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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...

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  • $\begingroup$ In what you have written the base space is the coset space, however. $\endgroup$ Commented Nov 26, 2012 at 10:44
  • $\begingroup$ @ Charles Matthews: My apologies for imprecise formulation. The base is, of course, the coset space. $\endgroup$
    – user29418
    Commented Nov 26, 2012 at 13:27

2 Answers 2

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

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  • $\begingroup$ Dear Robert, thank you very much for the extensive answer! One small question I still have though. What makes the fibration $SU(3)\rightarrow SU(4)\rightarrow S^7$ so special in comparison to $SU(2)\times SU(2)\rightarrow SU(4)\rightarrow \frac{SU(4)}{SU(2)\times SU(2)}$ that in the first case the the Poincaré polynomial factorizes into the Poincaré polynomials of the fiber and the base, while in the second case not. $\endgroup$
    – user29418
    Commented Nov 26, 2012 at 20:59
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    $\begingroup$ @unknown(google): To answer that, I have to talk about spectral sequences. The point is that the cohomology of the total space is computed from the cohomology of the base and the fiber by a spectral sequence that starts with the tensor product of the cohomology rings of the base and the fiber (technically, this is the $E_2$-term of the spectral sequence). In the first case, all the higher differentials are zero (trivially, for degree reasons), so the $E_2$-term is the answer. In the second case, one of the higher differentials isn't zero, which kills the extra terms in the tensor product. $\endgroup$ Commented Nov 26, 2012 at 21:44
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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 $\mathbb{R}^6$.)

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