Some properties associated to of $\mathbf{SO}(n,\mathbb C)$ and $\mathbf{Sp}(2n,\mathbb C)$

Question

Denote $\mathbf{SO}(n,\mathbb C):=\{g\in \mathbf{GL}(n,\mathbb C):\det g=1, gg^t=I\}.$ Is $\mathbf{SO}(n,\mathbb C)$ connected, path connected or simply connected? I am reading Kanpp's book where he has mentioned this Lie group. But the above question has not been addressed. In the same vain let me ask the same question for $\mathbf{Sp}(2n,\mathbb C):=\{g\in \mathbf{SL}(2n,\mathbb C):g^tJg=J\}$ where $J(x\oplus y)=y\oplus-x.$ One more question how to show that center of $\mathbf{SO}(n,\mathbb C)$ and $\mathbf{Sp}(2n,\mathbb C)$ have finite center. I could to for $\mathbf{SL}(n,\mathbb C).$

Answer 1

I am a beginner in the area of Lie groups , so please point out if I have made any mistakes. I find a way to prove this problem using the well known "Iwasawa Decomposition", and the solution is almost in the sense of "linear algebra", without any advanced techniques , but the idea behind it is not easy. I only show the case when $N=2m$ is an even integer.

Let's review the Iwasawa Decomposition for $\mathbf{SL}(2m,\mathbb{C})$ .

Let $\mathbf{K}=\mathbf{SU}(2m,\mathbb{C})$ to be the subgroup of $\mathbf{SL}(2m,\mathbb{C}) $ consisting of special complex unitary group.

Let $\mathbf{A}$ to be the subgroup of $\mathbf{SL}(2m,\mathbb{C}) $ consisting of diagonal matrices with positive real diagonal elements.

Let $\mathbf{N}$ to be subgroup of $\mathbf{SL}(2m,\mathbb{C})$ consisting of complex upper triangular unipotent matrices, that is having components 1 on the diagonal, arbitrary above the diagonal, and 0 below the diagonal.

Then the Iwasawa Decomposition tells us the following product map is a bijection: $$ \mathbf{K}\times \mathbf{A} \times \mathbf{N}\rightarrow \mathbf{SL}(2m,\mathbb{C})$$ $$(k,a,n)\longmapsto kan.$$ But, the following map is not a bijection $$ (\mathbf{K}\cap\mathbf{SO}(N,\mathbb{C}))\times (\mathbf{A}\cap\mathbf{SO}(N,\mathbb{C})) \times (\mathbf{N}\cap\mathbf{SO}(N,\mathbb{C}))\rightarrow \mathbf{SO}(N,\mathbb{C})$$

Fortunately, there exists a complex unitary matrix $S\in \mathbf{U}(2m)$ , such that the conjugation $\mathbf{G}=S^{-1}\mathbf{SO}(N,\mathbb{C})S$ of $\mathbf{SO}(N,\mathbb{C})$ satisfies: $$(\mathbf{K}\cap\mathbf{G})\times (\mathbf{A}\cap\mathbf{G}) \times (\mathbf{N}\cap\mathbf{G})\rightarrow \mathbf{G}$$ is bijective. Then the path-connectedness of $\mathbf{G}$, hence of $\mathbf{SO}(N, \mathbb{C})$, can be shown if we have proved that $\mathbf{K}\cap\mathbf{G}$, $\mathbf{A}\cap\mathbf{G}$, $\mathbf{N}\cap\mathbf{G}$ are path-connected .

So we need to construct such a matrix $S\in \mathbf{U}(2m)$.Let $E_{r,s}=(a_{p,q})$ to be the $N\times N$ matrix, with $a_{p,q}=1$, if $p=r$ and $q=s$,and else $a_{p,q}=0$. Let $Z_{r,s}=\frac{\sqrt{2}}{2}(iE_{r,s}+E_{r+1,s})$. Then we define $$S=\sum_{k=1}^{m}(Z_{2k-1,k}+\overline{Z_{2k-1,2m+1-k}}).$$ It can be easily verified that:

$\overline{S^{T}}S=I_N$

$S^T S=\mathcal{J}_N$, where $\mathcal{J}=\mathcal{J}_N=(b_{p,q})$, $b_{p,q}=1$ if $p+q=N+1$, and else $b_{p,q}=0$.

For $\gamma \in \mathbf{K}$, $\mathcal{J}\gamma^T \mathcal{J}$ also lies in $\mathbf{K}$.

For $\gamma \in \mathbf{A}$, $\mathcal{J}\gamma^T \mathcal{J}$ also lies in $\mathbf{A}$.

For $\gamma \in \mathbf{N}$, $\mathcal{J}\gamma^T \mathcal{J}$ also lies in $\mathbf{N}$.

So we can solve this problem in four steps:

Step.1: Show that $(\mathbf{K}\cap\mathbf{G})\times (\mathbf{A}\cap\mathbf{G}) \times (\mathbf{N}\cap\mathbf{G})\rightarrow \mathbf{G}$ is bijective.

Step.2: Show that $\mathbf{K}\cap\mathbf{G}$ is path-connected.

Step.3: Show that $\mathbf{A}\cap\mathbf{G}$ is path-connected.

Step.4: Show that $\mathbf{N}\cap\mathbf{G}$ is path-connected.

