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I have a question about the proof of Theorem 3.5 in Pollack's 2003 paper On the $p$-adic L-function of a Modular Form at a Supersingular Prime.

The setup is as follows. Fix an eigenform $f\in S_k(N,\epsilon)$ with $a_p=0$ and let $K$ be the completion of the field of Fourier coefficients for $f$ at some fixed prime above $p$. Let $L_p(f,\alpha_i,\psi,T)\in K(\psi,\alpha_i)[[T]]$ be the associated unbounded $p$-adic $L$-functions for $f$, where $\alpha_1,\alpha_2$ are the roots of the Hecke polynomial $X^2-a_pX+\epsilon(p)p^{k-1}$ and $\psi$ is some fixed tame character on $\mathbb{Z}_p^\times\times (\mathbb{Z}/M\mathbb{Z})^\times$. The first sentence of Theorem 3.5 in Pollack's paper defines the power series $$ \begin{align*} G_\psi^{+}&=\frac{L_p(f,\alpha_1,\psi,T)+L_p(f,\alpha_2,\psi,T)}{2}\\ G_\psi^{-}&=\frac{L_p(f,\alpha_1,\psi,T)-L_p(f,\alpha_2,\psi,T)}{2\alpha_1}.\\ \end{align*} $$ Pollack states that both these power series lie in $K(\psi)[[T]]$. Why is this true?

Here are my thoughts... We have an action of $G=\operatorname{Gal}(K(\psi,\alpha_1)/K(\psi))$ on power series $F=\sum_{i\geq 0}(a_i+\alpha_1b_i)T^i\in K(\psi,\alpha_1)[[T]]$, where $a_i,b_i\in K(\psi)$, by $$ \sigma F = \sum_{i\geq 0}(a_i+\sigma \alpha_1b_i)T^i. $$ for $\sigma\in G$. If $G$ is trivial then the claim is obvious, so suppose $G$ has order 2 and let $\sigma$ be the nontrivial element. In this case, Pollack claims (see the last sentence of the proof) that the two $p$-adic $L$-functions are conjugate, which I take to mean that $$ \sigma L_p(f,\alpha_1,\psi,T)=L_p(f,\alpha_2,\psi,T). $$ Assuming this, I can see why $G_\psi^\pm$ are rational: $\sigma$ clearly fixes $G_\psi^+$, and since $\alpha_1=-\alpha_2$ (recall $a_p=0$) we see that $\sigma$ also fixes $G_\psi^-$.

It's not clear to me why the $p$-adic $L$-functions need to be conjugate. Writing $L_p(f,\alpha_1,\psi,T)=\sum_{i\geq 0} (a_i+\alpha_1 b_i)T^i$ and $L_p(f,\alpha_2,\psi,T)=\sum_{i\geq 0} (c_i+\alpha_2 d_i)T^i$ for some $a_i,b_i,c_i,d_i\in K(\psi)$, it appears that we would need $a_i=c_i$ and $b_i=d_i$ for all $i$... Can someone clarify?

EDIT: More thoughts... Perhaps the fact that $L_p(f,\alpha_i,\psi,T)$ are conjugate can be derived from their definition in terms of $p$-adic distributions? The $p$-adic $L$-functions are defined as a pair of $p$-adic distributions $\mu_{f,\alpha_i}^\pm$ (linear maps on locally analytic functions $\mathbb{Z}_p^\times \rightarrow \mathbb{C}_p^\times$, assuming $M=1$ for simplicity) whose local values (on the open cover $a+p^n\mathbb{Z}_p$ for $(a,p)=1$) at polynomial functions $P$ of degree $\leq k-2$ are given by $$ \mu^{\pm}_{f,\alpha_i}(P; a+p^n\mathbb{Z}_p)=\frac{1}{\alpha^n}\lambda^\pm(f,P;a+p^n\mathbb{Z}_p)-\frac{\epsilon(p)p^{k-2}}{\alpha^{n+1}}\lambda^\pm(f,P;a+p^{n-1}\mathbb{Z}_p). $$ where the $\lambda$ terms are $K$-valued modular symbols. Then for $\sigma\in G$ as above we have $$ \sigma\mu^{\pm}_{f,\alpha_1}(P; a+p^n\mathbb{Z}_p)=\mu^{\pm}_{f,\alpha_2}(P; a+p^n\mathbb{Z}_p). $$ From this, is it correct to say that for a locally analytic character $\chi$ we then have $$ L_p(f,\alpha_2,\chi)=\mu_{f,\alpha_2}^{\text{sgn}(\chi)}(\chi)=\sigma\mu_{f,\alpha_1}^{\text{sgn}(\chi)}(\chi)=\sigma L_p(f,\alpha_1,\chi)? $$

