I am trying to connect the definition of $\pm$-modular symbols given in [Pollack, pg. 529] and [MTT,pg. 11] to those appearing in [Greenberg-Stevens, pg. 200 in #20 here], but I can't seem to reconcile a factor of $(-1)^{k-j}$ (see below) showing up. Can someone clarify?
Fix an eigenform $f\in S_k(N,\epsilon)$ and let $K_f$ be its associated number field. Following MTT, Pollack considers the modular integrals $$ \phi(f,P,r) := 2\pi i \int_{i\infty}^rf(z)P(z)dz $$ for $P\in \mathbf{C}[z]$ of degree $\leq k-2$, $r\in \mathbf{P}^1(\mathbf{Q})$, and states that one can find complex periods $\Omega_f^\pm$ such that $$ \lambda^\pm(f,P(z);a,m) := \frac{\pi i}{\Omega_f^\pm}\bigg(\int_{i\infty}^{a/m}f(z)P(mz-a)dz\pm\int_{i\infty}^{-a/m}f(z)P(mz+a)dz\bigg)\in K_f $$
Now let $V_g(\mathbf{C})$ be the space of homogeneous polynomials of degree $g$ in variables $X$ and $Y$ over $\mathbf{C}$. Greenberg-Stevens defines the modular symbol $\xi_f\in \operatorname{Hom}(\operatorname{Div}^0(\mathbf{P}^1(\mathbf{Q})),V_{k-2}(\mathbf{C}))$ by $$ \xi_f(\{r\}-\{s\}):=2\pi i\int_s^rf(z)(zX+Y)^{k-2}dz\in V_{k-2}(\mathbf{C}). $$ These modular symbols should be `dual' to the modular integrals defined above in the sense that the integrals should appear as coefficients of the modular symbol. In any case, it should be possible to recover at least $\lambda^\pm(f,z^j;a,m)$ from the modular symbol. The action of $\operatorname{GL}_2(\mathbf{Z})$ on maps $\xi:\operatorname{Div}^0(\mathbf{P}^1(\mathbf{Q}))\rightarrow V_{k-2}(\mathbf{C})$ is given by $$ (\xi\mid \gamma)(\{r\}-\{s\})=\xi(\{\gamma r\}-\{\gamma s\})\mid\gamma, $$ where the action on homogeneous polynomials $P\in V_{k-2}(\mathbf{C})$ is given by $ P(X,Y)\mid\gamma=P(dX-cY,-bX+aY). $ Since $\Gamma_1(N)$ is normalized by $\iota=\begin{pmatrix} -1 &0\\ 0&1\end{pmatrix}$, we have a decomposition $\xi_f=\xi_f^++\xi_f^-$ by orthogonal idempotents $(1\pm \iota)/2$. Let's now scale $\xi_f^\pm$ by the same periods $\Omega_f^\pm$ as above and consider the modular symbol $\varphi_f^\pm:=\xi_f^\pm/\Omega_f^\pm$. My hope had been that the coefficients of $\big(\varphi_f^\pm\mid \gamma \big)(D)$, where $\gamma=\begin{pmatrix} 1 & a\\ 0&m\end{pmatrix}$ and $D=\{\infty\}-\{0\}$, would allow us to recover the values $\lambda^\pm(f,z^j;a,m)$ of Pollack. This is nearly true: expanding everything out, we get that $$ \big(\varphi_f^\pm\mid \gamma \big)(D)=\sum_{j=0}^{k-2}\binom{k-2}{j}c_{f,j}^\pm(a,m)X^jY^{k-2-j} $$ where $$ c_{f,j}^\pm(a,m)=\frac{\pi i}{\Omega_f^\pm}\bigg(\int_{i\infty}^{a/m}f(z)(mz-a)^jdz\pm(-1)^{k-j}\int_{i\infty}^{-a/m}f(z)(mz+a)^jdz\bigg). $$ Thus, it appears that $c_{f,j}^\pm(a,m)$ only agree with $\lambda^\pm(f,z^j;a,m)$ when $(-1)^{k-j}=1$. Is it possible that Pollack is missing this factor in his definition? That is, one can define a right action of $\operatorname{GL}_2(\mathbf{Z})$ on the maps $\phi_f:=\phi(f,\_,\_):\mathbf{C}[z]\times \mathbf{P}^1(\mathbf{Q})\rightarrow \mathbf{C}$ by setting $$ (\phi_f\mid\gamma)(P,r) := \phi_f(P(\gamma^*z),\gamma r), $$ where $\gamma^*$ is the adjugate of $\gamma$. One can then scale and decompose $\phi_f$ by orthogonal idempotents and, doing this, it seems like the $\lambda^\pm$ may indeed be off by a sign... Can someone clarify/verify this?
