I know that there is this result of Milnor that all algebraic links are fibered. And the $(p,q)$-torus link is an algebraic link. But then I'm reading this paper of Baader and Graf (http://dx.doi.org/10.1016/j.exmath.2016.06.006), and they give an argument (Example 3.1) that the $(2,2n)$-torus link is not fibered.
In a bit more detail, they recall a result that the fiber of a fibered link is the minimal genus Seifert surface for the link. They then show that an unknotted ribbon with $n$-full twists, whose boundary is a $(2,2n)$-torus link, cannot be a fiber surface. Since this ribbon (i.e. annulus) clearly has minimal genus, its boundary cannot be fibered.
It's even more puzzling because the following example in the paper (Example 3.2) shows a way to fiber the complement of any $(p,q)$-torus link. A friend suggested this might have something to do with orientation of the link components. Can anyone fill in the details?
Indeed, Milnor's Fibration Theorem implies all torus links $T(p,q)$ are fibered. This is because the polynomial $f(z,w)=z^p+w^q$ satisfies the hypothesis that the zero set of the system $f(z,w)=0$, $f_z(z,w)=0$, and $f_w(z,w)=0$ is only at the origin. Hence, the function $$(z,w) \mapsto \frac{f(z,w)}{\lvert f(z,w)\rvert}$$ when restricted to the complement of the zero locus of $f(z,w)$ in the unit sphere $S^3\subset \mathbb{C}^2$ gives a map $S^3\setminus T(p,q) \to S^1$. This is a fibration. The preimage of each point of $S^1$ is a Seifert surface for $T(p,q)$.
That paper cites Kawauchi, "A survey of Knot Theory." The definition they give for a Seifert surface is
This is a little different from the usual definition, which insists that $S$ be connected, but it ends up not mattering in what follows. But, the key thing here is that $L$ is oriented. Sort of the point of Seifert surfaces is they geometrically represent a class in $H_2(S^3-\nu L, \partial(S^3-\nu L))$ that is Poincaré dual to the class in $H^1(S^3-L)$ that is Alexander dual to the link's orientation class in $H_1(L)$. That is to say, a Seifert surface lets you measure the orientation of a link by taking a meridian loop and computing the intersection number with the surface: whether it is $+1$ or $-1$ determines whether or not the meridian loop is oriented correctly with respect to the link component. That is to also say that the oriented boundary of the Seifert surface matches the orientation of the link.
Importantly, all Seifert surfaces for the same oriented link are homologous in $H_2(S^3-\nu L, \partial(S^3-\nu L))$ by the above dualities.
By the way, recall that classes in $H^1(S^3-L)$ are the same as homotopy classes of maps $S^3-L \to S^1$. Given a smooth representative map, the fiber at a regular value gives a Seifert surface. This is what connects Milnor fibers to Seifert surfaces.
Going back to $T(p,q)$ as an algebraic link, we haven't specified orientations! It turns out that there is a well-enough defined orientation of the algebraic link coming from orienting the zero-locus in $\mathbb{C}^2$ and intersecting it with $S^3$. The induced orientation is one where all the components are co-oriented with respect to the core axes of the Clifford torus.
According to the book, a link $L$ is a fibered link if there is a Seifert surface $S$ for $L$, called the fiber surface, such that $(E', \partial E')$ is homeomorphic to $(S_E, \partial S_E)\times [0,1]$, where $E=S^3-\nu L$ is the link exterior, $S_E=S\cap E$, and $E'$ is $E$ cut along $S_E$. The idea here is that you could glue $(S_E,\partial S_E)\times\{0\}$ to $(S_E,\partial S_E)\times\{1\}$ as a mapping torus of $(S_E,\partial S_E)$ to obtain the fibration; the homeomorphism of the surface that you use to do the gluing is known as the monodromy.
Fiber surfaces must be connected, so at this point the definition of Seifert surface agrees with the usual one.
Another way we could have defined a fibered link is a link $L$ such that its exterior has a fibration $S^3\setminus \nu L\to S^1$ such that one can extend this to be an open book decomposition of $S^3$. (What's this additional constraint? It's that the fiber surfaces define longitudinal curves. The exterior of a Hopf link is a thickened torus, and there are ways to fiber it such that fiber surfaces do not correspond to Seifert surfaces.) This map induces an orientation on the link. That is to say, a link is fibered if any of its orientations are fibered with respect to the above definition.
This is the result that the paper cites:
A catch-phrase you can derive from this is that "the fiber surface has the minimum genus in its homology class."
Going to the paper now, Example 3.1 discusses a $T(2,2n)$ torus link where one of the components has reversed orientation. This is the oriented boundary of a standard annulus with some number of twists, which is a genus-0 Seifert surface. They use their nice cord criterion to show this is not a fiber surface, so therefore the torus link with these orientations is not a fibered link.
They mention that the Alexander polynomial of this torus link is not monic. There is a characterization of the Alexander polynomial of a fibered knot as the characteristic polynomial of the monodromy's induced map on $H_1(S)$, and characteristic polynomials are monic. Alexander polynomials of links depend on the orientations of the components, and I think (but haven't verified) that the above orientation has Alexander polynomial $n(t-1)$, so unless $n=\pm 1$ it's not monic (Alexander polynomials are defined up to multiplication by $\pm t^{\pm 1}$, so take an appropriate definition of "monic" here). However, the orientation as an algebraic link has an Alexander polynomial that's a product of cyclotomic polynomials, so it's monic (and, indeed, fibered!)
After writing all this, I went back to the introduction of the paper, where in the second sentence they make it clear that all links are oriented :-)