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First time posting, so sorry if this is an uninteresting or overly long post!

The inspiration for this question was sparked by this answer given by Bruno Martelli in response to a question about horizontal surfaces in Seifert fibered spaces as the fiber of a fiber bundle over the circle with periodic monodromy.

When seeing Seifert fibered space and a surface bundle over the circle with periodic monodromy, torus knots in $S^3$ fit into this class and are well behaved. In Martelli's response however, he restricts to Seifert fibered spaces without boundary, so let's Dehn fill the torus knot, say $T(p,q)$ along boundary slope $\frac{r}{s}$.

Now if I remember correctly, we can't fill along $\frac{pq}{1}$ to get a closed Seifert fibered space, because it results in a connect sum of lens spaces from capping off the cabling annulus. Every other slope should yield a Seifert fibered space over $S^2$ with 3 singular fibers, a rather tricky class of Seifert fibered spaces. The surgered manifold should be a Seifert fibered space over the sphere as the underlying topological space for the torus knot's base orbifold was a disk, so Dehn filling the boundary should cap off the base orbifold's underlying topological space boundary to an $S^2$. Another way to see this is perhaps simply from the classification/construction of Seifert fibered spaces.

If $T(p,q)$'s exterior was $M_k$ we can give an easy van Kampen argument to yield a presentation for $\pi_1(M_k)$ as $<a,b |a^pb^q>$ where $a$ and $b$ are the classes of the singular fibers in the Seifert fibering of $M_k$. From here we should be able to give another Van Kampen argument to yield a presentation for the fundamental group of the the surgered manifold $M_k(\frac{r}{s})$, where we just add a relation to the previous presentation based on $\frac{r}{s}$. (EDIT, my original post had calculated $\pi_1(M_k(\frac{r}{s})) = <a,b |a^pb^q, a^rb^s>$, which I believe is wrong. I think mixed up the Heegard torus from the previous calculation with the knot's boundary torus so the new relation is wrong.)

I suppose geometrization in all its high powered glory may be able to tell us something about the geometry of $M_k(\frac{r}{s})$ based on this fundamental group calculation, if this presentation isn't too unpleasant. A more pedestrian understanding would be appreciated though, as I understand little about geometrization and how the geometry looks near the decomposing tori, and know nothing of how the geometries glue together coherently to produce whatever it is they produce.

So, I'm wondering if we can track through the construction of these Seifert fibered spaces through Dehn filling a trivial circle bundle over a pair of a pants with slopes $\frac{p}{q}$, $\frac{q}{p}$, and $\frac{r}{s}$, how the geometry changes, if we imagine that we start with a hyperbolic pair of pants $P$ with geodesic boundary. So onto the actual questions:

0.) Is there a way to see the geometry evolve as we fill $P \times S^1$?

1.) Of particular interest is the $\mathbb{H}^2 \times \mathbb{R}$ geometry that might arise, given that we can start with an $\mathbb{H}^2 \times \mathbb{R}$ geometry. Can the $\mathbb{H}^2 \times \mathbb{R}$ geometry ever show up in filled torus knots?

2.) It is known that torus knots are L-space knots, and so for every filling slope $\frac{r}{s} \geq pq-(p+q) = 2g-1$, where $g$ denotes the knot genus of $T(p,q)$, the surgered manifolds are L-spaces, is the class of possible geometries for the L-space slopes different from the class of geometries possible for the non-L-space slopes?

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    $\begingroup$ Just some pointers. Moser gave explicit descriptions of Dehn fillings of torus knots (Pacific J. Math. 38, 1971). (Infinitely many of them are lens spaces: all $pq\pm 1/n$ for $0 \neq n\in \mathbb{Z}$). There's also a very nice chapter on SFS in Fomenko and Matveev's book Algorithmic and computer methods for 3-manifolds. I think your geometric questions are already answered there, but additionally you can take a look at Scott's The geometry of 3-manifolds (Bull. LMS 15,1983). $\endgroup$ – Marco Golla Aug 19 '19 at 19:47
  • $\begingroup$ Those references are indeed helpful, thank you. I wasn't able to detect the geometric flavor in Fomenko and Mateev's book, Peter Scott's clearly discusses the geometry. I wasn't aware the geometry and base orbifold structure were so nicely related. I had previously only picked up on the topology from Hatcher's notes, and in passing heard that it underlies a fancier orbifold structure. I'll try to do better research before I ask a question next time. $\endgroup$ – Unlit Candle Aug 21 '19 at 21:30
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The general principle is that a generic filling of a geometric manifold belongs to the same geometry of the original manifold. This holds notably in hyperbolic geometry by Thurston's Dehn filling theorem. In the other geometries, this principle is somehow also true, but some care is needed to interpret it correctly: the manifold $P\times S^1$ can be equipped with two geometries, either $\mathbb H^2 \times \mathbb R$ or $\widetilde{SL_2}$, and both may appear in a filling - the latter being much more generic than the former.