I have a question about the proof of Theorem 3.5 in Pollack's 2003 paper On the $p$-adic L-function of a Modular Form at a Supersingular Prime.

The setup is as follows. Fix an eigenform $f\in S_k(N,\epsilon)$ with $a_p=0$ and let $K$ be the completion of the field of Fourier coefficients for $f$ at some fixed prime above $p$. Let $L_p(f,\alpha_i,\psi,T)\in K(\psi,\alpha_i)[[T]]$ be the associated unbounded $p$-adic $L$-functions for $f$, where $\alpha_1,\alpha_2$ are the roots of the Hecke polynomial $X^2-a_pX+\epsilon(p)p^{k-1}$ and $\psi$ is some fixed tame character on $\mathbb{Z}_p^\times\times (\mathbb{Z}/M\mathbb{Z})^\times$. The first sentence of Theorem 3.5 in Pollack's paper defines the power series $$ \begin{align*} G_\psi^{+}&=\frac{L_p(f,\alpha_1,\psi,T)+L_p(f,\alpha_2,\psi,T)}{2}\\ G_\psi^{-}&=\frac{L_p(f,\alpha_1,\psi,T)-L_p(f,\alpha_2,\psi,T)}{2\alpha_1}.\\ \end{align*} $$ Pollack states that both these power series lie in $K(\psi)[[T]]$. Why is this true?

Here are my thoughts... We have an action of $G=\operatorname{Gal}(K(\psi,\alpha_1)/K(\psi))$ on power series $F=\sum_{i\geq 0}(a_i+\alpha_1b_i)T^i\in K(\psi,\alpha_1)[[T]]$, where $a_i,b_i\in K(\psi)$, by $$ \sigma F = \sum_{i\geq 0}(a_i+\sigma \alpha_1b_i)T^i, $$ for $\sigma\in G$. If $G$ is trivial then the claim is obvious, so suppose $G$ has order 2 and let $\sigma$ be the nontrivial element. In this case, Pollack claims (see the last sentence of the proof) that the two $p$-adic $L$-functions are conjugate, which I take to mean that $$ \sigma L_p(f,\alpha_1,\psi,T)=L_p(f,\alpha_2,\psi,T). $$ Assuming this, I can see why $G_\psi^\pm$ are rational: $\sigma$ clearly fixes $G_\psi^+$, and since $\alpha_1=-\alpha_2$ (recall $a_p=0$) we see that $\sigma$ also fixes $G_\psi^-$.

It's not clear to me why the $p$-adic $L$-functions need to be conjugate. Writing $L_p(f,\alpha_1,\psi,T)=\sum_{i\geq 0} (a_i+\alpha_1 b_i)T^i$ and $L_p(f,\alpha_2,\psi,T)=\sum_{i\geq 0} (c_i+\alpha_2 d_i)T^i$ for some $a_i,b_i,c_i,d_i\in K(\psi)$, it appears that we would need $a_i=c_i$ and $b_i=d_i$ for all $i$... Can someone clarify?

