Status: Questions 2 and 4 answered in the negative. Questions 1 and 3 ARE STILL UNANSWERED, despite previous claims.

On the third page of Wolfang K. Seiler's paper "lambda-rings and Adams operations in algebraic K-Theory" (pp. 93-102 in Beilinson's Conjectures), it is claimed that the Grothendieck ring of representations of a given group $G$ over a given commutative ring $A$ modulo exact sequences is a special $\lambda$-ring. Here, "representations" means finitely-generated projective $A$-modules with a $G$-module structure. ("Special $\lambda$-ring" is what most people just call "$\lambda$-ring", as opposed to "pre-$\lambda$-ring".)

1. Does anybody know a proof of this? I don't understand the sketched proof in 1, and I don't have the reference (R. G. Swan, A splitting principle in algebraic $K$-theory, in: Representation theory of finite groups and related topics, Proc. Symp. Pure Math. 21 (1971), 155-159) either. I have been told that Atiyah-Tall has a proof, but I am very skeptical about it (and if it has a proof, then I am unable to find it amid all the geometry).

I know how this is proven for the case of a characteristic $0$ field (yes, this is in Atiyah-Tall, but it is trivial using characters). My problem is to show this over arbitrary commutative rings.

2. What if we take the Grothendieck ring modulo split exact sequences (i. e., modulo direct sums) rather than modulo all exact sequences? Is this still a special $\lambda$-ring? What if we work over some field? What if the group is finite?

Update on 2: This is easily seen to be wrong, even if the field is $\mathbb F_2$ and the group is the $2$-element group. In fact, applying Krull-Schmidt and considering the Jordan decomposition of the matrix $\left(\begin{array}{cc} 1&1\\ 0&1\end{array}\right)$, we easily see that the $\lambda^2\left(uv\right)=\left(\lambda^1\left(u\right)\right)^2\lambda^2\left(v\right)+\left(\lambda^1\left(v\right)\right)^2\lambda^2\left(u\right)-2\lambda^2\left(u\right)\lambda^2\left(v\right)$ equation is not satisfied if both $u$ and $v$ are the representation of the $2$-element group on $\mathbb F_2^2$ in which the generator acts as $\left(\begin{array}{cc} 1&1\\ 0&1\end{array}\right)$.

3. Can we hope for any reasonable results if we replace the group by a cocommutative bialgebra? Note that cocommutativity is required to define the exterior powers of an $H$-module (where $H$ is our bialgebra). What if the bialgebra is Hopf?

4. What if we remove the projectivity condition on representations of $G$? I know that removing the finite-generatedness condition is a very bad idea (in fact, it makes the Grothendieck group collapse because of the Eilenberg swindle), but I have seen nobody talking about the Grothendieck group of finitely generated but not-necessarily-projective modules. Is it that embarrassingly stupid?

Update on 4: Question 4 has now been answered by Ben Wieland in the comments to this post.

Oh, and I have already asked this question over the "field" $\mathbb F_1$. The answer is "no" to all three questions in this case. But I am interested in "real" rings and fields this time.

EDIT: I might be understanding what Seiler is doing in 1, but in this case, he is doing it wrong, so I think I am not understanding him.

If I am understanding him right, Seiler's somewhat cryptic formulation "where the module induced by $M$ itself has a one-dimensional quotient, given by the linear functions on $S\left(M\right)$" means that we consider the surjection $M\otimes S\left(M\right) \to S\left(M\right)$ which sends $m\otimes m_1m_2...m_k$ to $mm_1m_2...m_k$, and its kernel is then a projective $S\left(M\right)$-module with $\lambda$-dimension one less than that of $M\otimes S\left(M\right)$ (because for every exact sequence $0\to U\to V\to W\to 0$ and every $k\geq 0$, we have the equality $\left[\wedge^k V\right] = \sum\limits_{i=0}^k \left[\wedge^i U \otimes \wedge^{k-i} W\right]$ in $K_0$). Unfortunately $M\otimes S\left(M\right)$ and $S\left(M\right)$ are not quite $S\left(M\right)$-modules with $G$-representations, because $G$ does not act $S\left(M\right)$-linearly on them (unless I did something wrong). Also we need a device to conclude that two projective $R$-modules $V$ and $W$ with $G$-module structure are equal in $K_0$ if and only if the corresponding $S\left(T\right)$-modules $V\otimes S\left(T\right)$ and $W\otimes S\left(T\right)$ with $G$-module structure are equal in $K_0$, where $T$ is some projective $R$-module; this looks like it should follow from the gradedness of symmetric algebras, but I don't see how (mostly because I have no idea what equality in $K_0$ actually means in terms of modules).

