$\DeclareMathOperator\SL{SL}$As explained in [this question](https://mathoverflow.net/questions/109642/lambda-ring-structures-on-mathbb-a2) , there are two $\lambda$-ring structures on ${\mathbb Z}[x]$. In layman's terms, both come from a realization of ${\mathbb Z}[x]$ as the representation ring of an algebraic (semi)-group: it is either $R({\mathbb C},\times)$, or $R(\SL\_2({\mathbb C}))$.


**Question 1:** What are $\lambda$-ring homomorphisms between these? As we can choose any of the two structures for the range and the domain, it is four separate questions.


In particular, I am interested in the semi-group of endomorphisms of $R(\SL\_2({\mathbb C}))$. I can see 3 of its elements: zero, identity and
$$P: R(\SL\_2({\mathbb C}))\cong R(\operatorname{PSL}\_2({\mathbb C}))\hookrightarrow R(\SL\_2({\mathbb C})).$$
If one thinks of $x$ as the standard 2-dimensional representation of $\SL\_2$, they are given by
$$0: x\mapsto 2, \ \ 1:x\mapsto x, \ \ P:x\mapsto x^2-2.$$
**Question 2:** Is this semigroup cyclic? Note that $P$ will be the generator.



$\lambda$-ring endomorphisms of ${\mathbb Z}[x]$
No, it is not cyclic.


For each of the rings, the semigroup of endomorphisms consists of one element of degree $n$ for each positive integer $n$ plus one or two elements of degree $0$. For a positive integer, this is given by the unique product of Frobenius lifts of that degree, i.e. in $R(\mathbb C)$ by $x \mapsto x^n$ and in $R(SL\_2)$ by the $n$'th (normalized) Chebyshev polynomial of the first kind. That these are endomorphisms follows from the characterization of torsion-free lambda-rings in terms of commuting lifts of Frobenius.


The degree $0$ elements are $x\to 0, x\to 1$ for $R(\mathbb C)$ and $x\to 2$ for $R(SL\_2)$.


For maps from one ring to the other, the only choices are degree $0$, i.e. $x\to 2$ for $R(SL\_2) \to R(\mathbb C)$ and $x\to 0$ or $x\to 1$ for $R(\mathbb C) \to R(SL\_2)$.


To calculate these, we can use the following trick: Both rings embed into $\mathbb Z[t,t^{-1}]$, with the lambda-ring structure arising from viewing it as the representation ring of $\mathbb G\_m$, by the embeddings $x\to t$ and $x\to t+t^{-1}$. So it suffices to find the lambda-ring homomorphisms from each ring to $\mathbb Z[t,t^{-1}]$ and then throw away the ones whose image does not lie inside the desired target rings.


Since $\lambda^2(x) = 0 $ for $x$ the standard generator of $R(\mathbb C)$, any homomorphism $R(\mathbb C) \to \mathbb Z[t,t^{-1}]$ must send $x$ to an element $y$ with $\lambda^2(y)=0$. It's easy to see that such an element must have leading coefficient $1$ and no coefficients below the leading coefficient, i.e. must be a momomial $t^n$ for $n\in \mathbb Z$, or have no leading coefficients at all, since otherwise the leading term of $\lambda^2(y)$ would be nonvanishing. Then one can check that $x\to 0$ and $x \to t^n$ are in fact homomorphisms. The image of these homomorphisms is contained in $R(\mathbb C)$ if and only if $n\geq 0$, and contained in $R(SL\_2)$ only if $n=0$ or $x$ is sent to $0$.


Similarly, $R(SL\_2(\mathbb C))$ is generated by an element $x$ with $\lambda^2(x)=1$, so $x$ must be sent to an element $y$ with $\lambda^2(y)=1$. Such an element must have leading coefficient $1$ or $2$, with the second case only in degree $0$, or otherwise the leading term of $\lambda^2(y)$ is something other than $1$. In the first case, the leading term of $\lambda^2(y)$ comes from the next term beyond the leading term of $y$, so we must have $y = t^n + t^{-n} + $ lower order terms. and in the second case we have $y = 2+$ lower order terms. Symmetrically, we must have $y= t^n + t^{-n} + $ higher order terms or $y = 2+$ higher order terms, and this is only possible if $y= t^n+ t^{-n}$ for $n\geq 0$ (with $n=0$ encapsulating the $y=2$ case). One checks that $x \mapsto t^n + t^{-n}$ indeed gives a homomorphism for all $n$, and then observes that its image lies in $R(SL\_2)$ for all $n$ and lies in $R(\mathbb C)$ only if $n=0$.


