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Does a left translation of an automorphic form satisfy left $K$-finiteness?

Let $F$ be a number field and $G$ is an algebraic group. Let $\phi$ is an automorphic form on $G$. Let $K$ be a maximal compact subgroup of $G(A)$.

Then $\phi$ has $K$-finiteness.

For arbitrary $x$, let $\phi_x(g)=\phi(xg)$. Then I am wondering whether $\phi_x$ has also $K$-finiteness.

Decompose $K=K_{\infty}K^{\infty}$. I checked $K^{\infty}$-finiteness of $\phi_x$ but cannot check $K_{\infty}$-finiteness.

Does this hold? If so, how can we prove it?

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  • $\begingroup$ Though "left translation" appears in the title, it does not appear in the question, nor do you give a good context. Please amplify? $\endgroup$ Apr 3, 2020 at 23:00
  • $\begingroup$ @paul, oh I am very sorry. I edited my question. Would you see it again? Thank you! $\endgroup$
    – Monty
    Apr 3, 2020 at 23:11
  • $\begingroup$ I think, but am not sure, that the usual term is "is $K$-finite", not "has $K$-finiteness". Anyway, doesn't it just mean that the set of $K$-translates lies in a finite-dimensional space? Then testing the $K$-finiteness of $\phi_x$ is just testing the $x^{-1}K x$-finiteness of $\phi$, which we have already established. $\endgroup$
    – LSpice
    Apr 3, 2020 at 23:16
  • $\begingroup$ @LSpice, since $\phi_x$ is the left translate of $\phi$, I think it is not of the $x^{-1}Kx$-finiteness. $\endgroup$
    – Monty
    Apr 4, 2020 at 6:11
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    $\begingroup$ Is $\phi$ $K$-finite on the left? Or right? $\endgroup$ Apr 4, 2020 at 8:16

2 Answers 2

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Presumable (in an automorphic forms context with contemporary left-right conventions) you mean that $\varphi$ is right $K$-finite. This could apply to any (complex-valued) function $\varphi$ on a topological group $G$, with compact subgroup $K$. It means that the space of functions obtained by right translation $R_k$ by $k\in K$ is finite-dimensional. These are functions $R_k\varphi(g)=\varphi(gk)$.

Left translation $T_x$ by $x$ stabilizes that collection of functions, because left and right translation commute: $T_xR_k\varphi=R_kT_x\varphi$.

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  • $\begingroup$ I am very sorry for late reply and always thank you very much! I am learning much from your answers. $\endgroup$
    – Monty
    Apr 16, 2020 at 3:42
  • $\begingroup$ Dear Prof. @Paul, if you are fine, would you take a look my another question related to Harish-Chandra isomorphism? Thank you very much! mathoverflow.net/questions/357661/… $\endgroup$
    – Monty
    Apr 16, 2020 at 17:22
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The answer is yes at the finite places, but no at the infinite place.

Because each finite-dimensional complex representation of a totally disconnected group factors through a finite quotient, a vector is $K_\textrm{f}$-finite iff its stabilizer in $G(\mathbb{A}_\textrm{f})$ is open in that group, and this condition is obviously invariant under right-translation by $G(\mathbb{A}_\textrm{f})$.

On the other hand, at infinite places this is false: translating a $K_\infty$-finite vector by an element of $G(F_\infty)$ will generically give a non-$K_\infty$-finite vector. For example consider the action of $\mathrm{SL}_2(\mathbb{R})$ on $\left\{ f\colon\mathbb{R}^2\setminus{0}\to \mathbb{C} \vert f(rx) = r^{\frac12+it}f(x) \right\}$. A vector is $\mathrm{SO}(2)$-finite iff its restriction to the circle is a trigonometric polynomial. But it is easy to check that for essentially any $g\in\mathrm{SL}_2(\mathbb{R})\setminus\mathrm{SO}(2)$, if $f(x)$ is a non-constant trigonometric polynomial then $f(xg)$ isn't.

This is one of the motivation for passing to $(\mathfrak{g},K_\infty)$-modules: since $\mathfrak{g}$ is a finite-dimensional representation of $K_\infty$, its action on smooth vectors preserves $K$-finiteness.

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  • $\begingroup$ I am sorry for late reply but thank you very much for your kind reply. Since I didn’ t know the connection of modular form and automorphic form, I studied modular form for a while to understand your answer. But I couldn’t understand your last comment. Why the finite dimension of $\mathfrak{g}$ guarantee the $\mathfrak{g}$-action preserves the $K$-finiteness? $\endgroup$
    – Monty
    Apr 16, 2020 at 4:28
  • $\begingroup$ Let $v$ be a $K$-finite smooth vector in a representation $V$ of a Lie group $G$. Then $v$ lies in a $K$-invariant subspace $U$. If $X\in\frak{g}$ then $Xv$ lies in the image of a map $\frak{g}\otimes U\to V$, whose image is a f.d. rep. of $K$ since its domain is the tensor product of two such representations. $\endgroup$ Apr 25, 2020 at 16:56

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