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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 $J^2 \neq -1$.

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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.

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    $\begingroup$ 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. $\endgroup$
    – Victor Protsak
    Commented Jun 10, 2010 at 5:26
  • $\begingroup$ 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. $\endgroup$
    – Bo Peng
    Commented Jun 10, 2010 at 6:22
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    $\begingroup$ 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. $\endgroup$
    – José Figueroa-O'Farrill
    Commented Jun 10, 2010 at 8:20
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    $\begingroup$ 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 $J^2 = -4$. $\endgroup$
    – David E Speyer
    Commented Jun 10, 2010 at 13:43
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    $\begingroup$ 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. $\endgroup$
    – Tim Perutz
    Commented Jun 10, 2010 at 16:25

1 Answer 1

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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 $J$.

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