# HKR generalized character theory question regarding a certain construction

In that paper https://web.math.rochester.edu/people/faculty/doug/mypapers/hkr.pdf Hopkins-Kuhn-Ravenel introduced the idea of generalized character corresponding to a complex-oriented cohomology theory $$E^*$$. They construct an $$E^*$$-algebra $$L(E^*)$$, which is defined to be the colimit: $${\rm colim} E^*(B\mathbf{Z}^n_p)$$ (More details regarding the whole construction can be found in the link.)

Later on, they use this construction to define the invariant ring $$L(E^*)^{{\rm Aut}(\mathbf{Z}^n_p)}$$, where $${\rm Aut}(\mathbf{Z}^n_p)$$ acts as $$E^*$$-algebra homomorphisms ($$\mathbf{Z}_p$$ denotes the additive group of $$p$$-adic integers). To prove this they define $$L_r(E^*)=E^*(B\mathbf{Z}_{p^r})$$, and the natural $${\rm Aut}(\mathbf{Z}_{p^r})$$ action on gives $$L_r(E^*)^{{\rm Aut}(\mathbf{Z}_{p^r})}=p^{-1}E^*$$.

My question has to do with the proof of the above: What I understand is that the invariant rings $$L_r(E^*)^{{\rm Aut}(\mathbf{Z}_{p^r})}=p^{-1}E^*$$, induce a direct system of $$E^*$$-algebras and the colimit must be $$L(E^*)^{{\rm Aut}(\mathbf{Z}^n_p)}$$. However, they don't give any proof hence should be somehow straightforward why the colimit is the invariant ring $$L(E^*)^{{\rm Aut}(\mathbf{Z}^n_p)}$$ (not at all to me). Can you explain me please if my understanding makes sense? If yes, probably an explanation why the above colimit converges on the invariant ring $$L(E^*)^{{\rm Aut}(\mathbf{Z}^n_p)}$$ would be really helpful. If not, a sort of insight would be very appreciable!

I think what you understand is correct. Write $$A_r= L_r(E^*)$$, which fit into a direct system $$A_r\to A_{r+1}\to \cdots$$. Let $$G=\mathrm{Aut}(\mathbb{Z}_p^n)$$, which acts compatibly on every $$A_r$$ (it actually acts on $$A_r$$ through the quotient group $$\mathrm{Aut}((\mathbb{Z}/p^r)^n)$$).

If $$\cdots \to A_r\to A_{r+1}\to \cdots$$ is a sequence of injective maps between $$G$$-sets, then it is straightforward to show that $$\mathrm{colim} (A_r^G) \to (\mathrm{colim} A_r)^G$$ is an isomoprhism. In fact, you only need them to be injective for all $$r\geq R$$ for some $$R$$.

It remains to show that the $$A_r\to A_{r+1}$$ are injective. I don't see a proof of this in the paper, but it is surely true.

Here's a proof that $$A_r\to A_{r+1}$$ is injective if $$r\geq1$$. There's probably a better proof, but it's what comes to mind now. I use the description in the proof of [HKR, 6.5]: $$A_r = S_r^{-1} E^*B\Lambda_r,$$ where $$\Lambda_n=(\mathbb Z/p^r)^n$$, and $$S_r\subseteq E^*B\Lambda_r$$ is a certain multiplicatively closed subset, which can be defined as follows: $$S_r = \{ c(\alpha):=(B\alpha)^*(x)\; | \; \alpha\in \Lambda_r^*\smallsetminus\{0\} \},\qquad \Lambda_r^*:=\mathrm{Hom}(\Lambda_r, U(1)),$$ where $$x\in E^*BU(1)$$ is a chosen coordinate of the formal group. Although the set $$S_r$$ depends on the choice of $$x$$, the fraction ring $$A_r$$ doesn't, because any two coordinates differ by a unit in $$E^0BU(1)$$. Note that the direct system of $$A_r$$ comes from the inverse system $$\cdots\to\Lambda_{r+1}\to \Lambda_r\to\cdots$$

Note that $$c(\alpha^p)=[p](c(\alpha))= c(\alpha)f(c(\alpha))$$ where $$[p](x)$$ is the $$p$$-series of the formal group and $$f(x)=[p](x)/x$$ is a power series. This implies that if we invert $$c(\alpha^p)$$ then we automatically invert $$c(\alpha)$$ as well. For every $$\alpha\in \Lambda_r^*\smallsetminus\{0\}$$ with $$r\geq1$$, some $$\alpha^{p^k}$$ is in the image of $$\Lambda_1^*\to \Lambda_r^*$$, so in fact $$A_r = S_1^{-1}E^*B\Lambda_r,$$ where $$S_1$$ really means the image of $$S_1\subset E^*B\Lambda_1$$ under the map induced by $$\Lambda_r\to \Lambda_1$$. Since $$E^*B\Lambda_r\to E^*B\Lambda_{r+1}$$ is injective and $$S_1^{-1}E^*B\Lambda_1$$ is flat over $$E^*B\Lambda_1$$, the claim follows.

• Thanks for your effort Charles. I haven't read your answer through yet. Though I have a question from the very beginning. How would you explicitly describe that: $G$ acts compatibly on every $A_r$ through the quotient group ${\rm Aut}((\mathbf{Z}/p^r)^n)$? – user430191 Oct 21 '18 at 17:24
• My problem isn't the context in that paper. But rather, how you can write things up more explicitly. I understand that many things have been swept under the rug for simplicity. But it's not entirely obvious to me how this action can be written? – user430191 Oct 21 '18 at 17:27
• We have an action of $G=GL_n(\mathbb{Z}_p)$ on $\Lambda=\mathbb{Z}_p^n$, which passes to an action on every quotient $\Lambda_r=\Lambda/p^r\Lambda$. Everything comes from that. – Charles Rezk Oct 21 '18 at 19:40
• Charles, thanks for the answer. It was very illuminating indeed. A last question only. How you come up with "injective" morphisms $E^*B\Lambda_r \to E^*B\Lambda_{r+1}$? – user430191 Oct 22 '18 at 10:05