Consider a chevalley group a field $K$, with the right chevalley basis. Let $\alpha$ be a root. Let $x_{\alpha}(t)$ be the corresponding root space. Define $w_{\alpha}(t)=x_{\alpha}(t)x_{-\alpha}(-t^{-1})x_{\alpha}(t)$ then define $h_{\alpha}(t)=w_{\alpha}(t)w_{\alpha}(1)^{-1}$. Then chevalley proves in his exposition that the subgroup $H$ generated by the $h_{\alpha}(t)$ where $\alpha$ varies over the roots and $t$ varies over $K$ is isomorphic to the maximal torus.

But I don`t see this in the case of $Sp(4)$. Its Lie algebra is $C_2$. For example let $\alpha =e_1-e_2$ be the fundamental root. Then I take $x_{\alpha}(t)=1+t(E_{12}-E_{43})$. Then I get that $h_{\alpha}(t)$ is not a diagonal matrix. Where did I go wrong?

After Nick's comment my new question is

Does there exists a chevalley basis of $Sp(2n)$ such that the $h_{\alpha}(t)$ are diagonal matrices?

The chevalley basis I am using is in Cartar's book ' Simple groups of Lie type'. The definition of the root space that I am using is in the yale lectures of steinberg which is in math.ucla.edu/~rst/YaleNotes.pdf

My definitions are the following: Let $\Phi$ be the set of roots. Let $\mathfrak{g}=\mathfrak{H}+\sum_{r \in \Phi} \mathfrak{g}_r$. Then we choose $X_r\in \mathfrak{g}_r$ such that $\{h_s, s \in R;X_r, r \in \Phi \}$ is a chevalley basis where $R$ is a basis of the root system and $h_s$ is the fundamental coroot. Then we define $x_r(t)=exp(tX_r)$. This is my definition. In my case $X_{e_1-e_2}=E_{12}-E_{43}, X_{e_2-e_1}=-E_{21}+E_{34}$. So $x_{e_1-e_2}(t)=1+t(E_{12}-E_{43})$ and $x_{e_2-e_1}(t)=1+t(-E_{21}+E_{34})$. With this my calculation shows that $h_{e_1-e_2}(t)$ is not a diagonal matrix. This bring problem for me.

Thanks for help in advance.

conjugate toa set of diagonal matrices. Perhaps you've just chosen a basis for which the $h_\alpha(t)$ are not diagonal? (My guess for how to fix this would be to have $x_\alpha(t)=1+t(E_{12}-E_{34})$ -- i.e. swap the last two vectors in your basis -- but I don't have the wherewithal to do the calculation right now.) $\endgroup$ – Nick Gill Dec 4 '15 at 12:11