On the topological proof of the Fundamental Theorem of Algebra

[swampwiz]

Sorry for the misspelling, but this forum doesn't allow enough characters for the title. The title should be:

For the topological proof of the Fundamental Theorem of Algebra, what is the deal when the roots are at the same magnitude, either at different complex angles, or repeated roots?

I was looking at this explanation @ Youtube, which explains that if the polynomial variable is some value at some very large magnitude, the value of the function as per that value of the polynomial variable is a loop that must circle the origin with the number of times - i.e., the winding number - being the degree of the polynomial, and then as this magnitude is modulated continuous to smaller values, this loop will collapse onto the point that is the value of the free term of the polynomial (let's presume it is non-zero), with the winds of the loops crossing the origin, thus signifying that there exists a root with the magnitude of that root being the modulated radius of the circle, and such that further modulation of the radius to lower values will result in that wind of the loop never again winding around, thus reducing the winding number by 1. And so finally, since the loop that is simply the degenerate value of the free term has a winding number of 0, there must be the number of roots that is equal to the degree of the polynomial.

OK, so if the roots are all at different magnitudes, I can see this happening. However, if there a multiple roots at the same magnitude - and even more so if there are multiple roots (i.e., same magnitude and complex angle) - I am confused about how this schema handles this situation.

 

[Orodruin]

The only thing that is important is that you have a winding number, you do not actually have to shrink the circle as a circle, you can deform it in any continuous way you choose.

However, note that the version of the fundamental theorem that he shows is that every polynomial of degree $d \geq 1$ has at least one complex root. All you need is therefore that you have to pass zero. It is relatively easy to go from there to show that it has exactly $d$ roots by factoring out $z-z_0$ and getting a polynomial on the form $P_d(z) = (z-z_0)P_{d-1}(z)$.

 

[Orodruin]

As an example, here is what happens for $f(z) = z(z-i)^2$ as you shrink the radius from 1.3 (purple) to 0.7 (green). $z = 0$ is shown as a red star for clarity.

As you can see, what happens is that you get a small loop around zero which shrinks to zero size just when you pass the double root with your curve. For a multiplicity of $n$, this small loop will wind $n-1$ times around zero.

 

[wrobel]

Very nice explanation but I do not believe that it is a proof in mathematical sense. When he shrinks the circle, all the terms of the polynomial become to matter. Why the image of the circle can not be broken open during this shrinking? I think that to answer this question, one have to employ theorems of complex analysis again.

By the way, the true elementary proof is to show first that the function $z\mapsto |f(z)|$ attains its minimum in $\mathbb C$ This minimum is a root. Two simple assertions to prove.

 

[Orodruin]

Why the image of the circle can not be broken open during this shrinking?

This is where topology enters the picture. Essentially you need the continuity of $z \mapsto f(z)$. Of course he is not spelling out the proof in its entirety, it is more of a casual video after all.

 

[lavinia]

Very nice explanation but I do not believe that it is a proof in mathematical sense. When he shrinks the circle, all the terms of the polynomial become to matter. Why the image of the circle can not be broken open during this shrinking? I think that to answer this question, one have to employ theorems of complex analysis again.

By the way, the true elementary proof is to show first that the function $z\mapsto |f(z)|$ attains its minimum in $\mathbb C$ This minimum is a root. Two simple assertions to prove.

The proof is intuitive and is meant to illustrate an analogy to the Intermediate Value Theorem. The value of the winding number per se is not important but merely that for large enough circles the image curve surrounds the origin while for very small circles it does not.

 

[swampwiz]

As an example, here is what happens for $f(z) = z(z-i)^2$ as you shrink the radius from 1.3 (purple) to 0.7 (green). $z = 0$ is shown as a red star for clarity.
View attachment 240588
As you can see, what happens is that you get a small loop around zero which shrinks to zero size just when you pass the double root with your curve. For a multiplicity of $n$, this small loop will wind $n-1$ times around zero.

OK, so what is going on is that the winding itself is collapsing on the origin.

 

[swampwiz]

The only thing that is important is that you have a winding number, you do not actually have to shrink the circle as a circle, you can deform it in any continuous way you choose.

However, note that the version of the fundamental theorem that he shows is that every polynomial of degree $d \geq 1$ has at least one complex root. All you need is therefore that you have to pass zero. It is relatively easy to go from there to show that it has exactly $d$ roots by factoring out $z-z_0$ and getting a polynomial on the form $P_d(z) = (z-z_0)P_{d-1}(z)$.

So what you are saying is that as soon as you get that first zero, you lock it away and divide the polynomial by it to get a polynomial of -1 degree, and then start anew, thereby obviating any need to examine a multiple or even another root. This makes sense.

 

[Orodruin]

OK, so what is going on is that the winding itself is collapsing on the origin.

