When swapping roots of a polynomial, how to prove discriminant loops?

[swampwiz]

I was looking at this discussion of swapping roots of a polynomial causing the discriminant to loop around the origin.

https://www.akalin.com/quintic-unsolvability

Although it appears to be the case, has this mathematical fact ever been proven?

It seems that the formula for the discriminant is a product of squares of the difference between each combinatorial pair of roots, and thus the polar angle of the discriminant is simply twice the sum of the polar angle of the vector for each combinatorial pair of roots.

I can see how it works for the case of a quadratic polynomial, since there is only a single combinational pair, and the polar angle of the vector must make a net sweep of 180 degrees (i.e, in one of the directions), and so the polar angle for the discriminant must make a net sweep of 360 degrees. However, I don't see how this can be extended to the case of the cubic or higher polynomial.

 

[pasmith]

I don't think there is such a proof - partly because I believe the conjecture to be false, and partly because the topology of $\mathbb{C}$ is largely irrelevant to the theory of polynomials with complex coefficients and paths between points of the complex plane are just a neat way of illustrating graphically what is actually a discrete action of a symmetric group on the set of roots.

Root transpositions leave the discriminant unchanged, because the discriminant is a function of the coefficients, and the coefficients are symmetric functions of the roots.

A continuous function $f_{ij}: [0,1] \to \mathbb{C}^2$ with $f_{ij}(0) = (x_i,x_j)$ and $f_{ij}(1) = (x_j,x_i)$ defines paths taken by each root, and hence a path taken by the discriminant, which must necessarily be a closed loop because it starts and ends at the same point.

But the paths taken by the roots are otherwise entirely arbitrary. I'm sure you could find an $f_{ij}$ such that the discriminant doesn't loop the origin.

For example, in the case of a quadratic, if both roots are real then you can use paths which lie entirely on the real axis, and the discriminant would therefore be non-negative throughout, although the paths must intersect at some point, so the discriminant will touch the origin. In the case of a complex conjugate pair, you just have to ensure that the paths don't cross the real axis simultaneously, and the discriminant will never be real and positive and so can't loop the origin.

Of course since $f_{ij}$ has nothing to do with the theory of polynomials you can impose whatever symmetry conditions you like, and if you require that the paths be semi-circles of radius $\frac12|x_i - x_j|$ about $\frac12(x_i + x_j)$ then perhaps the discriminant must loop the origin.

 

[pasmith]

Of course since $f_{ij}$ has nothing to do with the theory of polynomials you can impose whatever symmetry conditions you like, and if you require that the paths be semi-circles of radius $\frac12|x_i - x_j|$ about $\frac12(x_i + x_j)$ then perhaps the discriminant must loop the origin.

In fact for this restricted choice of $f_{ij}$ it does, because the only factor of the discriminant which changes is $(x_i - x_j)^2$, and you can write $$f_{ij}(t) = \begin{pmatrix} f_1(t) \\ f_2(t) \end{pmatrix} = \frac{x_i + x_j}2\begin{pmatrix} 1 \\ 1 \end{pmatrix} + \frac{x_i - x_j}2e^{\mathrm{i} \pi t}\begin{pmatrix} 1 \\ -1 \end{pmatrix}$$ and thus if $x_i \neq x_j$ $$\begin{align*} \Delta(t) &= \Delta(0) \frac{(f_1(t) - f_2(t))^2}{(x_i - x_j)^2} \\ &= \Delta(0) \frac{(x_i - x_j)^2e^{2\pi \mathrm{i} t}}{(x_i - x_j)^2} \\ &= \Delta(0)e^{2\pi \mathrm{i} t} \end{align*}$$ which is a circle centered on the origin.

