Consider the ring $W(\mathbb{F}_p)$ of big Witt vectors of $\mathbb{F}_p$. This has a natural structure of a $\lambda$-ring (in the strong sense) since rings of big Witt vectors always do.

$\mathbb{Z}_p$ also has the structure of a $\lambda$-ring, given by the maps $\lambda^n(x)= \binom{x}{n}$.

There is a canonical quotient map (of rings) $W(\mathbb{F}_p) \to \mathbb{Z}_p$ from the big Witt vectors to the classical $p$-adic Witt vectors. Is this a $\lambda$-ring homomorphism? More generally, when (if ever) do we get a $\lambda$-ring structure on the smaller $p$-adic Witt vectors?

I know this quotient map admits an additive section. We could potentially use this to define $\lambda$ operators on the ring of $p$-adic Witt vectors, but it is not clear to me that this would actually give the structure of a $\lambda$-ring. And in my particular case, it is unclear if this would agree with the $\lambda$-ring structure I already have.


I think the answer is ``No'' for $p$ odd. Stronger yet, there exists no $\lambda$-ring structure on $\mathbb{Z}_{p}$ under which the canonical quotient map $W\left( \mathbb{F}_{p}\right) \rightarrow\mathbb{Z}_{p}$ becomes a $\lambda$-ring morphism. However, due to the manifold occasions for mistakes and misunderstandings in my argument, please check it thoroughly.

Let $p$ be an odd prime. Let $\Lambda$ be the Hopf algebra of symmetric functions over $\mathbb{Z}$ (in the indeterminates $x_{1},x_{2},x_{3},\ldots $). It is known that $\Lambda$ carries further structures, such as plethysm, inner multiplication and inner comultiplication. I am using the material of Michiel Hazewinkel's Witt vectors, part 1 here (although not always his notations: my $\Lambda$ is his $\mathbf{Symm}$, and my $x_{1},x_{2} ,x_{3},\ldots$ are his $\xi_{1},\xi_{2},\xi_{3},\ldots$).

Fix a commutative ring $\mathbf{k}$. The $\lambda$-ring of big Witt vectors $W\left( \mathbf{k}\right) $ is defined as the set $\operatorname*{Ring} \left( \Lambda,\mathbf{k}\right) $ of all ring homomorphisms from $\Lambda\rightarrow\mathbf{k}$, endowed with an addition that comes from the comultiplication of $\Lambda$, an additive inverse that comes from the antipode of $\Lambda$, a multiplication that comes from the inner comultiplication of $\Lambda$, and a $\lambda$-ring structure that comes from the plethysm of $\Lambda$.

We let $p_{1},p_{2},p_{3},\ldots$ be the power sum symmetric functions in $\Lambda$ (that is, $p_{n}=\sum_{i\geq1}x_{i}^{n}$). I am sorry for using the letter $p$ which already stands for a prime, but this is a standard (and the $p$ in ``$p_{1},p_{2},p_{3},\ldots$'' always comes with a subscript).

We let $w_{1},w_{2},w_{3},\ldots\in\Lambda$ be the ``Witt vector coordinate'' symmetric functions in $\Lambda$; these are recursively defined by the power-series equality

$\prod_{d\geq1}\left( 1-w_{d}t^{d}\right) ^{-1}=\prod_{i\geq1}\dfrac {1}{1-x_{i}t}\in\mathbb{Z}\left[ \left[ x_{1},x_{2},x_{3},\ldots\right] \right] \left[ \left[ t\right] \right] $.

(These $w_{1},w_{2},w_{3},\ldots$ are called $x_{1},x_{2},x_{3},\ldots$ in §9.63 of Hazewinkel's notes. For a quick overview of their properties, see Exercise 2.79 in Victor Reiner and YHS, Hopf Algebras in Combinatorics, arXiv:1409.8356v3.) Let $\Lambda^{\prime}$ be the subring of $\Lambda$ generated by $w_{p^{0}},w_{p^{1}},w_{p^{2}},\ldots$. Then, the restriction map $\operatorname*{Ring}\left( \Lambda,\mathbf{k}\right) \rightarrow \operatorname*{Ring}\left( \Lambda^{\prime},\mathbf{k}\right) $ is (equivalent to) the canonical quotient map $W\left( \mathbf{k}\right) \rightarrow W_{p}\left( \mathbf{k}\right) $ from the big Witt vectors $W\left( \mathbf{k}\right) $ to the $p$-typical Witt vectors $W_{p}\left( \mathbf{k}\right) $.