It is clear that $\mathbf{G}=S^{-1}\mathbf{SO}(N,\mathbb{C})S=\{g\in \mathbf{SL}(2m,\mathbb{C}): g^T\mathcal{J}g=\mathcal{J}\}$. Let $g=kan$ to be the Iwasawa Decomposition for $g\in \mathbf{G}$. So we must show $k^T\mathcal{J}k=\mathcal{J}$, $a^T\mathcal{J}a=\mathcal{J}$, $n^T\mathcal{J}n=\mathcal{J}$, from $n^T a^T k^T \mathcal{J} kan=\mathcal{J}$.Using $\mathcal{J}^2=I_N$, we have $$ (\mathcal{J}n^T\mathcal{J})(\mathcal{J}a^T\mathcal{J})(\mathcal{J}k^T\mathcal{J})kan= I_N$$ Let $n_1=\mathcal{J}n^T\mathcal{J}$, $a_1=\mathcal{J}a^T\mathcal{J}$, $k_1=\mathcal{J}k^T\mathcal{J}$. Then the equation becomes $n_1a_1k_1kan=I_N$, or $$(k_1k)(aa_1)(a_1^{-1}nn_1a_1)=I_N.$$ We know $(k_1,a_1,n_1)\in\mathbf{K}\times \mathbf{A} \times \mathbf{N}$ ,and $\mathbf{A}$ normalizes $\mathbf{N}$, then we see$$(k_1k,aa_1,a_1^{-1}nn_1a_1)\in\mathbf{K}\times \mathbf{A} \times \mathbf{N}$$ By the uniqueness part of Iwasawa Decomposition, we see $(k_1k,aa_1,a_1^{-1}nn_1a_1)=(I_N,I_N,I_N)$, i.e. $k^T\mathcal{J}k=\mathcal{J}$, $a^T\mathcal{J}a=\mathcal{J}$, $n^T\mathcal{J}n=\mathcal{J}$. So we have finished Step.1.

It is easy to show that $\mathbf{K}\cap\mathbf{SO}(N,\mathbb{C})=\mathbf{SO}(N,\mathbb{R})$, so $$\mathbf{K}\cap (S^{-1}\mathbf{SO}(N,\mathbb{C})S)=(S^{-1}\mathbf{K}S)\cap (S^{-1}\mathbf{SO}(N,\mathbb{C})S)=S^{-1}\mathbf{SO}(N,\mathbb{R})S\simeq\mathbf{SO}(N,\mathbb{R}),$$ which is path-connected.

It is easy to show that$$\mathbf{A}\cap\mathbf{G}=\{ a=\mathbf{diag}(a_k):a_k\in \mathbb{R}^+ , a_{k}a_{2m+1-k}=1, \forall 1\leq k \leq 2m \},$$ which is path-connected.

Let $\gamma =(c_{p,q}) \in \mathbf{N}\cap \mathbf{G}$, then we have $\forall 1\leq p < q\leq 2m$,$$c_{p,q}+c_{2m+1-q,2m+1-p}+\sum_{k=p+1}^{q-1}c_{p,k}c_{2m+1-q,2m+1-k}=0.$$ So we see if we have defined the elements $c_{p,q}$ with $q-p\leq d$, then it is free to define $c_{p,q}$ with $q-p= d+1,p+q\leq 2m$, and hence we define all $c_{p,q}$ with $q-p= d+1$. So we see $\gamma$ is uniquely determined by $c_{p,q}$ ,with $q-p\geq 1, q+p\leq 2m$, hence we see $\mathbf{N}\cap \mathbf{G}$ is path-connected.

I suggest you to check the matrix $S$ for $\mathbf{SO}(2m+1,\mathbb{C})$ and $\mathbf{Sp}(2m,\mathbb{C})$，and establish the similar Iwasawa Decomposition for such groups.And your final question will be clear by using such decomposition.

For example, we want to show $\mathbf{SO}(2m,\mathbb{C})$ have finite center. But $\mathbf{SO}(2,\mathbb{C})$ is in fact an abelian group. So we prove the claim when $m\geq 2$. A matrix $\gamma $ is in the center of $\mathbf{SO}(2m,\mathbb{C})$, $g=S^{-1}\gamma S $ is in the center of $\mathbf{G}$. Then You can check that the centralizer of $\mathbf{A}\cap\mathbf{G}$ in $\mathbf{G}$ is equal to the subgroup of diagonal matrices in $\mathbf{G}$, hence in particular, $g$ is a diagonal matrix, say $g=\mathbf{diag}[\lambda_1,...\lambda_m,\lambda_m^{-1},...,\lambda_1^{-1} ]$, with $\lambda_p\in \mathbb{C}^*$(We may not necessarily have $\lambda_p\in \mathbb{R}^+$). Then direct computation shows that such $\gamma=SgS^{-1}$ is also a blocked diagonal matrix, say, $\gamma=\mathbf{diag}[A(\lambda_1),...,A(\lambda_m)]$; where $A(\lambda_p)$ is ($\frac{\lambda_p+\lambda_p^{-1}}{2}$, $\frac{\lambda_pi-\lambda_p^{-1}i}{2}$, $\frac{\lambda_p^{-1}i-\lambda_pi}{2}$, $\frac{\lambda_p+\lambda_p^{-1}}{2}$ ). Take $\sigma \in \mathbf{SO}(2m,\mathbb{R}) $ to be the translation matrix corresponds to the permutation$(12)(34)$. (This needs $m\geq2$！) Check $\sigma\gamma=\gamma\sigma$ gives that $\lambda_1^2=1$, we have $\lambda_1=\pm 1$. Similarly we have $\lambda_p=\pm 1, \forall p$. Then similarly we see the center of $ \mathbf{SO}(2m,\mathbb{C}) $ is $\{\pm I_N\}$, when $m \geq 2$.