Lie algebra of O(n) – Random Thought

For the orthogonal set $O(n)$ , we can define the Lie algebra

$o(n) = \{X: \exp(tX) \in O(n), \forall t \in \mathbb{R}\}$

Note that $X \in o(n)$ iff $X=-X$ .

First note that $\exp(tX) \in O(n)\Rightarrow I=\exp(tX)^\top \exp(tX) = \exp(t X^\top) \exp(tX)$

And since $\exp(A)\exp(B) = exp(A+B)$ if $AB=BA$ , $\exp(A)\exp(-A) = 0 \Rightarrow exp(A)^{-1} = exp(-A)$ .

Therefore $X = -X\Rightarrow I=\exp(tX)^\top \exp(tX)$ .

Now for the opposite direction, since $I=\exp(t X^\top) \exp(tX)$ , then

$0 = \frac{d}{dt}(\exp(tX^\top)\exp(tX)) = X^\top \exp(t X^\top)\exp(tX)+\exp(tX^\top)X \exp(tX), \forall t$

Let $t=0 \Rightarrow 0=X^\top+X\Rightarrow X^\top=-X$ .

Remark: $O(n)$ is topologically closed. That is, it contains its limit.

Define $F$ as a mapping from any matrix $A$ to $A^\top A$ . Then $O(n) = F^{-1}(I)$ . Since $F$ is continuous, $\forall A_k \in O(n)$ , $I=\lim F(A_k) = F(\lim A_k)$ and thus $\lim A_k \in O(n)$ .

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