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In the paper "Smooth $S^1$ actions on homotopy complex projective spaces and related topics" By Ted Petrie (Bulletin of the AMS, 1972), a certain claim is made on page 149, in the proof of lemma 4.4. There is no proof of the claim in the paper as far as I can tell.

Before stating it, it will be convenient to set one notation. Consider $S^{3}$ as the unit sphere in $\mathbb{C}^{2}$ and let $f: S^{3} \rightarrow SU(2)$ be the diffeomorphism $$f(z) = \begin{bmatrix} z_{0} & z_{1} \\ -\bar{z_{1}} & \bar{z_{0}} \end{bmatrix} ,$$ for $z = (z_{0},z_{1}) \in S^3$.

The claim is as follows.

Claim: Consider two copies of $S^2 \times D^4$, glued along their boundary $S^{2} \times S^{3}$. The gluing function is $H: S^{2} \times S^{3} \rightarrow S^{2} \times S^{3}$ given by $H(u,z) = (uf(z),z)$. Then, the resulting manifold is diffeomorphic to $\mathbb{CP}^{3}$.

Question: How to prove the claim?

I am also interested to see how the gluing hypersurface $S^{2} \times S^{3} $ sits inside $\mathbb{CP}^{3}$ explicitly.

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Identifying complex 4-space $\mathbb{C}^4$ with the quaternionic plane $\mathbb{H}^2$ in the standard way we get an identification of the unit spheres

$$S^7\cong S(\mathbb{C}^4)\cong S(\mathbb{H}^2).$$

With this done we have Hopf fibrations

$$\gamma=\gamma^\mathbb{C}_3:S^7\rightarrow\mathbb{C}P^3,\qquad\lambda=\lambda^\mathbb{H}_1:S^7\rightarrow \mathbb{H}P^1\cong S^4$$

where we identify $\mathbb{H}P^1$ with $S^4$ in the usual way. Then the fibring $\gamma$ has the structure of an $S^1\cong S(\mathbb{C})$-principal bundle and $\lambda$ has the structure of an $S^3\cong S(\mathbb{H})\cong SU_2$-principal bundle. If we're careful when making our identifications we can arrange for things to make sense, so that under the isomorphism $\mathbb{C}^4\cong\mathbb{H}^2$ the $S^3=S(\mathbb{H})$ action on $\mathbb{H}^2$ becomes the standard $SU_2$-action on $\mathbb{C}^4$. Then the $S^1$-action on $\mathbb{C}^4$, as multiplication by complex numbers of modulus $1$, becomes the action of the subgroup $S^1\leq S^3$.

The point is that the subgroup inclusion now induces a map of orbit spaces

$$\pi:S^7/S^1\rightarrow S^7/S^3$$

which is a locally trivial fibration with fibre $S^3/S^1$. When we identify $\mathbb{C}P^3\cong S^7/S^1$ and $S^4\cong S^7/S^3$, as well as $S^3/S^1\cong S^2$, the map above becomes the fibre bundle

$$S^2\hookrightarrow \mathbb{C}P^3\xrightarrow\pi S^4.$$

With this we have given $\mathbb{C}P^3$ the structure of a sphere bundle over a sphere. Everything we did was actually smooth (all groups are Lie and all actions are smooth), so the standard theory of such gadgets in the smooth category gives rise to a diffeomorphism of $\mathbb{C}P^3$ with the clutching construction you describe in your question.

To make everything completely explicit you just need to find bundle charts for $\pi$ over the two hemispheres $D^4_\pm\subseteq S^4$. Since these are just disks, and hence contractible, bundle charts always exist. With your choice of local trivialisations there are induced maps from the pushout $(S^2\times D^4_+)\cup_{S^2\times S^3}(S^2\times D^4_-)$ to $\mathbb{C}P^3$. Chasing through this you find exactly how $S^2\times S^3$ sits inside $\mathbb{C}P^3$. Since $S^2\times 1\subseteq S^2\times S^3$ is just the fibre over the basepoint, this slice is easily seen to be a copy of $S^2\cong\mathbb{C}P^1$ sitting inside $\mathbb{C}P^3$ in the standard way.

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