Algebraic Topology – Group Structure on CP^infinity

at.algebraic-topologylie-groups

I was inspired by the following algebraic topology orals question:

"Is $S^1$ the loop space of another space?"

This is easy to see if you recognize that $S^1$ is a $K(\mathbb{Z},1)$, and the loop space of any $K(G,n)$ is a $K(G,n-1)$.

I then also remembered that the loop space functor is a functor from pointed topological spaces and continuous maps to the category of H-spaces and continuous homomorphisms. H-spaces being topological spaces that satisfy the axioms of a group up to homotopy (see Spanier, Chapter 1, Section 5).

I have three questions:

  1. Is there a useful criterion for when an H-space is actually a topological group?
  2. Seeing that $S^1$,$S^3$, and $S^7$ are the only spheres that support group structures, it doesn't seem coincidental that $S^1$ is a loop space, because it is in fact an H-space. Since $CP^{\infty}$ is the loop space of $K(Z,3)$ it too is an H-space, but is it known if it is a topological group?
  3. Even if not, is there a way (other than concatenation of loops) to "see" this structure on $CP^{\infty}$?

Thanks!

Best Answer

Here's a few thoughts on your questions.

  1. See algori's answer. (Incidentally, the "necessity is clear" step is because if $G$ is a topological group then it has a classifying space and then $G \simeq \Omega B G$, hence is homotopy equivalent to a loop space.)

  2. For $CP^\infty$, here's a construction that makes it a topological group. Take the unitary group on a Hilbert space, $U(H)$. This is contractible by Kuiper's theorem (MR0179792). The centre of this group is the circle, $S^1$, acting by diagonal operators. As this is normal, the quotient $PU(H) = U(H)/S^1$ is a topological group. Since $U(H)$ is contractible, this is a $K(\mathbb{Z},2)$ and hence "is" $CP^\infty$.

  3. The group structure on $CP^\infty$ can be viewed in a nice way using the fact that $CP^\infty$ represents $H^2(-,\mathbb{Z})$ and that $H^2(-,\mathbb{Z})$ classifies complex line bundles. Both of these views of $[-,CP^\infty]$ have obvious groups structures: for $H^2(-,\mathbb{Z})$ it is addition whilst for complex line bundles it is given by tensor product (the inverse operation is complex conjugation). (These two group operations are the same operation under the correspondence, by the way). As these are natural operations, they are represented by a group structure on the representing space, making $CP^\infty$ a group object in the $hTop$. That this is the "correct" group structure (aka, that coming from $CP^\infty \simeq K(\mathbb{Z},2) \simeq \Omega K(\mathbb{Z},3)$) is clearest from the characterisation of $CP^\infty$ as the representing space for $H^2(-,\mathbb{Z})$. The equivalence $CP^\infty \simeq \Omega K(\mathbb{Z},3)$ comes from the suspension isomorphism, $H^2(-,\mathbb{Z}) \cong H^3(\Sigma -, \mathbb{Z})$ which is additive and hence preserves the group structures on the representing spaces.

Related Question