On the Grassmannian Gr(2,5) and spheres

Question

${\bf Question:}$

We can write $U(N)\approx \prod_{k=1}^ N S^{2N-1} $ where "$\approx$" means "locally equal to", $S^m$ is $m-$ sphere and $U(N)$ is the unitary group of dimension $N^2$. I want to check whether we can use this to show that

$$Gr(2,5)\approx S^7\times S^5$$

where $Gr(2,5)$ being a complex Grassmannian manifold. However I am not even sure of this kind of local realization of the Grassmannian. We have a problem at hand which from a physics point of view implies that this should hold. Then we can proceed to use 3rd Hopf map to prove that there is more we can get. But mathematically we are not able to prove the above realization. Can we really do that?

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${\bf Motivation:}$

Well, thanks for the arguments. My initial motivation was to show that $\mathbf{Gr}(2,\mathbb{C}^5)$ could be expressed ${\it locally}$ as the product of two spheres appropriately. This is important in physics from many aspects in particular for people working on Quantum Hall Effects (a.k.a Landau problem) in higher dimensional spaces which can be associated with a non-trivial second homotopy group, cf. [1,2,3]. The importance can be most simply understood in the following way. The space $S^2$ has the coset realization $$S^2\equiv CP^2=SU(2)/U(1).$$ This is essentially another representation of the first Hopf fibration $$S^1\hookrightarrow S^3 \rightarrow S^2.$$ The quantum Hall effect on $S^2$ basically corresponds to electrons on $S^2$ under the influence of a "Dirac monopole" located at the center. In this picture, the wavefunctions of electrons are indeed sections of the principal $S(1)$-bundles over $S^2$ where the fiber represents an Abelian "Dirac monopole" in a physicist's sense (see, for example, [4]).

Now we are trying to find a similar connection between such monopoles in $\mathbf{Gr}(2,\mathbb{C}^5)$ and the 3rd Hopf fibration $$S^7\hookrightarrow S^{15} \rightarrow S^{8}.$$

After seeing that my first guess failed, from what I learned here I can guess the following, $$\mathbf{Gr}(2,\mathbb{C}^5)\approx S^4\times CP^4,$$ holds. This can be obtained from the use of "iterated extension" of all unitary groups involved in the coset representation of the Grassmannian, exactly the same way Allen did in a comment below, giving $$\mathbf{Gr}(2,\mathbb{C}^5)\approx (S^9 \times S^7)/(S^3 \times S^1),$$ and using the 2nd Hopf fibration to obtain $$\mathbf{Gr}(2,\mathbb{C}^5)\approx S^4\times (S^9 / S^1)\approx S^4 \times CP^4.$$

This is a big improvement however we are still not ${\it there}$ because I would rather have $CP^4$ replaced locally by $S^8$ to be able to make use of the 3rd Hopf fibration to justify the physics side of the deal, which works perfectly fine. I am not sure but this most probably will fail.

References

[1] http://arxiv.org/pdf/cond-mat/0306045v1.pdf

[2] http://arxiv.org/pdf/hep-th/0203264v1.pdf

[3] http://arxiv.org/pdf/1302.2754v2.pdf

[4] http://iopscience.iop.org/0305-4470/13/2/012

AB

Answer 1

That is definitely not true. In particular, $H^2(\text{Gr}(k,\mathbb{C}^n),\mathbb{Z})$ is isomorphic to $\mathbb{Z}$ for every pair of integers $n>1$ and $1<k<n$. Yet, by Künneth, $H^2(S^7\times S^5,\mathbb{Z})$ vanishes.

Edit.Your second guess is still false, for almost precisely the same reason as the first. If $\text{Gr}(k,\mathbb{C}^n)$ were homeomorphic (or even just homotopic) to $S^a\times \mathbb{C}P^b$, then the cohomology rings would be isomorphic. Since the socle degree of the cohomology ring equals the (real) dimension, first this implies that $2k(n-k)=a+2b$, so $a$ is even. Also, $H^2(\text{Gr}(k,\mathbb{C}^n)) = \mathbb{Z}x$, for a generator $x$. Yet $b_2(S^2\times \mathbb{C}P^b)$ equals $2$, not $1$. Hence, also $a\geq 4$. Then, by Künneth, $H^2(S^a\times \mathbb{C}P^b)$ is isomorphic to $H^0(S^a)\otimes H^2(\mathbb{C}P^b)$. Thus every generator is of the form $1\otimes y$, where $y$ is a generator of $H^2(\mathbb{C}P^b)$. In particular, we have $(1\otimes y)^{b+1}$ equals $0$. On the other hand, for the generator $x$ of $H^2(\text{Gr}(k,\mathbb{C}^n))$, $x^{k(n-k)}$ is nonzero. Since $2k(n-k)$ equals $a+2b$, in particular $k(n-k)$ is at least $b+1$. Thus we have a contradiction.

Answer 2

Even if you just have a fibration with one sphere as the base and the other as the fiber, you will have Euler characteristics $4$ which is not the Euler characteristics of $G(2,5)$.

The description of $G(2,5)$ that is closest to a fibration that I can come up with is by looking at the blowup of $G(2,5)$ at a $G(2,4)$ in it, which has the structure of a $\mathbb P^3$ bundle over a $\mathbb P^3$. Think of it as looking at a four-dimensional subspace $V_4$ of a five-dimensional space $V_5$ and then replacing the Grassmannian of dimension two subspaces of $V_5$ by the space of data "subspace $W$ of dimension two of $V_5$ and a choice of dimension one subspace of $W\cap V_4$". Then the fibration structure is the map to $\mathbb PV_4$ which records the above dimension one subspace.