EDIT: More thoughts... Perhaps the fact that $L_p(f,\alpha_i,\psi,T)$ are conjugate can be derived from their definition in terms of $p$-adic distributions? The $p$-adic $L$-functions are defined as a pair of $p$-adic distributions $\mu_{f,\alpha_i}^\pm$ (linear maps on locally analytic functions $\mathbb{Z}_p^\times \rightarrow \mathbb{C}_p^\times$, assuming $M=1$ for simplicity) whose local values (on the open cover $a+p^n\mathbb{Z}_p$ for $(a,p)=1$) at polynomial functions $P$ of degree $\leq k-2$ are given by $$ \mu^{\pm}_{f,\alpha_i}(P; a+p^n\mathbb{Z}_p)=\frac{1}{\alpha^n}\lambda^\pm(f,P;a+p^n\mathbb{Z}_p)-\frac{\epsilon(p)p^{k-2}}{\alpha^{n+1}}\lambda^\pm(f,P;a+p^{n-1}\mathbb{Z}_p). $$ where the $\lambda$ terms are $K$-valued modular symbols. Then for $\sigma\in G$ as above we have $$ \sigma\mu^{\pm}_{f,\alpha_1}(P; a+p^n\mathbb{Z}_p)=\mu^{\pm}_{f,\alpha_2}(P; a+p^n\mathbb{Z}_p). $$ From this, is it correct to say that for a locally analytic character $\chi$ we then have $$ L_p(f,\alpha_2,\chi)=\mu_{f,\alpha_2}^{\text{sgn}(\chi)}(\chi)=\sigma\mu_{f,\alpha_1}^{\text{sgn}(\chi)}(\chi)=\sigma L_p(f,\alpha_1,\chi)? $$

I have a question about the proof of Theorem 3.5 in Pollack's 2003 paper On the $p$-adic L-function of a Modular Form and a Supersingular Prime.

The setup is as follows. Fix an eigenform $f\in S_k(N,\epsilon)$ with $a_p=0$ and let $K$ be the completion of the field of Fourier coefficients for $f$ at some fixed prime above $p$. Let $L_p(f,\alpha_i,\psi,T)\in K(\psi,\alpha_i)[[T]]$ be the associated unbounded $p$-adic $L$-functions for $f$, where $\alpha_1,\alpha_2$ are the roots of the Hecke polynomial $X^2-a_pX+\epsilon(p)p^{k-1}$ and $\psi$ is some fixed tame character on $\mathbb{Z}_p^\times\times (\mathbb{Z}/M\mathbb{Z})^\times$. The first sentence of Theorem 3.5 in Pollack's paper defines the power series $$ \begin{align*} G_\psi^{+}&=\frac{L_p(f,\alpha_1,\psi,T)+L_p(f,\alpha_2,\psi,T)}{2}\\ G_\psi^{-}&=\frac{L_p(f,\alpha_1,\psi,T)-L_p(f,\alpha_2,\psi,T)}{2\alpha_1}.\\ \end{align*} $$ Pollack states that both these power series lie in $K(\psi)[[T]]$. Why is this true?

Here are my thoughts... We have an action of $G=\operatorname{Gal}(K(\psi,\alpha_1)/K(\psi))$ on power series $F=\sum_{i\geq 0}(a_i+\alpha_1b_i)T^i\in K(\psi,\alpha_1)[[T]]$, where $a_i,b_i\in K(\psi)$, by $$ \sigma F = \sum_{i\geq 0}(a_i+\sigma \alpha_1b_i)T^i. $$ for $\sigma\in G$. If $G$ is trivial then the claim is obvious, so suppose $G$ has order 2 and let $\sigma$ be the nontrivial element. In this case, Pollack claims (see the last sentence of the proof) that the two $p$-adic $L$-functions are conjugate, which I take to mean that $$ \sigma L_p(f,\alpha_1,\psi,T)=L_p(f,\alpha_2,\psi,T). $$ Assuming this, I can see why $G_\psi^\pm$ are rational: $\sigma$ clearly fixes $G_\psi^+$, and since $\alpha_1=-\alpha_2$ (recall $a_p=0$) we see that $\sigma$ also fixes $G_\psi^-$.

It's not clear to me why the $p$-adic $L$-functions need to be conjugate. Writing $L_p(f,\alpha_1,\psi,T)=\sum_{i\geq 0} (a_i+\alpha_1 b_i)T^i$ and $L_p(f,\alpha_2,\psi,T)=\sum_{i\geq 0} (c_i+\alpha_2 d_i)T^i$ for some $a_i,b_i,c_i,d_i\in K(\psi)$, it appears that we would need $a_i=c_i$ and $b_i=d_i$ for all $i$... Can someone clarify?

