establish the diffeomorphism between $S^2$ and $SO(4)/U(2)$ I was trying to do this using the idea of homogeneous space, but stuck on the point of how $SO(4)$ acts on $S^2$. Naturally $SO(3)$ acts on $S^2$ transitively, but how does the group action works for $SO(4)$?
Any help is appreciated.
 A: I think you are asking for the following isomorphism:
$$(SU(2)\times SU(2))/\pm(1,1)\cong SO(4) $$
In terms of quaternion multiplication, this says that every rotation of $\mathbb{R}^4 (=\mathbb{H})$ can be uniquely expressed as a map of the form:$$q\mapsto aqb^*,$$
for some pair of unit quaternions up to sign $\pm(a,b)\in \mathbb{H}^2$.
To see that the map sending $\pm(a,b)$ to the rotation $q\mapsto aqb^*$ is an isomorphism, note that it is injective, and the left hand side is compact, so its image is closed.  However both the LHS and RHS are 6-dimensional manifolds, so the image is open.  Finally note that $SO(4)$ is connected, so the image must be the whole of $SO(4)$.
Now we have: $$SO(4)/SU(2)\cong SU(2)/\pm1\cong SO(3)$$ and
$$
SO(4)/U(2)\cong SO(4)/((SU(2)\times S^1)/\pm1)\cong SO(3)/S^1\cong S^2
$$
Here we use the homomorphism embedding $U(2)\to (SU(2) \times SU(2))/\pm1\cong SO(4)$ mapping: $$A\mapsto \left(A\left(\begin{array}{cc}\sqrt{Det(A)}&0\\0&\sqrt{Det(A)}\end{array}\right)^{-1},\left(\begin{array}{cc}\sqrt{Det(A)}&0\\0&\overline{\sqrt{Det(A)}}\end{array}\right)\right)$$
A: $SO(4)$ acts on its (real) lie algebra, $\mathfrak{so}(4)$, by conjugation. This is the adjoint representation.
The elements of $\mathfrak{so}(4)$ are traceless antisymmetric matrices. This is a $6$D vector space. One can check directly that being antisymmetric and traceless are preserved by conjugating by an orthogonal matrix.
Another way to think about this is the exterior square $\Lambda^2\mathbb{R}^4$. The elements look like $v\wedge w$ (where $v,w\in\mathbb{R}^4$) or sums of these wedges. The wedge symbol itself is antisymmetric and bilinear. Note $\{e_i\wedge e_j\}$ (with $i<j$) is a basis, which has $\binom{4}{2}=6$ elements. Note $v\wedge w$ can be thought of as the operator $vw^T-wv^T$, which is a traceless and antisymmetric matrix. So, for instance, $e_1\wedge e_2$ is the operator which rotates $90^{\circ}$ in the $e_1e_2$-plane (from $e_1$ to $e_2$) and acts as $0$ on the $e_3e_4$-plane.
The (oriented) Grassmanian $\widetilde{\mathrm{Gr}}_2(\mathbb{R}^4)$ is the space of all (oriented) 2D subspaces of $\mathbb{R}^4$. Note that $v\wedge w$ doesn't depend on which oriented orthonormal basis of a 2D subspace on picks, since rotation gives
$$ (\cos\theta ~v+\sin\theta~w)\wedge(-\sin\theta~v+\cos\theta~w)=v\wedge w. $$
Thus, $\widetilde{\mathrm{Gr}}_2(\mathbb{R}^4)$ is a subset of $\Lambda^2\mathbb{R}^4$. The Hodge star $\ast$ turns a 2D subspace into its orthogonal complement (appropriately oriented). So for instance, $\ast(e_1\wedge e_2)=e_3\wedge e_4$. This turns out to extend to a linear operator on all of $\Lambda^2\mathbb{R}^4$. It has two 3D eigenspaces, corresponding to  eigenvalues $\pm1$, spanned respectively by elements of the form $e_i\wedge e_j\pm e_k\wedge e_{\ell}$ for even permutations $ijk\ell$ of $1234$.
There is an inner product $\Lambda^2\mathbb{R}^4$, defined by the Grammian determinant:
$$ \langle v_1\wedge v_2,w_1\wedge w_2\rangle = \det\begin{bmatrix} \langle v_1,w_1\rangle & \langle v_1,w_2\rangle \\ \langle v_2,w_1\rangle & \langle v_2,w_2\rangle \end{bmatrix}. $$
Geometrically, this is $\cos\phi$ where $\phi$ is the dihedral angle between the 2D planes $V=\mathrm{span}\{v_1,v_2\}$ and $W=\mathrm{span}\{w_1,w_2\}$ if they reside in a common 3D hyperplane. Otherwise, it measures the signed area distortion when orthogonally projecting from $V$ to $W$ or vice-versa (just as the inner product in $\mathbb{R}^n$ measures the signed length distortion when projecting from the 1D span of one vector to another).
These two eigenspaces are orthogonal with respect to this inner product (and indeed the bases I gave are orthonormal bases, up to a constant). Thus, $SO(4)$ acts on the $S^2$ within each 3D eigenspace.
