VizualBusiness: More New Modules from Old

Tuesday, September 25, 2012

More New Modules from Old

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

and

are two finite-dimensional

-modules, then I say we can put an

-module structure on the space $\hom(V,W)$ of linear maps from

. Indeed, we can identify $\hom(V,W)$ with $V^*\otimes W$ : if $\{e_i\}$ is a basis for

and $\{f_j\}$ is a basis for

, 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

and all the other basis elements of

. 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

we get a dual action on V^*

. And because we have actions on V^*

and

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

is $\hom(V,V)$ , and so it get an

-module structure like this.
The other case of interest is the space of bilinear forms on a module

. A bilinear form on

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

? 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

itself, where