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In a footnote to the 2018 Zerbes-Loeffler lecture notes from the Arizona Winter School, it's stated that Euler systems constructed from algebraic cycles "cannot give a full Euler system, only an anticyclotomic one." Why is this the case? It is not obvious to me.

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    $\begingroup$ The link does not work! I'm getting a $\sqrt{-404}$ error, which is pretty bad I believe. $\endgroup$ – RP_ Dec 24 '19 at 0:24
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    $\begingroup$ should be fixed now! $\endgroup$ – xir Dec 24 '19 at 3:33
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Let me explain a bit more what that footnote was supposed to mean.

As I'm sure you know, an Euler system for a Galois representation $V$ over a number field $K$ consists of a bunch of classes in $H^1(L, V)$, as $L$ varies over a suitable class of abelian extensions of $K$. If we're willing to temporarily forget about integrality, we can project to eigenspaces for the action of $Gal(L/K)$ and think of an Euler system equivalently as a collection of classes $z_\tau \in H^1(K, V(\tau))$, as $\tau$ varies over some collection of finite-order characters of the Galois group $G_K$ -- either all finite-order characters (a full Euler system) or only anticyclotomic ones, assuming $K$ is CM (an anticyclotomic Euler system).

Now, if your classes $z_\tau$ come from something geometric, they will have a special property: they will lie in the Bloch--Kato "$H^1_{\mathrm{g}}$" subspace. Up to a minor correction coming from local Euler factors, which will be trivial for almost all $\tau$ and thus can be ignored [*], this is the same as the Bloch--Kato "$H^1_{\mathrm{f}}$" subspace. Now, the Bloch--Kato conjecture predicts that the dimension of $H^1_{\mathrm{f}}(K, V(\tau))$ is the order of vanishing of $L(V^* \otimes \tau)$ at $s = 1$. So, the moral of that is the following: for an Euler system coming from geometry to exist, you need the L-values of all of these character twists of $V$ to vanish.

There are basically two possible mechanisms for forcing lots of L-values to vanish in a systematic way. One is poles of Archimedean $\Gamma$-factors (which is what gives the "trivial" zeroes of $\zeta(s)$ at negative even integers); the other is from sign considerations when $s = 1$ is the central value. (These are mutually exclusive, because Archimedean $\Gamma$-factors can only force vanishing when $s = 1$ is not the central value).

However, "sign-related" vanishing is somehow rather fragile: it only applies when $W = Ind_K^{\mathbb{Q}} V$ is Tate self-dual ($W = W^*(1)$). If you want to twist $V$ by a character $\tau$ while preserving this self-duality, you need $\tau$ to be self-dual in a suitable sense -- and this ends up forcing you to consider only anticyclotomic characters.

If your geometric classes come from algebraic cycles (i.e. from $K_0$ of algebraic varieties), then they correspond to an $L$-value at the centre of the functional equation (motivic weight $-1$) so the only possibility is sign-induced vanishing. This is why "algebraic cycle" Euler systems, like Heegner points and the more recent constructions of Cornut and of Jetchev, are always anticyclotomic. To get a full (non-anticyclotomic) Euler system, you have to use geometric classes coming from K-theory in positive degrees in a setting where there is "$\Gamma$-factor" vanishing, as in the case of cyclotomic units and Kato's Euler system.

[*] Historical remark: this gap between $H^1_{\mathrm{g}}$ and $H^1_{\mathrm{f}}$ is precisely where Flach's cohomology classes for the symmetric square of an elliptic curve live; and the fact that this gap closes up again when you twist by a character is why it is so hard to extend Flach's construction to a full Euler system.

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  • $\begingroup$ thanks for the great answer! there's a bit I'm still confused on, though, surely just because of my own ignorance - you're not saying the special reason cycles can only yield anticyclotomic ones is because their classes land in H^1_g, correct? this seems to be true of the other kind of euler system as well, made of units. $\endgroup$ – xir Dec 22 '19 at 22:40
  • $\begingroup$ rather, the special feature is that for whatever reason, the way you get suitable vanishing of the s=1 bloch-kato value tied to the existence of these cycles is via the sign of the functional equation, whereas in situations with you units, the way you get the suitable vanishing tied to the existence of the classes is via gamma factors (e.g. stark's conjectures, units for totally real characters are "seen" by the gamma factor). is this correct? if so, i'm not sure i understand why one expects to find the link cycles ~ sign of functional equation, units ~ gamma factors. $\endgroup$ – xir Dec 22 '19 at 22:42
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    $\begingroup$ The degree of $K$-theory that is relevant in Beilinson's conjecture is exactly $2\times$ the distance away from the centre of the functional equation. So $K_0$, i.e. algebraic cycles, corresponds to central values (as in BSD); $K_1$ corresponds to values $\tfrac{1}{2}$ away from the centre; etc. Thus "algebraic cycle" Euler systems are bound to be at the centre, so the only relevant vanishing is sign-induced vanishing. $\endgroup$ – David Loeffler Dec 22 '19 at 23:00
  • $\begingroup$ ah, how did i miss that?? it was even implicit in your comment about them being disjoint cases. many thanks! $\endgroup$ – xir Dec 22 '19 at 23:19

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