4
$\begingroup$

Is there a clean proof that the $L_n$, localization at $E(n)$, is simply rationalization (i.e. $L_0$) on Eilenberg-MacLane spectra? Eric Peterson asked this here, but I haven't seen an answer.

$\endgroup$
11
$\begingroup$

First, recall that any rational spectrum is a wedge of suspended copies of $H\mathbb{Q}$. It follows that $H\mathbb{Q}$ is a retract of $E(n)\mathbb{Q}$ and so is $E(n)$-local. From this we see that the canonical map $H\to H\mathbb{Q}$ factors uniquely through $L_nH$.

Over $\pi_*(E(n)\wedge H/p)$ we have a formal group law $F$ coming from $E(n)$, and a formal group law $G$ coming from $H$ (which is just $x+_Gy=x+y$). These are isomorphic, by the standard yoga of complex oriented cohomology theories. We have $[p]_G(x)=0$, and it follows easily that $[p]_F(y)=0$. However, we also have $[p]_F(y)=\sum^F_{0\leq k\leq n}v_ky^{p^k}$, so $v_k=0$ in $\pi_*(E(n)\wedge H/p)$ for all $k$. But $v_n$ is invertible, so $E(n)\wedge H/p=0$. It follows that $L_n(H/p)=0$. But $L_n(H/p)$ is the cofibre of $p$ times the identity on $L_nH$, so we see that $p.1_{L_nH}$ is an equivalence, so $L_nH$ is rational. It follows that the canonical map $H\to L_nH$ factors through a map $H\mathbb{Q}\to L_nH$. This is inverse to the map $L_nH\to H\mathbb{Q}$ that we exhibited earlier.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy