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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.

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  • $\begingroup$ 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

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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.

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