# $G$-bundles over $S^2$

In light of the fact that the only homogeneous (under a finite--dimensional Lie group action) $$S^2$$-bundle over $$S^2$$ is the trivial one. I would like to know if this fact is more general. i.e., is any homogeneous fiber bundle over $$S^2$$ and with a fiber that is connected and simply-connected is trivial?

I did a google search and I found the following in Encyclopedic dictionary of math. p. 572, the following:

"The set of the equivalence classes of principal $$G$$-bundles or $$G$$-bundles with fiber $$F$$ over the base $$S^n$$ is one-to-one correspondence with the set $$\pi_{n-1}(G)/\pi_0(G)$$ of equivalence classes under the operation of $$G$$ on $$\pi_{n-1}(G)$$..."

So for example in the (principal) Hopf-fibration $$SO(2)\hookrightarrow SO(3) \to SO(3)/SO(2)=S^2$$ we have $$\pi_1(SO(2))=\mathbb Z$$, so there are infinitely many different $$S^1$$-bundles over $$S^2$$. However, I didn't find an "actual" reference for the result claimed in the shaded box in the case of $$G$$-homogeneous bundles? But in case it is true, shouldn't be a condition on the effectivity of the group action as we can take our group to be simply connected (by going up to the universal covering)?

I have a hunch about the following situation:

Let $$J/H\hookrightarrow G/H\to G/J$$ be a fiber bundle consisting of a connected (finite--dimensional) Lie group $$G$$ and closed connected subgroups $$H\subset J$$ with base $$G/J\;\cong\; \mathrm{SU}(2)/S^1\;\cong \;S^2$$ and a compact fiber $$J/H\;\cong\;\mathrm{SU}(n)/T$$ where $$T$$ is a maximal torus in $$\mathrm{SU}(n)$$. In this case, the following can be shown:

(1) the total space $$G/H$$ can be given a structure of a (generalized) flag variety. Thus, $$G/H\cong \mathrm{SU}(m)/T_m$$.

(2) the $$\mathrm{SU}(2)$$ acting in the base is a normal subgroup of $$G$$.

Since maximal tori are conjugate to each other, then I can pick any maximal torus $$T_m$$ such that $$G/H \;\cong\; \mathrm{SU}(m)/ T_m\; \cong \; \Big(\mathrm{SU}(n)\times \mathrm{SU}(2)\Big)\Big/\Big(T\times S^1\Big)\; \cong \; \mathrm{SU}(n)/T \;\times \mathrm{SU}(2)/S^1$$

But even if my hunch is right, a general fact would be more interesting!

My question: Let $$G$$ be a connected and simply-connected finite dimensional Lie group and $$H\subset J$$ be closed subgroups of $$G$$ such that $$J$$ is connected and $$J/H$$ is simply-connected. Is the following fibration trivial? $$J/H\hookrightarrow G/H\to G/J\cong S^2$$

• Is $\pi_n$ here supposed to mean free homotopy classes? Otherwise, I don’t see how $\pi_0$ acts on it. – Connor Malin Jan 18 '20 at 14:27
• Since $G$ is a topological group, it is simple, so there is no real distinction between based and unbased homotopy classes of maps if it is connected. – Tyrone Jan 18 '20 at 14:58
• Amrat, assuming the shaded box is correct, then the set of equivlence classes of principal $G$-bundles over $S^2$ in bijective correspondence with $\pi_1(G)/\pi_0(G)$, and $\pi_1G$ need not be trivial (although, yes, $\pi_2G$ wil be). – Tyrone Jan 18 '20 at 15:01
• @Tyrone Could you explain what the action of $\pi_0$ is? – Connor Malin Jan 18 '20 at 15:02
• @ConnorMalin $\pi_0G\cong \pi_1BG$. It is the action of $\pi_1BG$ on the based homotopy set $[S^n,BG]=\pi_nBG=\pi_{n-1}G$, which yields the unbased homotopy set of of maps $<S^n,BG>$ as its quotient. The latter set is that of classifying maps for $G$-bundles over $S^n$. – Tyrone Jan 18 '20 at 15:03

It does not necessarily follow that the bundle is trivial.

In fact, a counterexamaple can be found in Kamerich's thesis

B. N. P. Kamerich, Transitive transformation groups of products of two spheres, Ph.D. thesis, Catholic University of Nijmegen, 1977.

He proves the following.

Theorem: Let $$G = SU(n+1)\times SU(2)$$. Let $$J = SU(n+1)\times S^1$$ and let $$H = SU(n)\times S^1_{p,q}$$, where the embedding $$J\rightarrow G$$ is the obvious one and the embedding $$H\rightarrow G$$ maps $$(A,z)\in SU(n)\times S^1_{p,q}$$ to the block diagonal $$\left( \operatorname{diag}(z^p A, z^{-np}), \operatorname{diag}(z^q, z^{-q})\right)$$ where $$p$$ and $$q$$ are relatively prime integers. Assume $$q|n$$, $$n$$ is even, $$n/q$$ is odd, and that $$p$$ is odd. Then $$G/H$$ is the (unique) non-trivial linear $$S^{2n+1}$$ bundle over $$S^2$$.

(If you want a specific example, take $$n = 6$$, $$q=2$$, and $$p = 1$$).