EDIT: More thoughts... Perhaps the fact that $L_p(f,\alpha_i,\psi,T)$ are conjugate can be derived from their definition in terms of $p$-adic distributions? The $p$-adic $L$-functions are defined as a pair of $p$-adic distributions $\mu_{f,\alpha_i}^\pm$ (linear maps on locally analytic functions $\mathbb{Z}_p^\times \rightarrow \mathbb{C}_p^\times$, assuming $M=1$ for simplicity) whose local values (on the open cover $a+p^n\mathbb{Z}_p$ for $(a,p)=1$) at polynomial functions $P$ of degree $\leq k-2$ are given by $$ \mu^{\pm}_{f,\alpha_i}(P; a+p^n\mathbb{Z}_p)=\frac{1}{\alpha^n}\lambda^\pm(f,P;a+p^n\mathbb{Z}_p)-\frac{\epsilon(p)p^{k-2}}{\alpha^{n+1}}\lambda^\pm(f,P;a+p^{n-1}\mathbb{Z}_p). $$ where the $\lambda$ terms are $K$-valued modular symbols. Then for $\sigma\in G$ as above we have $$ \sigma\mu^{\pm}_{f,\alpha_1}(P; a+p^n\mathbb{Z}_p)=\mu^{\pm}_{f,\alpha_2}(P; a+p^n\mathbb{Z}_p). $$ From this, is it correct to say that for a locally analytic character $\chi$ we then have $$ L_p(f,\alpha_2,\chi)=\mu_{f,\alpha_2}^{\text{sgn}(\chi)}(\chi)=\sigma\mu_{f,\alpha_1}^{\text{sgn}(\chi)}(\chi)=\sigma L_p(f,\alpha_1,\chi)? $$

I have a question about the proof of Theorem 3.5 in Pollack's 2003 paper On the $p$-adic L-function of a Modular Form at a Supersingular Prime.

The setup is as follows. Fix an eigenform $f\in S_k(N,\epsilon)$ with $a_p=0$ and let $K$ be the completion of the field of Fourier coefficients for $f$ at some fixed prime above $p$. Let $L_p(f,\alpha_i,\psi,T)\in K(\psi,\alpha_i)[[T]]$ be the associated unbounded $p$-adic $L$-functions for $f$, where $\alpha_1,\alpha_2$ are the roots of the Hecke polynomial $X^2-a_pX+\epsilon(p)p^{k-1}$ and $\psi$ is some fixed tame character on $\mathbb{Z}_p^\times\times (\mathbb{Z}/M\mathbb{Z})^\times$. The first sentence of Theorem 3.5 in Pollack's paper defines the power series $$ \begin{align*} G_\psi^{+}&=\frac{L_p(f,\alpha_1,\psi,T)+L_p(f,\alpha_2,\psi,T)}{2}\\ G_\psi^{-}&=\frac{L_p(f,\alpha_1,\psi,T)-L_p(f,\alpha_2,\psi,T)}{2\alpha_1}.\\ \end{align*} $$ Pollack states that both these power series lie in $K(\psi)[[T]]$. Why is this true?

Here are my thoughts... We have an action of $G=\operatorname{Gal}(K(\psi,\alpha_1)/K(\psi))$ on power series $F=\sum_{i\geq 0}(a_i+\alpha_1b_i)T^i\in K(\psi,\alpha_1)[[T]]$, where $a_i,b_i\in K(\psi)$, by $$ \sigma F = \sum_{i\geq 0}(a_i+\sigma \alpha_1b_i)T^i. $$ for $\sigma\in G$. If $G$ is trivial then the claim is obvious, so suppose $G$ has order 2 and let $\sigma$ be the nontrivial element. In this case, Pollack claims (see the last sentence of the proof) that the two $p$-adic $L$-functions are conjugate, which I take to mean that $$ \sigma L_p(f,\alpha_1,\psi,T)=L_p(f,\alpha_2,\psi,T). $$ Assuming this, I can see why $G_\psi^\pm$ are rational: $\sigma$ clearly fixes $G_\psi^+$, and since $\alpha_1=-\alpha_2$ (recall $a_p=0$) we see that $\sigma$ also fixes $G_\psi^-$.

