Let $G$ be simply connected chevalley group over a field $K$. I am following the notations as in 'Lectures on Chevalley group' by Steinberg (Yale lectures). Let $H$ be the subgroup generated by $\{h_{\alpha}(t), \alpha \in \Phi, t \in K\}$. Here $\Phi$ is the set of roots. Then in page $44$ after lemma $28$ the author says that every $h \in H$ can be written uniquely as $h=\prod_{i=1}^{i=l}h_{\alpha_i}(t_i)$ where $t_i \in K^{\times}$ and $\alpha_i$`s are basis elements of the roots.

The problem is that I do not understand its proof. The author states it as a corollary without any proof. Can someone write a proof for me and explain me why is it true?

Following the notations as in lemma $28$ I know that if $H_{\alpha}=\sum n_iH_i$ then $h_{\alpha}(t)=\prod_{i=1}^{i=l}{h_{\alpha_i}(t)}^{n_i}$. After this I do not see any further. How to use the information that $G$ is simply connected? Are these $n_i$`s equal to $1$ when $G$ is simply connected?

Thanks for your help and your effort for writing and explaining the proof.

  • $\begingroup$ Title: edit "spit" to "split". Please don't answer the comment, I'll erase it. $\endgroup$
    – YCor
    Commented Jan 12, 2016 at 12:39
  • $\begingroup$ Do you have a problem with the fact that it can be written in this fashion, or with the fact that it can be written uniquely in this fashion? Or both? $\endgroup$ Commented Jan 12, 2016 at 13:00
  • $\begingroup$ Both of the facts are unclear to me. $\endgroup$ Commented Jan 12, 2016 at 13:29
  • $\begingroup$ Isn't the existence what is stated in Lemma 28(b)? (Notice that each $h_i$ is multiplicative: $h_i(s) h_i(t) = h_i(st)$.) $\endgroup$ Commented Jan 12, 2016 at 13:45

1 Answer 1


By Lemma 28(b), $H$ is an abelian group generated by the $h_i(t)$'s (where $h_i = h_{\alpha_i}$), and since each $h_i$ is multiplicative (by Lemma 28(a)), the existence follows.

To prove uniqueness, it suffices to show that if $\prod_i h_i(t_i)=1$, then each $t_i=1$. By Lemma 28(c) now, we have $\prod_i t_i^{\langle \mu, \alpha_i \rangle} = 1$ for all $\mu \in L_V$, and the assumption that $G$ is simply connected tells you that $L_V = L_1$, i.e. the additive group generated by all the weights of all representations. In particular, this holds for any fundamental weight $\mu = \lambda_j$, for which, by definition, $\langle \lambda_j, \alpha_i \rangle = \delta_{ij}$. Since $\prod_i t_i^{\langle \lambda_j, \alpha_i \rangle} = t_j$, we get $t_j = 1$ for all $j$, so the result follows.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.