Second homotopy group of real Grassmannians $\textrm{Gr}(n,m)$, special case $n=m=2$ not clear.

Solution 1:

Here are a couple ways to see that $\widetilde{\mathrm{Gr}}(2,2)$ is $S^2\times S^2$.

Method One: Let $\mathrm{UT}S^3$ be the unit tangent bundle of $S^3$. This is a subbundle of the tangent bundle which only contains the unit tangent vectors. It can be defined as

$$ \mathrm{UT}S^3=\{(u,v):\in S^3\times S^3:u\perp v\}\subseteq\mathbb{R}^4\times\mathbb{R}^4 $$

Every ordered pair of perpendicular vectors induces an oriented plane (the one they span), in which case we get a map $\mathrm{UT}S^3\to\widetilde{\mathrm{Gr}}(2,2)$. The fibers are all pairs of orthogonal unit vectors in a given plane with the correct orientation; each such pair is determined by the first vector, so the fibers will simply be circles $S^1$.

On the other hand, if we treat $S^3$ as the unit quaternions, we have a trivialization

$$ \mathrm{UT}S^3\xrightarrow{\sim} S^3\times S^2: \quad (u,v)\mapsto (u,vu^{-1}). $$

Here we treat $S^2$ as a the set of unit imaginary quaternions. This is true because every tangent vector at $u$ will be of the form $wu$ where $w$ is a tangent vector at the identity.

What does the trivialization do to fibers? The elements of the fiber of $(u,v)\mapsto\mathrm{span}(u,v)$ are of the form $(e^{\theta w}u,e^{(\theta+\pi/2)w}u)$ where $w=vu^{-1}$, which turn into $(e^{\theta w}u,w)$. In other words the fibers become right cosets of the circle group $S^1=\exp(\mathbb{R}w)$ (in the first coordinate), which is a fiber of the map $(q,w)\mapsto (qwq^{-1},w)$ (orbit-stabilizer theorem), a fibration $S^3\times S^2\to S^2\times S^2$.

So by construction, the trivialization $\mathrm{UT}S^3\to S^3\times S^2$ is a diffeomorphism which restricts to diffeomorphisms on each fiber of their respective fibrations; thus, they determine a diffeomorphism of base spaces $\widetilde{\mathrm{Gr}}(2,2)\to S^2\times S^2$.

Method Two. With the Plucker embedding we can place $\widetilde{\mathrm{Gr}}(2,2)$ inside $\Lambda^2\mathbb{R}^4$. Every oriented plane with ordered orthonormal basis $\{u,v\}$ is identified with $u\wedge v$, which is easily checked independent of choice of basis for the plane.

The image may be identified as the solutions to the system

$$ x\wedge x=0, \qquad |x|^2=1. $$

To see this, plug $x=a\wedge b+c\wedge d$ into $x\wedge x=0$. The second equation normalizes $x$. (Note the image of this in $\mathbb{P}(\Lambda^2\mathbb{R}^5)=\mathbb{RP}^5$ is the so-called Klein quadric.) The inner product on $\Lambda^2\mathbb{R}^4$ is induced from the one on $\mathbb{R}^4$ in the standard way:

$$ \langle a\wedge b,c\wedge d\rangle=\det\begin{pmatrix} \langle a,c\rangle & \langle a,d\rangle \\ \langle b,c\rangle & \langle b,d\rangle \end{pmatrix}. $$

The act of "taking orthogonal complements" is an involution on $\widetilde{\mathrm{Gr}}(2,2)$ that extends to a linear operator on $\Lambda^2\mathbb{R}^2$ of order $2$, called the Hodge star $\ast$. Moreover, $\ast$ respects the inner product, and has two eigenvalues, $\pm1$. The $+1$ and $-1$ eigenspaces are comprised of elements of the form $a\wedge b+c\wedge d$ and $a\wedge b-c\wedge d$ respectively, where $\{a,b,c,d\}$ ranges over ordered orthonormal bases inducing the same orientation as the coordinate basis.

(Under the identification $\Lambda^2\mathbb{R}^4\cong\mathfrak{so}(4)$, where $a\wedge b$ acts as the linear operator which rotates $a$ to $b$ and annihilates the orthogonal complement of $a$ and $b$'s span, whenever $a,b$ are orthonormal, the $\pm1$ eigenspaces correspond to left and right isoclinic rotations, which in turn correspond to left and right multiplication of $\mathbb{H}\cong\mathbb{R}^4$ by imaginary quaternions.)

Call the $\pm1$-eigenspaces $\Lambda^2_L$ and $\Lambda^2_R$ respectively. They are perpendicular with respect to $\wedge$ and $\langle\cdot,\cdot\rangle$ (that is, $\nu_1\wedge \nu_2=\langle \nu_1,\nu_2\rangle=0$ whenever $\nu_1\in\Lambda^2_L$ and $\nu_2\in\Lambda^2_R$) and the bilinear form $\wedge$ has signatures $(3,0)$ and $(0,3)$ on them.

Write $x=\nu_1+\nu_2$. Then $x\wedge x=0$ corresponds to $|\nu_1|=|\nu_2|$ and $|x|^2=1$ corresponds to $|\nu_1|^2+|\nu_2|^2=1$, therefore $\widetilde{\mathrm{Gr}}(2,2)$ is the direct product of two spheres of radius $1/\sqrt{2}$ inside the eigenspaces of $\ast$, i.e. $S^2\times S^2\subset\Lambda^2_L\oplus\Lambda^2_R$.

The latter method gets to the heart of what's going on, in my opinion: every oriented plane (corresponding to $a\wedge b$) induces a pair of left and right forms (corresponding to $(a\wedge b\pm c\wedge d)/2$ where $\{c,d\}$ spans the orthogonal complement). What's not so obvious is that this is a one-to-one correspondence: every pair of (normalized) left/right forms determines an oriented plane, the left/right forms may be controlled independently of each other, and the (normalized) left/right forms each form a sphere. This is something magical that happens in $4$ dimensions.

Solution 2:

There is a correspondence between the oriented Grassmann manifold $\tilde{\mbox{Gr}}(2,2)$ and $S^2 \times S^2$. I guess this is kind of a "classical" fact, I know there is a proof in [H. Gluck and F. Warner, Great circle fibrations of the three-sphere] (see Lemma 5.2), maybe there is an intuitive proof but I don't know it off the top of my head.

In any case, the oriented Grassmann double covers the unoriented one, so the second homotopy groups are the same. So

$$ \pi_2(\mbox{Gr}(2,2)) = \pi_2(S^2 \times S^2) = \pi_2(S^2) \times \pi_2(S^2) = \mathbb{Z} \times \mathbb{Z}. $$