 

[swampwiz]

OK, I've figured it out. With respect to the discriminant, which is simply the product of the squares of the vectors of every combination of the roots, the vectors have either 0, 1 or both of the swapped roots. Those with 0 of them are simply constant. Those with 1 end up with the same vectors at the end (i.e., the swap), although there could be some fluctuation of the net polar angle depending on exactly the paths of the swapping. There is only one vector with both, and the result is that that vector does a 180 degree sweep, which means the square of it does a 360 degree sweep, thus the loop! Those with 1 could cause there to be additional loops, but in the end those loops will unwind, leaving on the single one.

 

[Office_Shredder]

I don't think there is such a proof - partly because I believe the conjecture to be false, and partly because the topology of $\mathbb{C}$ is largely irrelevant to the theory of polynomials with complex coefficients and paths between points of the complex plane are just a neat way of illustrating graphically what is actually a discrete action of a symmetric group on the set of roots.

There's an awesome proof that every complex polynomial has a complex root - Liouville's theorem says every nonconstant holomorphic function is unbounded on C, given a polynomial f 1/f is bounded and nonconstant unless f has a zero. So I think trying to construct analytic solutions to questions about polynomials is in general a reasonable approach. I don't think you should just throw up your hands and say this can't work because it's not algebra.

 

[akalin]

Hi, blog post author here! (I saw some traffic from here and figured it'd be an interesting discussion.) This is an interesting question! I don't know of any proof in a textbook, but it seems that with a little thought (and mild restrictions on the path) it can be proven.

I can see how it works for the case of a quadratic polynomial, since there is only a single combinational pair, and the polar angle of the vector must make a net sweep of 180 degrees (i.e, in one of the directions), and so the polar angle for the discriminant must make a net sweep of 360 degrees. However, I don't see how this can be extended to the case of the cubic or higher polynomial.

Your intuition is on the right track with thinking about the polar angle of the difference vector in the quadratic case. You basically want to calculate the winding number of the curve traced out by the discriminant around $0$, which must be a closed curve since any swap leaves the set of roots unchanged.

Now in order to talk about the winding number of a curve around $0$, we must stipulate that $0$ not be in the image of the curve; in other words, the polynomial must have no repeated roots, and furthermore any path chosen to swap two roots must not cause any root to coincide with any other root.

So cribbing from Wikipedia, one definition of the winding number around $0$ of a "nice" curve $𝛾 : [0, 1] → \mathbb{C}$, which here we take to mean that it's continuous and doesn't pass through $0$, is
$$n_𝛾 = \frac{𝜃_𝛾(1) - 𝜃_𝛾(0)}{2𝜋}\text{,}$$ where $𝜃_𝛾(t) := \operatorname{arg}(𝛾(t))$.

In particular, for a quadratic function with distinct roots $r_1$ and $r_2$, let $𝛾_1(t)$ and $𝛾_2(t)$ be the continuous curves swapping $r_1$ and $r_2$ such that $𝛾_1(t) ≠ 𝛾_2(t)$, and let $𝛥(t) = 𝛾_1(t) - 𝛾_2(t)$. Since $𝛥(1) = -𝛥(0)$, $𝜃_𝛥(1) - 𝜃_𝛥(0)$ must be $𝜋(1 + 2k)$ for some integer $k$, and thus for the curve $𝛥^2$ traced out by the discriminant, which must be a nice curve, $𝜃_{𝛥^2}(1) - 𝜃_{𝛥^2}(0) = 2𝜋(1 + 2k)$, and thus $𝛥^2$ must have winding number $1 + 2k$ for some integer $k$, i.e. an odd integer.

(You can generate any winding number by making $r_1$ and $r_2$ "circle" around each other $k$ times before swapping.)

In particular, $n_{𝛥^2} ≠ 0$, and thus the discriminant must loop around the origin at least once.

OK, I've figured it out. With respect to the discriminant, which is simply the product of the squares of the vectors of every combination of the roots, the vectors have either 0, 1 or both of the swapped roots. Those with 0 of them are simply constant. Those with 1 end up with the same vectors at the end (i.e., the swap), although there could be some fluctuation of the net polar angle depending on exactly the paths of the swapping. There is only one vector with both, and the result is that that vector does a 180 degree sweep, which means the square of it does a 360 degree sweep, thus the loop! Those with 1 could cause there to be additional loops, but in the end those loops will unwind, leaving on the single one.

