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Prove that $SO(n)$ is parallelizable.

How would I go about showing this? My supervisor could not help me with this problem, and I am stumped.

glS
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user166854
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2 Answers2

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We give an approach that does not just say $``$oh, it's a Lie group so it must be true.$"$ Here are a few standard facts we will assume (they are worthwhile exercises for the reader if one does not know how to show them):

  1. A manifold $M$ of dimension $n$ is parallelizable if and only if the tangent bundle $TM$ is trivial, i.e., diffeomorphic to $M \times \mathbb{R}^n$ with the diffeomorphism $\rho$ being linear in each fiber and satisfying $\pi \circ \rho^{-1}(p, \xi) = p$ for all $p \in M$ and $\xi \in \mathbb{R}^n$. Here $\pi: TM \to M$ is the canonical projection onto the base.
  2. Matrix multiplication and inversion are smooth on $SL(n, \mathbb{R})$.
  3. The tangent space to $SO(n)$ at the identity is precisely the set of $n \times n$ skew-symmetric matrices.

We first show that if $A \in T_\text{Id}(SO(n))$ and $M \in SO(n)$, then $MA \in T_M(SO(n))$. For any tangent vector $V \in T_M(SO(n))$, we know from $(3)$ above that we must have $V^\text{T}M + M^\text{T}V = 0$. Given $A \in T_\text{Id}(SO(n))$ and $M \in SO(n)$, we know $A^\text{T} = -A$ and $M^\text{T}M = \text{Id}$. Note that$$(MA)^\text{T} + M^\text{T}MA = A^\text{T}M^\text{T}M + M^\text{T}MA = A^\text{T} + A = 0,$$so $MA \in T_M(SO(n))$.

We now verify that the map $$\Psi: SO(n) \times T_\text{Id}(SO(n)) \to T[SO(n)],\text{ }(M, A) \mapsto (M, MA)$$is a diffeomorphism. The mapping $\Psi: (M, A) \mapsto (M, MA)$ essentially amounts to matrix multiplication, which is smooth. Also, $\Psi^{-1}: (M, MA) \mapsto (M, A)$ amounts to left multiplying by $M^{-1}$; since inverting a matrix and matrix multiplication are booth smooth, this is also smooth. Thus $\Psi$ is a diffeomorphism.

Since $T_\text{Id}(SO(n))$ is diffeomorphic to $\mathbb{R}^{n(n−1)/2}$ $($why?$)$, it follows that $T[SO(n)]$ and $SO(n) \times \mathbb{R}^{n(n−1)/2}$ are diffeomorphic $($and if we check, this diffeomorphism satisfies the nice properties necessary from $(1)$ above$)$. Thus $SO(n)$ is parallelizable.

user149792
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From Wikipedia:

Any Lie group $G$ is parallelizable, since a basis for the tangent space at the identity element can be moved around by the action of the translation group of $G$ on $G$ (any translation is a diffeomorphism and therefore these translations induce linear isomorphisms between tangent spaces of points in $G$).