Symplectic positive matrix.


by MathematicalPhysicist
Tags: matrix, positive, symplectic
MathematicalPhysicist
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#1
Mar13-12, 10:11 AM
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I need any help with the next question I posted in math.stackexchange, thanks.
http://math.stackexchange.com/questi...efinite-matrix
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quasar987
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#2
Mar13-12, 10:37 AM
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You can self-administer yourself hints by looking at p.45 of McDuff-Salamon's intro to sympolectic topology.
MathematicalPhysicist
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#3
Mar13-12, 12:02 PM
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I appreciate your help, but it doesn't help.

MathematicalPhysicist
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#4
Mar13-12, 12:24 PM
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Symplectic positive matrix.


I think I can see how to prove that this is valid for $\alpha =n$ where n is an integers. Because the eigenspace decomposition of $\mathbb{R}^{2n}$ in $P$ is the same as its decomposition by $P^{-1}$.

To extend this to rationals and then to real, is just taking common factors and reducing the rational case to integers. And for the real exponent to take the limit of rational sequence.

Am I right in this hand waving method? :-D
morphism
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#5
Mar13-12, 10:58 PM
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Provided you get the details right, I don't see why it wouldn't work.

How is ##A^\alpha## defined? Do you just write ##A=Q^T \text{diag}\{\lambda_1, \ldots, \lambda_{2n}\} Q## (possible because A is symmetric) and then let ##A^\alpha=Q^T \text{diag}\{\lambda_1^\alpha, \ldots, \lambda_{2n}^\alpha\} Q## (not problematic because each ##\lambda_i## is positive). It seems pretty clear then that ##A^\alpha## will preserve whatever symplectic form ##A## preserves (it's enough to check at eigenvectors, and for eigenvectors this isn't so bad).
MathematicalPhysicist
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#6
Mar13-12, 11:46 PM
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Yes, I thought along these lines too.

Thanks.
MathematicalPhysicist
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#7
Mar14-12, 12:12 PM
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I've got another question.

I want to show that for any two symplectic vector spaces of equal finite dimensions, W_i, and L_i are lagrange subspaces of [tex]W_i[/tex] (i=1,2), we have a symplectomorphism [tex]A:W_1\rightarrow W_2[/tex] s.t

[tex]A(L_1)=L_2[/tex].

I was thinking along the next lines, the dimensions of L_1 and L_2 are the same cause:
[tex] dim L_1+ dim L^{\omega}_1 = dim L_1 + dim L_1 =dim V_1=dim V_2= dim L_2+dim L^{\omega}_2 = 2dim L_2[/tex] so we must have an isomorphism between L_1 and L_2, I don't see how to extend it to W_1 and W_2.
quasar987
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#8
Mar14-12, 12:24 PM
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A linear map btw symplectic vector spaces is a symplectomorphism iff it sends a symplectic basis to a symplectic basis. (A symplectic basis for a symplectic bilinear map B is a basis (v_i) s.t. B(v_i,v_j) = the standard symplectic matrix -J.)

Observe then, that it suffices for you to prove that a basis of a lagrangian subspace L can always be completed to a symplectic basis of the whole space.

Note that lagrangian spaces are always of dimension half the dimension of the ambient space. So of course if W_1, W_2 are of equal dimension, say 2n, then L_1, L_2 are of equal dimension equal to n.
MathematicalPhysicist
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#9
Mar14-12, 12:36 PM
P: 3,170
OK, thanks. I see it's proved in the textbook you mentioned.

Cheers!
:-)


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