MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top


I was wondering if anybody can direct me to a paper or a book regarding the volume of $Gr(2,4) $ or generic complex Grassmanian manifolds of order $k$. My own heuristic method seems not to work! It is based on the adaption of the same procedure one has to follow for finding the volume of complex projective spaces $\mathbb{C}P^n$ using Hopf fibration $\mathbb{C}P^n\cong S^{2n+1}/S^1$. Here the volume can be roughly given by dividing the volume of $2n+1-$sphere by volume of $S^1$. Therefore in analogy with this example, we can estimate the volume of $Gr(k,n)$ by dividing the volume of $U(n)$ by that of $U(n-k) \times U(k)$ which gives me $12\pi^4 r^{16}$ for $Gr(2,4)$ where $r$ is the radius of $S^1$ and I don't like it because $Gr(2,4) $ is $8$ dimensional!

Thanks in Advance AB

share|cite|improve this question
up vote 9 down vote accepted

Check section 9.1.2 of these notes There I compute the volumes of real Grassmannians. A similar computation works in the complex case.

Update Using the description $\mathrm{Gr}\;(k, N)\cong U(N/U(k)\times U(N-k)$ and a bi-invariant metric on $U(N)$, this induces bi-invaraint metrics on $U(k),U(N-k)\subset U(n)$ and an invariant metric on $\mathrm{Gr}(k,N)$. The volume of $\mathrm{Gr}(k,N)$ with respect to this metric is

$$ {\rm vol} \mathrm{Gr}(k, N)= \frac{ {\rm vol}\; U(N)}{{\rm vol}\; U(k)\cdot {\rm vol}\; U(N-k)}. $$

The volume of a compact Lie group $G$ with respect to a bi-invariant metric $g$ was computed by I.G. Macdonald,

The volume of a compact Lie group, Invent. Math. 56(1980), no. 2, 93–95.

For the Lie group $U(n)$ this takes the form

$$ {\rm vol}\; U(n)=\frac{1}{(2P_n)^2(2\pi)^n}\times {\rm vol}\; T^n\times \prod_{k=1}^n {\rm vol}\;S^{2k-1}, $$

where ${\rm vol}\; T^n$ denotes the volume of the maximal torus of $U(n)$ equipped with the induced bi-invariant metric, and $P_n$ is the product of the lengths of the positive roots of $U(n)$.

share|cite|improve this answer
Well I found out that I made a mistake in calculating the radius part and the correct result is $12\pi^4 r^8$. The method you follow leads to a formula in proposition 9.1.12 which is very similar to that of mine above for the complex case. But I derived it by following the methodology I explained and I want to make sure fast if it is true! Could you explain if there is a quick way to reach the result for the complex case out of your computation? Or I have no choice but to spend much time to calculate Haar measure and stuff? – Alireza Apr 27 '13 at 17:08
Using the invariance it suffices to compute only the volume of $U(n)$, but you have to do that consistently. Here Weyl integration formula helps. – Liviu Nicolaescu Apr 27 '13 at 17:52

The volume of a Grassmanian can be computed using Wirtinger's theorem:

The volume of a $p$-dimensional complex submanifold $S$ of a complex Hermitian manifold $(X,\omega)$ is

$$ \frac{1}{p\!}\int_S\omega^p. $$

If $X=\mathbb{CP}^N$ the integral is equal to the degree of $S$ times the volume of $X$. Thus up to normalization factors, the volume of the Grassmanian $Gr(k,n)$ is its degree in the Plücker embedding $$Gr(k,n)\subset \mathbb{CP}^N, N=\binom{n}{k}-1.$$

share|cite|improve this answer
Thanks for bringing this into play. Since for $ B_r= \{z \in \mathbb{C}^p,|z|<r}\}$ the formula $Vol(B_r)=\frac{1}{p!}\int_{B_r}\omega^p$ gives $r^{2p}/{p!}$ we should introduce in it the normalization factor ${\pi}^{-p}$ by hand. Does this mean that the same normalization factor can be applied for the volume of any Grassmanian as well? – Alireza Apr 29 '13 at 2:04
The thing that I don't get is that for $k=2$ and $n=4$ this gives $Gr(2,4) \subset X=\mathbb{CP}^5$! Well no problem if your $N=5$ was $N=4$ from the embedding point of view yet your argument that the deg. of $Gr(2,4)$ times $Vol(\mathbb{CP}^5)$ gives the volume of the grassmanian seems not rational because one is $5$ complex dim and the other $4$ dim and the deg. of a submanifold of a Kaehlarian manifold is dimensionless, if I'm not mistaken. – Alireza Apr 29 '13 at 16:29

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.