The Unapologetic Mathematician

Mathematics for the interested outsider

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 Posted by | Algebra, Lie Algebras, Representation Theory | 19 Comments

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 Posted by | Algebra, Lie Algebras, Representation Theory | 2 Comments