Fibrations of $SU(4)$ I would like to apologize in advance if my question is too simple for mathematical community here: I am physicist by education.
It is well known that for a topological group $G$ acting transitively on a space $X$ and its subgroup $H \subset G$ one can construct a principal bundle whose fibers are homeomorphic to the coset space $G/H$. A typical example would be $SO(n-1)\rightarrow SO(n)\rightarrow S^{n-1}$.
$SU(2)\times SU(2)$ is a subgroup of $SU(4)$. Would it be possible to construct a corresponding fibration and what would be the base space? If the answer is 'yes' can we say something about the Betti numbers of the base based on this fact?
Thank you very much in advance...  
 A: I'm way late to the party, but with a bit more work, one can understand $SU(4)/SU(2)\times SU(2)$ even more precisely.  Namely

$SU(4)/SU(2)\times SU(2)$ is diffeomorphic to $T^1 S^5$, the unit tangent bundle of $S^5$.

The topology of the unit tangent bundles of spheres is well known.  In particular, using the Gysin Sequence, one easily sees that $H^\ast(T^1 S^5) \cong H^\ast(S^4\times S^5)$ (with integral coefficients), which recovers Robert's answer.
The "more work" is just a bit of representation theory.  There is a double covering map $$f:SU(4)\rightarrow SU(4)/\pm I \cong SO(6)$$ which is found as a real subrepresentation of $\Lambda^2 \mathbb{C}^4$, (where $\mathbb{C}^4$ denotes the standard rep of $SU(4)$).  Since $-I\in SU(4)$ lies in the subgroup $SU(2)\times SU(2)$ and $SU(2)\times SU(2)/-I = SO(4)$, $f$ induces a diffeomorphism between $SU(4)/SU(2)\times SU(2)$ and $SO(6)/SO(4)$.  For the block embedding $SO(4)\subseteq SO(6)$, the homogeneous space $SO(6)/SO(4)$ is diffeomorphic to T^1 S^5$.
Note that one gets lucky here:  there is a unique (up to conjugacy) subgroup of $SO(6)$ isomorphic to $SO(4)$.
(While we're at it, a similar argument shows $SU(4)/S(U(2)\times U(2))$ is diffeomorphic to $SO(6)/SO(4)\times SO(2)$, the Grassmanian of oriented $2$ planes in $\mathbb{R}^6$.)
A: From the homotopy exact sequence of the fibration
$$
\mathrm{SU}(2)\times \mathrm{SU}(2) \longrightarrow \mathrm{SU}(4)\longrightarrow 
\frac{\mathrm{SU}(4)}{\mathrm{SU}(2)\times \mathrm{SU}(2)} = Q
$$
and standard facts about $\pi_i\bigl(\mathrm{SU}(k)\bigr)$, one sees that $\pi_i(Q)=0$ for $i = 0, 1, 2, 3$ and that $\pi_4(Q)\simeq\mathbb{Z}$.  Thus, one knows, by the standard theorems, that $H_i(Q,\mathbb{R})=0$ for $i = 1,2,3$ while $H_4(Q,\mathbb{R})=\mathbb{R}$.  Now by Poincaré duality (since $Q$ has dimension $9$ and is connected and orientable), one has
$$
H_k(Q,\mathbb{R})=\mathbb{R}\qquad\text{for $k=0,4,5,9$}
$$
while
$$
H_k(Q,\mathbb{R})=0\qquad\text{for $k=1,2,3,6,7,8$}.
$$
The usual duality now determines the cohomology ring completely, and the Poincaré polynomial is $(1{+}x^4)(1{+}x^5)$.
