**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).

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