center of $O_n(\mathbb R)$

Question

How can I show center of $O_n(\mathbb R)=\{I,-I\}$ ?

I know $\det:O_n(\mathbb R) \rightarrow \{1,-1\}$ and so $O_n(\mathbb R)/SO_n(\mathbb R) \cong \{I,-I\}$

Answer 1

Denote $\mathfrak{S}_n$ the permutation group on $n$ elements. For any permutation $\sigma \in \mathfrak{S}_n$, denote $T_{\sigma} = (\delta_{i, \sigma(j)})_{1 \le i,j \le n}$. One checks that for all $\sigma, \tau \in \mathfrak{S}_n$, we have $T_{\sigma} T_{\tau} = T_{\sigma \circ \tau}$, and $T_{\sigma^{-1}} = \, ^t T_{\sigma}$, so $T_{\sigma} \in O_n(\mathbb{R})$.

If $A = (a_{i,j})_{1 \le i,j \le n}$ is an element of the center of $O_n(\mathbb{R})$, it must commute with all the $T_{\sigma}$ for $\sigma \in \mathfrak{S}_n$. So you get that for all $\sigma \in \mathfrak{S}_n$, and $1 \le i,j \le n$,

$$a_{i,j} = a_{\sigma(i), \sigma(j)}$$

If $i,j \in [\!|1,n|\!]$, there exists $\sigma \in \mathfrak{S}_n$ such that $\sigma(i) = j$, so $a_{i,i} = a_{j,j}$. This means all elements on the diagonal are equal. Similarly, if $i \ne j$ and $k \ne l$, there exists $\sigma \in \mathfrak{S}_n$ such that $\sigma(i) = k$ and $\sigma(j) = l$ so $a_{i,j} = a_{k,l}$. This means all elements outside of the diagonal are equal. So $A$ can be written as $A = aI + bH$, where $H = (1)_{1 \le i,j \le n}$ and $a,b \in \mathbb{R}$. But commutation of A with the matrix

$$\left(\begin{matrix}0 & -1 \\ 1 & 0\\ &&1 \\ &&& \ddots \\ &&&&1\end{matrix}\right)$$

Implies $b = 0$ (just look at the upper left $2 \times 2$ block). And because $(a_{i,j})$ is orthogonal, we get $a^2 = 1$, so $A = \pm I$.

Answer 2

Here is the hint: First try to find centre of $M_n(\mathbb R)$; this can be done by checking which matrices commutes with $E_{ij}:=i,j$ entry is 1, rest are zero. Then using orthogonal condition you can find centre of $O_n(\mathbb R)$.

Edit: Above does not answer the question. But here is the thought. Suppose we found centre of $SO_n(\mathbb R)= \{I_n,-I_n\}$. Then given data $O_n/SO_n={I,-I}$ we give us the answer. So let $n=2m$, and $A=R_{\theta_1,\cdots\theta_k}$ for some angles $\theta_i$'s and $A$ commutes with all $SO_n(\mathbb R)$. Now $A$ consists of diagonals $R_\theta:=$ rotation through an angle $\theta$. Note that $R_\theta$ does not commute with $B_2=e_1+(-1)e_2$ unless $sin \theta=0$ and $cos \theta =1 or -1$. Therefore, if $B_n$ is a matrix consisting of matrices $B_2$ on diagonal then $A$ will commute with $B_n$ only if $sin \theta_k=0$ and $cos \theta_k=1 or -1$. Thus matrix $A$ in centre of $SO_n$ has diagonal entries 1 or -1 and zeros elsewhere. If both 1 and -1 occur then we can construct a matrix with $R_\theta\in SO_n$ on diagonal at the position adjacet 1 and -1, otherwise 1's on the diagonal. So $A=I_n$ or $-I_n$

Similarly when $n=2m+1$ then above argument goes through except last step : $-I_{2m+1}$ does not belong to $SO_{2m+1}$, hence centre $SO_{2m+1}=I_{2m+1}$