This is a bin imprecise in language, but if you mean that you essentially get a smaller and smaller loop around the origin that when you pass the root collapses to the origin, then that is the intuitive picture if you have a multiple root.

If you just have two roots that happen to have the same magnitude, then it will just be two windings passing the origin at the same time.

 

[Orodruin]

This is where topology enters the picture. Essentially you need the continuity of $z \mapsto f(z)$. Of course he is not spelling out the proof in its entirety, it is more of a casual video after all.

To spell it out a little bit more: Assume that $f(z)$ is a continuous function without any zeros, then it can be seen as a map from $\mathbb S^1 \times \mathbb R$ to $\mathbb C \setminus \{0\}$, which is homeomorphic to $\mathbb S^1 \times \mathbb R$ and therefore has the same fundamental group ($\pi_1 = \mathbb Z$). Since $f(z)$ is continuous $f(\theta,r_1)$ and $f(\theta, r_2)$ are in the same homotopy class (must have the same winding number). However, since the winding number for $r \to \infty$ is $d$ and the winding number for $r \to 0$ is zero, this is not the case and $f(z)$ is therefore not continuous without any zeros. However, it is rather straightforward to show that it is continuous and so the assertion that fails is that it does not have any zeros.

 

[Orodruin]

while for very small circles it does not.

Unless $P(0) = 0$, but then the existence of a root is trivial.

 

[StoneTemplePython]

To spell it out a little bit more: Assume that $f(z)$ is a continuous function without any zeros, then it can be seen as a map from $\mathbb S^1 \times \mathbb R$ to $\mathbb C \setminus \{0\}$, which is homeomorphic to $\mathbb S^1 \times \mathbb R$ and therefore has the same fundamental group ($\pi_1 = \mathbb Z$). Since $f(z)$ is continuous $f(\theta,r_1)$ and $f(\theta, r_2)$ are in the same homotopy class (must have the same winding number). However, since the winding number for $r \to \infty$ is $d$ and the winding number for $r \to 0$ is zero, this is not the case and $f(z)$ is therefore not continuous without any zeros. However, it is rather straightforward to show that it is continuous and so the assertion that fails is that it does not have any zeros.

A direct form of the proof, which I prefer as it isn't set up as a contradiction, is, while building up the winding number, to focus on the key result that a value outside the function has winding number of zero (homotopic to a constant), so a winding number $\neq 0$ implies $y$ is in $f$ (specialized to a disk D in particular so I'll say $\in f(D)$).

Now, if roots of our polynomial exists, then (using one of many tools) we can bound them to be strictly inside some well chosen radius $r$. So construct a standard disk centered around the origin of $\mathbb C$ using $r$. We want the winding number computed on $r$, but this winding number is hard to compute, so look for an easy case: the winding number of $q(x) = x^n$ is easy to compute and we can apply an easy linear homotopoy to map from one polynomial to the other. The final detail is to show that the homotopy does not 'hit' the origin during the deformation (which is an easy inequality application with the right $r$) -- this tells us that our difficult polynomial and $q$ both wind around the origin the same number of times, which is non-zero (i.e. n), hence $0 \in p(D)$ and our polynomial has a root.

 

[lavinia]

A direct form of the proof, which I prefer as it isn't set up as a contradiction, is, while building up the winding number, to focus on the key result that a value outside the function has winding number of zero (homotopic to a constant), so a winding number $\neq 0$ implies $y$ is in $f$ (specialized to a disk D in particular so I'll say $\in f(D)$).

Now, if roots of our polynomial exists, then (using one of many tools) we can bound them to be strictly inside some well chosen radius $r$. So construct a standard disk centered around the origin of $\mathbb C$ using $r$. We want the winding number computed on $r$, but this winding number is hard to compute, so look for an easy case: the winding number of $q(x) = x^n$ is easy to compute and we can apply an easy linear homotopoy to map from one polynomial to the other. The final detail is to show that the homotopy does not 'hit' the origin during the deformation (which is an easy inequality application with the right $r$) -- this tells us that our difficult polynomial and $q$ both wind around the origin the same number of times, which is non-zero (i.e. n), hence $0 \in p(D)$ and our polynomial has a root.

You can also use Stokes Theorem on the pullback of the angular form $dθ$ if there are no zeros in the interior of the disk. The winding number would have to equal zero since

$∫_{C}P^{*}(dθ) = ∫_{D}dP^{*}(dθ) = 0$ since $P^{*}(dθ)$ is closed.

More generally, if there are zeros $z_{I}$inside the disk then the winding number of $P(C)$ around the origin is the sum of the winding numbers $P(C_{i})$ where $C_{i}$ is a small circle that surrounds $z_{i}$ and none of the other zeros. This again is by Stokes Theorem.

Your theorem follows that if the winding number is non-zero then there must zeros in the interior of the disk. The theorem works for any smooth function of the plane into the plane not just complex polynomials or analytic functions. This is a homology proof rather than homotopy proof.

