Why can Euler systems constructed from algebraic cycles only be anticyclotomic?

Question

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.

Answer 1

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.