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As far as I can understand, $E_\infty$ ring spaces are homotopy commutative H-spaces. If we have a space, and we have an action of an $E_\infty$-operad on it, then we say that it has a homotopy commutative multiplication?

This seems to be what people say, however, there must be more right? In the wikipedia article on $E_\infty$-operads, it is mentioned that what "$E_\infty$" gives us is a multiplication that is associative and commutative up to "all higher homotopies." It also says that an $E_2$ space is a homotopy commutative $A_\infty$ space, i.e. if all we want is a commutative multiplication, up to homotopy, then we're done after the second level. What is the significance of all these higher homotopies? I used to think the point was that we just wanted a way to get homotopy commutative spaces.

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Have a look at mathoverflow.net/questions/57589/…. Basically we want the homotopies themselves to satisfy certain relations, and so on, so on –  Juan S Nov 18 '11 at 9:40
    
Thanks. I guess my question is, I know what these relationships are, how they are parameterized essentially by discs, etc. but I don't know why we would want this. –  Jon Beardsley Nov 18 '11 at 11:49
    
Erm not discs actually, I was thinking of Associahedra, but the question remains. –  Jon Beardsley Nov 18 '11 at 11:55
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Another link - try en.m.wikipedia.org/wiki/Highly_structured_ring_spectrum –  Juan S Nov 19 '11 at 0:14
    
This is a fantastic article. Please post it as an answer so that I can mark it as the answer! –  Jon Beardsley Nov 19 '11 at 4:17
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up vote 2 down vote accepted

As per the comments this is explained very well on the wikipedia article http://en.m.wikipedia.org/wiki/Highly_structured_ring_spectrum.

Basically the notions of $ E_\infty$ and $A_\infty$ spectra give additional structure to the spectrum, that allow for useful constructions (e.g. topological Hochschild homology)

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