Is the non-trivial $S^3$ bundle over the torus $T^2$ an $H$-space?

Question

It follows from this MO post that for $n \ge 2$, the homotopy classes of maps from the torus $T^2$ to the classifying space of $SO(n + 1)$ is $$[T^2, BSO(n+1)] = \pi_2(BSO(n+1)) = \pi_1(SO(n+1)) = \mathbb{Z}_2.$$ Consequently, there exists precisely one non-trivial $S^n$ fiber bundle over $T^2$, with $SO(n+1)$-action.

Now, if we assume $n \le 3$, then we know $\rm{Diff}^+(S^n) \simeq SO(n+1)$, and so we have a classification of (oriented) $S^2$ and $S^3$-bundles over $T^2$. In this SE post, the non-trivial $S^2$ bundle was realized as a quotient of $S^2 \times T^2$.

Question: What is known about the non-trivial $S^3$-fiber bundle over $T^2$? In particular, is this space a Lie group, or maybe an $H$-space?

Any comment regarding this is highly appreciated. Cheers!

Answer 1

Here's one way to construct this manifold.

Note that $\pi_2(BSO(4)) \cong \pi_1(SO(4)) \cong \mathbb{Z}_2$, so up to isomorphism, there is only one non-trivial real rank four vector bundle $E \to S^2$ (note that such a bundle must be orientable as $S^2$ is simply connected). Namely, the bundle $E \cong \mathcal{O}(1)\oplus\varepsilon^2$. Let $f : T^2 \to S^2$ be a degree one map and consider the bundle $E' := f^*E \to T^2$. Note that $$\langle w_2(E'), [T^2]\rangle = \langle f^*w_2(E), [T^2]\rangle = \langle w_2(E), f_*[T^2]\rangle = \langle w_2(E), [S^2]\rangle = 1,$$

so $w_2(E') \neq 0$; in particular, $E'$ is (up to isomorphism) the unique non-trivial orientable real rank four vector bundle on $T^2$. The manifold you are asking about is the sphere-bundle of $E'$, call it $M$, with projection $p : M \to T^2$.

As $M$ is the sphere bundle of $E' \to T^2$, we have the following description of its tangent bundle:

$$TM\oplus\varepsilon^1 \cong p^*E'\oplus p^*T(T^2) \cong p^*E'\oplus\varepsilon^2.$$

So $w(TM) = w(p^*E') = p^*w(E') = 1 + p^*w_2(E')$. Applying the Gysin sequence to $p : M \to T^2$, we see that $p^* : H^2(T^2; \mathbb{Z}_2) \to H^2(M; \mathbb{Z}_2)$ is an isomorphism. Therefore $w_2(TM) = p^*w_2(E') \neq 0$ as $w_2(E') \neq 0$. In particular, $M$ is not parallelisable, so $M$ can not be equipped with a Lie group structure.