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, $Pic(X) \cong \mathbb{Z}/2\mathbb{Z}$.

  • $\begingroup$ How do you choose the symplectic and quadratic forms? $\endgroup$
    – Sasha
    Commented May 18, 2020 at 9:01
  • 1
    $\begingroup$ The standard symplectic form $\left(\begin{array}{cc} 0_n & I_n\\ -I_n & 0_n \end{array}\right)$ and the identity. $\endgroup$
    – Puzzled
    Commented May 18, 2020 at 9:09
  • $\begingroup$ 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 $H$. $\endgroup$ Commented May 18, 2020 at 17:36
  • $\begingroup$ I have restored your original question. Otherwise the reader cannot understand, what question was answered by Sasha. $\endgroup$ Commented May 18, 2020 at 17:47
  • $\begingroup$ Good idea, thank you very much. Is it obvious that the character group of the $H$ in the modified version of the question is $\mathbb{Z}/n\mathbb{Z}$? $\endgroup$
    – Puzzled
    Commented May 18, 2020 at 18:45

2 Answers 2


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 ${\rm Pic\,} X={\Bbb Z}/2{\Bbb Z}$.

  • $\begingroup$ I will compute $\widehat H$ in your case tomorrow. $\endgroup$ Commented May 18, 2020 at 20:07
  • $\begingroup$ That's too kind of you. Thank you very much. $\endgroup$
    – Puzzled
    Commented May 18, 2020 at 20:49
  • $\begingroup$ 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 $S^1$. $\endgroup$
    – Puzzled
    Commented May 20, 2020 at 8:17
  • $\begingroup$ @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 $L_2$! $\endgroup$ Commented May 20, 2020 at 11:37
  • $\begingroup$ @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 $-1$. $\endgroup$ Commented May 20, 2020 at 11:40

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, $\mathrm{Pic}(X) = \mathbb{Z}$.

  • $\begingroup$ 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 $X$. $\endgroup$
    – Puzzled
    Commented May 18, 2020 at 16:54
  • $\begingroup$ Sorry to bother you but a I have a question. I computed explicitly the isomorphism $GL(n)\rightarrow Sp(2n)\cap SO(2n)$ and I got a much more complicated form than $A\mapsto (A,A^{-1})$. $\endgroup$
    – Puzzled
    Commented Dec 22, 2020 at 18:00

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