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Milnor showed that if the Euler class of an $S^3$ bundle over $S^4$ is $\pm 1$, then the total space is a homotopy sphere. How many $S^3$ bundles over $S^4$ do we have with the total space is homotopic (i.e. homeomorphic) to $S^7$, if known? (or any related reference)

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we know $\pi_3(SO(4))$= Z + Z so there are infinitely many different such bundles, but are all the total spaces of these bundles homeomorphic to S7? i.e. how may I compute the homology groups? – wqr Jun 23 '12 at 13:08

$S^3$ bundles over $S^4$ (with structure group $SO(4)$) are classified, as you noted in the comments, by $\mathbb{Z}\oplus\mathbb{Z}$, that is, by two integers.

Milnor's gives an explicit construction as follows: Given $(h,j)\in\mathbb{Z}\oplus\mathbb{Z}$, the map $f_{j,h}:S^3\rightarrow SO(4)$ is given by $f_{j,h}(u)\cdot v = u^h v u^j$ where $u$ is intepreted as a unit quaternion and $v$ is intepreted as a quaternion.

Let $\xi_{h,j}$ denote the $S^3$ bundle over $S^4$ corresponding to $f_{h,j}$ above.

Milnor proves that the Pontrjagin class of $\xi_{h,j}$ is $\pm 2(h-j)\in \mathbb{Z}\cong H^4(S^4)$. I believe (though I admit I haven't checked the details) that a similar proof should show that the Euler class of $\xi_{h,j}$ is $(h+j)$.

So, how many bundles are there with Euler class $1$? Infinitely many, as, given any $h$, choose $j = 1-h$. Likewise, there are infinitely many with Euler class $-1$.


Given an fiber bundle over a "nice enough" base space $B$ with fiber a sphere of some dimension, there is an associated long exact sequence of cohomology groups called the Gysin sequence which related the cohomology of the total space with that of the base. One of the maps in the Gysin sequence is given by cupping with the Euler class of the bundle. In the case of an $S^3$ bundles over $S^4$, called $E$, one gets

$$...\rightarrow H^3(S^4)\rightarrow H^3(E)\rightarrow H^0(S^4)\rightarrow H^4(S^4)\rightarrow H^4(E)\rightarrow H^1(S^4)\rightarrow ...$$

or, filling in what one knows about the cohomology groups of $S^4$,

$$...\rightarrow 0 \rightarrow H^3(E)\rightarrow \mathbb{Z}\rightarrow \mathbb{Z}\rightarrow H^4(E)\rightarrow 0\rightarrow ...$$

where the map from $\mathbb{Z}$ to $\mathbb{Z}$ is cupping with the Euler class, and, in this case, can be identified with multiplication by the Euler class $e$.

Hence, we see that when $e = \pm 1$, the map in the middle is multiplication by $\pm 1$, which is an isomorphism. It follows that $H^3(E) = H^4(E) = 0$ in this case. (Using the Gysin sequence, one can see that $H^0(E) = H^7(E) = \mathbb{Z}$ and all others are $0$, regardless of $e$. So, in this case we get $H^*(E) = H^*(S^7)$.

On the other hand, when $e = 0$, we learn that $H^3(E) = H^4(E) = \mathbb{Z}$, so $H^*(E) = H^*(S^3\times S^4)$. Finally, if $|e|\geq 2$, then the map from $\mathbb{Z}$ to itself is injective. This shows that $H^3(E) = 0$ and $H^4(E) = \mathbb{Z}/|e|\mathbb{Z}$ in this case.

So, when $|e| \neq 1$, we see that $E$ does not have the cohomology groups of $S^7$.

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Thanks, but Milnor just considers the bundles $\xi_{hj}$ where $h+j=1$. Do we have any idea about the others? (though this does not change the result, infinitely many.) – wqr Jun 24 '12 at 12:37
Maybe I misunderstood your question. The space $\xi_{h,j}$ has the homotopy type of $S^7$ iff $h+j = \pm 1$. One can prove (and Milnor does) without the Poincare conjecture, that the space $\xi_{h,j}$ is homeomorphic to $S^7$ iff $h+j = \pm 1$ as well. – Jason DeVito Jun 24 '12 at 12:43
@wqr: I have added an explanation as to why $h+j = \pm 1$ is both necessary and sufficient. – Jason DeVito Jun 24 '12 at 12:59
thanx a lot.... – wqr Jun 24 '12 at 17:23

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