Someone once said to me, “$\mathbb{C}P^\infty$ is an abelian group”. Here is a summary of what I’ve learned in trying to make sense of that statement. Most of the ideas are from John Baez’s expository note on classifying spaces .

The Eilenberg-MacLane “space” $K(\mathbb{Z}, 2)$ is uniquely determined by any of several equivalent properties:

  • $\pi_2(K(\mathbb{Z}, 2)) = \mathbb{Z}$ and all other homotopy groups are trivial

  • The space of based loops in $K(\mathbb{Z}, 2)$ is homotopy equivalent to the circle: $\Omega K(\mathbb{Z}, 2) \simeq S^1$.

  • For any (reasonable) space $X$, homotopy classes of maps from X to $K(\mathbb{Z}, 2)$ are classified by elements of the second cohomology: $[X, K(\mathbb{Z}, 2)] = H^2(X; \mathbb{Z})$

  • It is the classifying space for principal circle bundles: $K(\mathbb{Z}, 2) \simeq BS^1$.

I put “space” in quotes because $K(\mathbb{Z},2)$ is not a space, but a homotopy class of spaces. Any space in this homotopy class satisfies all the above properties, and what one usually refers to as $\mathbb{C}P^\infty$ is just one of these spaces. So, while it is true that $K(\mathbb{Z}, 2)$ is an abelian topological group (see here, for example), it is not necessary that a given construction of $\mathbb{C}P^\infty$ should have a group structure. Or if it does, that this product should be the one induced by $\mathbb{Z}$ (or $S^1$).

According to the answers here, $\mathbb{C}P^\infty$ is homeomorphic to a topological abelian group, but I have no idea how to construct an actual map that gives a comprehensible product.

One way to construct $\mathbb{C}P^\infty$ is as the infinite symmetric product of $\mathbb{C}P^1$: define the finite symmetric product

\[\begin{aligned} SP^n := (\mathbb{C}P^1)^n / S_n, \end{aligned}\]

where the symmetric group $S_n$ acts in the obvious manner, and then defined the infinite symmetric product

\[\begin{aligned} SP := \text{colim}_n SP^n. \end{aligned}\]

We can think of elements of $SP^n$ as unordered tuples of elements of $\mathbb{C}P^1$. In fact $SP^n$ is isomorphic to $\mathbb{C}P^n$. Here’s how: for a collection $(\alpha_1, \ldots, \alpha_n)$ of points on $\mathbb{C}P^1$, define the polynomial $p(x) = \Pi_{\alpha_i\neq \infty} (x - \alpha_i)$ (where the empty product is the constant polynomial 1), and suppose $p(x) = \sum_i a_i x^i$. Then the map

\[\begin{aligned} SP^n &\xrightarrow{\sim} \mathbb{C}P^n\\ (\alpha_1, \ldots, \alpha_n) &\mapsto [a_0:\ldots:a_n] \end{aligned}\]

is an isomorphism because $\mathbb{C}$ is algebraically closed. It follows that the colimit $SP$ is isomorphic to $\mathbb{C}P^\infty$. An element of $\mathbb{C}P^\infty$ is then an unordered tuple of points on $\mathbb{C}P^1$, only finitely many of which are not $\infty$ (this is because all of these constructions are of based spaces, and I chose $\infty$ to be my basepoint without explicitly saying so). The multiplication of polynomials gives a natural product

\[\begin{aligned} SP^m \times SP^n &\to SP^{m+n}\\ (\alpha_1, \ldots, \alpha_m), (\beta_1, \ldots, \beta_n) &\mapsto (\alpha_1, \ldots, \alpha_m, \beta_1, \ldots, \beta_n) \end{aligned}\]

which is commutative because all tuples are unordered. This induces the same product on $\mathbb{C}P^\infty$ (where we just ignore all the $\infty$’s), and by defining $SP^0 := {\infty}$ we can endow $\mathbb{C}P^\infty$ with the structure of a commutative topological monoid. There is nothing special about $\mathbb{C}P^1$ in this construction: any infinite symmetric product has this property. This product, however, does not have an inverse, but might have one up to homotopy. I don’t know if it does. (Should it?)

Is the product at least induced by the product on $\mathbb{Z}$? The easiest way to check this, at least for me, is by using the fourth property, that $\mathbb{C}P^\infty$ is a classifying space for principal circle bundles. Consequently, it is also a classifying space for complex line bundles. So the product on $\mathbb{C}P^\infty$ must correspond to the tensor product of line bundles. It suffices to check this for the tautological bundle on $\mathbb{C}P^\infty$. What is this bundle? A consequence of the preceding construction of $\mathbb{C}P^\infty$ is that it can also be realized as the projectivization of $\mathbb{C}[t]$, and in this setting, the product $\mu:\mathbb{C}P^\infty\times \mathbb{C}P^\infty\to \mathbb{C}P^\infty$ is just multiplication of polynomials. The fiber of the tautological bundle $\mathcal{O}(-1)$ over any unordered tuple $(\alpha_1, \ldots, \alpha_n)$ is the one-dimensional space of all polynomials vanishing at those (and only those) points.

We can now restate our question as: is $\mu^\ast\mathcal{O}(-1) \simeq p_1^\ast\mathcal{O}(-1)\otimes p_2^\ast\mathcal{O}(-1)$, where $p_i$ is projection onto the $i^{th}$ factor? Yes, because if $p_j$ are polynomials which vanish at $(\alpha_1, \ldots, \alpha_n)$ and $q_j$ vanish at $(\beta_1, \ldots, \beta_m)$, then $\sum_j p_jq_j$ vanishes at $(\alpha_1, \ldots, \alpha_n, \beta_1, \ldots, \beta_m)$. So this product is one possible answer, modulo its lack of an inverse.

Baez uses a different construction of $K(\mathbb{Z}, 2)$ for his claim of an abelian group structure. Here, instead of taking the projectivization of $\mathbb{C}[t]$, we take that of $\mathbb{C}(x)$. An element of $\mathbb{C}(x)$ is determined up to scaling by its zeroes and poles, all of which live on $\mathbb{C}P^1$ ($\mathbb{C}(x)$ being the space of meromorphic functions on it), and so this new “$\mathbb{C}P^\infty$”, which I denote by $\Theta$, has a topology where two points are “close” if the corresponding zeroes and poles are close. $\mathbb{C}(x)$ has a product which descends to one on $\Theta$, as does the inverse. So $\Theta$ is a topological abelian group. But now the question is: is $\Theta$ homeomorphic to $\mathbb{C}P^\infty$, and if so, what does the induced product on $\mathbb{C}P^\infty$ look like? (By the way, how does one construct $\Theta$ as a CW complex?)

So, to summarize:

  • $K(\mathbb{Z}, 2)$ “is” an abelian group

  • $\mathbb{C}P^\infty$ has an obvious product, but that does not have an inverse

  • $\Theta$ is an abelian group, but I don’t know if $\Theta$ is homeomorphic to $\mathbb{C}P^\infty$

I will update this once I have more clarity on some of these ideas.