What is the diagonal $\mathfrak{g}$-action on $V \otimes V^*$?

Question

Let $\mathfrak{g}$ be a Lie algebra and $V$ a left $\mathfrak{g}$-module. Then the dual vector space $V^*$ is a right $\mathfrak{g}$-module with right $\mathfrak{g}$-action given by $(f.g)(v) = f(g.v)$.

What is the diagonal $\mathfrak{g}$-action on $V \otimes V^*$? Is the diagonal $\mathfrak{g}$-action on $V \otimes V^*$ given by the following formula \begin{align} & \mathfrak{g} \otimes V \otimes V* \to V \otimes V^*, \\ & g.(v, f) = g.v \otimes f + v \otimes f.g, \quad v\in V, f \in V^*, g \in \mathfrak{g}? \end{align} Any help will be greatly appreciated!

Answer 1

For a more general point of view: a (left) module over a Lie algebra $\mathfrak{g}$ is the same thing as a (left) module over the universal enveloping algebra $\mathbb{U}\mathfrak{g}$, an associative algebra defined as the quotient of the free associative algebra on $\mathfrak{g}$ quotiented out by an ideal: $$\mathbb{U}\mathfrak{g} = T(\mathfrak{g}) / \langle x \otimes y - y \otimes x - [x,y] \rangle.$$

Then one can check that $\mathbb{U}\mathfrak{g}$ is a Hopf algebra, with a coproduct defined on generators $x \in \mathfrak{g} \subset T(\mathfrak{g})$ by $$\Delta(x) = x \otimes 1 + 1 \otimes x$$ (be careful: this formula doesn't hold for a general element $y \in \mathbb{U}\mathfrak{g}$). This factors through the quotient and is a coassociative comultiplication, and there's also an explicit counit $\varepsilon(x) = 0$ and antipode $\sigma(x) = -x$ (both for $x \in \mathfrak{g}$).

Now whenever $H$ is a Hopf algebra with comultiplication $\Delta(x) = \sum_{(x)} x' \otimes x''$ (I'm using a kind of Sweedler notation here), if $M$ and $N$ are modules over $H$ viewed as an associative algebra, then $M \otimes N$ can be endowed with an $H$-module structure too (again, module over the associative algebra $H$) by setting: $$x \cdot (m \otimes n) := \sum_{(x)} (x' \cdot m) \otimes (x'' \cdot n).$$

Finally, there's a little trick: if $M$ is a right module over the Lie algebra $\mathfrak{g}$, then it can be given a left module structure by letting $x \cdot m := - m \cdot x$. In fact here I'm using the antipode: if $H$ is a Hopf algebra and $M$ is a right module over $H$ seen as an associative algebra, then setting $x \cdot m := m \cdot \sigma(x)$ turns $M$ into a left module over $H$. This comes from the fact that $\sigma$ is an anti-homomorphism of algebras.

So finally recalling $\Delta(x) = x \otimes 1 + 1 \otimes x$ when $x$ is the image of an element in $\mathfrak{g}$, you get, for $v \in V$ and $f \in V^*$: $$\begin{align} x \cdot (v \otimes f) & = (1 \cdot v) \otimes (x \cdot f) + (x \cdot v) \otimes (1 \cdot f) \\ & = v \otimes (- f \cdot x) + (x \cdot v) \otimes f \\ & = - v \otimes (f \cdot x) + (x \cdot v) \otimes f. \end{align}$$

So in fact you see that there is a sign missing from your formula. If you try checking the Jacobi relation you'll see that it's necessary.

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The same reasoning works for groups too (as Dietrich Burde mentions): for a group $G$, the group algebra $\Bbbk[G]$ is also a Hopf algebra, with $\Delta(g) = g \otimes g$ on generators $g \in G$. So you get $g \cdot (x \otimes y) = (g \cdot x) \otimes (g \cdot y)$.

Answer 2

If $V$ and $W$ are $\mathfrak{g}$-modules, then the usual formula you have applied defines the "diagonal" $\mathfrak{g}$-action on the tensor product $V\otimes W$, i.e., it gives the tensor product of two Lie algebra representations - see Definition $1.12$ here. I think the word "diagonal" comes from the group case, where the action is really $g(v_1\otimes g_2)=gv_1\otimes gv_2$. This definition would not work for Lie algebras (because it is not linear), but differentiating it it gives the correct strcuture - your formula (up to a sign).