Question

A lambda-ring $R$ is called "special" if it satisfies the $\lambda^i\left(xy\right)=...$ and $\lambda^i\left(\lambda^j\left(x\right)\right)=...$ relations, or, equivalently, if the map $\lambda_T:R\to\Lambda\left(R\right)$ given by $\lambda_T\left(x\right)=\sum\limits_{i=0}^{\infty}\lambda^i\left(x\right)T^i$ (where the $\sum$ sign means addition in $R\left[\left[T\right]\right]$, not addition in $\Lambda\left(R\right)$) is a morphism of lambda-rings. If you are wondering what the hell I am talking about, most likely you belong to the school of algebraists that denote only special lambda-rings as lambda-rings at all.

Anyway, let $A$ and $B$ be two special lambda-rings, and for every $i>0$, let $\Psi_A^i$ and $\Psi_B^i$ be the $i$-th Adams operations on $A$ and $B$, respectively. Let $f:A\to B$ be a ring homomorphism such that $f\circ\Psi_A^i=\Psi_B^i\circ f$ for every $i>0$. Does this yield that $f$ is a lambda-ring homomorphism, i. e. that $f\circ\lambda_A^i=\lambda_B^i\circ f$ for every $i>0$ ?

Note that this is clear if both $A$ and $B$ are torsion-free as additive groups (i. e., none of the elements $1$, $2$, $3$, ... is a zero-divisor in any of the rings $A$ and $B$), but Hazewinkel, in his text Witt vectors, part 1 (Lemma 16.35), claims the same result for the general case. I am writing a list of errata for his text, and I would like to know whether this should be included - well, and I'd like to know the answer anyway, as I am writing some notes on lambda-rings as well.

For the sake of completeness, here is a definition of Adams operations: These are the maps $\Psi^i:R\to R$ for every integer $i>0$ (where $R$ is a special lambda-ring) defined by the equation

$\sum\limits_{i=1}^{\infty} \Psi^i\left(x\right)T^i = -T\frac{d}{dT}\log\left(\lambda_{-T}\left(x\right)\right)$ in the ring $R\left[\left[T\right]\right]$ for every $x\in R$.

Here, even if the term $\log\left(\lambda_{-T}\left(x\right)\right)$ may not make sense (since some of the fractions $\frac{1}{1}$, $\frac{1}{2}$, $\frac{1}{3}$, ... may not exist in $R$), the logarithmic derivative $\frac{d}{dT}\log\left(\lambda_{-T}\left(x\right)\right)$ is defined formally by

$\displaystyle \frac{d}{dT}\log\left(\lambda_{-T}\left(x\right)\right)=\frac{\frac{d}{dT}\lambda_{-T}\left(x\right)}{\lambda_{-T}\left(x\right)}$.

Answer 1

Here is a more general point of view on Charles's example, which someone might find helpful.

Let $M$ be an abelian group, and let $\mathrm{Z}[M]$ denote $\mathrm{Z}\oplus M$ with the usual split ring structure $(a,m)(a',m')=(aa',am'+a'm)$. There is a simple description of the (special) $\lambda$-ring structures on such rings. For any prime number $p$, let $\theta_p$ denote the symmetric function $(\psi_p-e^p)/p$, where $e=x_1+x_2+\cdots$ and $\psi_p=x_1^p+x_2^p+\cdots$. Observe that $\theta_p$ has integral coefficients and therefore defines a natural operation on any $\lambda$-ring, and for any element $x$ in any $\lambda$-ring, we have $\psi_p(x)=x^p+p\theta_p(x)$. In particular, for any $\lambda$-structure on $\mathrm{Z}[M]$, the Adams operation $\psi_p$ satisfies $\psi_p(m)=p\theta_p(m)$ for all $m\in M$.

Exercise: Given a commuting family of additive maps $\theta_p:M\to M$, there is a unique $\lambda$-ring structure on $\mathrm{Z}[M]$ whose $\theta_p$ operators on $M$ are the given ones. Conversely, suppose $\mathrm{Z}[M]$ has a $\lambda$-ring structure. Then each $\theta_p$ preserves the ideal $M$, and resulting map $\theta_p:M\to M$ is additive.

Now it is a general fact that a ring map between two $\lambda$-rings is a $\lambda$-ring map if and only if it commutes with the $\theta_p$ operators. (This is because such symmetric functions and all their compositions under plethysm generate the whole ring of symmetric functions.) Thus, in our case, a $\lambda$-ring map $\mathrm{Z}[M]\to\mathrm{Z}[N]$ is the same as a linear map $M\to N$ that commutes with each $\theta_p$. But that is not the same as commuting with each $p\theta_p$, i.e. the Adams operations, and it is easy to make counterexamples. For instance Charles's counterexample is with $M=N=\mathrm{Z}/2\mathrm{Z}$, where $\theta_p$ on $M$ is the identity for all $p$, but $\theta_p$ on $N$ is the identity for all odd $p$ but is zero for $p=2$. The identity map $M\to N$ therefore commutes with each $p\theta_p$ and so is an Adams map. It commutes with each $\theta_p$ with $p$ odd, but it doesn't commute with $\theta_2$. Therefore it is not a $\lambda$-map.

Answer 2

I think the following gives an counterexample.

First, here are two Adams-ring structures on the truncated polynomial ring $Z[x]/(x^2)$. Let $A'=Z[x]/(x^2)$ with Adams operations defined by $\Psi_{A'}^m(x)=mx$ for all $m\geq1$. Let $B'=Z[x]/(x^2)$ with Adams operations defined by $\Psi_{B'}^m(x)=mx$ if $m$ is odd, and $\Psi_{B'}^m(x)=0$ if $m$ is even. Since $A'$ and $B'$ are torsion free, Wilkerson's congruence criterion says that $A'$ and $B'$ both admit unique lambda ring structures compatable with the given Adams operations (i.e., both satisfy $\Psi^p(x)\equiv x^p$ modulo $p$).

If I'm not mistaken, this gives $$ \lambda_{T,A'}(x) = 1 +xT-xT^2+xT^3-xT^4+\cdots $$ $$ \lambda_{T,B'}(x) = 1 +xT + xT^3+ xT^5+\cdots $$

Now let $A=A'/(2x)$ and $B=B'/(2x)$ as rings, so both are isomorphic as rings to $Z[x]/(x^2,2x)$. Clearly the Adams ring structures descend to $A$ and $B$, and there is an isomorphism $f: A\to B$ of Adams rings. If we can show that the $\lambda$-ring structures on $A'$ and $B'$ descend to $A$ and $B$, then it's clear that $f$ is not a morphism of $\lambda$-rings.

It's certainly true that $A$ is a $\lambda$-ring (for instance, it's isomorphic to $K^0RP^2$, the complex $K$-theory of real projective plane). I believe $B$ is also a $\lambda$-ring; it should suffice to check that $\lambda_{T,B'}(u)\equiv \lambda_{T,B'}(u+2k\,x)$ modulo $(2x)$ for all $u\in B'$ and integers $k$, whence $\lambda_{T,B'}$ descends to a well-defined function on $B$. Since $\lambda_{T,B'}(2x)=(1+x(T+T^3+T^5+\cdots))^2 = 1+2x(T+T^3+T^5+\cdots)$, this appears to be the case.