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... 



Fibrations of $SU(4)$
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 $(1{+}x^4)(1{+}x^5)$.


