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Let $(p,q)$ be a pair of coprime (positive) integers. Consider the torus knot $T_{p,q}$. What is the minimal genus of an (embedded) oriented Seifert surface for this knot?

It is not had to convince oneself that in the simplest case $p =2$, there is a Seifert surface of genus $(q-1)/2$. I do not know whether that is optimal.

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  • $\begingroup$ Yes. The slice genus for torus knots is $(p-1)(q-1)/2$ (this is the Milnor conjecture) and the Seifert genus is at least the slice genus. For torus knots this lower bound on the Seifert genus is sharp; you've verified the $p = 2$ case of this. $\endgroup$ – dvitek Feb 13 '15 at 2:34
  • $\begingroup$ This can be found on Wikipedia - see en.wikipedia.org/wiki/Torus_knot#Properties - so I am voting to close. $\endgroup$ – Sam Nead Feb 13 '15 at 12:08
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One of the "classical" proofs involves the Alexander polynomial.

The knot group $\pi_1(S^3\setminus T_{p,q})$ has a presentation $\langle x,y \mid x^p = y^q\rangle$, and using Fox calculus one can quickly compute the Alexander polynomial to be $$\Delta_{T_{p,q}}(t) = \frac{(t^{pq}-1)(t-1)}{(t^p-1)(t^q-1)}.$$ The degree of the Alexander polynomial gives a bound on the genus, so we get $2g(T_{p,q})\ge\deg\Delta_{T_{p,q}} = (p-1)(q-1)$. Since this lower bound agrees with the upper bound given by Seifert's algorithm, you're done.

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Here's another route: the standard picture of the torus knot is a positive braid, so applying Seifert's algorithm gives a minimal genus surface. There are (say) $p$ seifert circles and $q(p-1)$ crossings, so rearranging the Euler characteristic gives the genus as $(p-1)(q-1)/2$.

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