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Let $f$ be a Maass form on the upper half-plane with nebentypus $\chi$. It is known that \begin{align*} \text{$f$ is self-dual}&\qquad\Longleftrightarrow\qquad\text{$L(f\times f,s)$ has a pole at $s=1$};\\ \text{$f$ is dihedral}&\qquad\Longleftrightarrow\qquad\text{$L(\mathrm{sym}^2f,s)$ has a pole at $s=1$}. \end{align*} On the other hand, $$L(f\times f,s)=L(\mathrm{sym}^2f,s)L(\chi,s),$$ and the left-hand side can at most have a simple pole at $s=1$, so we conclude that $$\text{$f$ is self-dual}\qquad\Longleftrightarrow\qquad\text{$f$ is dihedral}\quad\text{or}\quad \text{$\chi$ is trivial}.$$

Let $f$ be a Maass form on the upper half-plane with nebentypus $\chi$. It is known that \begin{align*} \text{$f$ is self-dual}&\qquad\Longleftrightarrow\qquad\text{$L(f\times f,s)$ has a pole at $s=1$};\\ \text{$f$ is dihedral}&\qquad\Longleftrightarrow\qquad\text{$L(\mathrm{sym}^2f,s)$ has a pole at $s=1$}. \end{align*} On the other hand, $$L(f\times f,s)=L(\mathrm{sym}^2f,s)L(\chi,s),$$ and the $L$-functions on the right-hand side do not vanish at $s=1$, so we conclude that $$\text{$f$ is self-dual}\qquad\Longleftrightarrow\qquad\text{$f$ is dihedral}\quad\text{or}\quad \text{$\chi$ is trivial}.$$

Let $f$ be a Maass form on the upper half-plane with nebentypus $\chi$. It is known that $$\text{$f$ is self-dual}\qquad\Longleftrightarrow\qquad\text{$L(f\times f,s)$ has a pole at $s=1$};$$ $$\text{$f$ is dihedral}\qquad\Longleftrightarrow\qquad\text{$L(\mathrm{sym}^2f,s)$ has a pole at $s=1$}.$$ On the other hand, $$L(f\times f,s)=L(\mathrm{sym}^2f,s)L(\chi,s),$$ and the left-hand side can only have a simple pole at $s=1$, so we conclude that $$\text{$f$ is self-dual}\qquad\Longleftrightarrow\qquad\text{$f$ is dihedral}\quad\text{or}\quad \text{$\chi$ is trivial}.$$

Let $f$ be a Maass form on the upper half-plane with nebentypus $\chi$. It is known that \begin{align*} \text{$f$ is self-dual}&\qquad\Longleftrightarrow\qquad\text{$L(f\times f,s)$ has a pole at $s=1$};\\ \text{$f$ is dihedral}&\qquad\Longleftrightarrow\qquad\text{$L(\mathrm{sym}^2f,s)$ has a pole at $s=1$}. \end{align*} On the other hand, $$L(f\times f,s)=L(\mathrm{sym}^2f,s)L(\chi,s),$$ and the left-hand side can at most have a simple pole at $s=1$, so we conclude that $$\text{$f$ is self-dual}\qquad\Longleftrightarrow\qquad\text{$f$ is dihedral}\quad\text{or}\quad \text{$\chi$ is trivial}.$$

The second sentence is incorrect. The symmetric square $L$-function of a Maass form $f$ (on the upper half-plane) has a pole at $s=1$ if and only if $f$ is dihedral. On the other hand, most self-dual Maass forms are not dihedral. For example, the Maass forms of level $1$ are self-dual but not dihedral.

Let $f$ be a Maass form on the upper half-plane with nebentypus $\chi$. It is known that $$\text{$f$ is self-dual}\qquad\Longleftrightarrow\qquad\text{$L(f\times f,s)$ has a pole at $s=1$};$$ $$\text{$f$ is dihedral}\qquad\Longleftrightarrow\qquad\text{$L(\mathrm{sym}^2f,s)$ has a pole at $s=1$}.$$ On the other hand, $$L(f\times f,s)=L(\mathrm{sym}^2f,s)L(\chi,s),$$ and the left-hand side can only have a simple pole at $s=1$, so we conclude that $$\text{$f$ is self-dual}\qquad\Longleftrightarrow\qquad\text{$f$ is dihedral}\quad\text{or}\quad \text{$\chi$ is trivial}.$$