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> [!proposition] Proposition. ([[the orthogonal group is a Lie group]])
> The [[general orthogonal group|orthogonal group]] $\text{O}(n, \mathbb{R})=\{ A \in \text{GL}_{n}(n, \mathbb{R}) : A A ^{\top}= I \}$ has a [[smooth structure]] making it into a [[Lie group]] of dimension $n (n-1) / 2$. The [[Lie algebra of a Lie group|associated]] [[Lie algebra]] is the [[orthogonal Lie algebra]] $\mathfrak{o}_{n}$.
^proposition
Summary:
![[Pasted image 20250510231118.png|500]]
1. Recall that [[skew-symmetry exponentiates to orthogonality and vice versa, under suitable conditions]]
2. Namely, put $V_{0}:=\text{skew}(n) \cap \mathbb{B}_{\log{2}}(0) \subset \mathbb{R}^{n \times n}$. Restricting to this neighborhood of the origin means $\exp |_{V_{0}}:V_{0} \to \text{O}(n)$ is a diffeomorphism onto its image $U$. Thus, $\log:U \xrightarrow{\varphi} V_{0}$ is our first chart
3. We'll have one other chart per $g \in \text{O}(n)$. Specifically, take $gU$ to be the prospective coordinate neighborhood about $g$; then composing the maps $g ^{-1}(\cdot):gU \to U$ and $\varphi=\log:U \to V_{0}$ works
4. To gets that the lie algebra is $\text{skew}(n)$, differentiate a smooth path of $A(t)A(t)^{\top}=I$ with $A(0)=I$: $\dot{A}(0)I^{\top} + I \dot{A}(0)^{\top}=0 \implies X^{\top} + X=0, X:=\dot{A}(0).$
> [!proof]+ Proof. ([[the orthogonal group is a Lie group]])
> The [[Hausdorff space|Hausdorff]] and [[second-countable space|second-countable]] conditions are inherited since [[Hausdorff subspace is Hausdorff]] (same for second-countable) and $\text{O}(n, \mathbb{R}) \subset \mathbb{R}^{n^{2}}$. So we just need [[coordinate chart|charts]] that cover $\text{O}(n, \mathbb{R})$ and are [[transition map|smoothly compatible]]. Recall that [[skew-symmetry exponentiates to orthogonality and vice versa, under suitable conditions]] (the conditions are satisfied).
>
Consider the open neighborhood about zero $V_{0}=\{ B \in \mathbb{R}^{n \times n}: B^{\top}=-B \text{ and } |B|< \log 2\}$ of $\text{skew}(n, \mathbb{R}) \subset \mathbb{R}^{n(n-1)/2}$ with the [[subspace topology|subspace topology]] inherited from $\mathbb{R}^{n^{2}}$. Define $U := \{ \exp(V) : V \in V_{0} \} \subset \text{O}(n),$
We know that due to [[bijection|bijectivity]] of $\exp$ and $\log$ under our domain constraints and that they are [[continuous]], the map $\begin{align}
h: U & \to V_{0} \\
A & \mapsto \log A
\end{align} $
defines a [[homeomorphism]] $U \to V_{0}$. The let $U_{C}$ be the translates $U_{C}:=\{ CA : A \in U \},$
which are open (since [[open sets translate to open sets in topological groups]]).
Then define our charts as $\begin{align}
h_{C} : U_{C} &\to V_{0} \\
A & \mapsto \log (C ^{-1} A)
\end{align}$
('back-translate'). Clearly these cover $\text{O}(n)$ (e.g., since $C \in U_{C}$ for all $C \in \text{O}(n)$). Compatibility is verified as $h_{C_{2}} \circ h_{C_{1}}^{-1}(B)=h_{C_{2}}(C_{1}e^{B})=\log\big( C_{2}^{-1}C_{1}B \big)$
which is [[continuously differentiable|smooth]] as a composition of smooth maps. So we have a smooth structure on $\text{O}(n)$ *as a set*, inducing a topology on it as a [[smooth manifold]]. We already know $\text{O}(n)$ is a [[group]]; so the last thing to check to make it into a Lie group is that the group law and inversion are smooth. [[Lie group|It suffices]] to show that the function $F:\text{O}(n) \times \text{O}(n) \to \text{O}(n)$ defined by $F(A_{1},A_{2}):=A_{1}A_{2}^{-1};$
is a [[smooth maps between manifolds|smooth map]] between the product manifold $\text{O}(n) \times \text{O}(n)$ and $\text{O}(n)$ itself. To do this, we must pull back to local coordinates by composing with charts, and show that the map $F_{\text{loc}}=h_{A_{1}A_{2}^{-1}} \circ F \circ (h_{A_{1}} \times h_{A_{2}})^{-1}$ is smooth. We have $\begin{align}
F_{\text{loc}}(B_{1},B_{2}) & = h_{A_{1}A_{2}^{-1} } \big( F (h_{A_{1}}^{-1}(B_{1}), h_{A_{2}}^{-1}(B_{2})) \big) \\
&= \log [(C_{1}C_{2})^{-1} C_{1} e ^{B_{1}} (C_{2} e ^{B_{2}})^{-1}] \\
&= \log(C_{2} e^{B_{1}}e^{-B_{2}}C_{2}^{-1})
\end{align}$
and this is smooth as a composition of smooth maps.
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#### References
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