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Let $K$ be a finite extension of $\mathbf{Q}_p$ and $\mathcal{X}$ a smooth proper scheme over the ring of integers $\mathcal{O}_K$. For $i, j$ integers with $i \ne 2j$, there's a regulator map $$ H^i_{\mathrm{mot}}(\mathcal{X}, \mathbf{Q}(j)) \to H^1(K, H^{i-1}_{\mathrm{et}}(\overline{X}, \mathbf{Q}_p(j))),$$ and this is conjectured (and I think known in some cases) to land in the Bloch--Kato subspace $H^1_f \subset H^1$.

Now suppose $X$ is ordinary; I'm not sure I know what this means geometrically, but I want it to mean that $H^{i-1}_{\mathrm{et}}(\overline{X}, \mathbf{Q}_p(j))$ is ordinary as a Galois representation (i.e. can be written as an extension of 1-dimensional pieces). Then -- assuming the local $L$-factor at $p$ doesn't vanish -- we have $$ H^1_f(K, H^{i-1}_{\mathrm{et}}(\overline{X}, \mathbf{Q}_p(j))) = H^1(K, \operatorname{Fil}^1 H^{i-1}_{\mathrm{et}}(\overline{X}, \mathbf{Q}_p(j)))$$ where $\operatorname{Fil}^1$ means the unique subspace with all Hodge--Tate weights of the subspace $\ge 1$ and all weights of the quotient $\le 0$.

Let's define $\operatorname{Fil}^1 H^{i-1}_{\mathrm{et}}(\overline{X}, \mathbf{Z}_p(j)))$ to be the classes in $H^{i-1}_{\mathrm{et}}(\overline{X}, \mathbf{Z}_p(j))$ that land in $\operatorname{Fil}^1$ after inverting $p$. Then we know that the regulator map sends $H^i_{\mathrm{mot}}(\mathcal{X}, \mathbf{Z}(j))$ to $$ H^1(K, \operatorname{Fil}^1 H^{i-1}_{\mathrm{et}}(\overline{X}, \mathbf{Q}_p(j))) \cap H^1(K, H^{i-1}_{\mathrm{et}}(\overline{X}, \mathbf{Z}_p(j))).$$ However, this intersection isn't necessarily equal to $H^1(K, \operatorname{Fil}^1 H^{i-1}_{\mathrm{et}}(\overline{X}, \mathbf{Z}_p(j)))$, since the latter might not be saturated in $H^1(K, H^{i-1}_{\mathrm{et}}(\overline{X}, \mathbf{Z}_p(j)))$.

Does the regulator map actually land in $H^1(K, \operatorname{Fil}^1 H^{i-1}_{\mathrm{et}}(\overline{X}, \mathbf{Z}_p(j)))$?

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    $\begingroup$ This looks hard to me: I don't see how to rule out the possibility of torsion classes intervening (or equivalently, but from a different perspective, the case of the $L$-invariant being zero modulo $p$ seems hard to control a priori). $\endgroup$
    – Olivier
    Commented Oct 28, 2013 at 14:41
  • $\begingroup$ In fact, can you prove this in low dimensional cases (0 or 1)? This already looks hard to me. $\endgroup$
    – Olivier
    Commented Oct 30, 2013 at 9:28
  • $\begingroup$ Good point. The case $X = \operatorname{Spec} \mathcal{O}_K$, $i = j = 1$ says something about Kummer maps which is probably either obviously true or obviously false. I'll check. $\endgroup$
    – David Loeffler
    Commented Oct 30, 2013 at 12:19

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