It's not clear to me why the $p$-adic $L$-functions need to be conjugate. Writing $L_p(f,\alpha_1,\psi,T)=\sum_{i\geq 0} (a_i+\alpha_1 b_i)T^i$ and $L_p(f,\alpha_2,\psi,T)=\sum_{i\geq 0} (c_i+\alpha_2 d_i)T^i$ for some $a_i,b_i,c_i,d_i\in K(\psi)$, it appears that we would need $a_i=c_i$ and $b_i=d_i$ for all $i$... Can someone clarify?

EDIT: More thoughts... Perhaps the fact that $L_p(f,\alpha_i,\psi,T)$ are conjugate can be derived from their definition in terms of $p$-adic distributions? The $p$-adic $L$-functions are defined as a pair of $p$-adic distributions $\mu_{f,\alpha_i}^\pm$ (linear maps on locally analytic functions $\mathbb{Z}_p^\times \rightarrow \mathbb{C}_p^\times$, assuming $M=1$ for simplicity) whose local values (on the open cover $a+p^n\mathbb{Z}_p$ for $(a,p)=1$) at polynomial functions $P$ of degree $\leq k-2$ are given by $$ \mu^{\pm}_{f,\alpha_i}(P; a+p^n\mathbb{Z}_p)=\frac{1}{\alpha^n}\lambda^\pm(f,P;a+p^n\mathbb{Z}_p)-\frac{\epsilon(p)p^{k-2}}{\alpha^{n+1}}\lambda^\pm(f,P;a+p^{n-1}\mathbb{Z}_p). $$ where the $\lambda$ terms are $K$-valued modular symbols. Then for $\sigma\in G$ as above we have $$ \sigma\mu^{\pm}_{f,\alpha_1}(P; a+p^n\mathbb{Z}_p)=\mu^{\pm}_{f,\alpha_2}(P; a+p^n\mathbb{Z}_p). $$ From this, is it correct to say that for a locally analytic character $\chi$ we then have $$ L_p(f,\alpha_2,\chi)=\mu_{f,\alpha_2}^{\text{sgn}(\chi)}(\chi)=\sigma\mu_{f,\alpha_1}^{\text{sgn}(\chi)}(\chi)=\sigma L_p(f,\alpha_1,\chi)? $$

I have a question about the proof of Theorem 3.5 in Pollack's 2003 paper On the $p$-adic L-function of a Modular Form and a Supersingular Prime.

The setup is as follows. Fix an eigenform $f\in S_k(N,\epsilon)$ with $a_p=0$ and let $K$ be the completion of the field of Fourier coefficients for $f$ at some fixed prime above $p$. Let $L_p(f,\alpha_i,\psi,T)\in K(\psi,\alpha_i)[[T]]$ be the associated unbounded $p$-adic $L$-functions for $f$, where $\alpha_1,\alpha_2$ are the roots of the Hecke polynomial $X^2-a_pX+\epsilon(p)p^{k-1}$ and $\psi$ is some fixed tame character on $\mathbb{Z}_p^\times\times (\mathbb{Z}/M\mathbb{Z})^\times$. The first sentence of Theorem 3.5 in Pollack's paper defines the power series $$ \begin{align*} G_\psi^{+}&=\frac{L_p(f,\alpha_1,\psi,T)+L_p(f,\alpha_2,\psi,T)}{2}\\ G_\psi^{-}&=\frac{L_p(f,\alpha_1,\psi,T)-L_p(f,\alpha_2,\psi,T)}{2\alpha_1}.\\ \end{align*} $$ Pollack states that both these power series lie in $K(\psi)[[T]]$. Why is this true?

