Universal Chevalley group associated to $D_l$

Question

Consider the simple Lie algebra $D_l$. Consider the universal Chevalley group $G$ over a field $K$ associated to it. Then $G$ is a subgroup of the orthogonal group $O_{2l}(K, f)$ where $f$ is the quadratic form $x_1x_{-1}+\cdots +x_lx_{-l}$. Then it is shown in the book 'Simple groups of Lie type' by Carter p. 185 that $G$ is isomorphic to the commutator subgroup of $O_{2l}(K, f)$.

My question is :

What is the maximal torus of $G$? Can we write it explicitly? What is its dimension?. I think that its true that $G$ is contained in $SO_{2l}(K, f)$, which is the special orthogonal group. I know that the maximal torus of $SO_{2l}(K, f)$ is $Diag(t_1,t_2,\cdots , t_l, t_1^{-1},\cdots , t_l^{-1})$. So the maximal torus of $G$ is a subgroup of the above torus of $SO_{2l}(K, f)$. I want to write the maximal torus of $G$ explicitly like above. Thanks for help and reference. It will be also nice if someone also describes $G$ more clearly to me.

Answer 1

The question shows some confusion but also illustrates the difficulty of working with the Spin groups (the universal groups of types $B_\ell$ and $D_\ell$). It's easier to work with the single Lie algebra of type $D_\ell$, of course, but the associated groups are essential for many purposes including applications to particle physics. Chevalley (followed by Steinberg, Borel-Tits, and others) appreciated the possibility of working over arbitrary fields in the spirit of classical Lie group theory. But the literature and notation can easily cause some confusion.

First of all, type $D_\ell$ (of rank $\ell \geq 4$) leads to an algebraic or topological fundamental group of order 4, which is cyclic for $\ell$ odd (as in type $A_3$) but a Klein 4-group for $\ell$ even. So several simple algebraic groups of type $D_\ell$ tend to arise, or their analogues over smaller fields including finite fields. Each of the algebraic groups is equal to its derived group, and is generated by root groups corresponding to simple roots and their negatives; this much is true for any connected semisimple algebraic group. But over a field which is not algebraically closed, the abstract group structure may get more complicated: this was studied in detail by Dieudonne and others.

The most familiar algebraic group of type $D_\ell$ is $SO(2\ell)$, which in turn leads to the adjoint group $PSO(2 \ell)$ by factoring out a center of order 2. Here you use the first fundamental weight attached to the Lie algebra, to get a faithful irreducible representation in dimension $2 \ell$. However, your assertion (in the third sentence) that the "universal" group lives in the same dimension is false. So this confusion propagates to other fields as well. Instead, the universal covering group (simply connected algebraic group) of type $D_\ell$ is more elusive. To get a faithful representation of it you typically need dimension $2^\ell$, which involves study of an associated Clifford algebra. In turn, there is a "half-spin" group, arising from either of the fundamental weights attached to the two branched endpoints of the Dynkin diagram. Like the special orthogonal group this group has the Spin group as a two-fold universal covering group and projects onto the adjoint group. So even over an algebraically closed field the group theory gets complicated; it's more so over other fields.

Having said all of this, I think your main question has an easy answer: the diagonal subgroup of the universal (or other) group of type $D_\ell$ over an arbitrary field has a "maximal torus" which is a direct product of $\ell$ copies of the multiplicative group of the field. Of course, it doesn't have a "dimension" unless the field is algebraically closed (then the dimension is $\ell$), but if you start with the standard $2^\ell$-dimensional realization of the Spin group, such a torus (up to conjugacy) can be realized in the group of diagonal matrices. It's just not a convenient group to work with in matrix form. Working "downstairs" in $SO(2\ell)$ is usually more practical.