I have the following exercises:

(a) Verify directly that the group $SO(n)$ has a natural Lie group structure.

(b) Given this structure, prove that $SO(n) \subset GL(n;\mathbb{R})$ is a Lie subgroup.

(c) $GL(n;\mathbb{R})$ lies as an open subset in $M(n;\mathbb{R})$, the vector space of $n \times n$ matrices with real entries. Consequently, the tangent space $T_{e}GL(n;\mathbb{R})$ can be naturally identified with $T_{e}M(n;\mathbb{R}) \cong M(n;\mathbb{R})$. Using this identification, show that $T_{e}SO(n)$ is a subspace of $M(n;\mathbb{R})$.

So far this is what I've shown: First we consider the orthogonal group, $O(n)=(A \in GL(n;\mathbb{R}) : A^{T}A=I)$ where $(A^{T})_{ij}=A_{ji}$. Since we know that $(AB)^{T}=B^{T}A^{T}$, the set of orthogonal matrices forms a subgroup. We also see that $SO(n) \subset O(n)$ as the subgroup of matrices with determinant 1. It remains to be shown that $O(n), SO(n)$ have a manifold structure. We introduce $\mathrm{Sym}(n;\mathbb{R})$ to denote the vector spave of $n \times n$ symmetric matrices. This vector space has dimension $n(n+1)/2$ and is a linear subspace of $M(n;\mathbb{R})$. Thus, we may consider it as a local piece of $\mathbb{R}^{n(n+1)/2}$. We write $\phi: M(n;\mathbb{R}) \rightarrow \mathrm{Sym}(n;\mathbb{R})$ such that $A \rightarrow A^{T}A$. This map is quadratic in the entries of the matrix, so it is a smooth map between Euclidean spaces. Finally, we note that the identity matrix $I \in \mathrm{Sym}(n;\mathbb{R})$ is a regular value of $\phi$. Here we invoke the Implicit Function Theorem to see that $\phi^{-1}(I)$ is a submanifold of $GL(n;\mathbb{R})$. Thus, both $O(n), SO(n)$ have group and manifold structures. Hence, $SO(n)$ has a natural Lie group structure.

I guess I'm a little unsure of the distinction between (a) and (b). What must I further show? Have I given the natural structure or was this an overly complicated method? Finally, any help with (c) would be greatly appreciated.

  • $\begingroup$ You have exhibited $SO(n)$ as a closed subgroup of $GL_n(\mathbb{R})$. Any closed subgroup of a Lie group is a Lie subgroup, so $(b)$ is complete. I think you want $T_eSO(n)$ to be a vector subspace of $M_n(\mathbb{R})$ and not a subgroup. $\endgroup$ Feb 3 '14 at 4:50
  • $\begingroup$ Thanks! I made the change. $\endgroup$
    – user 3462
    Feb 4 '14 at 3:16
  • $\begingroup$ How might I proceed with (c)? $\endgroup$
    – user 3462
    Feb 4 '14 at 3:20
  • $\begingroup$ Having established that $SO(n)$ is a submanifold of $M_n(\mathbb{R})$, it follows that $T_{e}SO(n)$ is a vector subspace of $T_{e}M_n(\mathbb{R})$. However, the manifold structure on $M_n(\mathbb{R})$ comes from its vector space structure, meaning that $T_eM_n(\mathbb{R})$ is naturally isomorphic to $M_n(\mathbb{R})$. $\endgroup$ Feb 4 '14 at 14:50
  • $\begingroup$ Thanks, I worked it through. Do I need to say anything especially for (a)? It seems quite trivial to me... $\endgroup$
    – user 3462
    Feb 4 '14 at 21:36

For (c) let $X(t)$ be a curve in $O(n)$ such that $X(0)=e$. Then we have $X(t)X^T(t)=I$ differentiation gives $X'(0)+X'^T(0)=0$. And vice versa if you have a skew-symmetric matrix $A$ then $e^{tA}$ is a curve with tangent vector $A$.


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