# How commutative is Quillen's Plus-Construction?

This question is inspired by this question about the dependence of K-theory on the order of multiplication in the ring. I did not think long about it, so maybe the answer really lies on the surface; but I do not know.

Let $G$ be a discrete group, $G^{op}$ its opposite group (i.e. the one with reversed multiplication). Let $H \subset G$ be a subgroup, such that $H=[H,H]=[G,G]$. Note that $H^{op} \subset G^{op}$ has the same properties. I denote the Quillen Plus-Construction of a space with fundamental group $G$ with respect to $H$ by $X \mapsto X^+_H$

Question: Is there a homotopy equivalence between $BG_H^+$ and $B(G^{op})_{H^{op}}^+$, such that the induced map on $\pi_1$ is induced by identity $id: G \to G^{op}$.

Note that there is clearly a homotopy equivalence between $BG_H^+$ and $B(G^{op})_{H^{op}}^+$ such that the induced map on $\pi_1$ is induced by the inverse $inv: G \to G^{op}$; but that is not the one I am looking for.

However, this shows that one could also ask:

Question: Is there a homotopy equivalence between $BG_H^+$ and itself, such that the induced map on the abelian group $\pi_1(BG_H^+) = G/H$ is the inversion.

Given the motivation, any good answer in the case $G=GL_{\infty}(R)$ (for some ring) would be interesting too.

EDIT: Johannes Ebert suggested in a comment a strategy to give a negative answer to Question 2 in general. The Kan-Thurston construction gives a way of obtaining every (finite?) cell complex $X$ as the plus-construction on $BG$ with respect to some perfect subgroup $H$. It seems that the only missing piece is a finite cell-complex $X$ with $\pi_1$ abelian and no $\pi_1$-inversion-inducing self-homotopy equivalence. Then, $X = BG^+_H$ for some group $G$ and some perfect subgroup $H \subset G$. However, since $\pi_1 = G/H$ is abelian, we see that $[G,G] \subset H$ and hence, $[H,H] = [G,G]=H$ as required.

Question: Can anybody give an example of such a space?

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I think that the second question for $GL_\infty(R)$ follows directly from the fact that we have a connected $H$-space. –  Torsten Ekedahl Nov 19 '10 at 12:11
I see, every connected H-space has an inverse up to homotopy. This solves the problem for $BGL_{\infty}(R)^+$. Thanks for this comment. However, the $H$-space structure is not directly induced by the multiplication of $GL_{\infty}(R)$, so that there is probably no reason to assume that $BG^+_H$ will be an $H$-space in general. –  Andreas Thom Nov 19 '10 at 12:38
Of course not every plus-construction is an $H$-space, in fact, it seems to me that an $H$-space structure on $BG^+$ is a rather rare event. The Kan-Thurston theorem says that for any space $X$, there is a group $G$ and a homology isomorphism $BG \to X$, which will be a plus-construction. Take a simply connected $X$ that is manifestly not an $H$-space ($S^2$, for example) and you find a perfect group that does not have an $H$-space structure on its plus-construction. –  Johannes Ebert Nov 19 '10 at 16:23
In fact, I would use Kan-Thurston to construct a counterexample to question 2. 1. Find a space with abelian $\pi_1$ that does not admit a self-map inducing the inversion on $\pi_1$. I do not have a counterexample at hand, but would be \emph{really} surprised if such a space does not exist. Step 2. Apply Kan-Thurston to this space and check that the Kan-Thurston map is a plus construction (I believe that this is not obvious if there is some $H_1$ around). –  Johannes Ebert Nov 19 '10 at 16:29
@Johannes: The Kan-Thurston map as stated in their paper is in fact a plus construction on $\pi_1$, though indeed it's not always true that a map of spaces inducing a homology isomorphism and a surjection on fundamental groups is a plus construction. –  Tyler Lawson Nov 19 '10 at 18:42

To answer the final question, here is how you can construct a space with no self-equivalence inducing negation on (abelian) $\pi_1$.
If $X \to Y$ is a homotopy equivalence, then for any basepoint it induces isomorphisms $\pi_n(X) \to \pi_n(Y)$ commuting with the action of $\pi_1$. In particular, if we can construct abelian groups $A$ and $B$ with an action of $A$ on $B$ such that there is no isomorphism $\phi: B \to B$ such that ${}^a \phi(b) = \phi({}^{(-a)}b)$, then we can use these to construct the desired space; let $X$ be the homotopy orbit space $K(B,2) \times_A EA$, which has $\pi_1 = A$ and $\pi_2 = B$ with the given action.
If $B$ is cyclic, all endomorphisms commute and so it suffices to construct an action so that ${}^a b = {}^{(-a)}b$ does not hold for all $a, b$.
"Minimal" examples include the action of $\mathbb{Z}/4$ on $\mathbb{Z}/5$ with generator $x \mapsto 2x$ (not isomorphic to $x \mapsto 3x$) and the action of $\mathbb{Z}/3$ on $\mathbb{Z}/7$ with generator $x \mapsto 2x$ (not isomorphic to $x \mapsto 4x$).