While reading through some issues of Baez's (wonderful) "This Week's Finds in Mathematical Physics," I came across this statement (from week 149):

$K(\mathbb{Z},2)$ is a bit more complicated: it's infinite-dimensional complex projective space, $\mathbb{C}P^{\infty}$! [...] you can define $\mathbb{C}P^{\infty}$ as a "direct limit" of these [$\mathbb{C}P^n$] as $n$ approaches infinity, using the fact that $\mathbb{C}P^n$ sits inside $\mathbb{C}P^{n+1}$ as a subspace. Alternatively, you can take your favorite complex Hilbert space $\mathcal{H}$ with countably infinite dimension and form the space of all 1-dimensional subspaces in $\mathcal{H}$. This gives a slightly fatter version of $\mathbb{C}P^{\infty}$, but it's homotopy equivalent, and it's a very natural thing to study if you're a physicist: it's just the space of all "pure states" of the quantum system whose Hilbert space is $\mathcal{H}$.

This is fairly standard (I've seen similar statements in numerous other places, I just haven't thought much about it), but I'm just wondering: are these two realizations $\mathbb{C}P^{\infty}$ merely homotopy equivalent? Or can more be said (e.g. homeomorphic, diffeomorphic, etc.)?

  • 1
    $\begingroup$ Well John Baez says that the second description is "fatter." I doubt he said that for no reason. $\endgroup$ Commented Jan 14, 2016 at 14:32
  • $\begingroup$ Sure, I know. I suppose another way of phrasing what I'm getting at here would be something along the lines of: can someone be a bit more precise about how one is "fatter" than the other. $\endgroup$ Commented Jan 14, 2016 at 14:33
  • 2
    $\begingroup$ I believe the point is that the two local models are not locally homeomorphic: the first model being $\mathbb{C}^\infty = ($direct limit of $\mathbb{C} \subset \mathbb{C}^2 \subset ...)$ and the second being $\mathcal{H}$. But I could not find a reference. The closest thing I could find is a paper of Klee and Kakutani showing that they are not isomorphic as topological vector spaces, because the first is of countable dimension and has the "finite topology" generated by its finite subspaces, and the second has uncountable dimension and therefore (as they prove) cannot have the finite topology. $\endgroup$
    – Lee Mosher
    Commented Jan 14, 2016 at 16:21

1 Answer 1


The $CP^n$'s constructed as the direct limit are a $CW$ complex. Hence $CP^\infty$ is the countable union of closed subspaces with empty interior (in $CP^\infty$). However, a Hilbert space (and also Hilbert manifolds) are Baire. The countable union of closed subspaces with empty interior have empty interior hence cannot be the whole space. So these are not homeomorphic.

Another way of seeing the "fatness" is yet another model. We can also see $CP^\infty$ as the space of lines in $\mathbb{C}^\infty$. This is the vector space of sequences $(a_1,\ldots,)$ such that all but a finite number of $a_1$ are zero. This $\mathbb{C}^\infty$ can be seen to sit inside the Hilbert space (after choosing a basis). However, in the Hilbert space there are "more points" namely ones such that $\sum_{i=1}^\infty |a_i|^2<\infty$. One can argue that $C^\infty$ is "thin" inside the Hilbert space, hence the space of lines in $C^\infty$ is thin inside the space of lines in $H$.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .