# Picard group of symplectic group modulo orthogonal group

Let $$Sp(2n)$$ be the group of complex symplectic $$2n\times 2n$$ matrices, and $$O(2n)$$ the group of complex orthogonal $$2n\times 2n$$ matrices.

Consider $$Sp(2n)\cap O(2n)\subset Sp(2n)$$ and the quotient $$X=Sp(2n)/(Sp(2n)\cap O(2n))$$. How could one compute the Picard group of $$X$$?

EDIT. Consider the action of $$Sp(2n)$$ on the projective space $$\mathbb{P}^N$$ of $$2n\times 2n$$ matrices modulo scalar given by $$Sp(2n)\times\mathbb{P}^N\rightarrow\mathbb{P}^N$$, $$(A,Z)\mapsto AZA^t$$. The stabilizer $$H$$ of the identity is then given by those matrices in $$Sp(2n)$$ such that $$AA^t = \lambda I$$ for some $$\lambda\in\mathbb{C}^{*}$$.

Let $$X = Sp(2n)/H$$ be the orbit of the identity in $$\mathbb{P}^N$$.

How could one compute the Picard group of $$X$$?

Consider for instance the case $$n = 1$$. Since any $$2\times 2$$ symmetric matrix with non-zero determinant has a multiple that is symplectic the orbit $$X$$ is $$\mathbb{P}^2\setminus C$$ where $$C\subset\mathbb{P}^2$$ is the conic parametrizing matrices with zero determinant. So, in this case, $$Pic(X) \cong \mathbb{Z}/2\mathbb{Z}$$.

• How do you choose the symplectic and quadratic forms? – Sasha May 18 at 9:01
• The standard symplectic form $\left(\begin{array}{cc} 0_n & I_n\\ -I_n & 0_n \end{array}\right)$ and the identity. – F_L May 18 at 9:09
• The answer to both the original and the edited questions is ${\rm Pic}({\rm Sp}(2n)/H)=X(H)$, where $X(H)={\rm Hom}(H,{\Bbb G}_m)$ is the character group of $H$. – Mikhail Borovoi May 18 at 17:36
• I have restored your original question. Otherwise the reader cannot understand, what question was answered by Sasha. – Mikhail Borovoi May 18 at 17:47
• Good idea, thank you very much. Is it obvious that the character group of the $H$ in the modified version of the question is $\mathbb{Z}/n\mathbb{Z}$? – F_L May 18 at 18:45

With the suggested choice of the symplectic and orthogonal form, there is a direct sum decomposition of $$\mathbb{C}^{2n}$$ into the sum of two Lagrangian (with respect to the both forms) subspaces: $$L_1 = \langle e_k + ie_{n+k} \rangle_{k=1}^n, \qquad L_2 = \langle e_k - ie_{n+k} \rangle_{k=1}^n.$$ Moreover, the the pairings between $$L_1$$ and $$L_2$$ induced by the both forms are proportional. Therefore $$\mathrm{Sp}(2n) \cap \mathrm{O}(2n) \cong \mathrm{GL}_n$$ which acts on $$L_1 \oplus L_2$$ by $$A \mapsto (A,A^{-1})$$.

Using this, it is easy to see that $$X = \mathrm{LGr}(2n) \times \mathrm{LGr}(2n) \setminus D,$$ where $$\mathrm{LGr}(2n)$$ is the Lagrangian Grassmannian for the symplectic form, and $$D \subset \mathrm{LGr}(2n) \times \mathrm{LGr}(2n)$$ parameterizes pairs of intersecting Lagrangian subspaces. It is well known that $$\mathrm{Pic}(\mathrm{LGr}(2n)) = \mathbb{Z}$$ and it is easy to see that $$D$$ is a divisor of bidegree $$(1,1)$$. Therefore, $$\mathrm{Pic}(X) = \mathbb{Z}$$.

• Thank you very much for the answer. I slightly modiefied my question since I explaind mysefl incorrectly in the first version. Basically the only difference is that we must take into account matrices that are orthogonal up to scalars. I think the space you considered is a covering of the new $X$. – F_L May 18 at 16:54

Answer: $${\rm Pic\,} X={\Bbb Z}/2{\Bbb Z}$$; see Corollary 4 below.

Theorem 1. Let $$G$$ be a simply connected semisimple group over a field $$k$$ of characteristic 0. Let $$H\subset G$$ be an algebraic subgroup defined over $$k$$, not necessarily connected. Set $$X=G/H$$. Then there is a canonical isomorphism $${\rm Pic\,} X={\widehat H}(k)$$, where $${\widehat H}(k) ={\rm Hom}_k(H,{\Bbb G}_{m})$$ is the character group of $$H$$.

Proof. First assume that $$H$$ is connected. We deduce the theorem from results of the paper J.-J. Sansuc, Groupe de Brauer et arithmétique des groupes algébriques linéaires sur un corps de nombres. J. Reine Angew. Math. 327 (1981), 12–80.

By Proposition 6.10 of this paper, there is a natural exact sequence of abelian groups $${\widehat G}(k)\to {\widehat H}(k)\to{\rm Pic\,} X\to {\rm Pic\,} G.$$ Clearly we have $${\widehat G}(k)=0$$. By Sansuc's Lemma 6.9(iv), we have $${\rm Pic\,} G=0$$ (here Sansuc refers to a paper by Fossum and Iversen). We obtain an isomorphism $${\widehat H}(k)= {\rm Pic\,} X$$, as required.

Now we do not assume that $$H$$ is connected. We deduce Theorem 1 from a general result of M. Borovoi and J. van Hamel, Extended equivariant Picard complexes and homogeneous spaces. Transform. Groups 17 (2012), 51-86. Since $${\rm Pic\,} G_{\bar k}=0$$ and $$X$$ has $$k$$-points, by Theorem 2 (Theorem 7.1) of this paper there is a canonical isomorphism $${\rm Pic\,} X=H^1(k,[{\widehat G}({\bar k})\to {\widehat H}({\bar k})\rangle).$$ Here $${\bar k}$$ is an algebraic closure of $$k$$, $${\widehat H}({\bar k})={\rm Hom}_{\bar k}(H,{\Bbb G}_{m})$$, and similarly for $${\widehat G}({\bar k})$$. Further, $$[{\widehat G}({\bar k})\to {\widehat H}({\bar k})\rangle$$ denotes the complex of $${\rm Gal}({\bar k}/k)$$-modules $$\dots \to 0\to {\widehat G}({\bar k})\to {\widehat H}({\bar k})\to 0\to \dots$$ with $${\widehat H}({\bar k})$$ in degree 1, and $$H^1(k,[{\widehat G}({\bar k})\to {\widehat H}({\bar k})\rangle)$$ denotes the first Galois hypercohomology of this complex.

In our case $${\widehat G}({\bar k})=0$$, and therefore, $${\rm Pic\,} X=H^1(k,[0\to {\widehat H}({\bar k})\rangle)=H^0(k,{\widehat H}({\bar k}))={\widehat H}(k),$$ as required.

This looks like killing a fly with a bazooka, and there should be an elementary proof of Theorem 1.

Construction 2. The class in $${\rm Pic\,} X$$ corresponding to a character $$\chi\colon H\to{\Bbb G}_m$$ is described as follows. We consider the direct product $$G\times {\Bbb G}_m$$ and the injective homomorphism $$\iota_\chi\colon H\to G\times {\Bbb G}_m,\quad h\mapsto (h,\chi(h)).$$ Further, we consider the quotient $$Y_\chi:=(G\times {\Bbb G}_m)/\iota_\chi(H)$$ and the projection map $$\begin{gather*}\pi\colon\, Y_\chi=(G\times {\Bbb G}_m)/\iota_\chi(H)\,\longrightarrow\, G/H=X,\quad \\ [g,c]\,\mapsto\, [g]\quad \text{for }g\in G,\ c\in{\Bbb C}^\times.\end{gather*}$$ The group $${\Bbb G}_m$$ acts on the fibers of $$\pi$$ by $$c'\cdot [g,c]=[g,c'c]$$ for $$c'\in{\Bbb C}^\times$$. We see that $$\pi\colon Y_\chi\to X$$ is a principal $${\Bbb G}_m$$-bundle over $$X$$. To $$\chi$$ we associate the class of $$Y_\chi$$ in $${\rm Pic\,} X$$.

We compute the character group $$\widehat H$$ of the stabilizer $$H={\rm Sp}(2n)\cap{\rm GO}(2n)$$, where $${\rm GO}(2n)=\{A\in{\rm GL}(2n,{\Bbb C})\mid A^t A=\lambda_A I,\ \lambda_A\in{\Bbb C}^\times\}.$$

Proposition 3. For $$H={\rm Sp}(2n)\cap{\rm GO}(2n)$$ we have $${\widehat H}={\Bbb Z}/2{\Bbb Z}$$.