 

[StoneTemplePython]

More generally, if there are zeros $z_{I}$inside the disk then the winding number of $P(C)$ around the origin is the sum of the winding numbers $P(C_{i})$ where $C_{i}$ is a small circle that surrounds $z_{i}$ and none of the other zeros. This again is by Stokes Theorem.

Your theorem follows that if the winding number is non-zero then there must zeros in the interior of the disk. The theorem works for any smooth function of the plane into the plane not just complex polynomials or analytic functions. This is a homology proof rather than homotopy proof.

I think you are falling into the same trap as post 4. Differentiation (and smoothness) and Integration are optional here. As you said in post 6, this really should be thought of as the natural generalization of the Intermediate Value Theorem in the plane.

The issue is there are 3 different ways to define the winding number over closed paths --- (i.) in essence getting out your protractor and computing angles over sufficiently fine intervals, (ii.) as particular lift through the exponential map, (iii.) via a particular integral which (ignoring a normalizing constant) is in essence what you have in your post.

If you take (i) or (ii) as a definition you can get to these results via homotopy. If you want to use integration and Stokes, ok fine, but that really is optional.

It seems germane to point out that not so long ago, @mathwonk had a nice discussion of winding numbers here:

https://www.physicsforums.com/threa...unction-is-not-1-1.956650/page-2#post-6069870

 

[lavinia]

I think you are falling into the same trap as post 4. Differentiation (and smoothness) and Integration are optional here. As you said in post 6, this really should be thought of as the natural generalization of the Intermediate Value Theorem in the plane.

The issue is there are 3 different ways to define the winding number over closed paths --- (i.) in essence getting out your protractor and computing angles over sufficiently fine intervals, (ii.) as particular lift through the exponential map, (iii.) via a particular integral which (ignoring a normalizing constant) is in essence what you have in your post.

If you take (i) or (ii) as a definition you can get to these results via homotopy. If you want to use integration and Stokes, ok fine, but that really is optional.

It seems germane to point out that not so long ago, @mathwonk had a nice discussion of winding numbers here:

https://www.physicsforums.com/threa...unction-is-not-1-1.956650/page-2#post-6069870

The homology argument is also a topological and it gives you immediately the theorem you were relying on for the positive proof.

The idea of homology does not require differentiation but only continuity. The Stokes theorem argument shows that without zeros in the interior of the disk, the image curve is homologous to zero in the plane minus a point.

The fundamental group and the first integer homology group of the punctured plane are isomorphic but conceptually homology and homotopy are different.

 

[StoneTemplePython]

The homology argument is also a topological and it gives you immediately the theorem you were relying on for the positive proof.

Maybe we are talking past each other. I read your previous post to mean that knowledge of homology is necessary for said theorem -- it isn't. You may have been just saying it's sufficient, though as I read it again -- I'm still not sure to be honest.
- - - - - -
Really the underlying idea here is: look at an odd degree polynomial with real coefficients and truncate the real line to a suitable large closed and bounded interval (i.e. make use of compactness) with zero in middle. Choose wisely here and you'll have the polynomial being negative as a minimum and positive as a maxmium. Intermediate Value Theorem then guarantees a zero in its image.

For even degree poynomials this breaks, but we can resurrect it in the complex plane -- truncate to a suitably large compact region (i.e. a standard disk) centered at the origin and prove a non-zero winding number for the point at the origin. Any points that have been 'wound around' are in the image of the function inside the disk. Homotopy can be used to bring both of those points home. That seems like the simplest and most direct approach that I can think of.

(As far as choosing wisely for the radius when truncating: you can even reuse the same inequality for the first case in $\mathbb R$ and the second one in $\mathbb C$ as in both cases the underlying idea is for large magnitude $x$ the monic polynomial looks an awful lot like $x^n$ -- the inequality exists to sharpen this point.)

 

[lavinia]

Maybe we are talking past each other. I read your previous post to mean that knowledge of homology is necessary for said theorem -- it isn't. You may have been just saying it's sufficient, though as I read it again -- I'm still not sure to be honest.
- - - - - -
Really the underlying idea here is: look at an odd degree polynomial with real coefficients and truncate the real line to a suitable large closed and bounded interval (i.e. make use of compactness) with zero in middle. Choose wisely here and you'll have the polynomial being negative as a minimum and positive as a maxmium. Intermediate Value Theorem then guarantees a zero in its image.

For even degree poynomials this breaks, but we can resurrect it in the complex plane -- truncate to a suitably large compact region (i.e. a standard disk) centered at the origin and prove a non-zero winding number for the point at the origin. Any points that have been 'wound around' are in the image of the function inside the disk. Homotopy can be used to bring both of those points home. That seems like the simplest and most direct approach that I can think of.

