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The question is rather easy to formulate: when is the $L$-function of a pure motive over $\mathbb{Q}$ expected to have a holomorphic (as opposed to simply meromorphic) continuation to the complex plane? The Dedekind zeta function of a number field, which is a pure motive of weight 0, has a pole at $s=1$. Is this somehow the only case? (Meaning, I am just guessing, that for example if there are no Hodge classes then holomorphic continuation is expected?)

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    $\begingroup$ In the case of the pure motive $h^i(X)$ of weight $i$ (so the functional equation for $L(h^i(X),s)$ relates $s$ and $i+1-s$), if we look only at integers $s$, then the only possible pole is at the near-central point $s=i/2+1$. The order of this pole is given by the Tate conjecture, it is the dimension of $\mathrm{CH}^{i/2}(X)/\mathrm{CH}^{i/2}(X)_0$, the group of codimension $i/2$ algebraic cycles on $X$ modulo homological equivalence. This is explained in Nekovar, Beilison's conjectures, (6.2). $\endgroup$
    – François Brunault
    Oct 21, 2018 at 23:02
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    $\begingroup$ By Langlands, it is expected that motivic $L$-functions are automorphic, and it is also expected that automorphic $L$-functions belong to the Selberg class (not sure what is a reference for this). This would imply that the only possible pole (for all complex values $s$) is indeed at the near-central point. $\endgroup$
    – François Brunault
    Oct 21, 2018 at 23:24
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    $\begingroup$ @FrançoisBrunault: If we assume that the motive is irreducible then the group you mention must be 0. That seems to suggest that the OP's guess is right. $\endgroup$
    – Kapil
    Oct 22, 2018 at 3:21
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    $\begingroup$ @Kapil Hari Paranjape Do you mean irreducible with rank >1? The motive $\mathbb{Z}(-i/2)$ may appear (take the degree $i$ cohomology of $\mathbb{P}^n$ for example). $\endgroup$
    – François Brunault
    Oct 22, 2018 at 6:39
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    $\begingroup$ On the automorphic side, the $L$-function of a cuspidal automorphic representation of $\mathrm{GL}_n(\mathbb{A}_F)$ does not have a pole at $s = 1$ (unless you consider the trivial rep of $\mathrm{GL}_1$ to be cuspidal); see my answer here: mathoverflow.net/questions/285135/… $\endgroup$
    – Peter Humphries
    Oct 22, 2018 at 9:25

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