Timeline for Eckmann-Hilton for $A_{\infty}$-spaces?
Current License: CC BY-SA 3.0
11 events
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| Nov 12, 2011 at 16:40 | comment | added | Ulrich Pennig | Thank you for the quick answer and the proof. I feel a little stupid now: I actually knew that theorem, but somehow was too blind to think about the connectivity of $BG$ :-). | |
| Nov 12, 2011 at 13:28 | comment | added | Tyler Lawson |
$BG$ is always grouplike because it is connected. An H-space which is a CW-complex for which $\pi_0$ is a group is always grouplike. You can check this as follows. The "shear" map $H \times H \to H \times H$ given by $(x,y) \mapsto (xy,y)$ is a homotopy equivalence, because: (a) it is an isomorphism on $\pi_0$ (b) using the multiplication makes it suffices to check the identity component (c) using the projection to second factor, the shear map is a self-map of the fibration of connected spaces $H_e \times H_e \to H_e$ which is the identity on both base and fiber, hence is a weak equiv.
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| Nov 12, 2011 at 12:46 | comment | added | Ulrich Pennig | One last question about your second argument: You say, that both $BG$ and $BG_*$ are grouplike. But I only know this for G itself. Why is this true for $BG$ and $BG_*$? | |
| Nov 8, 2011 at 23:05 | comment | added | Ulrich Pennig | Thanks again. Eckmann-Hilton really seems to be a "buy one, get one free"-condition :-)! | |
| Nov 8, 2011 at 18:57 | comment | added | Tyler Lawson |
@Ulrich: Yes. An H-space structure on G is the same as a lift of the functor $X \mapsto [X,G]$ to monoids; it factors through a group-valued functor if and only if a homotopy inverse exists. Since you're assuming the H-space structure is the same on both ends, you do have that homotopy commutativity and a homotopy inverse exist for your $A_\infty$-structure.
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| Nov 8, 2011 at 18:52 | comment | added | Ulrich Pennig | Shouldn't it be possible to deduce the existence of homotopy inverses from the Eckmann-Hilton condition as well? We know that G is a group: Can we run through the Eckmann-Hilton clock (up to homotopy) to deduce that homotopy inverses exist for the H-space structure? | |
| Oct 31, 2011 at 16:37 | vote | accept | Ulrich Pennig | ||
| Oct 31, 2011 at 16:37 | comment | added | Ulrich Pennig | Thanks Tyler! I think the higher compatibility conditions also hold in the case I am considering. That counterexample is very enlightening. | |
| Oct 29, 2011 at 5:31 | history | edited | Tyler Lawson | CC BY-SA 3.0 |
counterexample
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| Oct 28, 2011 at 19:26 | history | edited | Tyler Lawson | CC BY-SA 3.0 |
i am an idiot
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| Oct 28, 2011 at 16:16 | history | answered | Tyler Lawson | CC BY-SA 3.0 |