  • 2
    $\begingroup$ 4: Finitely generated but not projective (or at least perfect) modules are not closed under derived tensor product. eg, $A=k[t]/t^2$, $M=k$, then $Tor_n(M,M)=k$ for all $n$. So you have a Grothendieck group but it isn't a ring, let alone a $\lambda$-ring. $\endgroup$ Mar 11, 2013 at 4:06
  • $\begingroup$ Hmm... What is derived tensor product? Why is it important? $\endgroup$ Mar 11, 2013 at 4:44
  • 2
    $\begingroup$ Tensor product is not exact, so it is not defined on the equivalence classes in $K$-theory. The main motivation for derived tensor product is to make it exact. One application is that if $A$ is regular noetherian, $K$ of fg $A$-modules is a ring. A more elementary way is to say that $K$ of fg equals $K$ of projective, where the latter has a ring structure by regular tensor product. (I'm not answering the question "what?") $\endgroup$ Mar 11, 2013 at 5:04
  • $\begingroup$ Ah!! This explains the focus on projectivity. One more question solved, thank you! $\endgroup$ Mar 11, 2013 at 5:29
  • $\begingroup$ This is probably a dumb question: If $M$ is a projective $A$-module, and $2$ is not invertible in $A$, how do I know that $\bigwedge^2 M$ is projective? $\endgroup$ Nov 6, 2015 at 16:48

1 Answer 1


After wondering about this question, I went and looked up Swan's paper. His argument is basically the one Seiler is trying to say; it also uses a paper by J. Burroughs: http://www.ams.org/mathscinet-getitem?mr=269719, which I have not looked at. According to Swan, his result only applies to the case when $A$ is noetherian for technical reasons, but that the general case would be discussed more generally elsewhere.

In the case of interest, suppose $C$ is the category of $G$ representations over $A$. Given an object $M$ of $C$, you are supposed to consider a category $C'$. In this case, this is a category of graded $S(M)$-modules with compatible $G$-action; that is the objects are graded $S(M)$-modules $Q$, together with a $G$-action on $Q$ such that the action map $S(M)\otimes Q\to Q$ is $G$-equivariant; here $G$ is acting in the evident non-trivial way on $S(M)$.

Seiler's (and Swan's) map $M\otimes S(M)\to S(M)$ is a morphism in $C'$, so you get a splitting relation of the form $[K]+[S(M)]=[M\otimes S(M)]$ in $K_0(C')$.

Swan then has an argument that the map $K_0(C)\to K_0(C')$, induced by tensoring up, is injective. I find it hard to parse, and don't really understand it. It involves a map of the form $$\theta: K_0(C')\to K_0(C)[[t]]/K_0(C)[t],$$ defined by sending $[Q]$ to $\sum [Q_k]t^k$. (It's necessary to quotient by $K_0(C)[t]$, in order to get something that takes exact sequences to sums.) The composite of $\theta$ with $K_0(C)\to K_0(C')$ is then shown to be injective, somehow using the fact that $\sum [S^n(M)]t^n$ has a multiplicative inverse in $K_0(C')$.

  • $\begingroup$ Thanks. The "compatible $G$-action" thing is what I was having troubles with; Seiler loses no words about it, so I assumed he thought it would be the trivial action. I hope that this compatible $G$-action does not interfer with taking exterior powers and tensor products. $\endgroup$ Oct 14, 2011 at 23:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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