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Consider the Hopf bundles $$S^1\rightarrow S^{2n+1}\rightarrow \mathbb{C}P^n$$ and $$S^3\rightarrow S^{4n+3}\rightarrow \mathbb{H}P^n.$$

In this question (and also here), it is shown that for any continuous map $f:X\rightarrow \mathbb{C}P^n$, there is a lift $\tilde{f}:X\rightarrow S^{2n+1}$ iff $f^\ast:H^2(\mathbb{C}P^2)\rightarrow H^2(X)$ is the $0$ map.

I'd like to know a similar characterization for the quaternionic Hopf bundles.

Suppose $f:X\rightarrow \mathbb{H}P^n$ is any continuous function. Is there a nice characterization of when there is a lift $\tilde{f}:X\rightarrow S^{4n+3}$?

The proof for $\mathbb{C}P^n$ relies on the fact that $\mathbb{C}P^\infty$ is a $K(\mathbb{Z},2)$, and so a map to $\mathbb{C}P^\infty$ is homotopically trivial iff the induced map on $H^2$ is $0$. This part of the proof fails for $\mathbb{H}P^n$.

Now, a necessary condition that $f$ have a lift is that $f^\ast$ be the $0$ map on $H^4$. But this condition is not sufficient, as the following example demonstrates:

By the long exact sequence in homotopy groups for the quaternionic Hopf fibration, we see that $\pi_5(\mathbb{H}P^n)$ surjects onto $\pi_4(S^3)\cong\mathbb{Z}_2$, so long as $n\geq 1$. In particular, $\pi_5(\mathbb{H}P^n)\neq 0$.

So, let $X= S^5$ and let $f:X\rightarrow \mathbb{H}P^n$ be homotopically nontrivial. Then $f^\ast$ is the $0$ map on $H^4$ for trivial reasons, but there is no lift $\tilde{f}$ because $f$ can't factor through a $5$-connected space.

This example shows that it is also necessary that $f$ induce the $0$ map on $\pi_5$. One can similarly argue that $f$ must induce the $0$ map on $\pi_6$. But this is as far as I can get.

Thank you!

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  • $\begingroup$ I think you're asking (a) what are the invariants you need to check that an $SU(2)$-bundle is trivial; and (b) what are the values of these invariants for the Hopf bundle over $\mathbb H P^n$? I think that, rationally, the invariants are just the Chern classes, which will be detected by $H^2$ and $H^4$. But integrally, I guess there are torsion invariants as well, the $2$-torsion obstruction coming from $\pi_5$ that you mention being one of these. $\endgroup$ – guy-in-seoul Sep 5 '14 at 23:50
  • $\begingroup$ Well, I'm not exactly sure what I'm asking ;-). That said, it seems conceivable to me that there could be a lift even if the pull back of the Hopf bundle to $X$ is not trivial. But perhaps not, because this can't happen in the $\mathbb{C}P^n$ case. $\endgroup$ – Jason DeVito Sep 6 '14 at 0:43
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The map $S^{4n+3} \to \mathbb H P^n$ realizes the source as an $SU(2)$-bundle over the target. By the universal property of pull-backs (fibre products), a morphism $f: X \to \mathbb H P^n$ lifts to a morphism $\tilde{f}: X \to S^{4n + 3}$ if and only if the pulled-back bundle $f^* S^{4n+3}$ (an $SU(2)$-bundle over $X$) admits a section.

So the question seems to break into two parts:

(a) what are the invariants that classify an $SU(2)$-bundle; and

(b) what are these invariants in the particular case of $S^{4n+3}$ over $\mathbb H P^n$.

Now the answer to these questions is at least partly related to Chern classes. If $f^* S^{4n+3}$ is trivial, then $f^*$ should at least kill the Chern classes of $S^{4n+3} \to \mathbb H P^n$. In general, for a rank $2$ bundle, there are two Chern classes, $c_1 \in H^2$ and $c_2 \in H^4$. When the bundle has structure group $SU(2)$, the class $c_1$ vanishes, and so the only interesting Chern class is $c_2 \in H^4$.

This fits well with the case of $\mathbb H P^n$, which has vanishing $H^2$, but whose $H^4$ is one-dimensional, generated by the $c_2$ of the Hopf bundle (I think).

So far, this analysis is completely in line with the corresponding $\mathbb C P^n$ case. But there is a key difference between the two:

For $S^1$-bundles, the first Chern class $c_1$ is a complete invariant (if we work with integral cohomology), so if it vanishes, the $S^1$-bundle is trivial. So in the case of a morphism $f: X \to \mathbb C P^n$, the triviality of $f^* H^2$ is not only necessary, but is also sufficient, for $f$ to lift to $\tilde{f}: X \to S^{2n+1}$.

For $SU(2)$-bundles, the second Chern class $c_2$ is not a complete invariant. My understanding (though I could be mistaken) is that it is a complete rational invariant. But there are additional torsion invariants, such as the class in $\pi_5$ that you found.

In fact, we have that $BSU(2) = \mathbb HP^{\infty}$, with $S^{\infty} \to \mathbb H P^{\infty}$ being the universal $SU(2)$-bundle. (See e.g. the discussion here on p.81.) Now $S^{\infty}$ is contractible, and so we see that $\pi_i(\mathbb H P^{\infty}) = \pi_{i-1}(S^3).$ In particular, if we work with rational homotopy, since $S^3$ has only one non-vanishing rational homotopy group, in degree $3$, we see that $\mathbb H P^{\infty}$ has only one non-vanishing rational homotopy group, in degree $4$. (And its generator is $c_2$ of the universal bundle; so I guess my suggestion above that $c_2$ is a complete rational invariant is correct.)

