Left translation of automorphic form satisfies $K$-finiteness?

Question

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?

Answer 1

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: $T_xR_k\varphi=R_kT_x\varphi$.

Answer 2

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.