# Understanding $\mathbb{C}P^2$

I am trying to understand $$\mathbb{C}P^2$$. Since I understand the Hopf fibration quite well, I like the following construction:

• Attach a $$\mathbb{D}^2$$ (2-cell) to a point $$\mathbb{D}^0$$ (0-cell) to get $$S^2$$ (thanks to @Leo Mosher for suggesting the more sensible order of attaching)
• Now attach $$\mathbb{D}^4$$ to $$S^2$$ by gluing the boundary $$\partial \mathbb{D}^4=S^3$$ to $$S^2$$ using the projection map P of the Hopf bundle $$S^1 \hookrightarrow S^3 \rightarrow S^2$$

I picture the result as a $$S^2$$ bundle over $$S^2$$ which I know is $$\textbf{wrong}$$*.

Here is my reasoning. Where did I go wrong?

1. As $$S^3 = \partial \mathbb{D}^4$$ is a $$S^1$$ bundle over $$S^2$$, the "interior" $$\mathbb{D}^4$$ is a $$\mathbb{D}^2$$ bundle over $$S^2$$ (by just "filling" every $$S^1$$)
2. Gluing the boundary $$\partial \mathbb{D}^4 = S^3$$ to $$S^2$$ via the projection of the Hopf fibraton amounts to gluing the boundary of every fiber (ie $$\partial \mathbb{D}^2 = S^1$$) together, giving us $$S^2$$ at every point.

I think that 2. probably does not work, ie that the boundaries of the smoothly over $$S^2$$ varying $$\mathbb{D}^2$$'s (making up $$\mathbb{D}^4$$) cannot be glued together to give smoothly varying $$S^2$$'s at every point, but that at one point this has to break. That would mean that $$\mathbb{C}P^2 - \{\text{point}\}$$ is a $$S^2$$ bundle.

Can someone help me clarify, where my reasoning is wrong? I would also gladly appreciate any other insight into $$\mathbb{C}P^2$$.

Thank you!

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* There are only two $$S^2$$ bundles over $$S^2$$: One being the trivial one, and the other one being a twisted one. Can someone maybe also confirm or reject my image of this twisted $$S^2$$ bundle:

• Imagine $$SO(3)$$ as $$S^1$$ bundle over $$S^2$$ (which is the unit tangent bundle of $$S^2$$ denoted $$T_1S^2$$).
• The interior is obtained by filling those $$S^1$$'s giving us a $$\mathbb{D}^2$$ bundle over $$S^2$$
• Now glue together the boundaries of every fiber $$\partial \mathbb{D}^2 = S^1$$ to obtain a sphere a $$S^2$$ at every point.

This gives a pretty "simple" visual of the twisted $$S^2$$ bundle over $$S^2$$.

$$\textbf{EDIT:}$$ I imagine the breaking via stereographic projection of $$S^3\subset \mathbb{R}^4$$ to $$\mathbb{R}^3$$: The projection gives tori filling $$\mathbb{R}^3$$. Every fiber $$S^1$$ becomes a Villarceau circle. One can move these circles to the points on $$S^2$$ they will end up under the Hopf map. The circle that ends up at the north pole is degenerated to a line (which is of course just an artifact of the projection). One cannot picture this fiber as an acual circle, without loosing the picture of the $$S^1$$'s varying smoothly. Now I can imagine filling all of these $$S^1$$'s to become $$\mathbb{D}^2$$'s including the degenerated one, which (if this is possible at all) has remain a degenerated line in order to vary smoothly (necessary not sufficient though). If I further glue the boundaries of these $$\mathbb{D}^2$$'s to get $$S^2$$'s at every point, I can (because $$\mathbb{C}P^2$$ is not a bundle) get smoothly varying $$S^2$$'s except for the degenerate $$\mathbb{D}^2$$ at the north pole somehow...

