# Maximal tori of $SO(n)$ and $Sp(n)$

I know that the maximal torus of the group $$GL_n (\mathbb{C})$$ is the set of all diagonal matrices $$\left( \begin{matrix} x_1 & 0 & \cdots & 0 \\ 0 & x_2 & \cdots & 0 \\ 0 & 0 & \ddots & 0 \\ 0 & 0 & \cdots & x_n \end{matrix} \right)$$ But what are the maximal tori of $$SO(n)$$ and $$Sp(n)$$? What about the maximal split tori?

• See wikipedia under examples and references. Aug 14, 2019 at 18:45
• But can elements of these tori be explicitly stated? I'm not sure what the elements of the tori look like. Aug 14, 2019 at 18:49
• Wikipedia is very explicit: "all block-diagonal matrices with $2\times 2$ diagonal blocks, where each diagonal block is a rotation matrix." Do you know, what a rotation matrix is? Aug 14, 2019 at 18:54
• Yes I do. So a matrix of the form $diag(a_1,\ldots,a_n,a_{1}^{-1},\ldots,a_{n}^{-1})$ would not be in either of those tori, yes? Aug 14, 2019 at 18:56

What a maximal torus looks like depends on your choice of bilinear form. For example, in the split group $$G = \operatorname{SO}_{2n}(\mathbb C)$$, we can choose the symmetric, nondegenerate bilinear form $$B: \mathbb C^{2n} \times \mathbb C^{2n} \rightarrow \mathbb C$$ given by

$$B(x_1, ... , x_{2n}, y_1, ... , y_{2n}) = x_1 y_{2n} + x_2y_{2n-1} + \cdots + x_{2n}y_1$$

By definition, $$G$$ is the group of invertible linear operators $$T$$ on $$\mathbb C^{2n}$$ satisfying $$B(Tv,Tw) = B(v,w)$$ for all $$v, w \in \mathbb C^{2n}$$.

In coordinates, if $$J_0$$ is matrix with $$1$$s on the antidiagonal, and zeroes elsewhere, and

$$J = \begin{pmatrix} & J_0 \\ J_0 & \end{pmatrix}$$

then

$$G = \{ g \in \operatorname{GL}_{2n}(\mathbb C) : \space ^tg Jg = J \}$$

Up to isomorphism, this does not depend on the choice of nondegenerate, symmetric bilinear form $$B$$ (equivalently, it does not depend on the choice of invertible symmetric matrix $$J$$). If you prefer another matrix $$J'$$ in place of $$J$$, you will define another special orthogonal group $$G'$$, and there will exist a $$g \in \operatorname{GL}_{2n}(\mathbb C)$$ such that $$gGg^{-1} = G'$$.

To find a maximal torus of $$G$$, we let $$D$$ be the group of diagonal invertible matrices in $$\operatorname{GL}_{2n}(\mathbb C)$$. It is easy to see that

$$T := D \cap G = \{ \begin{pmatrix} t_1 \\ & \ddots \\ & & t_n \\ & & & t_n^{-1} \\ & & & & \ddots \\ & & & & & & t_1^{-1} \end{pmatrix} \}$$

It is clear that $$T$$ is a torus in $$G$$. I claim it is moreover a maximal torus in $$G$$. If not, then the general theory of linear algebraic groups tells you that $$T$$ is contained in a maximal torus $$T_1$$ of $$G$$. Every element of $$T_1$$ must commute with every element of $$T$$. But there are no elements in $$\operatorname{GL}_{2n}(\mathbb C)$$ which commute with every element of $$T$$, except for diagonal matrices. Thus $$T_1$$ must consist of diagonal matrices, which forces $$T_1$$ to equal $$T$$.

• And for $Sp(2n)$, could one take $$B(x_1, \ldots, x_{2n}, y_1, \ldots, y_{2n}) = x_1 y_1 + \cdots + x_{2n} y_{2n}$$ and then obtain $T=diag(t_1, \ldots, t_n, t_{1}^{-1}, \ldots, t_{n}^{-1})$? Aug 17, 2019 at 13:41
• That bilinear form is not alternating, so you can't define $Sp(2n)$ that way. An alternating form needs to satisfy $B(v,v) = 0$.
– D_S
Aug 17, 2019 at 14:51
• Ah! Yes.. thank u! Aug 17, 2019 at 16:58
• @D_S Why must every element of $T_1$ commute with every element of $T$ ? Mar 14 at 15:06
• @rae306 because $T_1$ is a torus (and in particular, an abelian group), containing $T$ as a subgroup.
– D_S
Mar 16 at 22:42