# A gluing construction for complex projective space

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.

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.