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Every de Rham representation is Hodge--Tate, but the converse is not true -- there are Hodge--Tate representations which are not de Rham, and thus cannot appear in geometry. (Examples of these arise i …

For simplicity, let's take $K = \mathbf{Q}_p$. One of the few things about $\mathbf{B}_{\mathrm{cris}}$ that one can straightforwardly prove directly from its definition is that it contains $\widehat …

A p-adic L-function is expected to depend on (at least) two pieces of data: a family $V$ of representations of $G_{\mathbf{Q}}$ over some base space $X$ (which should be a p-adic formal scheme or rig …

No, triangulations are not in general unique. A simple way of seeing this is to consider the case when $K = \mathbf{Q}_p$, $V$ is 2-dimensional and crystalline with distinct Hodge–Tate weights, say $\ …

This is purely formal. If $V$ is crystalline, then $V \otimes \mathbf{B}_{\mathrm{cris}}$ has a basis as a $\mathbf{B}_{\mathrm{cris}}$-module in which the action of $G_K$ is trivial. Hence a fortiori …

This is a question about the base-rings appearing in the the theory of $(\varphi, \Gamma)$-modules in $p$-adic Hodge theory. Let $p$ be prime, $n \ge 1$, and let $$ \mathbf{A}_{\mathbf{Q}_p}^{\dagger …

We have $(V \otimes B_e)^{G_K} = D_{\mathrm{cris}}(V)^{\varphi = 1}$. So any representation which is crystalline, but such that $\mathbf{D}_{\mathrm{cris}}(V)$ has zero $\varphi$-invariants, is an exa …

Let me suppose (for simplicity) that $M$ has distinct elements. Choose a polynomial $P \in \mathbf{Q}_p[X]$ with distinct roots, all of which have the same valuation, with this common valuation being …

Here are two arguments for why such a representation $\rho$ cannot exist. Automorphic argument: Fontaine and Mazur have conjectured that any irreducible $n$-dimensional geometric representation $\rh …

Since $B^+_{dR} / t B^+_{dR} \cong \mathbf{C}_p$, we have an injection $$\frac{(B^+_{dR} \otimes V)^{G_K}}{(t B^+_{dR} \otimes V)^{G_K}} \hookrightarrow (V \otimes \mathbf{C}_p)^{G_K}.$$ Multiplying b …

It might be worth distinguishing here between two different but related constructions: Breuil--Kisin modules, which are finite free modules over a relatively simple base ring, namely $\mathfrak{S} = …

As I commented above, the question needs some adjustment for $n \ge 3$, since the Galois representation doesn't go into $\operatorname{GSp}_{2n}$ but rather into its $L$-group, which is $\operatorname …

Yes, this can be done. One can realise the universal character $\epsilon^{\mathrm{univ}}$ as an $R$-linear representation $G_{\mathbf{Q}} \to R^\times$, where $R$ is isomorphic to a direct product o …

The isomorphism class of the $G_{\mathbf{Q}_p}$-representation $V_f$ determines (up to scaling) a class in $H^1(\mathbf{Q}_p, \delta \epsilon^{-1})$. The condition that $\mathbf{D}_{\mathrm{cris}}(V_f …

Much more is true: the subring $B_{\mathrm{cris}}^{\varphi = 1}$ surjects onto $B_{\mathrm{dR}} / B_{\mathrm{dR}}^+$, so there is an exact sequence $$ 0 \to \mathbf{Q}_p \to B_{\mathrm{cris}}^{\varphi …