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?

  • $\begingroup$ See wikipedia under examples and references. $\endgroup$ Aug 14, 2019 at 18:45
  • $\begingroup$ But can elements of these tori be explicitly stated? I'm not sure what the elements of the tori look like. $\endgroup$
    – Britt K
    Aug 14, 2019 at 18:49
  • $\begingroup$ 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? $\endgroup$ Aug 14, 2019 at 18:54
  • $\begingroup$ 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? $\endgroup$
    – Britt K
    Aug 14, 2019 at 18:56

1 Answer 1


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

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


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

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 $T$.

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

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