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Let $f_1$ and $f_2$ be two cuspforms of weights $k_1$ and $k_2$ and nebentypus $\epsilon_1$ and $\epsilon_2$ respectively such that $f_1 \neq f_2 \otimes \chi$ for some Dirichlet character $\chi$ of order $n$.

Is it possible that ${\rm sym}^n(f_1)$ is same as ${\rm sym}^n(f_2)$?

In other words, I am looking for a necessary and sufficient condition for two cuspforms to produce the same symmetric $n$-th representation for some fixed $n$.

Any help regarding this would be really appreciated.

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    $\begingroup$ Are you assuming these are newforms? $\endgroup$
    – Kimball
    Commented Feb 15 at 22:59
  • $\begingroup$ @Kimball Yes. Let's assume they are newform. $\endgroup$
    – user15243
    Commented Feb 16 at 5:37
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    $\begingroup$ Do you know about the works of Ramakrishnan and Rajan? $\endgroup$
    – Kimball
    Commented Feb 16 at 14:22
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    $\begingroup$ Just for clarification: you're thinking about $GL_2$, rather than fancier things? $\endgroup$
    – paul garrett
    Commented Feb 26 at 20:43
  • $\begingroup$ @paulgarrett Yes. $\endgroup$
    – user15243
    Commented Feb 26 at 22:57

1 Answer 1

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This can happen. For example, the dihedral group $D$ of order $10$ has two irreducible (Galois conjugate) representations $U$ and $V$ of dimension $2$ which are not twists of each other. But $\mathrm{Sym}^3(U) = \mathrm{Sym}^3(V)$. Now just take $f$ to be associated to an Artin representation with image $D$ and let $g$ be the Galois conjugate and you have an example. Similar things happen with other representations with small image.

If the Zariski closure of the image of (either of the) Galois representations associated to $f_1$ or $f_2$ contains $\mathrm{GL}_2$ however then such an equality forces them to be twists. The idea in this case is that $\mathrm{Sym}^2(U) \mathrm{det}(U)^{-1}$ will be the unique $3$-dimensional constituent of $\mathrm{Sym}^n(U) \otimes \mathrm{Sym}^n(U)^{\vee}$. This reduces the problem to showing that an equality of irreducible representations $\mathrm{Sym}^2(U) \mathrm{det}(U)^{-1} = \mathrm{Sym}^2(V) \mathrm{det}(V)^{-1}$ implies that $U$ and $V$ are twists, which I assume since it is the first special case of your question you already know.

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  • $\begingroup$ Thanks for the answer. I tried to understand it but could not figure it out. Can you provide some references? $\endgroup$
    – user15243
    Commented Mar 14 at 20:38
  • $\begingroup$ claymath.org/library/proceedings/cmip013c.pdf Here, in page no 483, the author showed that two cusp forms can give same symmetric cube without being a twist of each other. So I'm not able to understand what you wrote in the second paragraph of your answer. $\endgroup$
    – user15243
    Commented Mar 15 at 18:41
  • $\begingroup$ Can we say that if $f$ is a non-CM form, then the Zariski closure of the Galois representation contains ${\mathrm{GL}}_2$? $\endgroup$
    – user15243
    Commented Mar 25 at 18:44
  • $\begingroup$ This can't be true, as the determinant of the image contains only $k-1 $-th powers (there are other much more subtle obstructions). $\endgroup$
    – Olivier
    Commented Nov 7 at 20:40
  • $\begingroup$ @Olivier you are conflating a subtle problem (whether the image is open in the $\mathbf{Z}_p$ points of some algebraic group) with an easy one. The image of $(k-1)$th powers (for $k \ge 2$) is an infinite set inside the one-dimensional variety $\mathbf{G}_m$ as so is certainly Zariski dense. $\endgroup$
    – user523984
    Commented Nov 8 at 22:40

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