5
$\begingroup$

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

$\endgroup$
  • $\begingroup$ How do you choose the symplectic and quadratic forms? $\endgroup$ – Sasha May 18 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. $\endgroup$ – F_L May 18 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 $H$. $\endgroup$ – Mikhail Borovoi May 18 at 17:36
  • $\begingroup$ I have restored your original question. Otherwise the reader cannot understand, what question was answered by Sasha. $\endgroup$ – Mikhail Borovoi May 18 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 $\mathbb{Z}/n\mathbb{Z}$? $\endgroup$ – F_L May 18 at 18:45
7
$\begingroup$

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

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ 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 $X$. $\endgroup$ – F_L May 18 at 16:54
5
$\begingroup$

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

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ I will compute $\widehat H$ in your case tomorrow. $\endgroup$ – Mikhail Borovoi May 18 at 20:07
  • $\begingroup$ That's too kind of you. Thank you very much. $\endgroup$ – F_L May 18 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 $S^1$. $\endgroup$ – F_L May 20 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 $L_2$! $\endgroup$ – Mikhail Borovoi May 20 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 $-1$. $\endgroup$ – Mikhail Borovoi May 20 at 11:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.