Bloch-Kato's dual exponential map in a simple setting I have a quick question on Bloch-Kato p-adic Hodge theory. I'm wondering about the following easy case:
Note that $H^1(Q_p,Q_p(1))$ is two-dimensional, isomorphic by the Kummer map to the completion of $Q_p^\times$.
We have $H^1_e(Q_p,Q_p(1)) = H^1_f(Q_p,Q_p(1)) = Z_p^\times \otimes Q_p$, and the Bloch-Kato logarithm is simply the usual p-adic $\log$ on $Z_p^\times$.
-My question is: I thought that the dual exponential on $H^1(Q_p,Q_p(1))$ would simply be the p-adic valuation $\mathrm{ord}_p$ on $Q_p^\times$, but now I see that perhaps not, because $H^1_g = H^1$ in this case, and $\exp^*$ vanishes on $H^1_g$, so the dual exponential is just the zero map on $H^1(Q_p,Q_p(1))$.
Is this is correct, or am I wrong and one actually has $\exp^* = \mathrm{ord}_p$? 
If I am right, can one recover the $\mathrm{ord}_p$ map on $H^1_g/H^1_f = Q_p^\times/Z_p^\times$ as part of the general theory of Bloch-Kato? I'm aware this is a rather special setting, as for many representations one has $H^1_f = H^1_g$.
 A: As Francois has very clearly explained, the Bloch--Kato dual exponential for $\mathbf{Q}_p(1)$ is indeed zero. Just as a footnote, let me say a few words about how one can "get at" the one-dimensional quotient $H^1 / H^1_{\mathrm{f}}$, given that $\exp^*$ doesn't suffice to do this.
The point is that there are two slightly different versions of the Bloch--Kato exact sequence: the first goes
$$ 0 \to \mathbf{Q}_p \to \mathbf{B}_{\mathrm{cris}}^{\varphi = 1} \to \tfrac{\mathbf{B}_{\mathrm{dR}}}{Fil^0} \to 0$$
and the second goes
$$ 0 \to \mathbf{Q}_p \to \mathbf{B}_{\mathrm{cris}} \to \mathbf{B}_{\mathrm{cris}} \oplus \tfrac{\mathbf{B}_{\mathrm{dR}}}{Fil^0} \to 0,$$
with the last map being $x \mapsto ( (1 - \varphi) x, x \bmod Fil^0)$. 
If you tensor the first exact sequence with some Galois rep $V$ and take cohomology, you get a boundary homomorphism 
$$ \exp_{\mathbf{Q}_p, V}: \tfrac{\mathbf{D}_{\mathrm{dR}}(V)}{Fil^0\mathbf{D}_{\mathrm{dR}} + \mathbf{D}_{\mathrm{cris}}^{\varphi = 1}} \cong H^1_{\mathrm{e}}(\mathbf{Q}_p, V).$$
which is the Bloch--Kato exponential. However, if you use the second version, you get a homomorphism
$$ \widetilde\exp_{\mathbf{Q}_p, V}: \frac{\mathrm{D}_{\mathrm{cris}}(V) \oplus \tfrac{\mathbf{D}_{\mathrm{dR}}(V)}{Fil^0}}{\{ ((1-\varphi)x, x \bmod Fil^0): x \in \mathrm{D}_{\mathrm{cris}}(V)\}} \cong H^1_{\mathrm{f}}(\mathbf{Q}_p, V).$$
If $\mathbf{D}_{\mathrm{cris}}(V)$ surjects onto $\tfrac{\mathbf{D}_{\mathrm{dR}}(V)}{Fil^0}$, e.g. if $V$ is crystalline, then this simplifies into an isomorphism
$$ \widetilde\exp_{\mathbf{Q}_p, V}: \frac{\mathrm{D}_{\mathrm{cris}}(V) }{ (1-\varphi) Fil^0} \cong H^1_{\mathrm{f}}(\mathbf{Q}_p, V).$$
(The quotient $\frac{\mathrm{D}_{\mathrm{cris}}(V) }{ (1-\varphi) Fil^0}$ is a rather natural object: one can see by easy explicit linear algebra that this quotient computes extensions of $\mathbf{Q}_p$ by $\mathrm{D}_{\mathrm{cris}}(V)$ in the category of filtered $\varphi$-modules.)
When $V = \mathbf{Q}_p$, the map $\exp$ is zero, but $\widetilde\exp$ is not. If we define $\widetilde\exp^*$ to be the dual of $\widetilde\exp$, then $\widetilde\exp^*$ gives an isomorphism $H^1(\mathbf{Q}_p, \mathbf{Q}_p(1)) / H^1_{\mathrm{f}} \to \mathbf{Q}_p$, which agrees with the $p$-adic order map $\operatorname{ord}: \mathbf{Q}_p^\times / \mathbf{Z}_p^\times \otimes \mathbf{Q}_p \to \mathbf{Q}_p$ up to a sign. 
See Proposition 2.5.5 of my paper "Local epsilon isomorphisms" with Venjakob and Zerbes (journal website, arxiv).
A: You're right that Bloch-Kato's dual exponential map on $H^1(\mathbf{Q}_p,\mathbf{Q}_p(1))$ is the zero map. By definition of the dual exponential map, it suffices to show that Bloch-Kato's exponential map
\begin{equation*}
\exp : D_{\mathrm{dR}}(\mathbf{Q}_p) \to H^1(\mathbf{Q}_p,\mathbf{Q}_p)
\end{equation*}
is the zero map. But $D_{\mathrm{dR}}(\mathbf{Q}_p) \cong \mathbf{Q}_p$ and the exponential map vanishes on $\mathrm{Fil}^0 D_{\mathrm{dR}}(\mathbf{Q}_p) = (B_{\mathrm{dR}}^+)^{\mathrm{Gal}(\overline{\mathbf{Q}}_p/\mathbf{Q}_p)} = \mathbf{Q}_p$, hence the result.
