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.



Maximal split torus of universal chevalley group
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.


