Calculating the Seifert framing for an exceptional fiber in a Seifert-fibered integer homology 3-sphere Let $Y=\Sigma(\alpha_{1},\dots,\alpha_{n})$ be a Seifert fibered homology 3-sphere, let $\pi:Y\to\Sigma$ denote the projection to the orbifold surface, and let $T\subset Y$ be the pre-image under $\pi$ of a small disk neighborhood of the singular point of order $\alpha_{i}$, which we identify with a standard fibered solid torus $T(\alpha_{i},p_{i})\cong S^{1}\times D^{2}$.
Here, $T(\alpha_{i},p_{i})$ is such that a regular fiber is identified with the subset $(t,e^{2\pi it p_{i}/\alpha_{i}}\cdot z)\subset S^{1}\times D^{2}$ for $z\in D^{2}\setminus\{0\}$, $p_{i}$ is such that $1\le p_{i}<\alpha_{i}$, $p_{i}\beta_{i}\equiv 1\pmod{\alpha_{i}}$, and $(b,(\alpha_{1},\beta_{1}),\dots,(\alpha_{n},\beta_{n}))$ are the Seifert invariants of $Y$, normalized so that $1\le \beta_{i}<\alpha_{i}$.
Now let $F_{i}\subset T(\alpha_{i},p_{i})$ denote the core of the Seifert fibered solid torus, corresponding to the exceptional fiber in $Y$ over the singular point of order $\alpha_{i}$ in $\Sigma$. Given this setup, we obtain a distinguished framing of $F_{i}$ given by $S^{1}\times \{z\}\subset S^{1}\times D^{2}\cong T(\alpha_{i},p_{i})$ for $z\in D^{2}\setminus\{0\}$, which we'll denote by $\lambda_{T}$. On the other hand, since $F_{i}$ is null-homologous there is also a canonical framing $\lambda_{S}$ of $F_{i}$ induced by any Seifert surface $S$ with $\partial S=F_{i}$.
My question is: how do the framings $\lambda_{T}$ and $\lambda_{S}$ compare? Furthermore, is the Seifert framing $\lambda_{S}$ determined locally by the tuple $(\alpha_{i},\beta_{i})$ corresponding to this fiber, or does it depend in an essential way on the global topology of $Y$? Thanks!
 A: Here's an answer to my own question (thanks to Matt Hedden for the approach):
It will be helpful to fix a presentation of $\pi_{1}(Y)$. Let $T_{1},\dots,T_{n}$ be regular neighborhoods of the exceptional fibers. We can write $Y'=Y\setminus(\cup_{j}T_{j})$ as a circle bundle over $S^{2}\setminus(\cup_{i}D_{i})$ with Euler class $b$. Let $\ell\in\pi_{1}(Y')$ correspond to the singular fiber, and for $j=1,\dots, n$ let $m_{j}\in\pi_{1}(Y')$ correspond to the meridians around the deleted disks, oriented so that $m_{j}=\partial D_{j}$. With respect to these generators we can write
$$\pi_{1}(Y')=\langle m_{j},\ell\;|\;[m_{j},\ell]=m_{1}\cdots m_{n}\ell^{-b}=1\rangle.$$
After Dehn filling to obtain $Y$, we obtain the presentation
$$\pi_{1}(Y)=\langle m_{j},\ell\;|\;[m_{j},\ell]=m_{1}\cdots m_{n}\ell^{-b}=m_{j}^{\alpha_{j}}\ell^{\beta_{j}}=1\rangle.$$
The meridians $\mu_{j}$ and longitudes $\lambda_{T_{j}}$ of the filling fibered solid tori $T_{j}$ are given by $\mu_{j}=\alpha_{j}m_{j}+\beta_{j}\ell$ and $\lambda_{T_{j}}=-p_{j}m_{j}+q_{j}\ell$, where $p_{j},q_{j}$ are the unique integers which satisfy $q_{j}\alpha_{j}+p_{j}\beta_{j}=1$ and $1\le p_{j}<\alpha_{j}$.
Now fix $1\le i\le n$ and write the Seifert longitude of the exceptional fiber $K_{i}$ as
$$\lambda_{S_{i}}=\lambda_{T_{i}}+N_{i}\mu_{i}\in H_{1}(\partial T_{i}).$$
The Seifert longitude is characterized by the property that it maps to zero under the inclusion $H_{1}(\partial T_{i})\hookrightarrow H_{1}(Y\setminus T_{i})$. The above fundamental group presentation for $Y$ induces a natural presentation
$$H_{1}(Y\setminus T_{i})=\mathbb{Z}\langle m_{j},\ell\rangle/(\alpha_{j}m_{j}+\beta_{j}\ell=0\text{ for all }j\neq i, \sum_{j}m_{j}=b\ell).$$
Solving for $\lambda_{S_{i}}=0$, we see that
$$0=\lambda_{S_{i}}=\lambda_{T_{i}}+N_{i}\mu_{i}=(N_{i}\alpha_{i}-p_{i})m_{i}+(N_{i}\beta_{i}+q_{i})\ell$$
$$=(N_{i}\alpha_{i}-p_{i})(b\ell-\sum_{j\neq i}m_{j})+(N_{i}\beta_{i}+q_{i})\ell$$
$$=(N_{i}\alpha_{i}-p_{i})(b\ell+\sum_{j\neq i}\frac{\beta_{j}}{\alpha_{j}}\ell)+(N_{i}\beta_{i}+q_{i})\ell$$
$$\implies (N_{i}\alpha_{i}-p_{i})(b+\sum_{j\neq i}\frac{\beta_{j}}{\alpha_{j}})+(N_{i}\beta_{i}+q_{i})=0.$$
Rearranging, we obtain
$$N_{i}=\frac{-q_{i}+p_{i}b+p_{i}\sum_{j\neq i}\frac{\beta_{j}}{\alpha_{j}}}{\beta_{i}+\alpha_{i}b+\alpha_{i}\sum_{j\neq i}\frac{\beta_{j}}{\alpha_{j}}}.$$
Finally using the relation $\sum_{j\neq i}\frac{\beta_{j}}{\alpha_{j}}=-\Big(\frac{\beta_{i}}{\alpha_{i}}+b+\frac{1}{\alpha_{1}\cdots\alpha_{n}}\Big)$ and rearranging, we obtain the explicit formula
$$N_{i}=\frac{\alpha_{1}\cdots\alpha_{n}}{\alpha_{i}}(q_{i}-bp_{i})-p_{i}\sum_{j\neq i}\frac{\alpha_{1}\cdots\alpha_{n}}{\alpha_{i}\alpha_{j}}\beta_{j}.$$
