1=d-pc,\quad e_2=c;$$ then $$X\cong{\Bbb Z}^2/\langle 2e_1\rangle\simeq {\Bbb Z}/2{\Bbb Z}\oplus{\Bbb Z}$$ with a generator $$[e_1]$$ of order 2 and a generator $$[e_2]=[c]$$ of infinite order.

We assume that $$n\ge 2$$. If $$k, the map $${\sf X}^*(G)\to {\sf X}^*(\GL_{n-k})$$ is bijective, and hence $$\operatorname{coker}\left[{\sf X}^*(G)\to {\sf X}^*(F)\right]\simeq {\sf X}^*(\GO_k)=X.$$ Thus for $$1\le k $$\Pic Y\simeq \begin{cases} {\Bbb Z} &\text{if k is odd;}\\ {\Bbb Z}/2{\Bbb Z}\oplus {\Bbb Z} &\text{if k is even.} \end{cases}$$

For $$k=n$$ we have $${\sf X}^*(F)=X$$ and $$\Pic Y\cong \operatorname{coker}\left[{\sf X}^*(G)\to X\right].$$ Now $$\operatorname{coker}\left[{\sf X}^*(G)\to X\right]={\Bbb Z}^2/\langle (1,0),(2,-n)\rangle\simeq {\Bbb Z}/n{\Bbb Z}$$ with the generator $$[c]$$ of order $$n$$.

EDIT: Our answers for $$k=1$$ and $$k=n$$ coincide with those of OP, but not for $$k=2$$, $$n=3$$. Below we compute $${\sf X}^*(F)$$ and $$\operatorname{coker}[{\sf X}^*(G)\to{\sf X}^*(F)]$$ is the case $$k=2$$, $$n>2$$.

Recall that $$\GO_2=\{A\in\GL_2\mid AA^T=c(A) I_2\}.$$ Elementary calculations show that $$\GO_2=\{U(a,b), V(a,b)\mid a^2+b^2\neq 0\},$$ where $$U(a,b)=\begin{pmatrix}a&b\\-b&a\end{pmatrix},\quad V(a,b)=\begin{pmatrix}a&b\\b&-a\end{pmatrix}.$$ We have $$U(a,b)\cdot U(a,-b)=(a^2+b^2)I_2$$ whence $$U(a,b)^{-1}=U(a,-b)/(a^2+b^2).$$ Set $$U=\{U(a,b)\},\quad V=\{V(a,b)\},\quad v=V(1,0)={\rm diag}(1,-1).$$ Then $$U$$ is a subgroup of $$\GO_2$$, and $$\GO_2=U\cup vU$$. The group $$\GO_2$$ is not abelian. Indeed, $$vU(a,b)v^{-1}U(a,b)^{-1}=U(a,-b)\cdot U(a,b)^{-1}=U(a,-b)^2/(a^2+b^2).$$ We obtain that the commutator subgroup $$(\GO_2,\GO_2)=U_1:=\{U(a,b)\mid a^2+b^2=1\}.$$ It follows that $${\sf X}^*(U)\cong {\Bbb Z}$$ with generator $$\omega$$ given by $$\omega(U(a,b))=a^2+b^2$$. Then $$\omega=d|_U=c|U$$ and $$\ker\omega=U_1$$. We have an isomorphism $$\omega_*\colon U/U_1\to {\Bbb G}_m,\quad U(a,b)\cdot U_1\mapsto a^2+b^2.$$ The map $$U\times\{1,v\}\to \GO_2,\quad (U(a,b), 1)\mapsto U(a,b), \ \, (U(a,b),v)\mapsto vU(a,b)=V(a,b)$$ induces an isomorphism $$U/U_1\times \{1,v\}\overset\sim\longrightarrow \GO_2/U_1.$$ Thus $$\GO_2/(\GO_2,\GO_2)\cong {\Bbb G}_m\times {\Bbb Z}/2{\Bbb Z}$$, and $${\sf X}^*(\GO_2)\cong {\Bbb Z}\oplus{\Bbb Z}/2{\Bbb Z}$$. We have \begin{align*}&d(U(a,b))=c(U(a,b))=a^2+b^2,\\ &c(V(a,b))=a^2+b^2,\ \text{ but }\ d(V(a,b))=-(a^2+b^2). \end{align*} Thus the character $$\zeta:=d/c$$ of $$\GO_2$$ takes the value 1 on $$U$$ and the value $$-1$$ on $$V$$. Clearly, $$\zeta\neq 1$$, but $$\zeta^2=1$$. Thus the character $$\zeta$$ is of order 2. Clearly, $${\sf X}^*(\GO_2)\cong {\Bbb Z}\oplus {\Bbb Z}/2{\Bbb Z}$$ with generator $$d$$ of infinite order and generator $$\zeta$$ of order 2.

Now we assume that $$n>2$$. We compute $$\operatorname{coker}[{\sf X}^*(G)\to{\sf X}^*(F)]$$. We have $${\sf X}^*(G)\cong {\Bbb Z}$$ with generator $$d_n$$, and $${\sf X}^*(F)={\sf X}^*(\GL_{n-k})\oplus{\sf X}^*(\GO_2)\cong {\Bbb Z}\oplus{\Bbb Z}\oplus{\Bbb Z}/2{\Bbb Z}$$ with generators $$d_{n-k}$$, $$d=d_2$$, and $$\zeta$$. The restriction map $${\sf X}^*(G)\to{\sf X}^*(F)$$ sends $$d_n$$ to $$(d_{n-k},d_2,0)$$. It follows that $$\operatorname{coker}[{\sf X}^*(G)\to{\sf X}^*(F)]\simeq {\Bbb Z}\oplus{\Bbb Z}/2{\Bbb Z}$$ with generator $$[d_2]$$ of infinite order and generator $$[\zeta]$$ of order 2. Thus $$\Pic Y\simeq {\Bbb Z}\oplus{\Bbb Z}/2{\Bbb Z}$$.

I construct explicitly a $${\Bbb G}_m$$-torsor over $$Y$$ of order 2 in the Picard group. We consider the homomorphism $$\zeta_*\colon F\to G\times{\Bbb G}_m,\quad f\mapsto (f,\zeta(f))\,\text{ for }f\in F,$$ where we extend the character $$\zeta$$ of $$\GO_2$$ to $$F$$ by $$\zeta\begin{pmatrix} A&B\\0&D \end{pmatrix} :=\zeta(A).$$ We consider the quotient $$Y_\zeta=(G\times{\Bbb G}_m)/\zeta_*(F)$$ and the projection map $$\pi\colon Y_\zeta\to Y,\quad (g,z)\cdot \zeta_*(F)\mapsto g\cdot F.$$ Then the class in $$\Pic Y$$ of the $${\Bbb G}_m$$-torsor $$(Y_\zeta,\pi)$$ is of order 2.

• 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 19:22
• 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 $X^{*}(F)$.
– user168611
Dec 17, 2020 at 19:38
• 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 21:35
• I agree with your answers, see my edited answer. Note that $F$ is not connected when $k<n$ is even. Dec 18, 2020 at 20:06
• 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 20:32