# Haar measure on the Grassmannian space

The grassmannian space $G(n,m)$ may be identified with the quotient space $O(n)/(O(m)\times O(n-m)$. As such, it is endowed with a natural invariant probability measure which I call "Haar measure on $G(n,m)$". Let $\gamma_{n,m}$ be this measure. Let us endow $G(n,m)$ with the usual metric $d(V,W)=\|P_V-P_W\|$, where $P_V$ is the orthogonal projection onto $V$.

Question. Is it true that there is some constant $C$ such that $\frac{1}{C} r^{m(n-m)} \leq \gamma_{n,m}(B(V,r)) \leq C\times r^{m(n-m)}$ for any $V \in G(n,m)$ and any $r>0$ small enough?

• Am I missing something, or doesn't this follow from the uniqueness (up to scalar multiples) of invariant measures? The metric is $O(n,\mathbb R)$ (I'm assuming real orthogonal groups, or else you'd not get a finite total measure) invariant, so its associated measure would be, too. Aug 6, 2016 at 22:09
• I suppose the norm appearing in the question represents the operator norm on $\text{End}(\mathbb{R}^n)$. You should replace it with the Hilbert-Schmidt norm and pay a constant, as all norms are equivalent. Then you view the Grassmanian as a submanifold of a Riemannian manifold. Argue that the restriction of the metric is comparable with a Riemmanian metric on the submanifold. Aug 7, 2016 at 4:59

Yes, this is true. The answer by Ben Webster gives the idea, but for completion let me give you a list of easy facts that combined together answer the question.

1. The inequalities you seek are correct for any Riemannian manifold with bounds on the curvature (upper/lower bounds for corresponding sides of the inequalities), once you take the Riemannian volume and Riemannian metric. In particular, this is the case for a compact Riemannian manifolds.

2. Our Grassmanian could be seen as a Riemannian manifold, by embedding it in $\text{End}(\mathbb{R}^n)$ as you do, and restrict the standard inner product to the tangent spaces. The obtained Riemannian structure will be $G$-invariant (for $G=\text{O}(n)$ of course).

3. For our manifold, up to normalization, the Riemannian volume coincides with the Haar measure, as both are $G$-invariant, by "uniqueness".

4. For a Riemannain manifold $M$ and a compact submanifold $N$ the two metrics on $N$, the restriction to $N$ of the Riemannian metric of $M$ and the Riemannian metric of $N$ associated with the induced Riemannian structure, are biLipischits (this is an easy excercise).

5. All norms on a Euclidean space are equivalent.

Of course, we apply 5 for whatever norm you had in mind in the formulation of the question and the standard inner product on $\text{End}(\mathbb{R}^n)$ and we apply 4 with $M=\text{End}(\mathbb{R}^n)$ and $N$ being the image of the Grassmannian.

• What's the difference between "the restriction to $N$ of the Riemannian metric of $M$" and "the induced Riemannian structure"? Oct 8, 2019 at 11:25
• @Călin, For $M=\mathbb{C}$ and $N=S^1$, the distance between $1$ and $-1$ is $2$ in the former and $\pi$ in the latter. Oct 8, 2019 at 18:50

The Haar measure is the volume induced by the unique (up to scalar) $O(n)$ invariant Riemannian metric on $G(n,m)$, and any volume induced by a Riemannian metric has this property (since the metric looks flatter and flatter as you zoom in).

EDIT: Uri's comment below points out that a Riemannian metric has this property for balls in that metric and I haven't addressed whether the metric mentioned by the OP is Riemmannian. Of course, the question isn't well-defined, since the OP didn't state what norm s/he is using on complex matrices. If one uses $L^2$ norm, then presumably that's the invariant Riemmanian metric. (EDIT: From Uri's comment above, I see that it's not. On all matrices, the $L^2$ norm gives the usual flat metric, and geodesics in that metric (i.e. straight lines) don't stay in the Grassmannian.) If one uses any other norm, then presumably not. However, these norms are all comparable (since any two norms on a finite dimensional $\mathbb{C}$ vector space are comparable), so if this property holds for one, it holds for all.

• Indeed. And, I think, with the explicit description of the metric, the associated measure is inevitably invariant, too, ... so there's more than one reason that the conclusion is true? Aug 6, 2016 at 22:57
• I think you should explain why you think the metric in the OP is Riemannian. Being invariant is just not enough. Aug 7, 2016 at 4:40
• @Ben, regarding your edit: I don't think the OP metric is Rimannian even if the norm is $L^2$. However, in this case it is not hard to see that it is bilipschitz with the unique invariant Riemannian metric. Aug 9, 2016 at 16:14