But $S^3$ has lots of torsion higher homotopy, and hence so does $\mathbb H P^{\infty}$.

At this point, I have to stop, since I'm not sure how to go on, other than noting certain explicit torsion obstructions, as in the OP.

Added: I think we are literally in the context of this wikipedia discussion about obstructions to lifting sections of principal bundles.

The class $c_2 \in H^4$ is the first obstruction, coming from $\pi_3(S^3)$. The next obstruction (if this one vanishes) will be a class in $H^5$ with coefficients in $\pi_4(S^3) = \mathbb Z/2;$ for the map $S^5 \to \mathbb H P^n$ that you discussed, this obstruction will be non-zero.

Since $S^3$ has many (!) other non-vanishing homotopy groups, there will be higher obstructions too (at least as $n$ grows).

For what it's worth, I think that the problem of computing these higher obstructions is also related to the problem of computing a Postinikov tower for $\mathbb H P^{\infty}$.

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  • $\begingroup$ Thanks for the answer (and +1 from me). I'll have to digest it a bit over time, but I did want to add the following. The Hopf bundle $S^3\rightarrow S^{4n+3}\rightarrow \mathbb{H}P^n$ is a principal $S^3$ bundle, and therefore, so is $f^\ast S^{4n+3}$. Now, a principal bundle is trivial iff it has a section, so it as you suggested in your comment on the question: A necessary and sufficient condition is that the pull back $f^\ast S^{4n+3}$ be trivial. Also, $\mathbb{H}P^\infty$ is a $K(\mathbb{Q},4)$, as you noted, and so rationally the argument for $\mathbb{C}P^n$ works. Thanks! $\endgroup$ – Jason DeVito Sep 6 '14 at 17:47
  • $\begingroup$ @JasonDeVito: You're welcome. Actually, your comment nicely summarizes my answer in just a couple of sentences! $\endgroup$ – guy-in-seoul Sep 6 '14 at 17:51
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Here are some general statements. Suppose $X$ is a space and $\pi : E \to B$ is a map of spaces. Given a map $f : X \to B$, we might want to know when $f$ admits a lift to $E$ up to homotopy (that is, a map $\tilde{f} : X \to E$ together with a homotopy $\pi \circ \tilde{f} \cong f$). The reason this is a more fun question to ask than the question of whether $f$ admits a lift is that it is a homotopically invariant question to ask, and so we can hope to get a homotopically invariant answer.

The general answer is the following. Assume for simplicity that $B$ is path-connected and pick a basepoint of it. Let $f$ be the homotopy fiber of $\pi$, so that $E$ is (up to homotopy) an $F$-fibration over $B$. Considering the homotopy pullback of the diagram $X \xrightarrow{f} B \xleftarrow{\pi} E$ shows that a lift up to homotopy iff the homotopy pullback $F$-fibration $X \times_B E$ over $X$ has a homotopy section.

Now I believe it is furthermore the case that if we replace $\pi$ with an actual fibration, then above we can replace "homotopy lift" with "lift," "homotopy pullback" with "pullback" and "homotopy section" with "section." But don't quote me on this.

In any case, in general obstructions to the existence of a homotopy section of an $F$-fibration over a space live in the (local) cohomology groups $H^{n+1}(X, \pi_n(F))$, where $\pi_n(F)$ denotes a local system and not just a group (you can ignore this if $X$ is simply connected). If $F$ has infinitely many nonzero homotopy groups then there are infinitely many of these obstructions, although if $X$ has finite cohomological dimension then all but finitely many of them automatically vanish. I may need to assume that the fibration has a Postnikov tower.

Things get simpler if $\pi$ is itself the homotopy fiber of a further map $\phi : B \to Y$ for some based space $Y$, so that we have a fiber sequence

$$F \cong \Omega Y \to E \xrightarrow{\pi} B \xrightarrow{\phi} Y.$$

In this case $E$ is now a homotopy "principal $\Omega Y$-bundle" over $B$ and the situation simplifies: by the universal property of the homotopy fiber, $f$ admits a homotopy lift iff the composite $\phi \circ f : X \to Y$ is nullhomotopic. In the nicest cases $Y$ is a product of Eilenberg-MacLane spaces so we immediately see that we have a complete set of obstructions given by various cohomology classes, but even if this isn't the case we can still get a complete set of obstructions by looking at a Postnikov tower of $Y$ if one exists; these will now live in $H^n(X, \pi_n(Y))$ (which is the thing one expects if $Y$ is a product of Eilenberg-MacLane spaces).

In this particular case we can take $Y = BSU(2)$ which has infinitely many nonzero homotopy groups, so there are infinitely many obstructions as already mentioned, although again all but finitely many of them can be ignored depending on the cohomological dimension of $X$.

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    $\begingroup$ I should mention something important about these obstructions, which is that the obstruction for $n+1$ is only well-defined provided that the obstruction for $n$ vanishes. $\endgroup$ – Qiaochu Yuan Sep 6 '14 at 19:15

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