I'm seeking help understanding this video. The author considers the equation

$ax^5+bx^4+cx^3+dx^2+ex+f = 0$

and shows both the coefficients $a, b$... and solutions $x_1, x_2$... in the complex plane. The author claims that if the coefficients are varied, moving them along short loops so they return to their original values and the solutions don't return to their original values, but instead exchange places, an expression for the solution cannot be found.

What is the significance of the solutions not returning to themselves?


1 Answer 1


I think the claim is a bit more complicated than that. For simplicity, let's look at a quadratic polynomial. If the roots are $r$ and $s$, and we impose the condition that the leading coefficient be $1$, then the polynomial is $$ (x-r)(x-s)=x^2-(r+s)x+rs=x^2+bx+c, $$ with $b=-(r+s)$ and $c=rs$. So we can find $b$ and $c$ if we know $r$ and $s$. The problem at hand, however, is the reverse: to find $r$ and $s$ given $b$ and $c$. To solve this in general means to find functions $f$ and $g$ such that $$ r=f(b,c),\qquad s=g(b,c). $$

The issue is that this is a somewhat paradoxical demand. The expressions for $b$ and $c$ in terms of $r$ and $s$ are symmetric under interchange of $r$ and $s$ (as they must be, since permuting the roots doesn't change the product $(x-r)(x-s)$). Because of the symmetry between $r$ and $s$, how can the function $f$ know which of $r$ and $s$ it is supposed to be finding?

This issue is put into sharp relief by the idea of setting $r$ and $s$ in motion. If $r$ and $s$ move around and then return to their starting values, but with $r$ and $s$ interchanged, $b$ and $c$ will return to their original values. Therefore $f(b,c)$ and $g(b,c)$ would seem to return to their original values, which were $r$ and $s$. But this appears to be wrong: since $r$ moved around and ended up at $s$, it seems that $f(b,c)$ should equal $s$ at the end of this motion, not $r$.

The resolution of the paradox is that, in complex analysis, functions can have multiple branches. So the value of the function $re^{i\theta}\mapsto\sqrt{r}e^{i\theta/2}$ does not return to its original value if $r$ is held fixed (say to $1$) and $\theta$ varies from $0$ to $2\pi$, but, instead, picks up a minus sign. A second circuit around the origin does return the value to the starting value.

This is exactly what is needed to allow $r$ and $s$ to move in such a way that $b$ and $c$ end up with their original values, but $f(b,c)$ and $g(b,c)$ end up swapping values. In fact we know that the quadratic formula makes use of the function $re^{i\theta}\mapsto\sqrt{r}e^{i\theta/2}$, which is what allows this magic to occur.

In summary, there is, in the final analysis, no problem with roots not returning to their original values at the same time that the coefficients in the polynomial do return to their original values.

So how does the proof in the video work? What it does is to try to construct very special paths so that, not only do $b$ and $c$ return to their original values, but any allowable formulas one might propose for $f(b,c)$ and $g(b,c)$ also return to their original values, while at the same time $r$ and $s$ swap values. ("Allowable formulas" means formulas that involve only taking roots and using the four arithmetic operations.) If such paths could be found, then we would have a proof by contradiction that formulas for $f(b,c)$ and $g(b,c)$ cannot exist.

We, of course, know this strategy must fail in the quadratic case, since the quadratic formula does, in fact, exist. But it is exactly this strategy that works in the quintic case. While in the quadratic case (and also in the cubic and quartic cases) any path the forces the values of $f(b,c)$ and $g(b,c)$ back to their starting values also ends up forcing $r$ and $s$ back to their original values (rather than swapping them), in the case of five roots, there are paths such that allowable formulas expressing the roots in terms of coefficients, $g(b,c,d,e,f)$, $h(b,c,d,e,f)$, ..., are all forced to return to their original values, but the roots, $r$, $s$, ..., are permuted. This produces the proof by contradiction.

Added: I hope I was clear about two points in my original answer: (1) the paradox described in the paragraph beginning "The issue is put into sharp relief..." is only apparent; (2) that paragraph does not contain the idea of the proof of Abel's theorem, although the correct idea is somewhat related to the apparent paradox described there.

So what is the apparent paradox? To make precise how the roots move, we imagine continuous functions $R(t)$, $S(t)$ of $t\in[0,1]$ such that $R(0)=r$, $S(0)=s$, $R(1)=s$, $S(1)=r$. Let the coefficients of the quadratic be $B(t)=-(R(t)+S(t))$, $C(t)=R(t)S(t)$. These are continuous functions of $t$ as well. Suppose that our hypothetical formulas for the roots are such that $f(B(0),C(0))=r$ and $g(B(0),C(0))=s$. The apparent paradox is that two different lines of reasoning give different values for $f(B(1),C(1))$.

One argument says that $B(1)=B(0)$ and $C(1)=C(0)$, and therefore $$ \begin{aligned} f(B(1),C(1))=f(B(0),C(0))&=r,\\ g(B(1),C(1))=g(B(0),C(0))&=s. \end{aligned} $$ (As discussed above, this argument is actually incorrect.)

The other argument is that $R(t)$ and $S(t)$ change continuously with $t$, and therefore so do $B(t)$, $C(t)$, $f(B(t),C(t))$, and $g(B(t),C(t))$. From this, we conclude that, since $f(B(0),C(0))=R(0)$, $f(B(t),C(t))$ remains equal to $R(t)$ throughout the entire motion, and therefore $f(B(1),C(1))=R(1)=s$. (The roots typically remain a finite distance apart throughout their motion, so the only way to end up with $f(B(1),C(1))=r$ is for a discontinuous jump to occur at some point, which cannot happen.)

If both arguments had been correct, we would have a proof by contradiction that the function $f$ (and $g$) cannot exist. But since the first argument is incorrect, this does not prove the nonexistence of $f$. In fact, $f$ does exist in the quadratic case. To prove the nonexistence of the analogous functions in the quintic case requires a more elaborate argument, as discussed above.

  • $\begingroup$ Can you clarify the paragraph that begins "This issue is put into sharp relief..."? I'm sorry. I still don't understand. $\endgroup$
    – Noah
    Jul 20, 2016 at 0:02
  • $\begingroup$ >"If r and s move around and then return to their starting values, but with r and s interchanged." If r and s return to their starting values, they are not interchanged. If they are interchanged they do not return to their starting values. Please clarify $\endgroup$
    – Bob Ueland
    Feb 10, 2023 at 9:11
  • $\begingroup$ @BobUeland I was trying to express the that at the end of some continuous motion the set consisting of the values of $r$ and $s$ returns to itself, but with the values of $r$ and $s$ interchanged. For more details, look at the paragraph in the "Added" part that begins "So what is the apparent paradox?" The sets $\{R(0),S(0)\}$ and $\{R(1),S(1)\}$ are the same, but $R(0)$ and $R(1)$ are not the same. Rather $R(0)=S(1)$ and $R(1)=S(0)$. $\endgroup$ Feb 10, 2023 at 15:38

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .