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Suppose I have an operad $P(n)$ in topological spaces in the sense of the book by Markl, Shnider and Stasheff, i.e. there is no space $P(0)$. In agreement with some of the answers to this question, I will call such an operad unreduced.

Let $H$ be an infinite dimensional separable Hilbert space. Then an example for $P$ would be

$$ P(n) = Iso(H^{\otimes n}, H) $$

the linear isometric isomorphisms from $H^{\otimes n}$ to $H$. There is no obvious candidate for $P(0)$ (since we insist on isomorphisms, $P(0) = Hom(\mathbb{C},H)$ will not work).

It is still possible to talk about unreduced $E_{\infty}$-operads in this case. But now it is less clear (at least to me) that I can deloop a space, which is an algebra over an unreduced $E_{\infty}$-operad. Hence my question:

Let $X$ be a space, which is an algebra over an unreduced $E_{\infty}$-operad $P$, such that $\pi_0(X)$ is a group. Is $X$ an infinite loop space?

If this is not true:

Are there conditions on the space or the operad (other than that it can be reduced) to ensure that $X$ is an infinite loop space?

Using the bar construction by May, I think this would correspond to the question, whether I can drop degeneracies and still get a delooping.

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Embed what you call unreduced operads into operads by taking $P(0)=\varnothing$. –  Fernando Muro Sep 4 '12 at 12:29
@Fernando: I can do that, but the delooping I know needs that $P(0)$ is a point, not empty. –  Ulrich Pennig Sep 4 '12 at 12:50

2 Answers 2

up vote 15 down vote accepted

The answer to your first question is no: for example, take $X$ to be any connected pointed space, and regard $X$ as a nonunital commutative monoid by saying that the product of any two points of $X$ is equal to the base point. This endows $X$ with the structure of an algebra over any operad $\mathcal{O}$ with $\mathcal{O}(0) = \emptyset$, but $X$ need not be an infinite loop space.

However, with a slightly stronger hypothesis, the answer is yes. Let $X$ be a nonunital $E_{\infty}$-space with the property that translation by any point $x \in X$ is a weak homotopy equivalence from $X$ to itself. Then $X$ is weakly homotopy equivalent to an infinite loop space.

Assuming that your operads are sufficiently nice (so that "algebras up to coherent homotopy" can be rectified), one has the following more precise statement: the homotopy category of $E_{\infty}$-spaces is equivalent to the subcategory of the homotopy category of "nonunital" $E_{\infty}$-spaces, whose objects are nonunital $E_{\infty}$-spaces $X$ for which $\pi_0 X$ contains a unit element $e_X$ such that translation by (any representative of) $e_X$ is a weak homotopy equivalence, and whose morphisms are maps of nonunital $E_{\infty}$-spaces $X \rightarrow Y$ which carry $e_{X}$ to $e_{Y}$.

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That is interesting. Just to clarify the terminology: Is a "non-unital $E_{\infty}$-space" the same thing as an algebra over an unreduced $E_{\infty}$-algebra? –  Ulrich Pennig Sep 4 '12 at 16:35
Where can I find a proof of this? –  Ulrich Pennig Sep 4 '12 at 16:36
By nonunital E-infinity space, I mean an algebra over some operad O where O(n) is contractible for n > 0, and O(0) is empty. There is a proof (in the homotopy coherent context) in my book "Higher Algebra": see Corollary –  Jacob Lurie Sep 4 '12 at 19:13
Another way of putting it is that in the absence of units the condition $\pi_0(X)$ is a group" should be replaced by the condition the shear map $(x,y) \mapsto (xy, y) :X \times X \to X \times X$ is an equivalence". –  Oscar Randal-Williams Sep 4 '12 at 19:55
Thank you for the reference. This really looks promising. Suppose I have a space $X$ that satisfies your condition. Is there an explicit construction of the delooping? –  Ulrich Pennig Sep 4 '12 at 20:20

Hey Ulrich, you sent me essentially that question by email, and I answered you just now. (Its earlier here, I just got up). So here publicly is the answer I sent you privately.

I know the situation. Let me refer you to Section 7 of a recently posted paper by Guillou and myself where certain operads P, Q, and R are defined and related. The operad R is analogous to your example and embedded in operad Q. The operad Q allows monos as well as isos.


If you can play a trick like that, and you have an analog of our Q that acts on your P-algebras, then you are fine. Otherwise I have no ideas.

Dealing with isomorphisms intrinsically gets you into that unreduced situation, and then you have no unit elements in your P-algebras, which means you can't do infinite loop space theory. I would bet that there are operads like P, maybe your P itself, which act on spaces that even have non-Abelian fundamental groups. Just a hunch.

Sorry not to be more helpful. Of course, I liked your question.

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