Action of $\mathfrak{sl}({V})$ in tensor spaces what is the natural action of $\mathfrak{sl}({V})$ in tensor spaces ?
 A: 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 $L$.
A: 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.
