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Any Eilenberg-MacLane space $K(A,n)$ for abelian $A$ can be given the structure of an $H$-space by lifting the addition on $A$ to a continuous map $K(A\times A,n)=K(A,n)\times K(A,n)\to K(A,n)$.

Does somebody know an explicit way to describe this structure in the cases $K({\mathbb Z}/2{\mathbb Z},1)={\mathbb R}P^\infty$ and $K({\mathbb Z},2)={\mathbb C}P^{\infty}$?

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3 Answers 3

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Look at $\mathbb R^\infty\setminus 0$ as the space of non-zero polynomials, which you can multiply. Pass to the quotient to construct the projective space and, from the multiplication, its $H$-space product.

The complex case is quite the same.

NB: Jason asks in a comment below if this is the same $H$-space structure that Hanno had in mind. To check, we can use the fact that Hanno's is characterised by the fact that if $\mu:K(\mathbb Z_2,1)\times K(\mathbb Z_2,1)\to K(\mathbb Z_2,1)$ is his product and $\alpha\in H^1(K(\mathbb Z_2,1), \mathbb Z_2)$ is the class represented by the identify map $K(\mathbb Z_2,1)\to K(\mathbb Z_2,1)$, then $\mu^\*(\alpha)=\alpha\times 1+1\times\alpha$. One should be able to check that this holds for the map given by multiplication of polynomials in a very small skeleton.

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    $\begingroup$ As a footnote, the construction does not extend to the quaternionic case since commutativity of multiplication of coefficients is needed in order for the multiplication of polynomials to be well defined modulo scalar multiplication. In the complex case, if you only factor out by scalar multiplication by numbers that are a real number times a p-th root of unity, you get an H-space structure on an infinite-dimensional lens space, a $K({\mathbb Z}_p,1)$. $\endgroup$
    – Allen Hatcher
    Commented Jan 8, 2010 at 7:48
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    $\begingroup$ Thank you, Mariano & Allen! This is really beautiful, and the structure you described is even strictly associative and unital. What about the (homotopy) inversion of this H-space structure - is there a nice way to describe it, too? $\endgroup$
    – Hanno Becker
    Commented Jan 8, 2010 at 10:14
  • $\begingroup$ Is it somehow obvious that this H-space structure and the one Hanno was talking about are the same? $\endgroup$
    – Jason DeVito - on hiatus
    Commented Jan 8, 2010 at 16:41
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    $\begingroup$ In the complex case, you can also use the fundamental theorem of algebra to replace $\mathbb{CP}^\infty$ to the infinite symmetric power of $\mathbb{CP}^1$. $\endgroup$
    – Reid Barton
    Commented Jan 8, 2010 at 17:17
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    $\begingroup$ The H-space structure in a $K(A,n)$ is unique up to homotopy since homotopy classes of maps $K(A,n)\times K(A,n) \to K(A,n)$ correspond bijectively with homomorphisms $A\times A \to A $, and the H-space condition says the homomorphism restricts to the identity on each factor so it is just the addition operation in the abelian group $A$. $\endgroup$
    – Allen Hatcher
    Commented Jan 8, 2010 at 17:57
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There's also a different way of writing down the $H$-space structure, that I like for its algebro-geometric flavor. (I'll talk about $\mathbb{C}P^\infty$ here, and $\mathbb{R}P^\infty$ should be analogous.)

Regarding $\mathbb{C}P^\infty$ as a classifying space for complex line bundles, we know that this $H$-space structure is supposed to implement "tensor product of line bundles". In a (not very explicit) sense this tells us the homotopy class of $\mathbb{C}P^\infty \times \mathbb{C}P^\infty \to \mathbb{C}P^\infty$: It represents the line bundle $\mathcal{O}(1,1) = p_1^* \mathcal{O}(1) \otimes p_2^* \mathcal{O}(1)$. We can use this description to write down a much more explicit (and classical) explicit representative.

First, let's recall what the analogous picture looks like for finite projective spaces. The line bundle $\mathcal{O}(1,1)$ determines (upon picking generating sections) the Segre map $\mathbb{C}P^n \times \mathbb{C}P^m \to \mathbb{C}P^{nm+n+m}$ which takes (in homogeneous coordinates)

$([X_0:\ldots:X_n] , [Y_0:\ldots:Y_m]) \mapsto[X_0 Y_0: \ldots : X_i Y_j: \ldots: X_n Y_m]$

where I'm choosing to be vague on the precise ordering of the coordinates. (In the end this won't matter up to homotopy, as the maps will become homotopic upon composing with $\mathbb{C}P^{nm+n+m} \hookrightarrow \mathbb{C}P^\infty$.)

The analogous formula with infinitely many homogeneous coordinate makes just as much sense, one just has to a good ordering of pairs of non-negative integers. Such an infinite Segre map gives another realization of the $H$-space structure.

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  • $\begingroup$ Thank you, Anatoly! I also thought of the Segre map, but didn't notice that the ordering of the factors doesn't matter up to homotopy, therefore struggled when trying to check associativity. $\endgroup$
    – Hanno Becker
    Commented Jan 8, 2010 at 10:27
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John Baez has a nice expository page about this called Classifying spaces made easy. About two thirds down the page he talks about multiplicative structure on $\mathbb{C}P^\infty$

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    $\begingroup$ Baez is talking about a different version of ${\mathbb C}P^\infty$ from the one topologists usually consider. Namely he takes nonzero rational functions with coefficients in ${\mathbb C}$ modulo scalar multiplication. The rational functions form an infinite dimensional vector space over ${\mathbb C}$, but of uncountable dimension since all the functions $1/(z+a)$ are linearly independent as $a$ ranges over ${\mathbb C}$. This gives a fatter version of ${\mathbb C}P^\infty$ that's actually an abelian group. $\endgroup$
    – Allen Hatcher
    Commented Jan 9, 2010 at 16:59

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