If I have a function $f(z,\alpha)$ (let's keep it a polynomial of order $\geq 2$ in $z$, for simplicity), what would be a necessary condition for there to be branch points for this function? A friend mentioned that $f(z,\alpha)$ and $f_{\alpha}(z,\alpha)$, where the second term is the partial derivative w.r.t. $\alpha$, should have a common root. I don't see this intuitively. Is this true? If so, why?
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$\begingroup$ What is the definition of "branch point" that you are using? $\endgroup$– pinakiCommented Mar 25, 2011 at 22:22
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$\begingroup$ I didn't know there were multiple definitions... I'm calling that point, for which going around it in an arbitrary loop results in a different value for the function, as the branch point $\endgroup$– doobCommented Mar 25, 2011 at 23:22
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$\begingroup$ You need to say more of what you mean. Polynomials themselves have no branch points. Or give us an example. $\endgroup$– Gerald EdgarCommented Mar 25, 2011 at 23:52
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1 Answer
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Your question is not very clear. However, I guess you are asking for the branch points of the cover of $\mathbb{C}$ defined by $f(z, \alpha)=0$.
In this case, let us assume for the sake of symplicity that $f(z, \alpha)$ is monic in $z$; then
$f(z, \alpha)=z^n + f_{n-1}(\alpha)z^{n-1}+ \cdots + f_0(\alpha)$.
The Riemann surface $X \subset \mathbb{C}^2$ defined by $f(z, \alpha)=0$ is a cover $X \to \mathbb{C}$ of degree $n$, defined by $(z, \alpha) \to \alpha$.
The branch points of this cover are precisely the points $\bar{\alpha}$ such that $f(z, \bar{\alpha})$ has a multiple root. This is equivalent to require that $f(z, \bar{\alpha})$ and $f_z(z, \bar{\alpha})$ have a common root.
In other words, the branch points are the roots of the equation
$\textrm{Disc}_z(f)=0,$
where $\textrm{Disc}_z(f)$ is the discriminant of $f$ with respect to $z$, i.e. the resultant of $f$ and $\partial f / \partial z$ computed with respect to $z$.
For instance, the branch points of the Riemann surface defined by
$z^3+f_1(\alpha)z+f_0(\alpha)=0$
are the roots of
$4 f_1(\alpha)^3+27 f_0(\alpha)^2=0$.
answered Mar 26, 2011 at 0:12