Clarification about adjoint representation and Lie Algebras

Question

Let $L = sl(2, F)$, and take the standard basis

$$x = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} \qquad y = \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix} \qquad h = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}$$

The multiplication table is $[xy] = h$, $[hx] = 2x$, $[hy] = -2y$.

So far so good, I understood why those are a basis and the multiplication table is correct. Not it says "notice that $x, y, h$ are eigenvectors for ad$(h)$, corresponding to eigenvalues $2, -2, 0$$...

And here I got stuck, because

• How can a matrix be an eigenvector? (got it)

• It says nothing on what ad$(h)$ is (only "adjoint representation") and I have no clue in how to "compute" it, if it's a thing one can do (?) (solved this part on my own)

Thank you for the help!

EDITS

I still don't get this part:

"Consider a generic element $ax + by + ch$; appling ad$(x)$ twice we get $-2bx$."

I tried to do the calculation, like using ad$(x) = 2x$ and applying it but I did not get that result.

Answer 1

Consider the structure equations of $sl(2, F)$:

$$[xy] = h \qquad [hx] = 2x \qquad [hy] = -2y$$

From here, let's say we search for ad$(x)$, then we have:

$$\text{ad}(x, x) = [xx] = 0$$ $$\text{ad}(x, y) = [xy] = h$$ $$\text{ad}(x, h) = [xh] = -2x$$

Which means

$$\text{ad}(x) = \begin{pmatrix} 0 & 0 & -2 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{pmatrix}$$

In a very similar way we find

$$\text{ad}(y) = \begin{pmatrix} 0 & 0 & 0 \\ -1 & 0 & 0 \\ 0 & 2 & 0 \end{pmatrix}$$

and so on.

Answer 2

I still don't get this part:

"Consider a generic element $ax + by + ch$; appling ad$(x)$ twice we get $-2bx$."

I tried to do the calculation, like using ad$(x) = 2x$ and applying it but I did not get that result.

Because $ad(x) =2x$ is not true, or rather, makes no sense. I do admit the notation is unfortunate, and I forgive you your rude "shut up" because I do remember it confused me too at the beginning.

The map you want to apply is $ad(x)$. You still have to apply that map, to $x$, to $y$ and to $h$, respectively. I will highlight it in red and put brackets around it so you see it. We have (check this!)

$\color{red}{(ad(x))} (x) = 0$ and $\color{red}{(ad(x))} (y) = h$ and $\color{red}{(ad(x))} (h) = -2x$.

Now let's just call that red map $f$. If you have any linear map $\color{red}{f}$ such that

$\color{red}{f} (x) = 0$ and $\color{red}{f} (y) = h$ and $\color{red}{f} (h) = -2x$,

then what is

$\color{red}{f}(ax+by+ch) = ?$

And then, what do you get if you apply that map $\color{red}{f}$ to the result again ? I.e. what is

$\color{red}{f} ( \color{red}{f}(ax+by+ch)) = ??$

If you do not get $-2bx$ for ??, show me what you got for ?.