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Let $G$ be a compact Lie group, and let $G^\mathbb{C}$ be its complexification. I am looking for a symplectic structure (without use of coordinates) on $$ Sym^kG^{\mathbb{C}}, $$ PS:Here $G^{\mathbb{C}}=T^*G$.(this equality is trivial by polar decomposition in the case, when $G$ is compact Lie group )

i.e. on the space of all symmetric tensors of order $k$ defined on $G^\mathbb{C}$.

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  • $\begingroup$ Do you have any reason to believe that such a thing exists? And I don't understand the notation; is this the $k^{th}$ symmetric power of the tangent bundle? $\endgroup$
    – Qiaochu Yuan
    Commented Nov 3, 2013 at 19:41
  • $\begingroup$ Dear Qiaochu , I edited it. $\endgroup$
    – user21574
    Commented Nov 3, 2013 at 19:46
  • $\begingroup$ It is not clear what you really want: the space $Sym^kG^{\Bbb C}$ is not smooth, and there are several different definitions of symplectic structures on singular spaces. $\endgroup$
    – Misha Verbitsky
    Commented Nov 3, 2013 at 19:48
  • $\begingroup$ Misha Verbitsky, I didn't know this fact.I know that $$ Sym^kG^{\mathbb{C}}=Sym^kT^*G, $$,do you have any referrence about symplectic structures on singular spaces. ? $\endgroup$
    – user21574
    Commented Nov 3, 2013 at 20:02
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    $\begingroup$ The notation is very strange. Usually if $X$ is a variety, $\text{Sym}^k X$ means the quotient of the Cartesian product $\prod^k X$ by the symmetric group on $k$ letters. $\endgroup$
    – Ben McKay
    Commented Nov 3, 2013 at 21:31

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Such a thing doesn't exist. The symmetric square of the cotangent bundle of a real $n$-dimensional manifold has dimension $n + {n+1 \choose 2}$, which is in particular odd whenever $n \equiv 2 \bmod 4$. So for example we can take $G = \text{SU}(2), G_{\mathbb{C}} = \text{SL}_2(\mathbb{C})$, which has real dimension $6$.

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  • $\begingroup$ If we take the bundle $B$ whose sections are the holomorphic symmetric 2-tensors on the complex manifold $X=G_{\mathbb{C}}$, then $X$ is Kaehler and so is $B$. Maybe that is what this is about? $\endgroup$
    – Ben McKay
    Commented Nov 3, 2013 at 21:34

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