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In Section 2 of Nevovar's paper "Beilinson's Conjectures", for a pure motive $M$ of the form $h^i(X)(n)$ where $X$ is a projective smooth variety over $\mathbb{Q}$ and $n$ is an integer such that the weight $w=i-2n$ is negative. Then Deligne's period map is defined to be \begin{equation} \alpha_M:M_B^+ \otimes \mathbb{R} \rightarrow (M_{dR}/F^0 M_{dR}) \otimes \mathbb{R} \end{equation} where $M_B$ is the Betti realisation of $M$, $M_B^+$ is the subspace of $M_B$ on which $\phi_{\infty}$ (induced from the action of conjugation on the complex-valued points $X(\mathbb{C})$ and on $\mathbb{Q}(n)=(2 \pi i)^n \mathbb{Q}$ ) acts as identity, $M_{dR}$ is the de Rham realisation of $M$ and $F^0$ is a subspace of $M_{dR}$ appears in the filtration. This homomorphism is induced from the standard comparison isomorphism $I_\infty$ followed by a quotient.

In the commutative diagram in section 2.2 (top of page 7), it claims that the following sequence is exact \begin{equation} 0 \rightarrow (F^0M_{dR} \oplus M_B^+)\otimes \mathbb{R} \rightarrow (M_B^-(-1) \oplus M_B^+ ) \otimes \mathbb{R} \rightarrow \text{Ker} (\alpha_{M^\vee(1)})^\vee \rightarrow 0 \end{equation} The weight of the pure motive $M^\vee(1)$ is $-w-2$, which might not be negative, but the period map is still defined to be \begin{equation} \alpha_{M^\vee(1)}:M^\vee(1)_B^+ \otimes \mathbb{R} \rightarrow (M^\vee(1)_{dR}/F^0M^\vee(1)_{dR}) \otimes \mathbb{R} \end{equation}

Question: How to prove the above short sequence is exact?

Two lines below the commuative diagram in page 7, it says that

"similarly, $\text{Coker}(\alpha_{M^\vee(1)}) \xrightarrow{\sim} \text{Ker} (\alpha_M)^\vee=0$"

which the author deduces from the commuative diagram. If assume $\text{Coker}(\alpha_{M^\vee(1)})=0$, I know how to show the above short sequence is exact. But without this assumption, I don't know how to prove the the short sequence is exact from the definition of $\alpha_{M^\vee(1)}$ directly!

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  • $\begingroup$ Just dualize the sequence and note that $F^0M_{dR}$ is the dual of $M^\vee(1)_{dR}/F^0M^\vee(1)_{dR}$. Your confusion may be because the standard indexing of the filtration on the dual ensures that $F^1M^\vee_{dR}$ is actually the annihilator of $F^0M_{dR}$. Therefore you need to twist $M^\vee$, which then makes $F^0M^\vee(1)_{dR}$ the annihilator instead. $\endgroup$
    – Keerthi Madapusi
    Apr 12, 2018 at 18:58
  • $\begingroup$ @KeerthiMadapusiPera Thank you. What still confuse me is that the injectivity of the left map of this short sequence is equivalent to $\alpha_{M^\vee(1)}$ being surjective, but I don't know how to prove $\alpha_{M^\vee(1)}$ is surjective! $\endgroup$
    – Wenzhe
    Apr 12, 2018 at 19:38
  • $\begingroup$ By duality, it's equivalent to $\ker \alpha_M$ being $0$; that is, to $M^+_B\otimes\mathbb{R}$ (which is contained in $M_{dR}\otimes\mathbb{R}$ under the comparison isomorphism) intersecting trivially with $F^0M_{dR}\otimes\mathbb{R}$. But in fact, $M_B\otimes\mathbb{R}$ intersects trivially with $F^0M_{dR}\otimes \mathbb{C}$ when $w<0$, as can be checked from the fact that the latter cannot contain both of any conjugate pair of weights. $\endgroup$
    – Keerthi Madapusi
    Apr 12, 2018 at 20:17
  • $\begingroup$ @KeerthiMadapusiPera Ifeel very sorry to ask you again, by which duality? Could you explain more carefully why it is equivalent to $\text{ker} \, \alpha_M=0$? $\endgroup$
    – Wenzhe
    Apr 12, 2018 at 20:23
  • $\begingroup$ The map $\alpha_{M^\vee(1)}$ is just the dual of the map $\alpha_M$, and so is surjective precisely when $\alpha_M$ is injective. $\endgroup$
    – Keerthi Madapusi
    Apr 12, 2018 at 20:27

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