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I am reading the book of Erdmann and Wildon. There is an excise in Chapter 10.

Let $L = sl(n, \mathbb{C}), n \ge2$ and let $H = span\{h\}$, where $h = e_{11}-e_{22}$. The book asked me to firstly find $L_0=C_L(H)$, and then determine the direct sum decomposition $$L=L_0 \oplus \bigoplus_{a \in H^*}L_a$$ with respect to $H$.

I am ok to find the centralizer of $H$, which is the diagonal matrices in $sl(n, \mathbb{C})$ (If i did it correctly)

Then I tried to find the eigenvalues of $ad_h$ and the corresponding weight space. However, the eigenvalues of $ad_h$ is getting weird when I calculated them.

It appears to me that $h$ has a matrix form with $a$ in position $e_{11}$ and $-a$ in position $e_{22}$. then

$[h,e_{1j}] = ae_{1j} \quad \forall j\neq1$

$[h,e_{2j}] = -ae_{2j} \ \forall j\neq2$

$[h,e_{i1}] = ae_{i1} \quad \forall i\neq1$

$[h,e_{i2}] = -ae_{i2} \ \forall i\neq2$

$[h,e_{ij}] = 2a \ \text{or} -2a$ if $i \neq j$ and $i,j = 1 \text{or}\ 2$

otherwise $[h,e_{ij}] = 0$

Then I am totally lost in finding the decompositions...Could someone help me please!

Thanks in advance! Any comments or hints would mean a lot to me

1 Answers1

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The list of commutators you computed basically looks fine except for the fact that there is some overlap between the first two lines an the last line, but you got the interpretation of the results wrong. Since $H$ has dimension $1$, you can simply identify $H^*$ with $\mathbb C$ by mapping each functional to its value on the matrix $h=e_{11}-e_{22}$. So you just have to care about eigenvalues of $ad_h$. The main mistake you made is that the centralizer of $H$ (which coincides with the space $L_0$ in the decomposition you are looking for) is much larger than just diagonal matrices. Indeed (as your computation shows) $C(H)$ not only contains all diagonal matrices but also all the elementary matrices $e_{ij}$ with $i,j\geq 3$ (so this looks like $\mathfrak{gl}(n-2,\mathbb C)\oplus\mathbb C$). Then you have verified that the other eigenvalues which occur are $\pm 2$ and $\pm 1$. The $\pm 2$ eigenspaces are $1$-dimensional and spanned by $e_{12}$ respectively $e_{21}$. For the eigenvalues $\pm1$ the eigenspaces have dimension $2n-4$, and are spanned by $e_{13},\dots,e_{1n}$ and $e_{32},\dots,e_{n2}$ respectively by $e_{23},\dots,e_{2n}$ and $e_{31},\dots,e_{n1}$.

If you know about these things, you can spot in this the decomposition of $\mathfrak{sl}(n,\mathbb C)$ as a representation of $\mathfrak{sl}(2,\mathbb C)$, which is embedded as the subalgebra corresponding to the first simple root.

Andreas Cap
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  • Thank you sir! I can understand mostly everything you said in the paragraph (with your $C(H)$ the dimension finally coincide on both sides of the decomposition.) It appears to me that your answer in the first paragraph has already solved the problem? However, could you please explain more to me the first simple root for $\mathfrak{sl}(2, \mathbb{C})$? –  Dec 05 '16 at 14:40
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    There is a general result saying that any root $\alpha$ of a semisimple Lie algebra $\mathfrak g$ gives rise to a subalgebra $\mathfrak{s}\alpha\subset\mathfrak g$ which is isomorphic to $\mathfrak{sl}(2,\mathbb C)$. If you do this for the first simple root in $\mathfrak g=\mathfrak{sl}(n,\mathbb C)$ then you get the subalgebra spanned by $h$, $e{12}$ and $e_{21}$. Via the adjoint action, this acts on $\mathfrak g$, and the exercise you are doing essentially decomposes this representation into irreducible components. – Andreas Cap Dec 05 '16 at 15:30