# Understanding Serre-Chevalley relations

I was studying Chevalley-Serre relations, which can be summed up to these

$$\tag{S1}\left[h_{i},\,h_{j}\right]=0$$

$$\tag{S2}\left[e_{i},\,f_{i}\right]=h_{i} \quad \left[e_{i},\,f_{j}\right]=0 \quad\text{for } i\neq j$$

$$\tag{S3} \left[h_{i},\,e_{j}\right]=A_{ij}e_{j} \quad \left[h_{i},\,f_{j}\right]=-A_{ij}f_{j}$$

$$\tag{S4} \text{ad}\left(e_{i}\right)^{1-A_{ij}}\left(e_{j}\right)=0 \quad\;\; \text{ad}\left(f_{i}\right)^{1-A_{ij}}\left(f_{j}\right)=0 \quad\text{for } i\neq j$$

where $$A_{ij}$$ are the coefficients of the Cartan matrix. Now it seems to me that relations (S1),(S2), and (S3) are really quite natural, but I don't fully understand relations in (S4). Does anybody has an insight on what does those relations mean?

The relations prescribe how the Lie algebra is supposed to decompose when considered as a module over the copy ${\mathfrak s}{\mathfrak l}_2(i)$ of ${\mathfrak s}{\mathfrak l}_2({\mathbb k})$ spanned by $\{e_i,f_i,h_i\}$. Namely, if you know that $\text{ad}(e_i)^{a+1}(e_j)=0$ but $\text{ad}(e_i)^{a}(e_j)\neq 0$, then the ${\mathfrak s}{\mathfrak l}_2(i)$-submodule of ${\mathfrak g}$ spanned by $e_j$ has dimension $a+1$ (note that $\text{ad}(f_i)(e_j)=0$, so $e_j$ is a lowest weight vector for the generated ${\mathfrak s}{\mathfrak l}_2(i)$ submodule).

If you look at the A2 root system of ${\mathfrak s}{\mathfrak l}_3({\mathbb C})$ for example, you see that if $\{\alpha,\beta\}$ is a basis of the root system, then the root string $\alpha, \alpha + \beta, ...$ has only length $2$, in accordance with the fact that the Cartan matrix is $\tiny\begin{pmatrix} 2 & -1 \\ -1 & 2\end{pmatrix}$. If, in contrast, you look at the G2 root system, you'll see one chain of length $4$ and one of length $2$, in accordance with the Cartan matrix $\tiny\begin{pmatrix} 2 & -3 \\ -1 & 2\end{pmatrix}$. The last Serre-Chevallley relation reflects these chain lengths (even the $2$'s on the diagonal make sense, because the ${\mathfrak s}{\mathfrak l}_2(i)$ submodule spanned by $e_i$ is just ${\mathfrak s}{\mathfrak l}_2(i)$ itself, so has dimension $3$; the sign is different because $e_i$ is a highest weight vector, though).

• Very interesting answer. What do you mean exactly fo ${\mathfrak s}{\mathfrak l}_2(i)$ – Dac0 Nov 7 '17 at 19:39
• I mean the ${\mathbb C}$-span of $e_i,h_i,f_i$ - it's a copy of ordinary ${\mathfrak s}{\mathfrak l}_2$, but should be labeled in some way to distinguish the various copies coming from the simple roots. – Hanno Nov 7 '17 at 19:40

It might help if we label the generators by the corresponding roots instead (since this is where they come from). When we do this, we get that if $[e_{\alpha_i},e_{\alpha_j}]$ is in the root space corresponding to $\alpha_i + \alpha_j$, so since there are only finitely many roots, at some point, if we keep taking brackets with the same $e_{\alpha_i}$, we need to get $0$.

That was the reason coming from actually looking at the Lie algebra we know we should get. But there is another way to look at this: What would happen if we left these out?

As it happens, if we leave out the relation, we no longer get a finite dimensional Lie algebra, so what the relation really does is make sure the Lie algebra stays small.

• the last relations as finiteness relations was indeed the only thing I came up with. Anyway I upvoted the answer since it had to be said. – Dac0 Nov 7 '17 at 19:37