In fact, not only is the bundle non-trivial, but Kamerich shows the total space is not even homotopy equivalent to $$S^{2n+1}\times S^2$$ - the second Stiefel-Whitney class is non-trivial for $$G/H$$ (but it is trivial for the parallelizable manifold $$S^{2n+1}\times S^2$$.)

Edit There are still examples where the total space has even dimension, simply by producting the total space and fiber of the above example with your favorite odd dimensional homogeneous space.

For example, given $$G,J,H$$ as above, consider $$G' = G\times SU(2)$$, $$J' = J\times SU(2)$$ (with $$J\subseteq G$$, $$SU(2)\subseteq SU(2)$$), and keep the same $$H$$ (still embedded in $$J$$ as above).

Then the homogeneous fibration is simply $$S^{2n+1}\times SU(2)\rightarrow (G/H)\times SU(2)\rightarrow S^2$$. The total space still has a non-trivial second Stiefel-Whitney class, so cannot be homotopy equivalent to $$S^{2n+1}\times SU(2)\times S^2$$.

However, I claim

Given a homogeneous fibration $$J/H\rightarrow G/H\rightarrow G/J$$ with $$G/J = S^2$$ and $$J/H$$ of non-zero Euler characteristic, then the bundle must be trivial.

Proof: Let's handle the case of $$G$$ simple first. Then the long exact sequence in homotopy groups associated to $$J\rightarrow G\rightarrow S^2$$ shows that the map $$\pi_3(J)\rightarrow \pi_3(G)\cong \mathbb{Z}$$ is trivial. In particular, it follows that $$J$$ is a torus. Since the higher homotopy groups of $$J$$ then vanish, the fact that $$\pi_k(S^2)$$ is torsion for all $$k > 4$$ implies the same of $$G$$. However, the rational homotopy groups of Lie groups are known, so this implies $$G = SU(2)$$ (up to finite cover), which then implies $$J = S^1$$. But then the only non-trivial $$H\subseteq J$$ of full rank (so that $$J/H$$ has non-zero Euler characteristic) is $$H = J$$. Thus, the fibration is the trivial fibration $$\{pt\}\rightarrow S^2\rightarrow S^2$$, so is obviously a product.

So we may assume (by passing to a finite cover) $$G = G_1 \times G_2 \times ...\times G_m \times T^n$$ is a product of semi-simple groups and a torus. Because the fiber has non-zero Euler characteristic, the same is true of $$G/H$$, so $$H$$ must have full rank in $$G$$. A theorem of Borel asserts then that $$H$$ splits as a product $$H = H_1\times H_2\times ...\times H_m\times T^n$$ with each $$H_i\subseteq G_i$$ of maximal rank. In particular, the torus factor of $$G$$ plays no role (since it's a common normal subgroup to both $$H$$, $$J$$, and $$G$$), so we may as well assume $$n = 0$$. In fact, if any $$H_i = G_i$$, then necessarily the action of $$G_i$$ on the homogeneous space $$G/H$$ and $$G/J$$ is trivial, so we may as well exclude it. That is, we may assume that $$H_i$$ is a proper subgroup of $$G_i$$ for each $$i$$.

Likewise, $$J$$ splits as $$J = J_1\times J_2\times...\times J_m$$. Now, $$G/J = (G_1/J_1)\times (G_2/J_2)\times ... \times (G_m/J_m)$$. Since $$\pi_1(G/J) = 0$$, $$\pi_1(G_k/J_k) = 0$$ for each $$k$$. This implies the Euler characteristic of $$G_k/J_k$$ is at least $$2$$ whenver $$J_k\neq G_k$$. Since $$\chi(S^2) = 2$$, we conclude that $$J_k = G_k$$ for all $$k$$ but one. By relabeling, we may as well assume that the "one" is $$k = 1$$. That is, $$J_k = G_k$$ for $$k = 2,..., m$$, and thus $$G_1/J_1 = S^2$$. By the simple case we did above, this means that $$G_1 = SU(2)$$ and $$J_1 = S^1$$.

What about $$H$$? Well, $$H_1\subseteq J_1 = S^1\subseteq G_2 = SU(2)$$, and $$H_1$$ must have full rank. Thus $$H_1 = J_1 = S^1$$. The other $$H_i\subseteq G_i$$ are arbitrary, except for having full rank.

It now follows that the homogeneous fibration really looks like $$(S^1/S^1)\times (G_2/H_2)\times ... \times (G_m/H_m)\rightarrow (SU(2)/S^1)\times (G_2/H_2)\times ....\times (G_m/H_m)\rightarrow (SU(2)/S^1)\times (G_2/G_2)\times ...\times (G_m/H_m)$$ with the obvious projection.

Canceling out any common normal subgroups and writing $$S^2 = SU(2)/S^1$$, $$F = (G_2/H_2)\times ...\times (G_m/H_m)$$, this is simply $$F\rightarrow S^2\times F\rightarrow S^2$$ with the obvious projection. $$\square$$

• I have extended the answer to address this comment. Short answer: easy counterexamples to what you asked, but no counterexamples if you assume $J/H$ has non-zero Euler characteristic. – Jason DeVito Jan 21 '20 at 14:55
• Thank you so much! – user328669 Jan 24 '20 at 12:46