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My question is about the extention of kirillov's symplectic structure on coadjoint orbits. The most remarkable feature of the coadjoint representation is the fact that all coadjoint orbits possess a canonical G-invariant symplectic structure. Kirillov defined the coadjoint orbit by the natural way as follows,

Let $G$ be a Lie Group and $\mathfrak{g}$ be its lie algebra,and also $\mathfrak{g^*}$ be the dual of Lie algebra

$\mathfrak{G}=\{Ad^*(g)F, g\in G\}$ where $F\in\mathfrak{g^*}$.

In fact, Kirillov introduced an antisymmetric bilinear form $B_F$ on $\mathfrak{g}$ by

$B_F(X,Y)=<F,[X,Y]>$ and showed that $B_F$ is invariant under $Stab(F)=\{g\in G: Ad^*(g)F=F \}$ and by using this fact, he introduced a $G$-Invariant symplectic structure $\omega_{\mathfrak{G}}(F)(ad^*(X)F,ad^*(Y)F)=B_F(X,Y)$ on $\mathfrak{G}$ which is now known as Kirillov-Kostant-Souriau Theorem .

Now, I am trying to extend this result for $\mathfrak{g}\oplus \mathfrak{g^*} $ instead of $\mathfrak{g^*} $and try to find a G-invariant symplectic structure. The fact is that $\mathfrak{g}\oplus \mathfrak{g^*}$ is exactly "equal"(in algebraic and geometric sense) to $Lie(T^*G)$, and we have a symplectic structure on $T^*G$ so we will have a symplectic structure on $Lie(T^*G)$, because we just can restrict the symplectic structure of $(T^*G)$ to $T_e(T^*G)$ and get a symplectic structure on $Lie(T^*G)$.(Also we can define a bilinear symmetric and antisymmetric form on $\mathfrak{g}\oplus \mathfrak{g^*}$ and define a generalized complex structure on it and so we will have a symplectic, poisson, kahler structure on $\mathfrak{g}\oplus \mathfrak{g^*}$ by this way. ) But it would be a good question , if we define the action of $g\in G$ on $X+F\in \mathfrak{g}\oplus \mathfrak{g^*} $ by $g.(X+F)=Ad(g)X+Ad^*(g)F$. and define an orbit as same method of kirillov by $\mathfrak{O}=\{Ad(g)X+Ad^*(g)F, g\in G\}$.

So, by this definition,we will have $Stab(X+F)=\{g\in G: Ad^*(g)F=F , gX=Xg \}$, and then $G$ is a fibre bundle over the base $\mathfrak{O}$.

Now, if we construct the symplectic structure

$\omega_{X+F}:T_{X+F}{\mathfrak{O}}\times T_{X+F}{\mathfrak{O}}\to R$

$\omega_{X+F}([Y,X]+ad^*(Y)F,[Z,X]+ad^*(Z)F)$=? THE FACT IS "?" (I mean right hand sight of equality) should be an invariant bilinear form $B_{X+F}(Y,Z)$ and we should find it. So, my question is how can we find this guy?

PS:I edited my question after comments of Ben and Mariano .

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you can see the definition 3 for bilinear form of… but just for action $G$ on $\mathfrak{g^*}$ – baba ab dad Sep 12 '13 at 16:50
Are you sure you want $B_{X+F}$ to be invariant under $G$? Note that neither $X$ nor $F$ have been required to be invariant under $G$. So if we are going to produce a formula for $B_{X+F}$ which works universally for all Lie groups and all $X$ and $F$, it seems unlikely that the formula could make any use of $X$ or $F$. As you can see, Woit doesn't arrive at a $G$-invariant bilinear form; his form is only invariant under the appropriate stabilizer subgroup. Moreover, if $G$ is not reductive then there is no $G$-invariant nondegenerate inner product on $\mathfrak{g}$. – Ben McKay Sep 12 '13 at 16:59
And why don't you explain that in the question? Questions are more interesting (and therefore more likely to attract attention...) if you make them real, if you provide them with a context. You asked a question which has «The zero bilinear form» as an answer, and apparently you have in mind something considerably more interesting than that: sell your question! – Mariano Suárez-Alvarez Sep 12 '13 at 17:28
@BenMcKay: Your last phrase is true only for positive-definite inner products. If you allow indefinite metrics, then there are plenty of nonreductive groups with ad-invariant inner products on their Lie algebras. In a way, they generalise the semidirect product $\mathfrak{g} \ltimes \mathfrak{g}^*$, with $\mathfrak{g}^*$ abelian, with inner product given by the natural dual pairing. – José Figueroa-O'Farrill Sep 12 '13 at 23:18
@HassanJolany, I am not a moderator of MO and I did not downvote your question. – Mariano Suárez-Alvarez Sep 13 '13 at 17:22
up vote 5 down vote accepted

There is no hope for such a construction. The orbits of $G$ on $\frak{g}\oplus\frak{g}^\ast$ are not always of even dimension, so they will not support symplectic structures.

For example, take $G=\mathrm{SO}(3)$. Because $G$ is simple, $\frak{g}^\ast$ is isomorphic to $\frak{g}$ as a $G$-module, so the action of $G$ on $\frak{g}\oplus\frak{g}^\ast$ is just the diagonal action of $G$ on $\frak{g}\oplus\frak{g}$. Now, if $x, y\in\frak{g}$ are linearly independent, then the stabilizer of $(x,y)\in \frak{g}\oplus\frak{g}$ is trivial, so its orbit is just a copy of $G$, which has dimension $3$.

A similar phenomenon happens for any odd-dimensional compact simple Lie group.

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Robert Bryant@ thanks a lot for your good answer. – baba ab dad Sep 14 '13 at 9:56
We know that $G\times G$ acts on $T^*G$ by following $(g_1,g_2).(g,F)=(g_1gg_2^{-1},Ad^*(g_2)F)$. So this action make sense. But what can we say about the same question , if we change $G$ to $G\times G$ – baba ab dad Sep 16 '13 at 13:57
Well, of course, $G\times G$ acts on $G$ on the left and right and therefore induced an action of $G\times G$ on $T^\ast G$, which is already a symplectic manifold. In particular, one can perform symplectic reduction on $T^\ast G$ under this group action and get symplectic manifolds. The action you describe on $G\times\frak{g}^\ast$ is equivalent to this action under the appropriate identification of this space with $T^\ast G$, so there is symplectic reduction. Frankly, I'm not sure what you are trying to accomplish, since essentially all homogeneous symplectic manifolds are coadjoint orbits. – Robert Bryant Sep 16 '13 at 16:25
Well, I constructed an orbit as same as coadjoint orbit, but your answer showed that it does not functional , and I wanted to construct a symplectic structure on this orbit. But now, if we change $G$ to $G\times G$, and construct an orbit on it, then can we construct a symplectic 2-form on it like Kirillov's symplectic 2-form ? – baba ab dad Sep 16 '13 at 17:42
Are you considering the orbits of $G\times G$ on $T^\ast G$ or on $\frak{g}\times\frak{g}^\ast$? In either case, if you want to construct a homogeneous symplectic manifold (which is what the coadjoint orbits are), then you won't get anything beyond a coadjoint orbit or a covering of a coadjoint orbit anyway because that's all the homogeneous symplectic manifolds there are. Even if you could construct something by your approach it wouldn't be a new homogeneous symplectic manifold. – Robert Bryant Sep 16 '13 at 18:41

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