Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

For an ideal $I \subset R$ with relative K-groups $K_i(R,I)$ we have an exact sequence

$K_2(R) \to K_2(R,I) \to K_2(R/I) \to K_1(R) \to K_1(R,I) \to K_1(R/I)$

$\to K_0(R) \to K_0(R,I) \to K_0(R/I)$

This is reminiscent of the relative homotopy exact sequence and suggests - perhaps not alone - that we are dealing with homotopy groups. Let's say $K_i(R) = \pi_i(X_R)$ for some space $X_R$.

There also exist 'multiplication' maps $K_0 \times K_0 \to K_0$, $K_1 \times K_0 \to K_1$, $K_0 \times K_1 \to K_1$, $K_1 \times K_1 \to K_2$, $K_2 \times K_0 \to K_2$, $K_0 \times K_2 \to K_2$ (as given in Milnor's book), which suggests that $\pi_{\bullet}(X_R)$ should have a graded ring structure.

Now, if $X_R$ is an H-space, then its multiplication induces a ring structure on its homology, but does this relate to the homotopy as well?

How might one reason that $X_R$ ought to be an H-space, given that $\pi_{\bullet}(X_R)$ gives a graded ring?


What sort of classes of 'nice' spaces naturally have a graded ring structure on their homotopy?

share|improve this question
In fact the $K$-groups of a ring are naturally modelled as the homotopy groups of a spectrum. (If you want to stick with spaces, you can in turn model this as an infinite loop space, i.e. a space which can be expressed as the form $\Omega^nX_n$ for some space $X_n$, for every $n$.) When you're starting with a ring, the spectrum you get is always a ring spectrum, which is the spectrum version of an $H$-space. I don't actually know why this is true, but I'd guess it pops out of one of the various constructions of this spectrum. –  Paul VanKoughnett Mar 22 '13 at 2:32
That isn't really an answer to your questions, so for now let me just make an insinuation: what could you mean by a space 'naturally' having a multiplication on its homotopy groups, other than this being induced by a multiplication on the space itself? –  Paul VanKoughnett Mar 22 '13 at 2:34
Detail, Paul. You want a commutative ring for the second statement. I'm afraid I haven't time to give the full intuition, but a short answer is that infinite loop space machines take symmetric monoidal categories (like projective modules over a ring under direct sum) to spectra and take symmetric bimonoidal categories (like projective modules over a commutative ring under sum and tensor product) to commutative ring spectra. –  Peter May Mar 22 '13 at 2:40
@Peter: Thanks for the nod. May you please elaborate when you do find the time? –  Joshua Seaton Mar 22 '13 at 2:46
add comment

1 Answer

up vote 5 down vote accepted

For based spaces $X$ and $Y$ there is a map $\pi_i(X)\times \pi_j(Y)\to \pi_{i+j}(X\wedge Y)$ given by $$ (a:S^i\to X,b:S^j\to Y)\mapsto (a\wedge b:S^i\wedge S^j\to X\wedge Y).$$ This function of $a$ (resp. $b$) is a homomorphism if $i$ (resp. $j$) is positive. Thus a space-level multiplication of the form $X\wedge X\to X$ gives a graded ring structure to $\pi_\ast(X)$ (except maybe on $\pi_0$ and $\pi_1$).

In particular this is so if $X=\Omega^\infty E$ is the infinite loopspace of a spectrum and you have a spectrum map $m: E\wedge E\to E$. In this case you have abelian groups $\pi_i(E)$ for all $i\in \mathbb Z$, not just $i\ge 2$, and $m$ makes this a graded ring.

Note: smash product, not cartesian product.

Edit: Here is the promised elaboration. This is extremely sketchy. It would probably be pointless to try to avoid mentioning spectra here.

When you have a spectrum you have a very special sort of $H$-space, an infinite loopspace. If $X$ is a loopspace, say $X=\Omega Y$, then this gives a group structure to $\pi_0X$ and makes the group $\pi_1X$ abelian (and has lots of other consequences, too, for the homotopy type of $X$). If $X$ is a double loopspace, the loopspace of a loopspace, then it is an even more special kind of loopspace. $\pi_0X$ is now abelian. In fact, the multiplication $X\times X\to X$ is now commutative up to homotopy. When you have a spectrum $E$, you have a sequence of spaces $E_n$ in which $E_n$ is (at least weakly homotopy equivalent to) $\Omega E_{n+1}$. The homotopy groups of the spectrum are defined by $\pi_kE=\pi_{k+n}E_n$. The space $E_0$ is then homotopy-commutative in a very strong sense.

You can think of infinite loopspaces as a generalization of topological abelian groups. Let's think of the 'multiplication' in such an $H$-space as addition, because we now want to think of the possibility of another binary operation called multiplication that makes this generalized abelian group into a generalized ring. A ring spectrum is a spectrum $E$ together with a map of spectra $E\wedge E\to E$. The smash product $E\wedge F$ of two spectra is somewhat awkward to define. Its homotopy groups can be described by saying that $\pi_k(E\wedge F)$ is the direct limit of $\pi_{k+m+n}(E_m\wedge F_n)$. A map $E\wedge F\to G$ gives you a map $\pi_iE\otimes \pi_jF\to \pi_{i+j}G$. The homotopy groups of a ring spectrum form a graded ring. As a very special case, suppose that $A$ is a simplicial abelian group, i.e. a simplicial set with a suitable addition law. It is an infinite loopspace: you can deloop it by explicitly writing down another such object; if you replace simplicial abelian groups by nonnegatively graded chain complexes using the Dold-Kan equivalence, then it's just a matter of regrading, shifting the chain groups by one dimension. (The homology groups of the chain complex are the homotopy groups of the realization of the simplicial set.)

As a very special case: If $A$ is not just a simplicial abelian group but a simplicial ring, then the spectrum becomes a ring spectrum. The multiplication in homotopy groups can be described either by writing down a multiplication on the chain complex level or by noting that the multiplication $A\times A\to A$ takes $A\times 0\cup 0\times A$ to $0$ and so gives a map $|A|\wedge |A|=|A\wedge A|\to |A|$. (Now look at the beginning of my answer above.)

More generally, if $X$ is the infinite loopspace $E_0$ of a ring spectrum $E$, then you always get a map $E_0\wedge E_0\to E_0$.

The algebraic $K$-groups of a ring are in fact the homotopy groups of an infinite loopspace, or of a spectrum. You make the spectrum by using a category of $R$-modules. If $R$ is commutative then the spectrum is a ring spectrum. The multiplication is related to tensor product of $R$-modules.

share|improve this answer
Thanks a lot Tom. I'm unfamiliar with spectra. Is there an H-space lurking in the background somewhere? –  Joshua Seaton Mar 23 '13 at 20:37
I'll try to explain a little in an edit to the answer. –  Tom Goodwillie Mar 23 '13 at 21:11
Great, that would be much appreciated. –  Joshua Seaton Mar 25 '13 at 1:07
@Tom: A friendly reminder: please update when you can. –  Joshua Seaton Mar 26 '13 at 14:18
@Tom: Thanks for taking the time to write this out. It's helped a great deal. –  Joshua Seaton Mar 26 '13 at 22:27
show 6 more comments

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.