1

I am absolutely boggled by the notation $ad_x$ as used to discuss the adjoint representation of a Lie Algebra. A few things I do understand:

I understand what a Lie algebra is in general, including the commutator bracket I feel like I understand that a homomorphism is any mapping of the form $f(a*b) = f(a) f(b)$. I feel like I understand that any representation, in general, can be reduced to an adjoint representation by evaluating the representation at zero.

But what totally blows me away is the notation $ad_x$, which is the same as $ad(X)$ (which is equally confusing). In the wikipedia article on adjoint representations of Lie Algebras, I see statements like:

$$ad_x(y) = [x,y]$$

and people saying that the above equation is a homomorphism. I completely fail to see how the statement above is a homomorphism. So:

is $ad_x$ some kind of operator? Is it a function notation like $f(x)$? Does $ad_x$ itself have some value that can be inserted (i.e. is it a matrix of some kind)?

Berci
  • 90,745
the_photon
  • 209

1 Answers1

4

I think the formula you copied is pretty telling: for any $x$ in your Lie algebra $L$, $ad_x$ is the function $$ad_x: L\to L$$ which sends $y\in L$ to $[x,y]\in L$.

It is clear that $ad_x$ is a linear function, so $ad_x\in \operatorname{End}(L)$. Now when we see $\operatorname{End}(L)$ as a Lie algebra, we usually write it as $\mathfrak{gl}(L)$. So we have a function $ad: L\to \mathfrak{gl}(L)$ which sends an element $x\in L$ to the linear map $ad_x\in \mathfrak{gl}(L)$.

Now the magic part is that $$ad_{[x,y]}=[ad_x,ad_y]$$ which can be rewritten as $$[[x,y],z] = (ad_x\circ ad_y-ad_y\circ ad_x)(z) = [x,[y,z]]-[y,[x,z]]$$ for all $z\in L$. (And this is just the Jacobi axiom.)

This means that $ad: L\to \mathfrak{gl}(L)$ is actually a Lie algebra morphism. I would guess this is the statement you're referring to in your question.

Captain Lama
  • 25,743