The geometry of the fillings of $P\times S^1$ can be described easily. Use the (oriented) boundaries of $P$ and the fiber $S^1$ as meridians and longitudes. Represent the fillings with respect to this basis as coprime pairs $$(p_1,q_1), \qquad (p_2,q_2), \qquad (p_3,q_3).$$ If $p_i=0$ for some $i$, you get a sum of lens spaces. If not, you get a Seifert manifold, whose geometry is nicely determined by the invariants $$\chi = -1 + \frac 1{p_1} + \frac 1{p_2} + \frac 1{p_3}$$ $$e = \frac{q_1}{p_1} + \frac{q_2}{p_2} + \frac{q_3}{p_3}$$ If $p_i=1$ for some $i$, you get a lens space, that has the $S^3$ geometry.

Here $\chi$ is the Euler characteristic of the base orbifold. Generically you get $\chi < 0$, and in this case you get the geometry $\mathbb H^2 \times \mathbb R$ if $e=0$ and $\widetilde{SL_2}$ if $e\neq 0$.

You get a torus knot by filling the first two components with parameters $(p_1,q_1)$ and $(p_2,q_2)$ such that $p_1q_2+p_2q_1=1$. In this case $$e = \frac{1}{p_1p_2} + \frac{q_3}{p_3}.$$ Therefore every torus knot has precisely one Dehn filling with $e=0$, obtained by taking $(p_3,q_3)=(-p_1p_2,1)$. This gives a manifold of geometry $\mathbb H^2 \times \mathbb R$ as soon as $\chi < 0$. The latter condition should be satisfied by all torus knots except the trefoil knot, where $p_1,p_2,p_3 = 2,3,6$ and you get a flat manifold.

Conclusion: Every torus knot except the trefoil has precisely one filling with geometry $\mathbb H^2 \times \mathbb R$. Generically, its Dehn fillings are of the geometry $\widetilde{SL_2 \mathbb R}$, but as noted by Marco there are also infinitely many fillings that are lens spaces, obtained by taking $p_3=1$: these have $\chi>0$, and are not generic in some natural sense.

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  • $\begingroup$ Thanks for the answer, this helps a lot! The remark that you get a torus knot by filling two of the components with $(p_1,q_1)$ and $(p_2,q_2)$ such that $p_1q_2 + p_2q_1=1$, is sufficient to arrive at a torus knot, not necessary correct? I could fill with $(p_1,q_1)$ and $(q_1,p_1)$ to get the $T(p_1,q_1)$ complement right? Also, if one imagines doing these two fillings first, followed by the third playing the role of filling the torus knot, is the original framing from $P x S^1$ compatible with the standard framing coming from the homologically trivial slope playing the role of longitude? $\endgroup$ – Unlit Candle Aug 21 '19 at 20:44
  • $\begingroup$ Given that there exists a unique filling of $T(p,q)$ that yields a closed orientable SFS with Euler number 0, I am now confused about [your answer to this question][1]. That answer says for an orientable SFS without boundary, Euler number 0 is equivalent to the existence of a horizontal surface. $T(p,q)$ has infinitely many fillings yielding a closed orientable SFS with horizontal foliations. Each leaf of such a foliation being a horizontal $\pi_1$ injective non-compact surface. Does your usage of horizontal discount this sort of problem? [1]: mathoverflow.net/a/338530 $\endgroup$ – Unlit Candle Aug 23 '19 at 18:39
  • $\begingroup$ A horizontal surface is usually meant to be compact. $\endgroup$ – Bruno Martelli Aug 23 '19 at 21:40

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