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Let $(V,\psi)$ be a symplectic vector space over some field $K$ (or just over the complex numbers, if you prefer), i.e., $\psi$ is a non-degenerate alternating form $\psi\colon \bigwedge^2 V \rightarrow K$. Then one defines the symplectic group $G := \text{Sp}(V,\psi)$ as the group of automorphisms $g$ of $V$ with $g^* \psi = \psi$. This is a Lie group (or an algebraic group), so it makes sense to speak of its Lie algebra $\text{Lie}(G)$, which carries a $G$-representation called the adjoint representation.

Question: Is there description of the adjoint representation of $G$ as some tensor construction applied to the $G$-representation $V$?

Some motivation: For a quadratic space $(V,Q)$ (assume the characteristic of the base field is $\neq 2$), the map $\bigwedge^{2} V \rightarrow \text{Lie}(\text{SO}(V,Q))$ given by $v \wedge w \mapsto Q(v,x)w - Q(w,x)v$ is easily seen to be an isomorphism of $\text{SO}(V,Q)$-representations.

Attempted partial answer: There is an obvious injective $G$-equivariant map $\text{Sym}^2 V \rightarrow \text{Lie}(G)$ given by sending $vw$ to the map $x \mapsto \psi(v,x)w + \psi(w,x)v$, where I've identified $\text{Lie}(G)$ with the space of endomorphisms $M$ of $V$ satisfying $\psi(Mv,w) + \psi(v,Mw) = 0$. However, the dimension of $\text{Sym}^2 V$ is still too low by at least $\dim V^2$.

EDIT: Andreas Cap pointed out in the comments that the attempted partial answer is an answer by virtue of the fact that the dimensions of $G$ and $\text{Sym}^2 V$ actually do coincide. That is, $\text{Lie}(G) \cong \text{Sym}^2 V$.

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    $\begingroup$ The map $Sym^2V\to Lie(G)$ you construct is an isomorphism. I don't know why you think that the dimension of $Sym^2V$ is too low (even less how it could be too low by $\dim(V)^2$). $\endgroup$ – Andreas Cap Apr 12 '17 at 7:37
  • $\begingroup$ Through thorough use of doublethink. I simultaneousy believed $\dim V = g$ and $\dim V = 2g$. I'm ever so slightly embarrassed. Thank you. $\endgroup$ – EstonianFootballerJoelLindpere Apr 12 '17 at 9:12

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