Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

I found something strange when I was working on some other problems.

I thought the triple intersection description of the unitary group said that any two of $(g, \omega, J)$ determines the third pointwisely. Then I found I was very wrong: when one tries to find a $J$ from some $(g,\omega)$ , one need not have $g(Jx, Jy)=g(x, y)$ for the resulting $J$, and hence may not have $J^2=−1$... very strange.

Consider the usual inner product on $\mathfrak{gl}(n)$:

$g(X,Y) = tr(XY)$

and this "symplectic structure"

$\omega(X,Y)=tr(A X A^{-1} Y - A Y A^{-1} X)$

where $A$ is a fixed constant element in $GL(n)$.

Now the $J$ corresponding to it seems to be

$J(X) = AXA^{-1} - A^{-1}XA$

but then $J^2 \neq -1$.

================================

Now the above symplectic form is degenerate, but I thought the real cause shall be something different.

Define $\omega$ on $\mathfrak{g} \times \mathfrak{g}$ as follows:

$\omega((x_1, x_2),(y_1, y_2)) = tr(x_1 y_2 - y_1 x_2 + Ad_g x_1 \cdot y_1 - Ad_g y_1 \cdot x_1 + Ad_h x_2 \cdot y_2 - Ad_h y_2 \cdot x_2)$

where $(g,h) \in G \times G$ is fixed, $(x_1,x_2) \in \mathfrak{g} \times \mathfrak{g}$, $(y_1,y_2) \in \mathfrak{g} \times \mathfrak{g}$.

One still have the $J$ issue.

share|improve this question
2  
This isn't a symplectic structure: the identity matrix is orthogonal to everything. Moreover, if $n$ is odd then there can be no symplectic structure at all. –  Victor Protsak Jun 10 '10 at 5:26
    
Actually I knew it is degenerate, but I thought the real cause shall be something different. Define $\omega$ on $\mathfrak{g} \times \mathfrak{g}$ as follows: $\omega((x_1, x_2),(y_1, y_2)) = tr(x_1 y_2 - y_1 x_2 + Ad_g x_1 \cdot y_1 - Ad_g y_1 \cdot x_1 + Ad_h x_2 \cdot y_2 - Ad_h y_2 \cdot x_2)$ where $(g,h) \in G \times G$ is fixed, $(x_1,x_2) \in \mathfrak{g} \times \mathfrak{g}$, $(y_1,y_2) \in \mathfrak{g} \times \mathfrak{g}$. One still have the $J$ issue. –  Bo Peng Jun 10 '10 at 6:22
4  
Why are you surprised that $J^2 \neq -1$? If $g$ is an inner product on a vector space and $\omega$ a symplectic structure, then in general the endomorphism $J$ defined by $\omega(x,y) = g(Jx,y)$, say, need not be a complex structure. All one can say is that it is a nondegenerate skewsymmetric endomorphism. –  José Figueroa-O'Farrill Jun 10 '10 at 8:20
3  
Indeed, a much simpler example. On $\mathbb{R}^2$, let $g$ be the standard inner product, and let $\omega$ be TWICE the standard symplectic form. Then $J^2 = -4$. –  David Speyer Jun 10 '10 at 13:43
2  
In $GL(2n,\mathbb{R})$, if you intersect $Sp(2n,\mathbb{R})$ with $O(2n)$ you'll get $U(n)$. If you intersect $Sp(2n,\mathbb{R})$ with some conjugate $g O(2n)g^{-1}$ (i.e., the orthogonal group of an arbitrary inner product on $\mathbb{R}^{2n})$ you'll get ... nothing that deserves a name besides $g O(2n)g^{-1}\cap Sp(2n,\mathbb{R})$, in general. –  Tim Perutz Jun 10 '10 at 16:25
show 3 more comments

1 Answer

It is true that any two of $(g,\omega,J)$ determine the third. What is not true is that arbitrary choices of these three ingredients give rise to a third.

For example, given a metric $g$ and an almost complex structure $J$, you won't get a symplectic form out of these two unless $J$ is skew (equivalently, $J$ is an isometry for $g$). In short, your two ingredients must have some kind of compatibility. Again, given $\omega$ and $J$, you will need $J^*\omega=\omega$ before you have even a chance of defining $g$ (and even then you need to worry about whether $g$ is positive definite).

I leave it as an exercise to determine the compatibility of $g$ and $\omega$ required to define $J$.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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