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!



Calculating the Seifert framing for an exceptional fiber in a Seifert-fibered integer homology 3-sphere
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


$$N\_{i}=\frac{\alpha\_{1}\cdots\alpha\_{n}}{\alpha\_{i}}(q\_{i}-bp\_{i})-p\_{i}\sum\_{j\neq i}\frac{\alpha\_{1}\cdots\alpha\_{n}}{\alpha\_{i}\alpha\_{j}}\beta\_{j}.$$