Proof. We compute the group $$H$$. We write the equations for $$A\in H$$: $$A^t A =\lambda_A I,\qquad A^t J A=J, \qquad\text{where } J= \begin{pmatrix} 0 & I_n\\ -I_n &0 \end{pmatrix}.$$ We obtain $$\lambda_A A^{-1} J A=J, \quad\text{whence } \lambda_A J A=AJ.$$ Let $$x$$ be an eigenvector of $$J$$ with eigenvalue $$\mu$$. Then $$Jx=\mu x,$$ whence $$AJx=\mu Ax,\qquad \lambda_A JAx=\mu Ax,\qquad Jy=\lambda_A^{-1} \mu y, \text{ where }y=Ax.$$ We see that $$y$$ is an eigenvector of the matrix $$J$$ with eigenvalue $$\lambda_A^{-1}$$. Thus $$\lambda_A^{-1}\mu$$ is an eigenvalue of $$J$$ as well. Since our matrix $$J$$ has only two eigenvalues $$i$$ and $$-i$$, we conclude that $$\lambda_A$$ can take values only $$1$$ and $$-1$$. Thus we obtain a homomorphism $$\lambda\colon H\to \mu_2,\quad A\mapsto \lambda_A.$$

Consider the matrix $$S=i\begin{pmatrix} 0 & I_n \\ I_n & 0\end{pmatrix}.$$ An easy calsulation shows that
$$S^t S=S^2=-I,\qquad S^t J S=SJS=J.$$ Thus $$S\in H$$, $$\lambda_S=-1$$. We obtain a short exact sequence $$1\to H_1\to H\to \mu_2\to 1,$$ where $$H_1={\rm Sp}(2n)\cap{\rm SO}(2,n)$$ and where the homomorphism $$\lambda\colon H\to\mu_2$$ is surjective because $$\lambda_S=-1$$. We have $$H=H_1\cup S\cdot H_1$$.

The group $$H_1$$ was computed by Sasha in his answer: it is isomorphic to $${\rm GL}(n,{\Bbb C})$$ acting on $$V=L_1\oplus L_2$$ by $$B\mapsto (B,B^{-1})$$. The linear operator $$S$$ permutes the subspaces $$L_1$$ and $$L_2$$, and it acts on the normal subgroup $$H_1$$ of $$H$$ as follows: $$S\cdot (B,B^{-1}) \cdot S^{-1}=(B^{-1},B).$$ Hence $$S\cdot (B,B^{-1}) \cdot S^{-1}\cdot (B,B^{-1})^{-1}=(B^{-2},B^2).$$ It follows that the commutator subgroup $$(H,H)$$ of $$H$$ is $$H_1$$. Thus $${\widehat H}=\widehat{H/H_1}=\widehat{\mu_2}={\Bbb Z}/2{\Bbb Z},$$ as required. The nontrivial element of the character group $${\widehat H}$$ is the character $$\lambda\colon H\to \mu_2\hookrightarrow{\Bbb G}_m,\quad A\mapsto \lambda_A\in {\Bbb C}^\times.$$

Corollary 4. For $$X={\rm Sp}(2n)/({\rm Sp}(2n)\cap {\rm GO}(2n))$$ we have $${\rm Pic\,} X={\Bbb Z}/2{\Bbb Z}$$.

• I will compute $\widehat H$ in your case tomorrow. – Mikhail Borovoi May 18 at 20:07
• That's too kind of you. Thank you very much. – F_L May 18 at 20:49
• Thank you very much for your answer. There is just a thing that is not clear to me. Let us look for instance at the case $n = 1$. If I want a matrix $\left(\begin{array}{cc} a & b \\ c & d \end{array}\right)$ to act as the multiplication by a scalar $\lambda$ on $L_1$ we must have $c = -b, a = d$ and then $\lambda = a+ib$. Now matrices of the form $\left(\begin{array}{cc} a & b \\ -b & a \end{array}\right)$ act via the multiplication by $\frac{1}{\lambda} = a-ib$ on $L_2$. Since we want these matrices to be symplectic and orthogonal we must have $a^2+b^2 = 1$. This is $S^1$. – F_L May 20 at 8:17
• @F_L: I don't want a matrix $\begin{pmatrix} a & b \\ c& d \end{pmatrix}$ to act as the multiplication by scalar on $L_1$! I want it to permute $L_1$ and $L_2$! – Mikhail Borovoi May 20 at 11:37
• @F_L: Take the following matrix: $S= \begin{pmatrix} 0 & i \\ i& 0 \end{pmatrix}$. Then $S\in H$ and $S$ permutes $L_1$ and $L_2$. Moreover, $S$ multiplies the quadratic form $x^2+y^2$ by $-1$. – Mikhail Borovoi May 20 at 11:40