5
$\begingroup$

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 $\mathrm{Z}(\lambda\mathrm{Ring})$?

$\endgroup$
1
  • $\begingroup$ 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). $\endgroup$ Dec 3, 2011 at 2:29

1 Answer 1

4
$\begingroup$

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 $Z(\mathbb{Q}\backslash\lambda\mathrm{Ring})=\mathbb{N}$.

Added.

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.

$\endgroup$
7
  • $\begingroup$ 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 . $\endgroup$ Dec 3, 2011 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 $\endgroup$ Dec 3, 2011 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. $\endgroup$ Dec 3, 2011 at 6:11
  • $\begingroup$ Ah, I see. Primitivity wrt $\Delta$ and group-likeness wrt $\Delta^{\prime}$ also follow from stuff done in Hazewinkel. $\endgroup$ Dec 3, 2011 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! $\endgroup$ Dec 3, 2011 at 6:23

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.