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!


* 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...

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

2 Answers 2


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.

  • 1
    $\begingroup$ 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. $\endgroup$
    – Thorgott
    Feb 16 at 15:52
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    $\begingroup$ 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. $\endgroup$
    – Thorgott
    Feb 16 at 15:55
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    $\begingroup$ (+1) IIRC, the last sentence (about the punctured complex projective plane) is true only by reversing orientation. Holomorphically, and as an oriented real four-manifold, manifold, the punctured plane is the total space of the hyperplane bundle, dual to the tautological bundle. $\endgroup$ Feb 16 at 16:16
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    $\begingroup$ @AndrewD.Hwang beat me to the catch. I noticed it a few hours ago but was waiting to be at my computer to write the comment. :) $\endgroup$ Feb 16 at 17:52
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    $\begingroup$ Thank you for the suggestions! Will definitely check it all out. $\endgroup$
    – Thomas
    Feb 19 at 9:36

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 $1$.


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