0
$\begingroup$

I am trying to prove that the unitary group associated to an $\omega$-compatible complex structure $J$ acts transitively on the Lagrangian Grassmannian $\mathcal{L}(V)$.

I know that for a symplectic space $(V, \omega)$, one can find an $ \omega$-compatible complex structure $J$. Then, a hermitian structure on $V$ is defined by \begin{equation*} \langle \cdot, \cdot \rangle = g_J(\cdot, \cdot)+ \mathrm{i} \omega(\cdot, \cdot), \end{equation*} where $g_J(x, y)= \omega(x,Jy)$ is an inner product.

The unitary group $U(V)$ consists of linear transformations $T\in GL(V)$ which preserve the Hermitian structure. Then the intersection of $Sp(V)$ and $O(V)$ equals $U(V)$.

Thus, for any $L_1,L_2\in \mathcal{L}(V)$, we begin with an orthogonal transformation $A:L_1\to L_2$. The proof from the book is as follows. From $A:L_1\to L_2$, we construct a symplectromorphism from $L_1\oplus L_1^*$ to $L_2\oplus L_2^*$. Then we can generate a unitary transformation $L_1\oplus JL_1\to L_2\oplus JL_2$ which maps $L_1$ to $L_2$.

I cannot understand the proof. What I know is that $L_i^*$ is isomorphic to $JL_i$. But I have no idea why we consider $L_1\oplus JL_1\to L_2\oplus JL_2$?

Any advice on explaining the proof or new ideas are appreciated. Thanks in advance.

$\endgroup$
  • $\begingroup$ I realize that for $x_1+y_1, x_2+y_2 \in L_1\oplus JL_1$, we have $g_J(x_1+y_1,x_2+y_2)=\omega(x_1+y_1,J(x_2+y_2))=g_J(x_1,x_2)+g_J(y_1,y_2)$. Moreover, $\omega(x_1+y_1,x_2+y_2)=\omega(x_1,y_2)-\omega(x_2,y_1)$ which is associated with the symplectic form on $L\oplus L^*$. I think that I misunderstood the meaning of orthogonal. The orthogonality is associated with the inner product $g_J$. Then we construct $A\otimes A^{-1}*$ which is the symplectromorphism. Thus, we construct a tranformation which preserves $g_J$ and $\omega$. Am I right? $\endgroup$ – Hu ju yuan Mar 25 at 10:23

Your Answer

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

Browse other questions tagged or ask your own question.