I am presently studying Rosenberg book: "The Laplacian on a Riemannian Manifold" and I came across the following exercise:
Show that the space of all positive definite inner products on $\mathbb{R}^{n}$ is in one-to-one correspondence with the set $M = GL(n; \mathbb{R}) / O(n)$. Hint: Such an inner product is determined by a basis of $\mathbb{R}^{n}$ which is orthonormal with respect to the inner product. However, two bases which differ by an orthogonal transformation (with respect to this inner product) determine the same inner product.
It also says that because $O(n)$ is a closed subgroup of the Lie group $GL(n; \mathbb{R})$, $M$ is a manifold. I understand how $O(n)$ can be a closed subgroup by construction a sequence converging to a point in $GL(n; \mathbb{R})$ and this point must necessarily be in $O(n)$ but it is not clear how this makes $M$ a manifold.
Thank you for clearing these for me.
To see that $GL(n)/O(n)$ is a manifold, you can show the general statement: if $G$ is a Lie group and $H$ is a Lie subgroup of $G$, then $G/H$ is a manifold. Hint: show the action of $H$ is smooth, free and proper.
To see why this quotient is in bijection with the set of all inner-products: let $I$ be the set of all inner-products of $\mathbb{R}^n$. Let $f : I \to GL(n)/O(n)$ be the map associating to an inner-product $p$ the set of all orthonormal bases.
This is a difficult and abstract point of view, but it is a common thing to visualise the set of all possible structures on a set as a quotient space. For example, the set of all complex structures on $\mathbb{R}^{2n}$ is $GL(2n,\mathbb{R})/GL(n,\mathbb{C})$.