Constructing positive definite inner products

59 Views Asked by At

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.

1

There are 1 best solutions below

0
On

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.

  1. Show $f$ is well defined. To do this, indentify an (ordered) orthonormal basis with the matrix where the rows are the vectors of the basis, and show that if $B$ and $B'$ are two orthonormal bases for $p$, then there exists $O\in O(n)$ with $B=OB'$.
  2. Show it is injective, that is if two inner products have some common orthonormal basis then they are equal
  3. Show it is surjective, that is if you have an orbit of basis under $O(n)$ then you can construct an inner-product on $\mathbb{R}^n$ for which everything works

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 $GL(2n,\mathbb{R})/GL(n,\mathbb{C})$.