# Action of $\mathfrak{sl}({V})$ in tensor spaces

what is the natural action of $\mathfrak{sl}({V})$ in tensor spaces ?

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If $L$ is a Lie algebra and $V$ and $W$ are representations of $L$, then an action of $L$ on $V\otimes W$ is given by $x.(v\otimes w) = (x.v) \otimes w + v \otimes (x.w)$. It is a nice little exercise to check that this makes $V\otimes W$ a representation of $L$.
Here's the standard explanation for why Tobias' answer is the natural action to use on the tensor product. First, the natural action of the group $G = SL(V)$ on a tensor product $U \otimes W$ (where $U$ and $W$ are $G$-representations) is $g \cdot (u \otimes w) = (g \cdot u) \otimes (g \cdot w)$.
Since $\mathfrak{g} = \mathfrak{sl}(V)$ is the tangent space at the identity of $G$ and $U$ and $W$ become $\mathfrak{g}$-representations by differentiating the $G$-representation, the same should be true of $U \otimes W$. So let $c(t) : (-\epsilon, \epsilon) \rightarrow G$ be a curve in $G$ passing through the identity (i.e. $c(0) = \text{id}$). Then $X = c'(0) \in \mathfrak{g}$ and the action of $X$ on $u \otimes w$ is obtained by differentiating the action of $c(t)$ on $u \otimes w$ and evaluating at $0$. So \begin{align*} X \cdot (u \otimes w) & = \left[ c(t) \cdot (u \otimes w) \right]'(0) \\ &= \left[ (c(t) \cdot u) \otimes (c(t) \cdot w) \right]'(0) \\ &= \left[ (c'(0) \cdot u) \otimes (c(0) \cdot w) \right] + \left[ (c(0) \cdot u) \otimes (c'(0) \cdot w) \right] \\ &= (X \cdot u) \otimes w + u \otimes (X \cdot w). \end{align*}
In the calculation, we used the product rule to evaluate the derivative and we used that $c(0) = \text{id}$ and $c'(0) = X$. Of course, everything here is done over either the real numbers or complex numbers, which are the primary fields we use for our geometric intuition.
Nice answer! Once can also derive the induced action on the Lie algebra from the identity $\Phi(e^X) = e^{\phi(X)}$ for $X \in \mathfrak{g}$, $\Phi : G \to \text{GL}(V)$ and $\phi : \mathfrak{g} \to \textrm{gl}(V)$. $G$ is some Lie group and $\mathfrak{g}$ the Lie algebra of $G$. – user38268 Jan 19 '13 at 3:30