# Volume of Gr(2,4)

Hello

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!

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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 $U(n)$.

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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 $$Gr(k,n)\subset \mathbb{CP}^N, N=\binom{n}{k}-1.$$

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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