(As far as choosing wisely for the radius when truncating: you can even reuse the same inequality for the first case in $\mathbb R$ and the second one in $\mathbb C$ as in both cases the underlying idea is for large magnitude $x$ the monic polynomial looks an awful lot like $x^n$ -- the inequality exists to sharpen this point.)

OK.

I was only trying to observe that your theorem that the disk must contain zeros if the winding number is not zero follows almost immediately from Stokes theorem. You also made a comment that defining the winding number using calculus was "optional" - not sure exactly what "optional" signifies - but in any case I thought it might mean that the definitions of winding number that use integrals are homological rather than homotopy based and that may be what you were referring to. It seems interesting that the homology definition - using an integral - gives a series of equivalent ways to think about the winding number. One possibility is the total angle traversed. But in general integrating any closed 1 form whose integral over the circle is 1 will work.

For instance if you use a bump function whose support on the unit circle is a small open interval and whose integral over the circle is 1, then the integral will calculate the number of times the projection of the curve onto the circle passes through the support interval - with opposite directions having opposite sign. It will behave more like a counter than as a total angular rotation calculator. If there are no zeros in the disk the total count will be zero. I suppose in the limit one can think of a dirac delta 1 form that is a pure counter - not sure if this makes sense but it would be a somewhat different definition of the winding number.

There is another aspect to this winding number approach to the Fundamental Theorem which I am not sure of.

A complex polynomial is a holomorphic mapping of the sphere into itself. Its mapping degree - viewing it as a smooth function - is equal to the winding number around the origin of the image of the large circle. This seems to say more than the Fundamental Theorem of Algebra which only asserts that the polynomial is a surjective mapping of the sphere into itself. It also says that it is not null homotopic.

Any mapping of the sphere into itself that misses a point is null homotopic and any that has non-zero mapping degree is not. So non-zero mapping degree certainly suffices to show that a map is surjective. But a priori it might be possible for a surjective map to have mapping degree zero. So non-zero mapping degree for polynomials seems to be a stronger statement than the Fundamental Theorem of Algebra.

The reason thoughts led to this is that there is at least one topological proof of the Fundamental Theorem that does not seem to use winding number and only shows that $P$ is surjective. My understanding of this proof is that it only uses that $P$ is a smooth map of the sphere into itself and that $P$ has only finitely many critical points. The proof follows as an almost trivial application of the Inverse Function Theorem and the compactness of the sphere. (See Milnor's Topology from the Differentiable Viewpoint.)

Note: The mapping degree of smooth map from the sphere into itself is similar to the winding number of a curve around a point in the plane. It is the number of times that the map wraps the sphere around itself. A complex polynomial of degree $n$ wraps the sphere around itself $n$ times. Just as winding numbers classify equivalence classes of homotopic loops in the puntured plane, the mapping degree classifies equivalence classes of homotopic maps of the sphere into itself. For smooth maps the mapping degree is the sum over points in the inverse image of any regular value of $+1$ if the differential is orientation preserving and $-1$ is it is orientation reversing. This sum is independent of the regular value. For a complex polynomial the sign is always $+$ because complex polynomials are orientation preserving at every non-singular point and the number of points in the inverse image of a regular value is $n$ if the polynomial has degree $n$.

The mapping degree defined in terms of regular values gives yet another way to define a winding number. One just counts the number of points - with signs accounted for - in the inverse image of a regular value of the projection of the image curve onto the circle.

 

[mathwonk]

here is an easy topological proof of the FTA; "elementary complex analysis" shows that a non constant holomorphic function is an open mapping. Since P^1 is compact, a non constant polynomial map extends to a holomorphic map of P^1 to itself, which is thus both open and closed. Since P^1 is connected it is also surjective.

This proof is somewhat similar to that of Milnor mentioned by Lavinia. Milnor proves that a smooth map from P^1 to itself with only finitely many critical points, is surjective. His argument is slightly more difficult than the one here, but he has not proved his map is open. Both Milnor's and the argument in this post, use the (smooth) inverse function. I.e. for holomorhic functions that theorem (plus an argument that a non constant holomorphic function factors locally through z^n and an invertible function) implies openness, and in Milnor's argument it implies non critical values are regular.

Probably a covering map argument would prove Lavinia's assertion that in Milnor's case too, the smooth map is open, since it should be topologically equivalent to a branched cover of P^1(complex).

 

[lavinia]

here is an easy topological proof of the FTA; "elementary complex analysis" shows that a non constant holomorphic function is an open mapping. Since P^1 is compact, a non constant polynomial map extends to a holomorphic map of P^1 to itself, which is thus both open and closed. Since P^1 is connected it is also surjective.

More generally, I think that if $n>1$, any smooth map of an $n$ dimensional domain into $R^{n}$ that has only isolated singularities is an open mapping.