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Let $\mathfrak{h} \subset \mathfrak{sl}(3,\mathbb{C})$ be the set of diagonal matrices. Then for $A = \begin{pmatrix} a_1 & 0 & 0 \\ 0 & a_2 & 0 \\ 0 & 0 & a_3 \end{pmatrix} \in \mathfrak{h}$ and $1 \leq i \leq 3$, define $L_i A = a_i$. One can see that the $L_i - L_j$ for $i \neq j$ are the weights with eigenvectors $E_{ij}$ ($1$ in position $(i,j)$, $0$ everywhere else). Then, in Fulton and Harris page 163, they give us this picture

enter image description here

I don't know what is going on here. In the weight space decomposition of $\mathfrak{sl}(3,\mathbb{C})$ we get the spaces $\mathfrak{g}_{ij}$ are the eigenvectors of $L_i - L_j$ and are spanned by $E_{ij}$.... but what does the picture below even mean?

I know this should be obvious but I'm quite confused now.

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The image is an angle preserving depiction of a certain Euclidian space together with some of its elements:

The Euclidean space in question is the real part of the dual ${\mathfrak h}^{\ast}$ of the Cartan subalgebra ${\mathfrak h}$ of ${\mathfrak s}{\mathfrak l}_3({\mathbb C})$ of diagonal matrices, equipped with the restriction of the dual of the Cartan-Killing-Form of ${\mathfrak s}{\mathfrak l}_3({\mathbb C})$. The depicted elements are combinations of the projections $L_i: {\mathfrak h}\to{\mathbb C}$.


General Theory: If ${\mathfrak g}$ is a semisimple complex Lie algebra, then its Cartan-Killing form $\kappa_{\mathfrak g}$ is non-degenerate, and its restriction $\kappa_{\mathfrak g}|_{{\mathfrak h}\times{\mathfrak h}}$ is non-degenerate on ${\mathfrak h}$. In particular, it induces an isomorphism ${\mathfrak h}\cong{\mathfrak h}^{\ast}$ along which it can itself be pulled back to a non-degenerate form on ${\mathfrak h}^{\ast}$. It can be proved that the restriction of this form to the ${\mathbb R}$-span ${\mathfrak h}_{\mathbb R}^{\ast}$ of the roots of $({\mathfrak g},{\mathfrak h})$ is even positive definite, hence providing ${\mathfrak h}_{\mathbb R}^{\ast}$ with the canonical structure of a Euclidean vector space.


In the example, the Killing form on ${\mathfrak s}{\mathfrak l}_3({\mathbb C})$ is (up to scalar) given by $(A,B)\mapsto \text{tr}(AB)$, and hence its restriction to ${\mathfrak h}\subset{\mathbb C}^3$ is (up to scalar) the standard bilinear form with $e_1,e_2,e_3$ as an orthonormal basis. In particular, the $L_i-L_j$ are (up to scalar) the duals of $e_i - e_j$, so we e.g. get the cosine $$\left\langle \frac{L_1-L_2}{\|L_1-L_2\|},\frac{L_2-L_3}{\|L_2-L_3\|}\right\rangle = -\frac{1}{2}$$ of the angle between $L_1-L_2$ and $L_2-L_3$ - the angle is therefore 120° as depicted.

Note also that $L_1 + L_2 + L_3 = 0$, so that e.g. $L_1 = \frac{1}{3}\left((L_1 - L_2) + (L_1 - L_3)\right)$, fitting with the picture.

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  • $\begingroup$ Isn't the killing form $(A,B) \mapsto tr(ad(A) ad(B))$? $\endgroup$ – nigel Apr 19 '15 at 22:26
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    $\begingroup$ Right! But up to a scalar it turns out to agree with the usual trace form. $\endgroup$ – Hanno Apr 19 '15 at 22:49
  • $\begingroup$ @nigelvr: Do you have any more questions? $\endgroup$ – Hanno Apr 21 '15 at 4:40

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