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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!

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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.

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  • $\begingroup$ 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 ${\rm Aut}((\mathbf{Z}/p^r)^n)$? $\endgroup$ – user430191 Oct 21 '18 at 17:24
  • $\begingroup$ 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? $\endgroup$ – user430191 Oct 21 '18 at 17:27
  • $\begingroup$ 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. $\endgroup$ – Charles Rezk Oct 21 '18 at 19:40
  • $\begingroup$ Charles, thanks for the answer. It was very illuminating indeed. A last question only. How you come up with "injective" morphisms $E^*B\Lambda_r \to E^*B\Lambda_{r+1}$? $\endgroup$ – user430191 Oct 22 '18 at 10:05

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