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Jan 8, 2010 at 17:57 comment added Allen Hatcher 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 $A$.
Jan 8, 2010 at 17:17 comment added Reid Barton In the complex case, you can also use the fundamental theorem of algebra to replace $\mathbb{CP}^\infty$ to the infinite symmetric power of $\mathbb{CP}^1$.
Jan 8, 2010 at 17:14 history edited Mariano Suárez-Álvarez CC BY-SA 2.5
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Jan 8, 2010 at 16:41 comment added Jason DeVito - on hiatus Is it somehow obvious that this H-space structure and the one Hanno was talking about are the same?
Jan 8, 2010 at 10:29 vote accept Hanno Becker
Jan 8, 2010 at 10:15 vote accept Hanno Becker
Jan 8, 2010 at 10:27
Jan 8, 2010 at 10:14 comment added Hanno Becker 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?
Jan 8, 2010 at 7:48 comment added Allen Hatcher 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 $K({\mathbb Z}_p,1)$.
Jan 8, 2010 at 7:38 history edited Mariano Suárez-Álvarez CC BY-SA 2.5
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Jan 8, 2010 at 7:31 history answered Mariano Suárez-Álvarez CC BY-SA 2.5