Finding the $J$ for a symplectic vector space - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T22:24:49Z http://mathoverflow.net/feeds/question/27642 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/27642/finding-the-j-for-a-symplectic-vector-space Finding the $J$ for a symplectic vector space Bo Peng 2010-06-10T05:03:40Z 2010-06-10T18:04:36Z <p>I found something strange when I was working on some other problems.</p> <p>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.</p> <p>Consider the usual inner product on $\mathfrak{gl}(n)$:</p> <p>$g(X,Y) = tr(XY)$</p> <p>and this "symplectic structure"</p> <p>$\omega(X,Y)=tr(A X A^{-1} Y - A Y A^{-1} X)$</p> <p>where $A$ is a fixed constant element in $GL(n)$.</p> <p>Now the $J$ corresponding to it seems to be</p> <p>$J(X) = AXA^{-1} - A^{-1}XA$</p> <p>but then $J^2 \neq -1$.</p> <p>================================</p> <p>Now the above symplectic form is degenerate, but I thought the real cause shall be something different.</p> <p>Define $\omega$ on $\mathfrak{g} \times \mathfrak{g}$ as follows:</p> <p>$\omega((x_1, x_2),(y_1, y_2)) = tr(x_1 y_2 - y_1 x_2 + Ad_g x_1 \cdot y_1 - Ad_g y_1 \cdot x_1 + Ad_h x_2 \cdot y_2 - Ad_h y_2 \cdot x_2)$</p> <p>where $(g,h) \in G \times G$ is fixed, $(x_1,x_2) \in \mathfrak{g} \times \mathfrak{g}$, $(y_1,y_2) \in \mathfrak{g} \times \mathfrak{g}$.</p> <p>One still have the $J$ issue.</p> http://mathoverflow.net/questions/27642/finding-the-j-for-a-symplectic-vector-space/27725#27725 Answer by Fran Burstall for Finding the $J$ for a symplectic vector space Fran Burstall 2010-06-10T18:04:36Z 2010-06-10T18:04:36Z <p>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.</p> <p>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).</p> <p>I leave it as an exercise to determine the compatibility of $g$ and $\omega$ required to define $J$.</p>