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.



Minimal genus of Seifert surface of torus knot
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.


