proof was based on a trace formula. Our approach is based on an integral representation, which was pioneered in [@G], and a method of ruling out poles which was pioneered in [@GJ].

Note that the action of ${{\mathrm{GL}}}_n$ on the space of $n\times n$ matrices may be regarded as the tensor product of the standard representation with its own dual. It decomposes as the direct sum of our adjoint representation, and the one dimensional span of the identity matrix equipped with trivial action. It follows that $$\label{Ad = RS/zeta}
L( s, \pi, {{\mathrm{Ad}}}\times \chi) = \frac{L( s, \pi \times {\widetilde}\pi\times \chi )}{L(s, \chi)},$$ where ${\widetilde}\pi$ is the contragredient of $\pi.$ From this perspective, Conjecture \[conj: adjoint GL3 is entire\] may be viewed as saying that $L(s, \pi \times {\widetilde}\pi\times \chi)$ should be “evenly divisible” by $L(s, \chi).$ More concretely, it says that $L(s, \pi \times {\widetilde}\pi)$ must vanish at each zero of $\zeta(s),$ to at least the same order. (Cf. [@Flicker p. 234].)

As explained in [@JngR], Conjecture \[conj: adjoint GL3 is entire\] is expected to be related to the conjecture that $\zeta_K(s)/\zeta_F(s)$ is entire for a field extension $K/F$. As explained in the introduction of [@BumpGinzburg], since $\zeta_K(s)/\zeta_F(s)$ is 