• In your opening construction it is better to say Attach $\mathbb D^2$ to $\mathbb D^0$ than the other way around. Feb 16 at 13:52
• $\mathbb D^4$ is not a bundle over $S^2$. It is not true, that $\mathbb D^4$ is obtained by filling each $S^1$-fiber of $S^3\to S^2$ Feb 16 at 13:54

This is best explained in the language of vector bundle, where you are observing a bunch of crucial phenomena. The Hopf fibration $$S^1\rightarrow S^3\rightarrow S^2$$ is the sphere bundle of the tautological (complex line) bundle $$\mathbb{C}\rightarrow H\rightarrow\mathbb{CP}^1$$, granted we identify $$\mathbb{CP}^1$$ and $$S^2$$ via stereographic projection. The idea of "filling every fiber" works and yields the disk bundle $$D^2\rightarrow D(H)\rightarrow\mathbb{CP}^1$$ of the tautological bundle. Its total space $$D(H)$$ is not a $$D^4$$, so point 1. is wrong. The issue, geometrically, is that filling every fiber with a disk results in having one "origin" in each fiber, whereas $$D^4$$ only has one "origin" total. This informs us that the relationship between $$D(H)$$ and $$D^4$$ is that the latter is obtained from the former by identifying all these origins. In proper language, this means that collapsing the zero section $$s$$ of the disk bundle yields $$D(H)/s(\mathbb{CP}^1)=D^4$$. In general, collapsing the zero section of a vector bundle's disk bundle yields the cone over its sphere bundle (at least if the base is compact). The converse of this process, taking a $$D^4$$ in a $$4$$-manifold and replacing it with $$D(H)$$ (i.e. replacing the origin with a whole copy of $$\mathbb{CP}^1$$) is an instance of what's called a "blow-up".
Now, having corrected point 1., point 2. actually becomes salient. Forming the adjunction space $$E:=D(H)\cup_{S^3}S^2$$ along the Hopf map corresponds to collapsing the boundary in each fiber separately, so yields a fiber bundle $$S^2\rightarrow E\rightarrow S^2$$, which turns out to be the non-trivial $$S^2$$-bundle over $$S^2$$ (note this really is the same as the description you give in your post). Now, collapsing the zero section of this bundle (the order does not make a difference here) yields $$\mathbb{CP}^2=D^4\cup_{S^3}S^2$$. This can be interpreted as saying that $$E$$ is the blow-up of $$\mathbb{CP}^2$$ along a point. Note also that removing this point of $$\mathbb{CP}^2$$ demonstrates that $$\mathbb{CP}^2-\{pt\}$$ is a $$S^2-\{pt\}$$-bundle over $$S^2$$ (rather than an $$S^2$$-bundle). In fact, you can show that the projection $$\mathbb{C}^3\rightarrow\mathbb{C}^2$$ onto the last coordinate yields induces a bundle $$\mathbb{CP}^2-\{[0\colon0\colon1]\}\rightarrow\mathbb{CP}^1$$ isomorphic to the dual of the tautological bundle.
• Thanks for the instructive question! I think intuition ultimately comes from playing around with examples. To grok these "fiber-wise" constructions, I found it helpful to understand how they're described in terms of a fiber bundles structure group and transition functions. This naturally leads to translating everything back-and-forth to and from principal bundles (that's also why I know the $S^2$-bundle I give is non-trivial, by the way). For a recommendation in this direction, I like these notes. Feb 16 at 15:52
• The constructions like disk/sphere bundles and also the projectivization of a vector bundle are good examples to familiarize oneself with. There's a nice proof that $\mathbb{RP}^{2n+1}$ is null-bordant by "filling in disks" along the bundle $\mathbb{RP}^{2n+1}\rightarrow\mathbb{CP}^n$ that is worth learning (there's a subtlety: every sphere-bundle bounds topologically, but not necessarily smoothly). The last point in my answer is describing what is called the "Thom space" of $H$, which is also worth looking up. This stuff should in parts be in "Characteristic Classes" by Milnor & Stasheff. Feb 16 at 15:55
Your reasoning goes wrong in your point number $$1$$. Just because you have fibered $$S^3$$ by disjoint $$S^1$$'s, you cannot assert without justification that you can extend this to a fibering of $$\mathbb D^4$$ by disjoint $$\mathbb D^2$$'s. And if you attempted to justify this assertion, you would run into this problem: if $$D_1,D_2 \subset \mathbb D^4$$ are two properly embedded discs with boundary circles $$C_1,C_2 \subset S^3$$, then the linking number of $$C_1$$ and $$C_2$$ in $$S^3$$ is $$0$$; but the linking number of any two fibers of the Hopf fibration is $$1$$.