Induced $\frak{g}$-action on exterior power and symmetric power? What to show.
Let $\frak{g}$ be a Lie algebra and $V$ a $\frak{g}$-representation. I am supposed to show that for $r \geq 0$ there exists a unique action of $\frak{g}$ on the exterior power $\bigwedge ^r V$ such that  the canonical projection $V^{\otimes r} \rightarrow \bigwedge ^r V$ is a $\frak{g}$-intertwiner. Here, we consider the tensor product of representations on the left-hand side. Afterwards, I am asked to show the above statement, where the exterior power is replaced by the symmetric power $S^rV$.
What I think.
Consider the exterior power.
Uniqueness seems easy: The canonical projection is supposed to be a $\frak{g}$-intertwiner. This forces us to define on equivalence classes of homogenous tensors $x.[v_1 \otimes ... \otimes v_r] := [x. (v_1 \otimes ... \otimes v_r)]$ for $x \in \frak{g}$. As homogenous tensors span $V^{\otimes r}$ there image under the canonical projection spans $\bigwedge ^r V$. Hence, the above definition determines the action uniquely. What remains to show is that this map is well-defined, and indeed gives a $\frak{g}$-action. That is where I am struggling. I have no experience in working with exterior powers. I don’t know what relations I can use.
Questions.
Any hints? Are well-definedness and the two action properties (bilinearity and the “Lie algebra action property”) the only thing left to show? Many thanks!
 A: I would formulate this in the following way: the data of a $\mathfrak{g}$-representation on $V$ is exactly the data of a Lie algebra homomorphism $\rho : \mathfrak{g} \to \operatorname{End}(V)$. (If you are unfamiliar with this perspective, here $\operatorname{End}(V)$ is just the set of linear maps $V \to V$, which is a Lie algebra because it is an associative $\mathbb{k}$-algebra.)
So, given a map $\rho:\mathfrak{g} \to \operatorname{End}(V)$ you are supposed to produce maps $\bigwedge^r \rho : \mathfrak{g} \to \operatorname{End}\left(\bigwedge^r V\right)$ and $\operatorname{S}^r \rho : \mathfrak{g} \to \operatorname{End}\left(\operatorname{S}^r V\right)$.
Let's focus on a more basic construction first, the $r$th tensor power: given $\rho$, we can define a linear map $\bigotimes^r \rho : \mathfrak{g} \to \operatorname{End}(\bigotimes^r V)$ by
$$
\bigotimes{}^r \rho := \rho \otimes 1 \otimes \cdots \otimes 1 + 1 \otimes \rho \otimes \cdots \otimes 1 + \cdots + 1 \otimes 1 \otimes \cdots \otimes \rho,
$$
i.e. defined on a pure tensor by
$$
(\bigotimes{}^r \rho)(x) : v_1 \otimes \cdots \otimes v_r \mapsto \rho(v_1) \otimes v_2 \otimes \cdots \otimes v_r + v_1 \otimes \rho(v_2) \otimes \cdots v_r + \cdots + v_1 \otimes v_2 \otimes \cdots \otimes \rho(v_r).
$$
This new map $\bigotimes{}^r \rho : \mathfrak{g} \to \operatorname{End}(\bigotimes^r V)$ gives a Lie algebra representation of $\mathfrak{g}$ on $\bigotimes^r V$ if we can show that $\bigotimes{}^r \rho$ preserves the Lie bracket. Exercise 1: Check this. Then we have a quotient map $\pi_\wedge : \bigotimes^r V \to \bigwedge^r V$ which exists just by the definition of the $r$th exterior power as a quotient of the $r$th tensor power. So, for each fixed $x \in \mathfrak{g}$ the we get a map $(\pi_\wedge \circ \bigotimes^r \rho)(x) : \bigotimes{}^r V \to \bigwedge^r V$. This map descends to a map $\bigwedge^r V \to \bigwedge^r V$ (i.e. factors through $\pi_\wedge$) if it is an alternating map. Exercise 2: Check that this is the case (is it for "easy" reasons?). Thus for each fixed $x \in \mathfrak{g}$ we have a recipe to construct a linear map $\bigwedge^r V \to \bigwedge^r V$. We just finally need to check that the map taking each $x \in \mathfrak{g}$ to each such associated map is a Lie algebra homomorphism, but I claim Exercise 3: Check that this follows directly from the same fact for the tensor algebra (i.e. Ex 1).
By replacing the $r$th exterior power with the $r$th symmetric power and "alternating" with "symmetric" you should obtain a construction for the structure of a $\mathfrak{g}$-representation on the symmetric powers of $V$ as well.
Attendum: Another more abstract (and clearer) way to see what is going on here can be described as follows. Let $I$ be the kernel of the quotient map $\pi_\wedge : \bigotimes^r V \to \bigwedge^r V$. Then note (check) that $I \subset \bigotimes^r V$ is in fact a $\mathfrak{g}$-subrepresentation of $\bigotimes^r V$. We can then define (the $\mathfrak{g}$-representation) $\bigwedge^r V$ to just be the quotient $\bigotimes^r V / I$ of $\mathfrak{g}$-representations. Symmetric powers are completely analogous---the point is that the constructions you care about are just special cases of the notion of a quotient of $\mathfrak{g}$-representations.
