Second homotopy group of real Grassmannians $\textrm{Gr}(n,m)$, special case $n=m=2$ not clear. I have been considering real Grassmanians
$$\textrm{Gr}(n,m)=O(n+m)/O(n)\times O(m)$$
appearing in certain condensed matter physics context (space of real flat-band Hamiltonians $Q(k)$ with $n$ occupied and $m$ unoccupied bands, the reality comes from commutation with antiunitary time-reversal squaring to $1$), and I am interested in their second homotopy group.
If I understand correctly, this can be derived from the exact sequence (https://en.wikipedia.org/wiki/Fibration#Long_exact_sequence_of_homotopy_groups) 
$$\pi_2[O(n+m)]\to\pi_2[\textrm{Gr}(n,m)]\to \pi_1[O(n)\times O(m)]\to\pi_1[O(n+m)].$$
The result depends on the choice of $n$ and $m$. If I didn't do any mistake, one can order all the results into a neat table:
$$\begin{array}{c||c|c|c}
        & n=1        & n=2                        & n \geq 3  \\ \hline
m=1     & 1          & \mathbb{Z}                 & 1         \\
m=2     & \mathbb{Z} & \mathbb{Z}\times\mathbb{Z} & \mathbb{Z}\\
m\geq 3 & 1          & \mathbb{Z}                 & \mathbb{Z}_2
 \end{array}$$
where $1$ is the trivial one-element group. 
I think I do understand most of these entries. The $\color{blue}{\textrm{blue}}$ ones in here
$$\begin{array}{c||c|c|c}
        & n=1        & n=2                        & n \geq 3  \\ \hline
m=1     & 1          & \color{blue}{\mathbb{Z}}                & 1         \\
m=2     & \color{blue}{\mathbb{Z}} & \mathbb{Z}\times\mathbb{Z} & \mathbb{Z}\\
m\geq 3 & 1          & \mathbb{Z}                 & \mathbb{Z}_2
 \end{array}$$
come from the fact that
$$\textrm{Gr}(1,2)\cong \textrm{Gr}(2,1) \cong \mathbb{R}P^2$$
is the space of lines through $0$ in 3D, and topologically it looks like a half-sphere. One can obviously wrap $S^2$ around it an integer number of times, thus the second homotopy group are integers. 
For the $\color{LimeGreen}{\textrm{green}}$ entries here
$$\begin{array}{c||c|c|c}
        & n=1        & n=2                        & n \geq 3  \\ \hline
m=1     & 1          & \mathbb{Z}                 & 1         \\
m=2     & \mathbb{Z} & \color{LimeGreen}{\mathbb{Z}}\times\mathbb{Z} & \color{LimeGreen}{\mathbb{Z}}\\
m\geq 3 & 1          & \color{LimeGreen}{\mathbb{Z}}                 & \color{LimeGreen}{\mathbb{Z}_2}
 \end{array}$$
I also know an explanation, although only in physics terms: On the $S^2$ one defines a "north pole" $\mathrm{N}$ and a "south pole" $\textrm{S}$, and chooses a set of paths $\gamma(\theta)$ ($\theta \in[0,\pi]$) such that


*

*$\gamma(\theta)$ depends continuously on $\theta$,

*$\gamma(0) = \textrm{N}$ and $\gamma(\pi) = \textrm{S}$ are just single points,

*and for $0<\theta<\pi: \textrm{N},\textrm{S}\notin \gamma(\theta)$.


Analogy with parallels on a globe might be helpful. Obviously $\cup_\theta \gamma(\theta)=S^2$. The physics interpretation has to do with a Wilson loop operator (I think this corresponds to parallel transport in mathematics -- not 100% sure though) on a closed path $\gamma$
$$W(\gamma) = \overline{\prod_{k\in\gamma}} P_k = \overline{\exp}\left[-\int_\gamma \mathcal{A}(k)\right]$$
where $P_k = \sum_{a=\textrm{occ}} u_{k,a}^{\phantom{\top}} u_{k,a}^\top$ is the projector onto the occupied (negative eigenvalue) or alternatively unoccupied (positive eigenvalue) eigenvectors of the Hamilotnian $Q(k)$ at $k$, the horizontal bar "$\,\overline{\phantom{\exp}}\,$" indicates path ordering, $\mathcal{A}$ is the (Wilczek-Zee-)Berry connection, and the points $k$ lie along $\gamma$ (a proper limit with number of points $N\to\infty$ is assumed).
Now the interpretation of the $\color{LimeGreen}{\textrm{green}}$ homotopy groups: It can be shown that $W(\gamma)$ is a gauge invariant $O(n)$ or $O(m)$ matrix (depending on whether one focuses on positive or negative eigenvalues). We choose the smaller one, i.e. $O(\min\{n,m\})$. We now look at the continuous function $W(\gamma(\theta))\equiv W(\theta)$. Because of the conditions listed above, $W(0) = W(\pi)$ is just the unit matrix $1\in SO(\min\{n,m\})$, and we trace some closed path in $SO(\min\{n,m\})$ for intermediate values of $\theta$. Thus we constructed a topological invariant related to $\pi_1 [SO(\min\{n,m\})]$ which is well-understood to be $\mathbb{Z}$ for $SO(2)$, and $\mathbb{Z}_2$ for all larger arguments. This gives the $\color{LimeGreen}{\textrm{green}}$ entries.
But there is one more non-trivial entry, the $\color{red}{\textrm{red}}$ one
$$\begin{array}{c||c|c|c}
        & n=1        & n=2                        & n \geq 3  \\ \hline
m=1     & 1          & \mathbb{Z}                 & 1         \\
m=2     & \mathbb{Z} & \mathbb{Z}\color{red}{\times\mathbb{Z}} & \mathbb{Z}\\
m\geq 3 & 1          & \mathbb{Z}                 & \mathbb{Z}_2
 \end{array}$$
that I have no clue about. 
I was wondering if anyone has some insight into or even suitable visualization of $\textrm{Gr}(2,2)$ to help me get over this last one. Does any one of you have an idea what's the meaning of that $\mathbb{Z}\times\mathbb{Z}$? Or perhaps some accessible reference into the topology of real Grassmannians? I would appreciate any hint.
 A: 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.
A: 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
$$
\pi_2(\mbox{Gr}(2,2)) = \pi_2(S^2 \times S^2) = \pi_2(S^2) \times \pi_2(S^2) = \mathbb{Z} \times \mathbb{Z}.
$$
