Submodule of a space of 2-vector fields

Question

I am reading an article about differential geometry. The author considers a Lie algebra $\mathfrak{g} = \mathrm{Lie}(G)$ and the submodule of $\Lambda^2 \mathfrak{g}$ given by the kernel of the map $$L\colon \Lambda^2\mathfrak{g}\rightarrow \mathfrak{g}, $$ where $L$ is induced by the Lie bracket $[{}\cdot{},{}\cdot{}]$. My doubt is on the word 'submodule': is it such with respect to the ring of constant functions on $G$? That would be my most qualified guess, based on the formula $$[fX,gY] = fg[X,Y]+fX(g)Y-gY(f)X, \quad f,g \in C^{\infty}(G),X,Y \in \mathfrak{g}.$$

Answer 1

My guess would be that $\Lambda^2 \mathfrak g$ is to be viewed as a module over the Lie algebra $\mathfrak g$. (When I say a "module over $\mathfrak g$", I mean a "representation of $\mathfrak g$". This is standard terminology.)

To specify this $\mathfrak g$-module structure of $\Lambda^2 \mathfrak g$, I need to tell you how an element of $\mathfrak g$ acts on an element of $\Lambda^2 \mathfrak g$ by left-multiplication. I will define the action of an element $x \in \mathfrak g$ on an element $y \wedge z \in \Lambda^2 \mathfrak g$ as follows: $$ x(y \wedge z) = [x, y] \wedge z + y \wedge [x,z].$$ In order for this to define a genuine $\mathfrak g$-module structure on $\Lambda^2 \mathfrak g$, we require the compatibility condition $$ ([w,x])(y \wedge z) = w(x(y \wedge z)) - x(w(y \wedge z))$$ to hold. You can verify that this does hold, using the Jacobi identity.

Note that the copy of $\mathfrak g$ that appears as the image of $L : \Lambda^2 \mathfrak g \to \mathfrak g$ can also be thought of a $\mathfrak g$-module, where the action of $\mathfrak g$ on itself is given by the Lie bracket.

$L$ is then not only a linear map, but also a $\mathfrak g$-module map, i.e. it satisfies the compatibility relation: $$ x(L(y \wedge z)) = L(x(y \wedge z)).$$ Indeed, recalling that $x$ acts on $L(y \wedge z)$ by the Lie bracket, the left-hand side is $$ x(L(y \wedge z)) = [x, [y,z]],$$ while the right-hand side is $$ L(x(y \wedge z)) =[[x,y],z] + [y, [x,z]],$$ and the two are equal by the Jacobi identity.

Finally, having established that $L:\Lambda^2 \mathfrak g \to \mathfrak g$ is a $\mathfrak g$-module map, the kernel $\ker(L)$ is guaranteed to possess a natural $\mathfrak g$-module structure, i.e. the kernel $\ker(L)$ is a $\mathfrak g$-submodule of $\Lambda^2 \mathfrak g$.