It is well-known that $p_{n}=\sum_{d\mid n}dw_{d}^{n/d}$ for every positive integer $n$. In particular, $p_{p^{k}}=\sum_{i=0}^{k}p^{i}w_{p^{i}}^{p^{k-i}}$ for every $k\in\mathbb{N}$. From this, it is easy to see that for every $k\in\mathbb{N}$,

(1) there exists a positive integer $N$ such that $Nw_{p^{k}}$ is a polynomial in $p_{p^{0}},p_{p^{1}},\ldots,p_{p^{k}}$ with no constant term.

We now define two ring homomorphisms $\alpha:\Lambda\rightarrow\mathbf{k}$ and $\beta:\Lambda\rightarrow\mathbf{k}$ by setting

$\alpha\left( f\right) =f\left( 1,0,0,0,\ldots\right) $ and $\beta\left( f\right) =f\left( -1,0,0,0,\ldots\right) $ for every $f\in\Lambda$

(where the $1$ and the $-1$ are taken in $\mathbf{k}$). (Incidentally, $\alpha$ is the internal counit of $\Lambda$.)

Now, $\alpha+\beta$ is the ring homomorphism $\Lambda\rightarrow\mathbf{k}$ satisfying

$\left( \alpha+\beta\right) \left( f\right) =f\left( 1,-1,0,0,0,\ldots \right) $ for every $f\in\Lambda$.

Thus, it is easy to see that $\left( \alpha+\beta\right) \left( p_{i}\right) =0$ for every odd $i\geq1$. In particular, $\left( \alpha +\beta\right) \left( p_{p^{i}}\right) =0$ for every $i\in\mathbb{N}$. Thus, from (1) we see that $\left( \alpha+\beta\right) \left( w_{p^{k} }\right) =0$ for every $k\in\mathbb{N}$ as long as all positive integers are invertible in $\mathbf{k}$. But we can remove the ``all positive integers are invertible in $\mathbf{k}$'' condition (because proving $\left( \alpha+\beta\right) \left( w_{p^{k}}\right) =0$ for $\mathbf{k}=\mathbb{Z}$ is sufficient to ensure that $\left( \alpha +\beta\right) \left( w_{p^{k}}\right) =0$ holds for every $\mathbf{k}$). Hence, we obtain $\left( \alpha+\beta\right) \left( w_{p^{k}}\right) =0$ for every $k\in\mathbb{N}$. In other words, the canonical quotient map $W\left( \mathbf{k}\right) \rightarrow W_{p}\left( \mathbf{k}\right) $ sends $\alpha+\beta$ to the zero of the ring $W_{p}\left( \mathbf{k}\right) $.

On the other hand, recall that the $\lambda$-ring structure on $W\left( \mathbf{k}\right) $ comes from plethysm on $\Lambda$. More precisely,

$\left( \lambda^{n}\gamma\right) \left( f\right) =\gamma\left( e_{n}\left[ f\right] \right) $ for every $n\geq1$, $\gamma\in W\left( \mathbf{k}\right) $ and $f\in\Lambda$,

where $e_{n}\left[ f\right] $ denotes the plethysm of $f$ into $e_{n}$ (denoted by $e_{n}\circ f$ in Hazewinkel's notes). Applying this to $n=2$, $\gamma=\alpha+\beta$ and $f=p_{1}$, we obtain

$\left( \lambda^{2}\left( \alpha+\beta\right) \right) \left( w_{1}\right) =\left( \alpha+\beta\right) \left( \underbrace{e_{2}\left[ w_{1}\right] }_{=e_{2}}\right) =\left( \alpha+\beta\right) \left( e_{2}\right) $

$=e_{2}\left( 1,-1,0,0,0,\ldots\right) =-1$.

