David Roberts geometry on S^5

Some notes on stuff on the 5-sphere

• There is only one nontrivial $SU(2)$-bundle on $S^5$.

• There is only one nontrivial $SO(3)$-bundle on $S^5$, and it lifts to the previous bundle.

• Any $SU(2)$-bundle on $\mathbb{CP}^2$ trivialises on pulling back along $\pi\colon S^5 \to \mathbb{CP}^2$. Here $S^5$ is the total space of an $S^1$-bundle

  • This is because a) any classifying map $\mathbb{CP}^2\to BSU(2)$ factors through $\mathbb{CP}^2 \to \mathbb{CP}^2/\mathbb{CP}^1 \simeq S^4$, the collapse of a generator of $\pi_2$; and b) the resulting map $S^5\to \mathbb{CP}^2\to S^4$ is null-homotopic.

• Any $SO(3)$-bundle on $\mathbb{CP}^2$, when pulled back to $S^5$, lifts to an $SU(2)$-bundle.

• Given any fibre $S^1\subset S^5$ and classifying map $S^5 \to BSU(2)$ there is a canonical lift of $S^1 \to BSU(2)$ through $B\mathbb{Z}/2\to BSU(2)$. This is because $S^1$ maps to a point in $BSO(3)$ and so we get an up-to-homotopy unique map to the homotopy fibre of $BSU(2) \to BSO(3)$.

• $SO(3)$-bundles on $\mathbb{CP}^2$ are classified by $w_2\colon\mathbb{CP}^2 \to B^2\mathbb{Z}/2$ and $p_1\colon\mathbb{CP}^2 \to K(\mathbb{Z},4)$, where the latter is subject to the condition $p_1 = \mathfrak{P}(w_2) (mod 4)$, for $\mathfrak{P}\colon B^2\mathbb{Z}/2 \to K(\mathbb{Z}/4,4)$ the Pontryagin square operation. Here “$mod 4$” means under the map $K(\mathbb{Z},4) \to K(\mathbb{Z}/4,4)$.

  • I guess this means that the 4-type of $BSO(3)$ is $K(\mathbb{Z}/2,2) \times_{K(\mathbb{Z}/4,4)} K(\mathbb{Z},4)$. I need to get at the 5-type, most notably the next $k$-invariant, $k^5$.

  • Antieau and Williams show this is not trivial, and identify the other independent elements, namely $w_2^3$ and $(Sq_1 w_2)^2$. In particular, $k^5$ is not either of those.

  • Might it be something like $w_2 p_1 (mod 2)$? Note that by the description of $H^*(K(\mathbb{Z}/2,2),\mathbb{Z}/2)$, it can’t come from a class there, like the other two basis elements for $H^6(BSO(3),\mathbb{Z}/2)$ do.

• Note that $w_2\in \mathbb{Z}/2$ when working on $\mathbb{CP}^2$. And so one has $p_1 = 0, 1 (mod 4)$.

• So if I have an $SO(3)$-bundle on $\mathbb{CP}^2$ which lifts to an $SU(2)$-bundle, i.e. $w_2=0$, then its pullback to $S^5$ trivialises.

• I would like to know what happens when I pull back an $SO(3)$-bundle with $w_2=1$ - will it never trivialise? Or depending on parity? Or always?

  • I think the answer must have something to do with cohomology operations, in particular the $k$-invariant $k^5\in H^6(K(\mathbb{Z}/2,2) \times_{K(\mathbb{Z}/4,4)} K(\mathbb{Z},4),\mathbb{Z}/2)$, or equivalently a map

$$K(\mathbb{Z}/2,2) \times_{K(\mathbb{Z}/4,4)} K(\mathbb{Z},4) \to K(\mathbb{Z}/2,6)$$

• The moduli space of $SO(3)$-instantons of $p_1=k$ on $\mathbb{CP}^2$ has dimension $2k - 3$, and we can only have $k=0,1 (mod 4)$, so it is inahbited only for $k\geq 4$ (cf Buchdahl Instantons on $\mathbf{CP}^2$). In particular, one must take $k=5$ to get a non-liftable $SO(3)$-bundle.

• If the nontrivial $SO(3)$-bundle on $S^5$ arises as a pullback from $\mathbb{CP}^2$ of a bundle with $p_1\geq 5$, then there are $SO(3)$ instantons on $S^5$, and in particular, $SU(2)$ instantons, since the connection clearly lifts to an $SU(2)$ connection on the (unique) lifted bundle.

• Every $SO(3)=PU(2)$-bundle on $\mathbb{CP}^2$ can be lifted to a $U(2)$-bundle. The obstruction to restricting to an $SU(2)$-bundle is identical to the obstruction to lift the original bundle to such a thing.

• Then $w_2$ is the mod 2 reduction of the first Chern class of the $U(2)$ bundle (and presumably $p_1$ is $c_2$).

• We could then repeat the analysis using $U(2)$-bundles pulled back to $S^5$.

• The tangent bundle of $S^5$ is in some sense detected by the secondary cohomology class $\Phi_{(2)}$ of Petersen and Stein (_Annals_ 1962). In particular, it can see that $TS^5$ reduces to a rank-4 bundle (hence the $SU(2)$-bundle we know and love, but it can’t see this as far as I can tell), but doesn’t reduce to a rank-3 bundle.

• However, they only show that $\Phi_{(2)}$ is nontrivial by appealing to known results about non-existence of nonzero sections of the resulting rank-4 bundle, rather than calculating $\Phi_{(2)}$.

• Also, $\Phi_{(2)}(P) \in H^5(S^5,\mathbb{Z}/2)$ is the pullback along $S^5 \to BString(3)$ of the extension along $K(\mathbb{Z},4) \hookrightarrow BString(3)$ of $K(\mathbb{Z},4) \to K(\mathbb{Z}/2,4) \stackrel{Sq^2}{\to} K(\mathbb{Z}/2,6)$. This doesn’t seem like it could detect a nontrivial $w_2$ down on $\mathbb{CP}^2$!

• One thing Nick suggested is to see if one can build an arbitrary $U(2)$-bundle on $\mathbb{CP}^2$ by deleting a point and gluing two trivial bundles along a ball surrounding it, twisted in some sense by the (nontrivial) line bundle given by $S^5 \to \mathbb{CP}^2$, which is the extension of the Hopf fibration, and what $c_1$ detects (and which survives to give a nontrivial $w_2$, as an $SO(3)$-bundle).

• Given such an explicit realisation, one could then pull this back, along with the gluing data, to $S^5$ and see if it becomes trivial there.

• The nontrivial $U(2)$-bundle on $S^5$ is $U(3) \to S^5$.

• One could ask whether there is descent data for this bundle to $\mathbb{CP}^2$, for instance depending on an integer, which would become (if this worked) $c_2$ of the descended bundle.

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Two options: either a bundle survives, in which case I know there is an instanton, or no bundle survives, in which case if I find an instanton it would be potentially a new result about the moduli space of such on $S^5$.