Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

I've been reading Hatcher's survey "A Short Exposition of the Madsen-Weiss Theorem". In it, he discusses the Barratt-Priddy-Quillen theorem, which says that the homology of the infinite symmetric group is the same as the homology of one component of $\Omega^{\infty} S^{\infty}$. My question : what space exactly is being referred to in this notation? It would appear that he intends to take the limit of $\Omega^n S^n$ as $n$ goes to infinity; however, I don't see a natural inclusion $\Omega^n S^n \subset \Omega^{n+1} S^{n+1}$, so I don't know how to make sense of this. Another possible interpretation would be to take the limit of $\Omega^n S^{\infty}$ as $n$ goes to infinity, but again I don't know a natural inclusion $\Omega^n S^{\infty} \subset \Omega^{n+1} S^{\infty}$.

share|improve this question

1 Answer 1

up vote 13 down vote accepted

Your first statement is correct: it is taking a colimit of $\Omega^n S^n$. The inclusion from one to the next is called the "suspension" map. The points of $\Omega^n S^n$ are basepoint-preserving functions $S^n \to S^n$, and the suspension map takes such a function $f$ and sends it to $f \wedge id: S^n \wedge S^1 \to S^n \wedge S^1$.

This is easier to describe in terms of another model. $S^n$ is homeomorphic to the space obtained by taking $[0,1]^n$ and identifying the boundary to a single point. Under this identification, you could describe a function $S^n \to S^n$ using coordinates as a function $$(x_1,\ldots,x_n) \mapsto f(x_1,\ldots,x_n),$$ and the suspension map sends this to the function $$(x_1,\ldots,x_n,x_{n+1}) \mapsto (f(x_1,\ldots,x_n), x_{n+1})$$ Of course, you have to check that this preserves the equivalence relation if $f$ does and so on.

Most models of $S^\infty$ are contractible, and hence so are $\Omega^n S^\infty$; these are definitely not what you want. The notation is not to be taken literally; at best, $S^\infty$ should denote an "infinite suspension" operator which doesn't actually produce spaces as output.

share|improve this answer
    
Thanks! This is very helpful! –  George Oct 27 '11 at 5:00

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.