Here are my thoughts... We have an action of $G=\operatorname{Gal}(K(\psi,\alpha_1)/K(\psi))$ on power series $F=\sum_{i\geq 0}(a_i+\alpha_1b_i)T^i\in K(\psi,\alpha_1)[[T]]$, where $a_i,b_i\in K(\psi)$, by $$ \sigma F = \sum_{i\geq 0}(a_i+\sigma \alpha_1b_i)T^i. $$ for $\sigma\in G$. If $G$ is trivial then the claim is obvious, so suppose $G$ has order 2 and let $\sigma$ be the nontrivial element. In this case, Pollack claims (see the last sentence of the proof) that the two $p$-adic $L$-functions are conjugate, which I take to mean that $$ \sigma L_p(f,\alpha_1,\psi,T)=L_p(f,\alpha_2,\psi,T). $$ Assuming this, I can see why $G_\psi^\pm$ are rational: $\sigma$ clearly fixes $G_\psi^+$, and since $\alpha_1=-\alpha_2$ (recall $a_p=0$) we see that $\sigma$ also fixes $G_\psi^-$.

It's not clear to me why the $p$-adic $L$-functions need to be conjugate, hence my main question is:

If $\alpha_1\notin K(\psi)$, why are the two $p$-adic $L$-functions $L_p(f,\alpha_1,\psi,T)$ conjugate?

Writing $L_p(f,\alpha_1,\psi,T)=\sum_{i\geq 0} (a_i+\alpha_1 b_i)T^i$ and $L_p(f,\alpha_2,\psi,T)=\sum_{i\geq 0} (c_i+\alpha_2 d_i)T^i$ for some $a_i,b_i,c_i,d_i\in K(\psi)$, it appears that we would need $a_i=c_i$ and $b_i=d_i$ for all $i$... Can someone clarify?

EDIT: More thoughts... Perhaps the fact that $L_p(f,\alpha_i,\psi,T)$ are conjugate can be derived from their definition in terms of $p$-adic distributions? The $p$-adic $L$-functions are defined as a pair of $p$-adic distributions $\mu_{f,\alpha_i}^\pm$ (linear maps on locally analytic functions $\mathbb{Z}_p^\times \rightarrow \mathbb{C}_p^\times$, assuming $M=1$ for simplicity) whose local values (on the open cover $a+p^n\mathbb{Z}_p$ for $(a,p)=1$) at polynomial functions $P$ of degree $\leq k-2$ are given by $$ \mu^{\pm}_{f,\alpha_i}(P; a+p^n\mathbb{Z}_p)=\frac{1}{\alpha^n}\lambda^\pm(f,P;a+p^n\mathbb{Z}_p)-\frac{\epsilon(p)p^{k-2}}{\alpha^{n+1}}\lambda^\pm(f,P;a+p^{n-1}\mathbb{Z}_p). $$ where the $\lambda$ terms are $K$-valued modular symbols. Then for $\sigma\in G$ as above we have $$ \sigma\mu^{\pm}_{f,\alpha_1}(P; a+p^n\mathbb{Z}_p)=\mu^{\pm}_{f,\alpha_2}(P; a+p^n\mathbb{Z}_p). $$ From this, is it correct to say that for a locally analytic character $\chi$ we then have $$ L_p(f,\alpha_2,\chi)=\mu_{f,\alpha_2}^{\text{sgn}(\chi)}(\chi)=\sigma\mu_{f,\alpha_1}^{\text{sgn}(\chi)}(\chi)=\sigma L_p(f,\alpha_1,\chi)? $$

I have a question about the proof of Theorem 3.5 in Pollack's 2003 paper On the $p$-adic L-function of a Modular Form and a Supersingular Prime.