As for the general case, you're also on the right track in considering the vectors between every root pair. The key tool in this case is additivity of winding numbers in $\mathbb{C}$ around $0$, namely that if $n_𝛼$ and $n_𝛽$ are winding numbers around $0$ of nice curves $𝛼$, $𝛽$, then $𝛼𝛽$ is also a nice curve with winding number $n_{𝛼𝛽} = n_𝛼 + n_𝛽$ around $0$. (See this StackExchange question for a proof.)

So given a polynomial with distinct roots $r_1, \dotsc, r_n$ where we want to swap $r_1$ and $r_2$, we can ignore the vectors from $r_i$ to $r_j$ with $i, j > 2$, since those contribute only a constant factor to the discriminant. The winding number of the square of the curve of the vector from $r_1$ to $r_2$ is odd, by the above. So all that remains is to consider the curves traced out from a root $r_i$ with $i > 2$ to $r_1$ and $r_2$.

Letting $𝛾_1$ and $𝛾_2$ be the curves swapping $r_1$ and $r_2$ as above, let $𝛾_i(t) = (r_i - 𝛾_1(t))(r_i - 𝛾_2(t))$, which is thus a nice curve. Therefore, it has some winding number $n_𝛾$ around $0$. But the curve traced out by the discriminant has a factor of $𝛾_i^2$, and thus by additivity $n_{𝛾_i^2} = 2n_{𝛾_i}$ is an even integer.

Therefore, again by additivity, the discriminant curve is a nice curve with winding number around $0$ equal to the sum of an odd integer and some number of even integers, i.e. an odd integer, and thus the discriminant must loop around the origin at least once.

(I'll address pasmith's and Office_Shredder's comments in another post. This one is already long!)

 

[akalin]

I don't think there is such a proof - partly because I believe the conjecture to be false...

For example, in the case of a quadratic, if both roots are real then you can use paths which lie entirely on the real axis, and the discriminant would therefore be non-negative throughout, although the paths must intersect at some point, so the discriminant will touch the origin.

So you're correct in that the conjecture as stated is false -- if at any point the paths are such that two roots coincide, the discriminant touches the origin, and so the winding number isn't well-defined. However, once you remove that case, the conjecture becomes true. (Unless, I've done something wrong above, which is a real possibility! :) )

In the case of a complex conjugate pair, you just have to ensure that the paths don't cross the real axis simultaneously, and the discriminant will never be real and positive and so can't loop the origin.

This isn't quite true; for a complex conjugate pair, you're looking at the vector of their difference, and so you need to ensure that the paths don't cross the same horizontal line simultaneously. It's straightforward to see that if you have continuous paths, this must happen at least once.

...and partly because the topology of $\mathbb{C}$ is largely irrelevant to the theory of polynomials with complex coefficients and paths between points of the complex plane are just a neat way of illustrating graphically what is actually a discrete action of a symmetric group on the set of roots.

Now I want to address this second part. I'd encourage you to keep an open mind about the relationship between various fields of mathematics; there are many beautiful parts of mathematics which arise from deep connections between seemingly-unrelated fields. :)

Office_Shredder already mentioned the Fundamental Theorem of Algebra, which is actually a theorem of analysis. Indeed, the "purely algebraic" proofs of the theorem need a small amount of analysis. In fact, my favorite proof of the FTA boils down to a single topological fact: removing a single point from the real line disconnects it, but removing a finite set of points from the complex plane leaves it connected. (My other blog post has more details.)

As for Galois theory, the application of topological methods to it is called topological Galois theory, pioneered by V.I. Arnold. It doesn't seem like a large field by any means, but the connection is there! (The idea of monodromy and monodromy group is also related, i.e. the Galois group of the generic $n$th degree polynomial over $\mathbb{C}$ can be realized as a monodromy group.)