The Unapologetic Mathematician

Mathematics for the interested outsider

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])

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September 21, 2012 - Posted by | Algebra, Lie Algebras, Representation Theory

2 Comments »

  1. […] 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

  2. Reblogged this on Mathesis universalis.

    Comment by mathesisuniversalis | April 6, 2013 | Reply


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