Recall that the center $\mathrm{Z}(C)$ of a category $C$ is the monoid of endomorphisms of $\mathrm{id}_C$. Thus $\eta \in \mathrm{Z}(C)$ is given by a familiy of endomorphisms $\eta_x : x \to x$, where $x \in C$, such that for all morphisms $x \to y$ the obvious diagram commutes. The center of the category of rings is trivial (see here).

In the category of special $\lambda$-rings $\lambda\mathrm{Ring}$, we have for every $q \in \mathbb{N}_{\geq 1}$ the Adams-Operation $\Psi^q$ which is known to be a $\lambda$-ring endomorphism for every special $\lambda$-ring and is compatible with $\lambda$-ring homomorphisms (see these notes by Darij Grinberg). Besides, we have $\Psi^p \circ \Psi^q = \Psi^{pq}$ and $\Psi^1 = \mathrm{id}$. This shows that there is a homomorphism of monoids

$(\mathbb{N}_{\geq 1}, *) \to \mathrm{Z}(\lambda\mathrm{Ring})$, $q \mapsto \Psi^q$

It is easy to see that it is injective. It is also surjective? If not, what do we have to add to get $\mathrm{Z}(\lambda\mathrm{Ring})$?

  • $\begingroup$ Charles' answer suggests a new question: What is the center of the category of $\lambda$-rings in which $n\cdot 1=0$, where $n$ is a given integer? The case of $n$ being a prime is probably the easiest one. John R. Hopkinson's PhD thesis ( dspace.mit.edu/bitstream/handle/1721.1/34544/… ) seems to suggest that we should somehow rule out what he calls the $\theta_p$'s (his Section 2.3). $\endgroup$ – darij grinberg Dec 3 '11 at 2:29

Your map is surjective too.

The free (special) $\lambda$-ring on one generator is a polynomial algebra of the form $F=\mathbb{Z}[\lambda^1(x),\lambda^2(x),\lambda^3(x),\dots]$. (This is well-known; I think Donald Yau proves it in his book on $\lambda$-rings.) The set of endomorphisms of the forgetful functor $\lambda\mathrm{Ring}\to \mathrm{Set}$ corresponds to the underlying set of $F$, so $Z(\lambda\mathrm{Ring})\subset F$.

One way to proceed from here is to use the fact that $A\mapsto A\otimes \mathbb{Q}$ is a functor $\lambda\mathrm{Ring}\to \mathbb{Q}\backslash\lambda\mathrm{Ring}$, and that $\lambda$-rings containing $\mathbb{Q}$ are nothing more that commutative $\mathbb{Q}$-algebras equipped with Adams operations. It should be easy to see that $Z(\lambda\mathrm{Ring})\subseteq Z(\mathbb{Q}\backslash \lambda\mathrm{Ring})$, since $F\subseteq F\otimes\mathbb{Q}$, and that $Z(\mathbb{Q}\backslash\lambda\mathrm{Ring})=\mathbb{N}$.


Let me write $F\{x_1,x_2,\dots\}$ for the free (special) $\lambda$-ring on generators $x_1,x_2,\dots$. Then $F\{x_1,x_2\}\approx F\otimes F$, since coproducts in $\lambda$-rings are tensor products. There is a comultiplication $\Delta\colon F\{x\}\to F\{x_1,x_2\}$ (i.e., $F\to F\otimes F$) defined by sending $x\mapsto x_1+x_2$; it makes $F$ into a Hopf algebra. The map $\Delta$ encodes how polynomials in $\lambda$-operations act on sums, so we see that $\psi^k(x)\in F$ is primitive: $\Delta(\psi^k(x))=\psi^k(x_1)+\psi^k(x_2)$.

The subgroup $P\subset F$ corresponds precisely to the set of polynomials $f(\lambda^1,\lambda^2,\dots)$ such that $f(x+y)=f(x)+f(y)$ in any $\lambda$-ring. I want to identify $P$ with the $\mathbb{Z}$-linear span of the $\psi^k(x)$. It is a little easier to identify the subgroup of primitives in $F\otimes \mathbb{Q} \approx \mathbb{Q}[\psi^1(x),\psi^2(x),\psi^3(x),\dots]$ as the $\mathbb{Q}$-linear span of the $\psi^k(x)$ (for instance, using structure theorems for Hopf algebras; $F\otimes\mathbb{Q}$ is primitively generated as a Hopf algebra.).

There is a second comultiplication $\Delta'\colon F\{x\}\to F\{x_1,x_2\}$ sending $x\mapsto x_1x_2$, which encodes how operations act on products. We want the elements inside $P$ (or just $P\otimes \mathbb{Q}$) which are grouplike with respect to $\Delta'$, (i.e., $\Delta'(u)=u\otimes u$). We already know that the $\psi^k(x)\in P$ have this property, so we just need to show that if a linear combination of $\psi^k(x)$'s is grouplike wrt to $\Delta'$, then it is just a single $\psi^k(x)$, which is relatively elementary.

  • $\begingroup$ To make the reference to Donald Yau's book more precise: it's his Proposition 1.38, avaliable online at worldscibooks.com/etextbook/7664/7664_chap01.pdf . $\endgroup$ – darij grinberg Dec 3 '11 at 2:31
  • $\begingroup$ I have troubles checking that $Z\left(\mathbb Q \backslash \lambda\text{Ring}\right) = \mathbb N$. Could you give some details $\endgroup$ – darij grinberg Dec 3 '11 at 3:11
  • $\begingroup$ Thanks for clearing things up (though it took me half an hour to understand why our element must be primitive wrt $\Delta$ and group-like wrt $\Delta^{\prime}$; does it follow from abstract nonsense?). As for computing the primitives of $P$ (i. e., showing that they are the $\mathbb Z$-linear span of the $\psi^k\left(x\right)$), this was done in Hazewinkel's "Witt vectors, part 1" ( arxiv.org/abs/0804.3888 ) §10.15. $\endgroup$ – darij grinberg Dec 3 '11 at 6:11
  • $\begingroup$ Ah, I see. Primitivity wrt $\Delta$ and group-likeness wrt $\Delta^{\prime}$ also follow from stuff done in Hazewinkel. $\endgroup$ – darij grinberg Dec 3 '11 at 6:22
  • $\begingroup$ One final remark: the fact that "if a linear combination of $\psi^k(x)$'s is grouplike wrt to $\Delta'$, then it is just a single $\psi^k(x)$" is indeed an elementary fact of Hopf algebra theory, stating that any set of grouplike elements of a bialgebra over a field must be linearly independent. Nice answer! $\endgroup$ – darij grinberg Dec 3 '11 at 6:23

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