V.I. Arnold's Topological Proof of the Fifth Degree I have been playing around with some topological ideas in Galois Theory, and I find that it is easy to force a swapping of the roots in a quadratic equation if I slowly drag one of the coefficients around the origin, returning it to where I started. But I can't easily do the same with a cubic. I can take an equation like x^3 + 7x -11 = 0, I can drag either the 7 or the 11 in a big circle about the origin, but I can't thereby swap the two complex roots (or any other pair of roots). Sometimes I can force a cyclic permutation, and sometimes the roots meander around and return to where I began (depending on what cubic equation I start with). But I haven't been able to swap two roots. Is there some reason for this?
 A: We can write out quite explicitly what a path through the space of cubics with distinct roots that transposes two roots looks like, even a path that maintains a zero quadratic term: take the three roots to be $r_1(t) = e^{it}, r_2(t) = - e^{it}, r_3(t) = 0$ where $t \in [0, \pi]$ (so that $r_1(0) = r_2(\pi) = 1, r_2(0) = r_1(\pi) = -1$). This gives the cubic polynomial
$$(x - r_1)(x - r_2)(x - r_3) = x^3 - e^{2it} x$$
which has a linear term going in a circle around the origin and a zero constant term.
Getting an example where the constant term is nonzero is a bit more annoying if you insist on keeping the quadratic term zero (it's easy without this, just translate the example above); this amounts to requiring that $\frac{1}{r_1(t)} + \frac{1}{r_2(t)} + \frac{1}{r_3(t)} = 0$ which makes the formulas less nice. We can take
$$\frac{1}{r_1(t)} = 2 + e^{it}$$
$$\frac{1}{r_2(t)} = 2 - e^{it}$$
$$\frac{1}{r_3(t)} = -4$$
which gives the cubic polynomial
$$(x - r_1)(x - r_2)(x - r_3) = x^3 - \left( \frac{4}{4 - e^{2it}} - \frac{1}{4} \right) x - (16 - 4e^{2it}).$$
So the constant term travels in a circle. The linear term does also, because it's an inversion of a circle. If you take the path to be $t \in \left[ \frac{\pi}{2}, \frac{3 \pi}{2} \right]$ you even get an example where the two roots being swapped start out complex.
I don't know what happens if you only make the constant and linear terms go around in circles separately. What might be going on there is the following: if you move both terms freely you are working in the space of (monic) cubics with distinct roots and zero quadratic term, which can be identified with the complement in $\mathbb{C}^2$ of the zero locus of the discriminant, which for a cubic polynomial $x^3 + px + q$ with zero quadratic term takes the form $\Delta = - 4p^3 - 27 q^2$. I believe this complement is homotopy equivalent to the complement of the trefoil knot; in any case there's a somewhat complicated locus of points you can't cross, and if you don't take sufficiently complicated loops you may not be able to generate the entire fundamental group (which is famously the braid group $B_3$, also the fundamental group of the entire configuration space $C_3(\mathbb{C})$, also known as the space of (monic) cubics with distinct roots).
I just found a quite nice page that explains the connection to configuration spaces and the trefoil knot (with pictures and a video!) here.
