# The Unapologetic Mathematician

## New Modules from Old

There are a few standard techniques we can use to generate new modules for a Lie algebra $L$ from old ones. We’ve seen direct sums already, but here are a few more.

One way is to start with a module $M$ and then consider its dual space $M^*$. I say that this can be made into an $L$-module by setting $\displaystyle\left[x\cdot\lambda\right](m)=-\lambda(x\cdot m)$

for all $x\in L$, $\lambda\in M^*$, and $m\in M$. Bilinearity should be clear, so we just check the defining property of a module. That is, we take two Lie algebra elements $x,y\in L$ and check \displaystyle\begin{aligned}\left[[x,y]\cdot f\right](m)&=-f([x,y]\cdot m)\\&=-f(x\cdot(y\cdot m)-y\cdot(x\cdot m))\\&=-f(x\cdot(y\cdot m))+f(y\cdot(x\cdot m))\\&=\left[x\cdot f\right](y\cdot m)-\left[y\cdot f\right](x\cdot m)\\&=-\left[y\cdot(x\cdot f)\right](m)+\left[x\cdot(y\cdot f)\right](m)\\&=\left[x\cdot(y\cdot f)-y\cdot(x\cdot f)\right](m)\end{aligned}

so $[x,y]\cdot f=x\cdot(y\cdot f)-y\cdot(x\cdot f)$ for all $f\in M^*$, as desired.

Another way is to start with modules $M$ and $N$ and form their tensor product $M\otimes N$. Now we define a module structure on this space by $\displaystyle x\cdot m\otimes n=(x\cdot m)\otimes n + m\otimes(x\cdot n)$

We check the defining property again. Calculate: \displaystyle\begin{aligned}{}[x,y]\cdot m\otimes n&=([x,y]\cdot m)\otimes n+m\otimes([x,y]\cdot n)\\&=(x\cdot(y\cdot m)-y\cdot(x\cdot m))\otimes n+m\otimes(x\cdot(y\cdot n)-y\cdot(x\cdot n))\\&=(x\cdot(y\cdot m))\otimes n-(y\cdot(x\cdot m))\otimes n+m\otimes(x\cdot(y\cdot n))-m\otimes(y\cdot(x\cdot n))\end{aligned}

while \displaystyle\begin{aligned}x\cdot(y\cdot m\otimes n)-y\cdot(x\cdot m\otimes n)=&x\cdot((y\cdot m)\otimes n+m\otimes(y\cdot n))-y\cdot((x\cdot m)\otimes n+m\otimes(x\cdot n))\\=&x\cdot((y\cdot m)\otimes n)+x\cdot(m\otimes(y\cdot n))-y\cdot((x\cdot m)\otimes n)-y\cdot(m\otimes(x\cdot n))\\=&(x\cdot(y\cdot m))\otimes n+(y\cdot m)\otimes(x\cdot n)+(x\cdot m)\otimes(y\cdot n)+m\otimes(x\cdot(y\cdot n))\\&-(y\cdot(x\cdot m))\otimes n-(x\cdot m)\otimes(y\cdot n)-(y\cdot m)\otimes(x\cdot n)-m\otimes(y\cdot(x\cdot n))\\=&(x\cdot(y\cdot m))\otimes n+m\otimes(x\cdot(y\cdot n))-(y\cdot(x\cdot m))\otimes n-m\otimes(y\cdot(x\cdot n))\end{aligned}

These are useful, and they’re only just the beginning.

September 17, 2012 - Comment by Tom Gregory | December 27, 2014 | Reply Comment by John Armstrong | December 28, 2014 | Reply