Given $V$ a finite dimensional normed vector space and $\beta: V\times V \to \mathbb{R}$ a non-degenerate symmetric or anti-symmetric bilinear form, we define the orthogonal group $O(V,\beta)$ by the matrices such that $AA^* = I$. Therefore, since $I$ is a regular value of the map $A\mapsto AA^*$ we can give $O(V,\beta)$ a smooth manifold structure using the implicit function theorem.

However, using the Cayley Transform $$\mathcal{C}(A) = (1-A)(1+A)^{-1}$$ we can define another smooth manifold structure on $O(V,\beta)$. First, note that if $A\in O(V,\beta)$ then $C(A)\in \mathfrak{so}(V,\beta)$ (matrices $B$ such that $B^* = -B$). Let $\mathcal{R}$ be the set of matrices such that $1+A$ is invertible, this set is clearly open. Define $U_A$ to be the translation by $A$ of the set $\mathcal{R}$ in the orthognal group; that is $U_A := \{AB\mid B\in \mathcal{R}\cap O(V,\beta)\}$. Let $\phi_A: U_A \to \mathcal{R}\cap\mathfrak{so}(V,\beta)$ by $B\mapsto \mathcal{C}(A^{-1}B)$ which is a homeomorphism sice it has a clear inverse. It is not difficult to show that $\{(U_a,\phi_a) : A\in O(V,\beta)\}$ is an atlas on $O(V,\beta)$.

My question is: Do these two differentiable structures coincide? Why?

  • 1
    $\begingroup$ There is at most one smooth structure on a topological group making it a Lie group, so yes. $\endgroup$ – Mariano Suárez-Álvarez Oct 13 '17 at 4:42
  • $\begingroup$ There is an excellent discussion here. $\endgroup$ – fredgoodman Oct 25 '17 at 21:39

That should be a general property of submanifolds. Check the definition of a submanifold, and how you get a smooth structure on it, from different charts that make it appear locally as a linear subspace of $\mathbb{R}^n$. You do not need any Lie group theory for this.

  • $\begingroup$ I believe that given that the inclusion is differentiable the structures must coincide, but why is that? $\endgroup$ – user263732 Oct 13 '17 at 5:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy