Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I was reading the Mathematica Guidebook to Symbolics by Michael Trott, and saw the following expression to evaluate the branch points of a Riemann surface given by $f(x,a)=x^3+x^2+ax-1/2$.

Eliminate[{#==0},D[#,x]==0},x]&[eq=x^3+x^2+ax-1/2]

This gives the result $36a-4a^2+16a^3+19=0$. The roots of this equation are the branch points of $f(x,a)$.

My question is: Why is this so? How does eliminating $x$ from $f(x,a)=0$ and $\dfrac{df(x,a)}{dx}=0$ lead us to the solution?

NOTE: I'm not asking for an explanation of the Mathematica command... I don't get the math behind it.

share|improve this question

2 Answers 2

up vote 2 down vote accepted

Imagine the plane $(x,a)$ where the equation $f(x,a)=0$ gives a curve $C$. We project the curve onto the $a$-axis. Branch points are the points on the $a$ axis over which the projection "is not so nice". You can imagine e.g. the points over which the projection "folds", e.g. if $f(x,a)=x^2-a$ then $a=0$ is such a point.

Now you have to imagine, though, that $a$ and $x$ are complex numbers (and hence $a$ is a point in a plane, and $C$ is 2-dimensional). Over a general point $a$ the equation $f(x,a)=0$ has $n$ different solutions (for $x$), where $n$ is the degree of $f$ as a polynomial in $x$ (in your case $n=3$). We have $n$ different solutions iff $f(x,a)=0$ and $df(x,a)/dx=0$ (for this given $a$) don't have a common root. Over such points $C$ covers the plane of $a$'s in $n$ sheets. When, however, $f(x,a)=0$ and $df(x,a)/dx=0$ do have a common root, the number of roots of $f(x,a)=0$ is less than $n$, so over these points $a$ some of the sheets meet. These are the branch points. In fact, what happens, if you go around such a point then you would see that you get from one sheet (also called one "branch") to another. That's why these points are called branching.

(I skipped some unimportant technical details)

share|improve this answer
    
But number of roots less than $n$ could happen even in the case of a double root, right? Needn't necessarily be a branch point. How does eliminating $x$ in this case help? –  user7815 Mar 23 '11 at 21:04
    
The common roots of $f(x,a)=0$ and of $df(x,a)/dx=0$ (for a fixed $a$) are precisely the multiple roots of $f(x,a)=0$. (In those skipped technical details I assumed e.g. that $f$ is irreducible as a polynomial in 2 variables - that implies that a multiple root happens only for finitely many $a$'s). In principle the fact that $f(x,a)$ has a multiple root (equivalently, has less than $n$ roots) is only a necessary condition for $a$ being a branch point. (again a technical detail) It means that the algorithm you described is not quite right - it might some false positives. –  user8268 Mar 23 '11 at 22:55
    
The point I don't understand is how does eliminating $x$ from $f(x,a)$ and its derivative help? What does the resulting function $g(a)$ mean w.r.t. the original function? –  user7815 Mar 23 '11 at 23:34
    
The question is: for which values of $a$ does the polynomial $f(x,a)$ (as a polynomial in $x$) have multiple roots (ie. less than $n$ roots)? A root is multiple iff it's also a root of $df(x,a)/dx$. Hence you want to know whether $f(x,a)=0$ and $df(x,a)/dx=0$ have a common solution (for a fixed $a$). That's what "elimination of $x$" does: it gives an equation for $a$ s.t. you have a common root of $f$ and $df/dx$ exactly for the solutions of that equation –  user8268 Mar 24 '11 at 6:17

Branch points will occur at values of $x$ which are double or higher roots of $f(x,a)=0$, that is, where $f(x,a)$ and $f_x(x,a)$ have common zeros. "Elimination of x from $f(x,a)$ and $f_x(x,a)$" amounts to running the Euclidean algorithm on $f$ and $f'$, where the field of coefficients is rational functions of $a$ with complex coefficients. The expression the algorithm produces is the (non-zero, constant) GCD, unless $f(x,a)$ and $f_x(x,a)$ have a common factor for every $a$, as in $f(x,a)=(x-a)^2$. Avoiding that kind of degeneracy, for only finitely-many values of $a$ that GCD vanishes, which shows a non-trivial GCD.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.