# Maximal torus of Chevalley group $Sp(4)$

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.

• I don't quite follow your calculations, but perhaps the problem is that a split torus in $Sp_4(K)$ need not consist of diagonal matrices, it just needs to be conjugate to a 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.) – Nick Gill Dec 4 '15 at 12:11
• I think that the element you mentioned is not an element of $Sp(4)$. So my new question is : Does there exists a chevalley basis of $Sp(2n)$ such that the $h_{\alpha}(t)$ are diagonal matrices? – MathStudent Dec 4 '15 at 15:29
• I did the computation and did find a diagonal matrix: $h_\alpha (t) ={\rm Diag}(t,t^{-1}, t^{-1},t)$. – Paul Broussous Dec 4 '15 at 16:20
• @ paul. what is your $x_{\alpha}(t)$? It will be helpful if you please elaborate your comment when you have time. – MathStudent Dec 4 '15 at 17:01
• Who is Nill?  – Alex B. Dec 4 '15 at 17:28

The question is not well-formulated (for instance, it's not clear what you mean by "the right chevalley basis", and Chevalley is a proper name). Most important, your third sentence doesn't make sense: "Let $x_\alpha(t)$ be the corresponding root space." The elements here should be in a related matrix group, which gets complicated when there are longer root strings than in special linear types. That's the main problem here, I think, since your definition of $x_\alpha(t)$ for the case of type $C_2$ isn't precise. Here the long simple root $\alpha$ you start with has a root string involving $\alpha + \beta, \alpha +2\beta$. where $\beta$ is the short simple root. So the expression for $x_\alpha(t)$ gets more complicated. (You should specify what sources you are following, since there are some differences in notation and terminology.)
In Chevalley's original 1955 construction of adjoint Chevalley groups, his starting point is certainly a Chevalley basis of the corresponding complex simple Lie algebra (which is not unique but involves delicate sign choices). Once that is in place, the group elements imitate characteristic 0 exponentials of root vectors. But this gets tricky, especially over fields of small characteristic, since the exponentials have summands with $n!$ in the denominator. The Chevalley basis helps to circumvent this problem in arbitrary characteristic.
[Note too that Steinberg in his 1967-68 Yale lectures broadened the framework to allow for arbitrary representations in characteristic 0; here you build the exponentials into the Kostant basis for the universal enveloping algebra. People working in algebraic K-theory subsequently drew more conclusions from Steinberg's formalism in order to study what happens over $\mathbb{Z}$ or other rings of "integers" in local and global fields.]
ADDED: At first I assumed you were working as Chevalley originally did, with the adjoint representation. Here that involves $10 \times 10$ matrices, so direct calculation is lengthy. But apparently your problem arises when the representation used is the 4-dimensional one, in which case you apparently made a computational error. As Paul comments, the $4 \times 4$ matrix $h_\alpha(t)$ is indeed diagonal. Aside from computational glitches, your term "root space" certainly doesn't apply to the matrix $x_\alpha(t)$. But the matrix notation depends on specifying a (faithful) representation of the Lie algebra you begin with. In Steinberg's approach there is a "Chevalley group" attached to each such representation of the given simple Lie algebra.
• @dulal: The choice of a Chevalley basis doesn't change the fact that $h_\alpha(t)$ is diagonal, but your definition of $x_\alpha(t)$ in the special case isn't correct. How are you defining $x_\alpha(t)$ in general? – Jim Humphreys Dec 4 '15 at 22:15
• @humphreys: I added the definition in the question. This is the definition in the steinberg's yale lectures. But i still do not understand why the $h_{\alpha}(t)$ is not diagonal. Thanks again for your help. – MathStudent Dec 5 '15 at 10:35