Understanding an eigenspace of semi-simple elements in Lie algebra

$\mathfrak{gl}(n, \Bbb{C})$ is not semisimple, but let us look at $L = \mathfrak{sl}(n, \Bbb{C})$. Let's say $n=4$ to have a very concrete example.

Choose $H$ to be the diagonal matrices, a general element being

$$h = \pmatrix{h_1 & 0 & 0 & 0\\ 0 & h_2 & 0 & 0\\ 0 & 0 & h_3 & 0\\ 0 & 0 & 0 & h_4}.$$

_{(With the extra condition $\sum_1^4 h_i = 0$ to ensure we are in $\mathfrak{sl}(4,\mathbb{C})$, which will not make any actual difference in what follows; see remark at the very end.)}

Now let us see what happens if we let this operate (via $ad$) on a completely arbitrary element of $L$, say, $$x = \pmatrix{d_1 & e_{12} & e_{13} & e_{14}\\ f_{21} & d_2 & e_{23} & e_{24}\\ f_{31} & f_{32} & d_3 & e_{34}\\ f_{41} & f_{42} & f_{43} & h_4}.$$ We have:

(*)$\qquad ad(h) (x) = [h,x] = \pmatrix{0 & (h_1-h_2)\,e_{12} & (h_1-h_3)\,e_{13} & (h_1-h_4)\,e_{14}\\ (h_2-h_1)\,f_{21} & 0 & (h_2-h_3)\,e_{23} & (h_2-h_4)\,e_{24}\\ (h_3-h_1)\,f_{31} & (h_3-h_2)\,f_{32} & 0 & (h_3-h_4)\,e_{34}\\ (h_4-h_1)\,f_{41} & (h_4-h_2)\,f_{42} & (h_4-h_3)\,f_{43} & 0}.$

I claim that this, in this concrete example, is nothing else than the famous $$(**) \qquad L = H\oplus \bigoplus_{\alpha\in \Phi} L_\alpha$$ where $\Phi \subset Hom(H,\Bbb{C})$ is the root system, and the $L_\alpha$ are the root spaces you ask about. This is a decomposition of $L$ as $H$-representation (where $H$ operates via $ad$); $L_\alpha$ is the subspace on which $H$ operates via the character $\alpha$. (It turns out that such an $L_\alpha$ is always one-dimensional, but that is a non-trivial result in general.)

Let's check how equation (*) "is" equation (**) applied to our concrete example. The roots are linear forms (characters, one-dimensional representations) of $H$. In this case, they are all of the form $$\alpha\left(\pmatrix{h_1 & 0 & 0 & 0\\ 0 & h_2 & 0 & 0\\ 0 & 0 & h_3 & 0\\ 0 & 0 & 0 & h_4}\right) = h_i-h_j$$ for certain $1\le i\neq j\le 4$.

Actually, let's call $\alpha_1$ the one sending $h$ to $h_1-h_2$. You will notice in (*) that there is exactly one matrix entry which gets multiplied by $h_1-h_2$ when the whole matrix is acted on by $h$: The one at position 12. Well, that means the root space to that root $\alpha_1: h\mapsto h_1-h_2$ is $$L_{\alpha_1} = \pmatrix{0 & * & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0}.$$

Similarly, if $\alpha_2$ is the one that sends $h$ to $h_2-h_3$, we have

$$L_{\alpha_2} = \pmatrix{0 & 0 & 0 & 0\\ 0 & 0 & * & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0}$$

and $\alpha_3:h \mapsto h_3-h_4$ is the one that rules over $$L_{\alpha_3} = \pmatrix{0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & *\\ 0 & 0 & 0 & 0}$$

Now I chose these three for a purpose; namely, all the others can be written as combinations of them. For example, what about the $\alpha$ that sends $h$ to $h_1-h_4$, the one that operates on the upper right corner? Well, that is $\alpha_1+\alpha_2+\alpha_3$ (because it sends $h$ to $(h_1-h_2)+(h_2-h_3)+(h_3-h_4) = h_1-h_4$). Or the one that operates on the entry $f_{31}$? That is $-\alpha_1-\alpha_2$, which sends $h$ to $h_3-h_1$.

We can rewrite our whole equation (*) as:

$$ad(h) (x) = \pmatrix{0 & \alpha_1(h)\,e_{12} & (\alpha_1+\alpha_2)(h)\,e_{13} & (\alpha_1+\alpha_2+\alpha_3)(h)\,e_{14}\\ -\alpha_1(h)\,f_{21} & 0 & \alpha_2(h)\,e_{23} & (\alpha_2+\alpha_3)(h)\,e_{24}\\ (-\alpha_1-\alpha_2)(h)\,f_{31} & -\alpha_2(h)\,f_{32} & 0 & \alpha_3(h)\,e_{34}\\ (-\alpha_1-\alpha_2-\alpha_3)(h)\,f_{41} & (-\alpha_2-\alpha_3)(h)\,f_{42} & -\alpha_3(h)\,f_{43} & 0}.$$

_{(This might look a bit awkward and not really simplifying in this example, but it is something that generalises and is of great importance to all root systems: $(\alpha_1, \alpha_2, \alpha_3)$ is a basis or set of simple roots of our $\Phi$.)}

It turns out that in our case, $\Phi$ consists of the twelve roots $$\{\pm \alpha_1, \quad \pm\alpha_2, \quad \pm \alpha_3, \quad \pm(\alpha_1+\alpha_2), \quad \pm(\alpha_2+\alpha_3), \quad \pm(\alpha_1+\alpha_2+\alpha_3\}.$$ And you can explicitly see those root spaces and that they are one-dimensional. Just one more example: $$L_{-\alpha_2-\alpha_3} = \pmatrix{0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & * & 0 & 0}$$ etc.

Their sum is $$\bigoplus_{\alpha\in \Phi} L_\alpha = \pmatrix{0 & * & * & *\\ * & 0 & * & *\\ * & * & 0 & *\\ * & * & * & 0}$$ and this should clarify equation ($**$). Finally, the $H$ in ($**$) is sometimes also written as $L_0$, and that is totally consistent, because in (*) you see how on the diagonal, $H$ actually acts via the constant eigenvalue zero ("$\alpha_0 = 0$"), and $$L_0 = H = \pmatrix{* & 0 & 0 & 0\\ 0 & * & 0 & 0\\ 0 & 0 & * & 0\\ 0 & 0 & 0 & *}$$ is the only space in the decomposition which is not one-dimensional.

_{All this works for $\mathfrak{gl}(4,\mathbb{C})$ as well. The extra condition in $\mathfrak{sl}(4, \Bbb{C})$ that all elements $h$ satisfy $\sum h_i=0$ basically "mods out" the centre of $\mathfrak{gl}(4) =$ the scalar matrices $\pmatrix{h & 0 & 0 & 0\\ 0 & h & 0 & 0\\ 0 & 0 & h & 0\\ 0 & 0 & 0 & h}$ and thus ensures that for every root $\alpha$, there are elements in $H$ on which $\alpha$ is not zero. The fact that the roots span the whole dual space $H^* = Hom(H, \Bbb{C})$ is useful in the abstract theory, so here $H$ should better be $3$-dimensional.}