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 John Armstrong | Algebra, Lie Algebras, Representation Theory

2 Comments »

[…] an illustration of how interesting these are, consider the modules we looked at last time. What are the invariant linear maps from one module to another ? We consider the action of on a […]

Pingback by The Submodule of Invariants « The Unapologetic Mathematician | September 21, 2012 | Reply
Reblogged this on Mathesis universalis.

Comment by mathesisuniversalis | April 6, 2013 | Reply

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