# Two equivalent measures on the real Grassmannians?

Denote by $G(n,k)$ the real Grassmannian, the set of $k$-dimensional subspaces of $\mathbb{R}^n$. It is a topological space, even metrizable (see A metric for Grassmannians), and so it is a measurable space with the Borel $\sigma$-algebra.

Here are two ways to furnish a probability measure on $G(n,k)$:

1) Pass the Haar probability measure on $O(n)$ to $G(n,k)$ via its natural action on $G(n,k)$ (see Measure on real Grassmannians),

2) Let $R = \mathbb{R}^{n-k} \times \{0\}^k$ be the subspace consisting of vectors ending in $k$ zeros (in the standard basis), and let $G^*(n,k) = \{V \in G(n,k) \mid V \cap R = \{0\} \}$. Consider the (standard?) coordinate chart whereby a real $(n-k) \times k$ matrix $M$ is associated with the element of $G^*(n,k)$ spanned by the columns of the matrix $\left(\begin{array}{c} M \\ \hline \text{Id}_{k \times k} \end{array}\right)$. Finally, pass your favorite probability measure on $\mathbb{R}^{(n-k)k}$ which is equivalent to the Lebesgue measure (say, the Gaussian measure) to $G^*(n,k)$ via this chart, and extend it to $G(n,k)$ by declaring that $G(n,k) \setminus G^*(n,k)$ has zero measure.

I believe these two measures on $G(n,k)$ are equivalent (in the usual sense that they share the same null sets), but I don't know how to show it. Any ideas would be much appreciated.

## 1 Answer

Yes, they are. The point is that any smooth manifold carries a natural smooth measure class (as transition maps obviously preserve the Lebesgue measure class on coordinate charts), which, in particular, contains the volume measure of any Riemannian metric.

Now, both measures you are talking about belong to the smooth measure class. For the first one it follows, for instance, from the fact that it is proportional to the Riemannian volume of any rotation invariant metric on the Grassmannian. The second measure belongs to the smooth class by its construction.

• Thanks R W; can you provide a reference? – Dan Glasscock Feb 26 '14 at 14:08
• References to what? - I think my explanation is pretty much self-contained – R W Feb 27 '14 at 13:29
• Your explanation is self-contained, but I'd like more details. Do you know of a good source (book, notes, etc...) for information on the smooth measure class of a smooth manifold? – Dan Glasscock Feb 27 '14 at 16:34
• I don't think there is anything specific as the notion is pretty straightforward. In principle this should be just a comment after the definition of a smooth manifold, but since geometers are not usually interested in measures, they rarely make this comment. – R W Feb 28 '14 at 17:13