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])

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

3 Comments »

  1. Hi John, don’t know whether it’s worth posting here but your blog has been mentioned in the list of mathematics blogs here-> http://www.talkora.com/science/List-of-mathematics-blogs_112 (look for entry #5 in the list)

    Comment by Will | January 3, 2013 | Reply

  2. Will you be posting anymore for the rest of time?

    Cheers,
    NS

    Comment by notedscholar | April 13, 2013 | Reply

  3. Reblogged this on Observer.

    Comment by Kamran | July 14, 2013 | Reply


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