5
$\begingroup$

I post this naive question on Math stackexchange, but have got no reply, so I decide to bother the mathoverflow community.

https://math.stackexchange.com/questions/2350436/rational-structures-of-cokernel-of-linear-maps

In Nekovar's paper on Beilinson Conjecture,

http://math.stanford.edu/~conrad/BSDseminar/refs/BeilinsonintroII.pdf

I got stuck with the rational structures on cokernel, see sections(2.1) and (2.2), and the question could be stated as follows,

$M_B^+$ and $M_{dR}/F^0$ are both rational vector spaces and $\alpha_M$ is an injetive linear map, \begin{equation} \alpha_M: M_B^+ \otimes_{\mathbb{Q}} \mathbb{R} \rightarrow (M_{dR}/F^0) \otimes_{\mathbb{Q}} \mathbb{R} \end{equation} From the paper the rational structures on $M_B^+$ and $M_{dR}/F^0$ define a natural rational structure on $\text{det(Coker}~\alpha_M)$ where $\text{det}(V)$ means the highest exterior power of the vector space $V$, but I don't know how?

If $\alpha_M$ is an isomorphism, then $\text{Coker}~\alpha_M$ is trivial, from the paper $\text{det(Coker}~\alpha_M)$ now is canonically isomorphic to $\mathbb{R}$, and the rational structure is $\text{det}(\alpha_M)^{-1}\mathbb{Q}$, which seems very confusing, could someone clarify this?

$\endgroup$

1 Answer 1

4
$\begingroup$

For every short exact sequence of $k$-vector spaces \begin{equation*} 0 \to V_1 \xrightarrow{\alpha} V \xrightarrow{\beta} V_2 \to 0 \end{equation*} there is an isomorphism $\det(V) \cong \det(V_1) \otimes \det(V_2)$. Looking at the special case $V_2=0$, it is natural to define $\det(0)=k$. Now if $k_0$ is a subfield of $k$ and if $V_1$ and $V$ are equipped with a $k_0$-rational structure then $\det(V_1)$ are $\det(V)$ are also equipped with a $k_0$-rational structure and it is natural to equip $\det(V_2)$ with the rational structure $\det(\alpha)^{-1} \cdot k_0$. In this way the isomorphism above respects the rational structures.

Actually when formulating the Beilinson conjecture for a motive $M$, under the assumptions of Nekovar's paper, it is convenient to introduce the fundamental $\mathbf{Q}$-line \begin{equation*} \Xi(M) = \det(H^1_f(M))^{-1} \otimes \det(H^1_f(M^*(1))^*) \otimes \det(M_B^+)^{-1} \otimes \det(M_{\mathrm{dR}}/\mathrm{Fil}^0) \end{equation*} (see The equivariant Tamagawa number conjecture by Flach). The conjecture says that there is a canonical isomorphism $\Xi(M) \otimes \mathbf{R} \cong \mathbf{R}$ and that the $\mathbf{Q}$-rational structure $\Xi(M)$ corresponds to the leading term of the $L$-function $L^*(M)$. Note that the rational structure defined by $\Xi(M)$ is equal to $\det(\alpha_M)$ times the determinant of the height pairing, and that the problem of defining $\det(0)$ does not appear anymore.

$\endgroup$
3
  • $\begingroup$ In the second last sentence of the first paragraph, shoud $\text{det}(\alpha)k_0$ be $\text{det}(\alpha)^{-1}k_0$? $\endgroup$
    – Wenzhe
    Commented Jul 10, 2017 at 9:56
  • $\begingroup$ Yes, you're right, thanks for pointing it out. I've not been very careful about the signs. $\endgroup$ Commented Jul 10, 2017 at 10:53
  • $\begingroup$ Thank you very much for your answer, this is really helpful. $\endgroup$
    – Wenzhe
    Commented Jul 10, 2017 at 11:06

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .