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Let $\pi$ be an automorphic representation of $GL(2)/\mathbb{Q}$. For simplicity, you can take it to be a Maass form for $SL(2,\mathbb Z)$. Kim, Shahidi, Gelbart-Jacquet prove that $$L(s, \pi, Sym^m)$$ is automorphic on $GL(m+1)$ for $m=2,3,4$. And therefore the analytic property (being entire) of $L(s, \pi, Sym^m)$ is known for $m=2,3,4$.

What is known for $m=5$ or higher? I think we know that for $m=5,6,7,8$, $L(s, \pi, Sym^m)$ is meromorphic and satisfies functional equation. Do we know that it is holomorphic on any (small) area on $\{1-\delta<\Re s\leq 1\}$ for some small $\delta$? Kim-Shahidi's DUKE paper can do it for $\Re s\geq 1$.

PS: Since Langlands-Shiahidi method could yield $Sym^4 \pi\times \pi'$ with $\pi'$ on GL(4), I can't believe the same method fails to say anything about $Sym^5\pi$.

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    $\begingroup$ You could take Rankin-Selberg of $\text{Sym}^2$ and $\text{Sym}^3$, which would factor as $\text{Sym}^5$ times $\text{Sym}^3$ times $\pi$. Since the Rankin-Selberg is holomorphic, one should get holomorphy of $\text{Sym}^5$ in any zero free region for $\pi$ and $\text{Sym}^3$. The usual arguments would give a small zero-free region to the left of $1$ for these two $L$-functions, and hence a small region of holomorphy for $\text{Sym}^5$. Is this what you were looking for, or did you need a larger region? $\endgroup$
    – Lucia
    Commented Jun 30, 2014 at 5:23
  • $\begingroup$ I do need a larger region. And moreover the standard zero free region of $Sym^3$ does not say anything about the real line. $\endgroup$
    – 7-adic
    Commented Jun 30, 2014 at 5:25
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    $\begingroup$ You might want to clarify the exact question then. Presently you only ask for holomorphy on "any small area to the left of $1$". $\endgroup$
    – Lucia
    Commented Jun 30, 2014 at 5:26

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