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Let us call a group object $G$ in a category $\mathcal C$ rigid, if it has the following property: For every group object $X$ in $\mathcal C$, every morphism $G\to X$ in $\mathcal C$ respecting the unit sections is already a morphism of group objects.

As a formal consequence, rigid group objects are commutative (because the inversion is a group object morphism).

This definition is motivated by the following fact: If $\mathcal C$ is the category of algebraic varieties over $\mathbb C$ (say), then the rigid group objects in $\mathcal C$ are precisely the complex abelian varieties. Indeed, that abelian varieties enjoy the rigidity property is a "standard fact", and the converse follows quickly from Chevalley's structure theorem (every algebraic group is an extension of an abelian variety by an affine group).

If $\mathcal C$ is the category of sets or of hausdorff topological spaces, there are no interesting rigid group objects. I am puzzled with the case where $\mathcal C$ is the category whose objects are the topological spaces which are homotopy equivalent to CW-complexes and morphisms are continuous maps up to homotopy. A group object in this category is commonly called $H$--group.

Is it true that the rigid $H$--groups are exactly the products of circles?

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Every space maps canonically (in the homotopy category) into some topological abelian group whose homology groups are the homotopy groups of the space. (Simplicial free abelian group generated by a simplicial set.) If this map is a homomorphism up to homotopy I wonder what restrictions that places on the homotopy type. –  Tom Goodwillie May 25 '11 at 13:25
I don't believe that even the product of two or more circles is rigid in this sense. –  Tom Goodwillie May 25 '11 at 13:56
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2 Answers

up vote 4 down vote accepted

Even the circle is not rigid in this sense! With such a weak notion of "group up to homotopy" you can make a multiplication on $G=K(\mathbb Z,1)\times K(\mathbb Z,2)$ such that there is no nontrivial homomorphism from $K(\mathbb Z,1)$.

Edit: Let me explain and generalize, showing that a point is the only example.

For $n>0$ and $F$ a field, take the product $G=K(F,n)\times K(F,2n)$ and define a multiplication $\mu:G\times G\to G$ as follows. Let $a\in H^n(G;F)$ and $b\in H^{2n}(G;F)$ be the canonical elements so that maps $f:X\to G$ correspond bijectively with pairs $(f^*a\in H^n(X;F),f^*b\in H^{2n}(X;F))$. Define $\mu$ by

$\mu^*a=a\times 1+1\times a$

$\mu^*b=b\times 1+1\times b +a\times a$.

This is unital. It is associative up to homotopy since both $\mu\circ (\mu\times 1)$ and $\mu\circ (1\times \mu)$ pull back $a$ to

$a\times 1\times 1+1\times a\times 1+1\times 1\times a$

and pull back $b$ to

$b\times 1\times 1+1\times b\times 1+1\times 1\times b+a\times a\times 1+a\times 1\times a+1\times a\times a$.

An inverse map exists because the shearing map $(x,y)\mapsto (x,\mu(x,y))$ is a homotopy equivalence.

Now if $(X,m:X\times X\to X)$ is any group up to homotopy and $\alpha\in H^n(X;F)$ any element, then the map $f:X\to G$ given by $f^*a=\alpha$ and $f^*b=0$ cannot be a homomorphism unless $\alpha=0$, since $\mu\circ (f\times f)$ pulls back $b$ to $f^*b\times 1+1\times f^*b+f^*a\times f^*a=\alpha\times \alpha$ while $f\circ m$ pulls back $b$ to $0$.

So $X$ cannot have any cohomology at all in positive dimensions, for any coefficient field. Since $\pi_1(X)$ must be abelian, this makes $X$ homotopically discrete, which as Neil pointed out makes it trivial.

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Let $A$ be any discrete abelian group. The classifying space $BA$ is then a topological abelian group. If $A=\mathbb{Z}^n$ then $BA$ is an $n$-torus, as in the question. For any connected based space $X$, we have $[X,BA]=H^1(X;A)=\text{Hom}(\pi_1(X),A)$. Now suppose that $X$ is an $H$-space. For any based map $f:X\to BA$ we can define $g:X^2\to BA$ by $g(x,y)=f(xy)f(x)^{-1}f(y)^{-1}$. This corresponds to a class in $H^1(X^2;A)$ that vanishes on $X\times\{1\}$ and on $\{1\}\times X$, but any such class is zero by the Kunneth Theorem, so $g$ is nullhomotopic, so $f$ is a homomorphism up to homotopy. Thus, $BA$ is rigid.

Conversely, suppose that $G$ is rigid and connected, and also that $G$ is associative up to coherent higher homotopies, so it has a classifying space $BG$ with $G\simeq \Omega BG$. Let $X$ be the topological group freely generated by $S^1$ modulo the relation that the basepoint is the identity. It is standard that this is weakly equivalent to the James construction, and thus that the suspension $\Sigma X$ is $\bigvee_{k=0}^\infty S^{k+1}$. We then have $$ [X,G] = [X,\Omega BG] = [\Sigma X,BG] = \prod_{k=0}^\infty \pi_{k+1}BG = \prod_{k=0}^\infty \pi_kG. $$ One can describe $[X^2,G]$ in similar terms and use it to understand which maps from $X$ to $G$ are homomorphisms up to homotopy. As $G$ is assumed to be rigid and connected, we can conclude that $\pi_kG=0$ for $k\neq 1$, so $G\simeq B(\pi_1(G))$. It is also standard that $\pi_1(G)$ is commutative for any $H$-space $G$, so we have a $BA$ as in the first paragraph.


As Tom pointed out, the question that I answered was the opposite way around from the question that was originally asked. So suppose instead that every map from $G$ to an $H$-space $X$ is a homomorphism up to homotopy. By taking $X=\mathbb{Z}$ we see easily that $G$ must be connected. Taking $X$ to be an Eilenberg-MacLane space $K(A,m)$, we see that every class in $H^m(G;A)$ is primitive. Now take $A=\mathbb{Q}$. As $H^{\ast}(G;\mathbb{Q})$ is a Hopf algebra, it is a tensor product of polynomial algebras and exterior algebras, and it is easy to check that the product of any two primitive elements is not primitive. From this one can show that $H^{\ast}(G;\mathbb{Q})$ is either $\mathbb{Q}$ or an exterior algebra on a single generator. Similarly, $H^{\ast}(G;\mathbb{Z}/p)$ is a tensor product of polynomial algebras, exterior algebras, and algebras of the form $\mathbb{Z}/p[x]/x^{p^m}$. Again, the only way that everything can be primitive is if $H^*(G;\mathbb{Z}/p)$ is either $\mathbb{Z}/p$ or an exterior algebra on a single generator. We can now use the universal coefficient theorem to see that $\tilde{H}_{\ast}(G)$ is a single copy of $\mathbb{Z}$, in degree $d$ say. Also $\pi_1(G)$ is abelian (because $G$ is an $H$-space) and so is isomorphic to $H_1(G)$. We now see using the Hurewicz theorem that $\pi_d(G)=\mathbb{Z}$ and that any generator of $\pi_d(G)$ is a homotopy equivalence $S^d\to G$. The only spheres that admit an associative H-space structure are $S^1$ and $S^3$ (ignoring the nonassociative structure on $S^7$). I don't believe that $S^3$ is rigid but I don't immediately see a proof.

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But the question was about rigidity in a dual sense, concerned with maps out of $G$ not into $G$! –  Tom Goodwillie May 25 '11 at 15:53
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