Assume now that $\mathbf{k}\neq0$. Thus, $\left( \lambda^{2}\left( \alpha+\beta\right) \right) \left( w_{1}\right) =-1\neq0$. Hence, the canonical quotient map $W\left( \mathbf{k}\right) \rightarrow W_{p}\left( \mathbf{k}\right) $ sends $\lambda^{2}\left( \alpha+\beta\right) $ to a nonzero element of the ring $W_{p}\left( \mathbf{k}\right) $.

Thus, the canonical quotient map $W\left( \mathbf{k}\right) \rightarrow W_{p}\left( \mathbf{k}\right) $ sends $\alpha+\beta$ to zero, but $\lambda^{2}\left( \alpha+\beta\right) $ to a nonzero element. Therefore, there exists no $\lambda$-ring structure on $W_{p}\left( \mathbf{k}\right) $ under which this canonical quotient map $W\left( \mathbf{k}\right) \rightarrow W_{p}\left( \mathbf{k}\right) $ becomes a $\lambda$-ring morphism. Now, applying this to $\mathbf{k}=\mathbb{F}_{p}$ and recalling that $W_{p}\left( \mathbb{F}_{p}\right) \cong\mathbb{Z}_{p}$, we see that there exists no $\lambda$-ring structure on $\mathbb{Z}_{p}$ under which the canonical quotient map $W\left( \mathbb{F}_{p}\right) \rightarrow \mathbb{Z}_{p}$ becomes a $\lambda$-ring morphism. As I said, this is all under the assumption that $p$ is odd, but I would not expect even $p$ to behave much different.

  • $\begingroup$ Hi Darij. In fact, your original statement is true for all p, including 2. The big Witt ring W(F_p) can be expressed as a product of copies of Z_p where the index set is the set of integers n>1 and prime to p, the projection to Z_p being the projection onto the first component. Further the n-th Adams operation takes the n-th component to the first one. s if Z_p has a lambda-structure, then it has an $\endgroup$ – JBorger Oct 2 '15 at 18:26
  • $\begingroup$ n-th Adams operator in particular. If the projection from W(F_p) is a lambda-ring map, then it has to commute with the n-th Adams operator. Looking at the effect on the idempotent describing the n-th component, you see the n-th Adams operator on Z_p would have to send 1 to 0 (if n>1!), which is impossible. $\endgroup$ – JBorger Oct 2 '15 at 18:30
  • $\begingroup$ @JBorger: Nice, and thanks for reminding me of the direct-product description! (Though I think your n>1 should be $n\geq 1$, and you mean "0 to 1" instead of "1 to 0".) $\endgroup$ – darij grinberg Oct 2 '15 at 19:33
  • $\begingroup$ Interesting. Follow-up question: Suppose I have a map of rings from a $\lambda$-ring $R$ to $\mathbb{F}_p$, under what circumstances does it lift to a map of $\lambda$-rings from $R$ to $\mathbb{Z}_p$? For example if $R$ is $\mathbb{Z}$ and I take the quotient mod $p$ map to $\mathbb{F}_p$ it lifts to the inclusion of $\mathbb{Z}$ in $\mathbb{Z}_p$ which is a map of $\lambda$-rings, what is it about this example that gives me this property? $\endgroup$ – Nate Oct 2 '15 at 20:45
  • $\begingroup$ A map from such an $R$ to $\mathbb{F}_p$ lifts to a unique $\lambda$-map from $R$ to $W(\mathbb{F}_p)$. Since you can identify $\mathbb{Z}_p$ with the subring of $W(\mathbb{F}_p)$ on which all prime-to-$p$ Adams operators are trivial, the $\lambda$-ring map $R\to W(\mathbb{F}_p)$ has image in $\mathbb{Z}_p$ if and only if the map $f:R\to\mathbb{F}_p$ satisfies $f(\psi_n(r))=f(r)$ for all $n$ prime to $p$. For example, in the case $R=\mathbb{Z}$, all Adams operations are the identity; so it does indeed work there. $\endgroup$ – JBorger Oct 4 '15 at 19:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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