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Unitary group acts transitively on the Lagrangian Grassmannian

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 $Tin GL(V)$ which preserve the Hermitian structure.
Then the intersection of $Sp(V)$ and $O(V)$ equals $U(V)$.

Thus, for any $L_1,L_2in mathcal{L}(V)$, we begin with an orthogonal transformation $A:L_1to L_2$.
The proof from the book is as follows.
From $A:L_1to L_2$, we construct a symplectromorphism from $L_1oplus L_1^*$ to $L_2oplus L_2^*$.
Then we can generate a unitary transformation
$L_1oplus JL_1to L_2oplus JL_2$ 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 $L_1oplus JL_1to L_2oplus JL_2$?

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

linear-algebra symplectic-geometry symplectic-linear-algebra

share|cite|improve this question

asked Mar 25 at 9:37

Hu ju yuanHu ju yuan

99110

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I realize that for $x_1+y_1, x_2+y_2 in L_1oplus JL_1$, 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 $Loplus L^*$. I think that I misunderstood the meaning of orthogonal. The orthogonality is associated with the inner product $g_J$. Then we construct $Aotimes A^{-1}*$ 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
  

  
  
  
  

add a comment | 

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 $Tin GL(V)$ which preserve the Hermitian structure.
Then the intersection of $Sp(V)$ and $O(V)$ equals $U(V)$.

Thus, for any $L_1,L_2in mathcal{L}(V)$, we begin with an orthogonal transformation $A:L_1to L_2$.
The proof from the book is as follows.
From $A:L_1to L_2$, we construct a symplectromorphism from $L_1oplus L_1^*$ to $L_2oplus L_2^*$.
Then we can generate a unitary transformation
$L_1oplus JL_1to L_2oplus JL_2$ 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 $L_1oplus JL_1to L_2oplus JL_2$?

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

linear-algebra symplectic-geometry symplectic-linear-algebra

share|cite|improve this question

asked Mar 25 at 9:37

Hu ju yuanHu ju yuan

99110

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I realize that for $x_1+y_1, x_2+y_2 in L_1oplus JL_1$, 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 $Loplus L^*$. I think that I misunderstood the meaning of orthogonal. The orthogonality is associated with the inner product $g_J$. Then we construct $Aotimes A^{-1}*$ 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
  

  
  
  
  

add a comment | 

$begingroup$

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 $Tin GL(V)$ which preserve the Hermitian structure.
Then the intersection of $Sp(V)$ and $O(V)$ equals $U(V)$.

Thus, for any $L_1,L_2in mathcal{L}(V)$, we begin with an orthogonal transformation $A:L_1to L_2$.
The proof from the book is as follows.
From $A:L_1to L_2$, we construct a symplectromorphism from $L_1oplus L_1^*$ to $L_2oplus L_2^*$.
Then we can generate a unitary transformation
$L_1oplus JL_1to L_2oplus JL_2$ 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 $L_1oplus JL_1to L_2oplus JL_2$?

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

linear-algebra symplectic-geometry symplectic-linear-algebra

share|cite|improve this question

asked Mar 25 at 9:37

Hu ju yuanHu ju yuan

99110

$endgroup$

share|cite|improve this question

asked Mar 25 at 9:37

Hu ju yuanHu ju yuan

99110

share|cite|improve this question

asked Mar 25 at 9:37

Hu ju yuanHu ju yuan

99110

asked Mar 25 at 9:37

Hu ju yuanHu ju yuan

99110

asked Mar 25 at 9:37

Hu ju yuanHu ju yuan

99110