The setup is as follows. Fix an eigenform $f\in S_k(N,\epsilon)$ with $a_p=0$ and let $K$ be the completion of the field of Fourier coefficients for $f$ at some fixed prime above $p$. Let $L_p(f,\alpha_i,\psi,T)\in K(\psi,\alpha_i)[[T]]$ be the associated unbounded $p$-adic $L$-functions for $f$, where $\alpha_1,\alpha_2$ are the roots of the Hecke polynomial $X^2-a_pX+\epsilon(p)p^{k-1}$ and $\psi$ is some fixed tame character on $\mathbb{Z}_p^\times\times (\mathbb{Z}/M\mathbb{Z})^\times$. The first sentence of Theorem 3.5 in Pollack's paper defines the power series $$ \begin{align*} G_\psi^{+}&=\frac{L_p(f,\alpha_1,\psi,T)+L_p(f,\alpha_2,\psi,T)}{2}\\ G_\psi^{-}&=\frac{L_p(f,\alpha_1,\psi,T)-L_p(f,\alpha_2,\psi,T)}{2\alpha_1}.\\ \end{align*} $$ Pollack states that both these power series lie in $K(\psi)[[T]]$. Why is this true?

Here are my thoughts... We have an action of $G=\operatorname{Gal}(K(\psi,\alpha_1)/K(\psi))$ on power series $F=\sum_{i\geq 0}(a_i+\alpha_1b_i)T^i\in K(\psi,\alpha_1)[[T]]$, where $a_i,b_i\in K(\psi)$, by $$ \sigma F = \sum_{i\geq 0}(a_i+\sigma \alpha_1b_i)T^i. $$ for $\sigma\in G$. If $G$ is trivial then the claim is obvious, so suppose $G$ has order 2 and let $\sigma$ be the nontrivial element. In this case, Pollack claims (see the last sentence of the proof) that the two $p$-adic $L$-functions are conjugate, which I take to mean that $$ \sigma L_p(f,\alpha_1,\psi,T)=L_p(f,\alpha_2,\psi,T). $$ Assuming this, I can see why $G_\psi^\pm$ are rational: $\sigma$ clearly fixes $G_\psi^+$, and since $\alpha_1=-\alpha_2$ (recall $a_p=0$) we see that $\sigma$ also fixes $G_\psi^-$.

It's not clear to me why the $p$-adic $L$-functions need to be conjugate. Writing $L_p(f,\alpha_1,\psi,T)=\sum_{i\geq 0} (a_i+\alpha_1 b_i)T^i$ and $L_p(f,\alpha_2,\psi,T)=\sum_{i\geq 0} (c_i+\alpha_2 d_i)T^i$ for some $a_i,b_i,c_i,d_i\in K(\psi)$, it appears that we would need $a_i=c_i$ and $b_i=d_i$ for all $i$... Can someone clarify?

EDIT: More thoughts... Perhaps the fact that $L_p(f,\alpha_i,\psi,T)$ are conjugate can be derived from their definition in terms of $p$-adic distributions? The $p$-adic $L$-functions are defined as a pair of $p$-adic distributions $\mu_{f,\alpha_i}^\pm$ (linear maps on locally analytic functions $\mathbb{Z}_p^\times \rightarrow \mathbb{C}_p^\times$, assuming $M=1$ for simplicity) whose local values (on the open cover $a+p^n\mathbb{Z}_p$ for $(a,p)=1$) at polynomial functions $P$ of degree $\leq k-2$ are given by $$ \mu^{\pm}_{f,\alpha_i}(P; a+p^n\mathbb{Z}_p)=\frac{1}{\alpha^n}\lambda^\pm(f,P;a+p^n\mathbb{Z}_p)-\frac{\epsilon(p)p^{k-2}}{\alpha^{n+1}}\lambda^\pm(f,P;a+p^{n-1}\mathbb{Z}_p). $$ where the $\lambda$ terms are $K$-valued modular symbols. Then for $\sigma\in G$ as above we have $$ \sigma\mu^{\pm}_{f,\alpha_1}(P; a+p^n\mathbb{Z}_p)=\mu^{\pm}_{f,\alpha_2}(P; a+p^n\mathbb{Z}_p). $$ From this, is it correct to say that for a locally analytic character $\chi$ we then have $$ L_p(f,\alpha_2,\chi)=\mu_{f,\alpha_2}^{\text{sgn}(\chi)}(\chi)=\sigma\mu_{f,\alpha_1}^{\text{sgn}(\chi)}(\chi)=\sigma L_p(f,\alpha_1,\chi)? $$