# The Unapologetic Mathematician

## The Submodule of Invariants

If $V$ is a module of a Lie algebra $L$, there is one submodule that turns out to be rather interesting: the submodule $V^0$ of vectors $v\in V$ such that $x\cdot v=0$ for all $x\in L$. We call these vectors “invariants” of $L$.

As an illustration of how interesting these are, consider the modules we looked at last time. What are the invariant linear maps $\hom(V,W)^0$ from one module $V$ to another $W$? We consider the action of $x\in L$ on a linear map $f$: $\displaystyle\left[x\cdot f\right](v)=x\cdot f(V)-f(x\cdot v)=0$

Or, in other words: $\displaystyle x\cdot f(v)=f(x\cdot v)$

That is, a linear map $f\in\hom(V,W)$ is invariant if and only if it intertwines the actions on $V$ and $W$. That is, $\hom_\mathbb{F}(V,W)^0=hom_L(V,W)$.

Next, consider the bilinear forms on $L$. Here we calculate \displaystyle\begin{aligned}\left[y\cdot B\right](x,z)&=-B([y,x],z)-B(x,[y,z])\\&=B([x,y],z)-B(x,[y,z])=0\end{aligned}

That is, a bilinear form is invariant if and only if it is associative, in the sense that the Killing form is: $B([x,y],z)=B(x,[y,z])$

September 21, 2012

## More New Modules from Old

There are a few constructions we can make, starting with the ones from last time and applying them in certain special cases.

First off, if $V$ and $W$ are two finite-dimensional $L$-modules, then I say we can put an $L$-module structure on the space $\hom(V,W)$ of linear maps from $V$ to $W$. Indeed, we can identify $\hom(V,W)$ with $V^*\otimes W$: if $\{e_i\}$ is a basis for $V$ and $\{f_j\}$ is a basis for $W$, then we can set up the dual basis $\{\epsilon^i\}$ of $V^*$, such that $\epsilon^i(e_j)=\delta^i_j$. Then the elements $\{\epsilon^i\otimes f_j\}$ form a basis for $V^*\otimes W$, and each one can be identified with the linear map sending $e_i$ to $f_j$ and all the other basis elements of $V$ to $0$. Thus we have an inclusion $V^*\otimes W\to\hom(V,W)$, and a simple dimension-counting argument suffices to show that this is an isomorphism.

Now, since we have an action of $L$ on $V$ we get a dual action on $V^*$. And because we have actions on $V^*$ and $W$ we get one on $V^*\otimes W\cong\hom(V,W)$. What does this look like, explicitly? Well, we can write any such tensor as the sum of tensors of the form $\lambda\otimes w$ for some $\lambda\in V^*$ and $w\in W$. We calculate the action of $x\cdot(\lambda\otimes w)$ on a vector $v\in V$: \displaystyle\begin{aligned}\left[x\cdot(\lambda\otimes w)\right](v)&=\left[(x\cdot\lambda)\otimes w\right](v)+\left[\lambda\otimes(x\cdot w)\right](v)\\&=\left[x\cdot\lambda\right](v)w+\lambda(v)(x\cdot w)\\&=-\lambda(x\cdot v)w+x\cdot(\lambda(v)w)\\&=-\left[\lambda\otimes w\right](x\cdot v)+x\cdot\left[\lambda\otimes x\right](w)\end{aligned}

In general we see that $\left[x\cdot f\right](v)=x\cdot f(v)-f(x\cdot v)$. In particular, the space of linear endomorphisms on $V$ is $\hom(V,V)$, and so it get an $L$-module structure like this.

The other case of interest is the space of bilinear forms on a module $V$. A bilinear form on $V$ is, of course, a linear functional on $V\otimes V$. And thus this space can be identified with $(V\otimes V)^*$. How does $x\in L$ act on a bilinear form $B$? Well, we can calculate: \displaystyle\begin{aligned}\left[x\cdot B\right](v_1,v_2)&=\left[x\cdot B\right](v_1\otimes v_2)\\&=-B\left(x\cdot(v_1\otimes v_2)\right)\\&=-B\left((x\cdot v_1)\otimes v_2\right)-B\left(v_1\otimes(x\cdot v_2)\right)\\&=-B(x\cdot v_1,v_2)-B(v_1,x\cdot v_2)\end{aligned}

In particular, we can consider the case of bilinear forms on $L$ itself, where $L$ acts on itself by $\mathrm{ad}$. Here we read $\displaystyle\left[x\cdot B\right](v_1,v_2)=-B([x,v_1],v_2)-B(v_1,[x,v_2])$

September 21, 2012