Sign up ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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?

share|cite|improve this question
What is the definition of "branch point" that you are using? – auniket Mar 25 '11 at 22:22
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 – doob Mar 25 '11 at 23:22
You need to say more of what you mean. Polynomials themselves have no branch points. Or give us an example. – Gerald Edgar Mar 25 '11 at 23:52

1 Answer 1

up vote 2 down vote accepted

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


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


are the roots of

$4 f_1(\alpha)^3+27 f_0(\alpha)^2=0$.

share|cite|improve this answer

Your Answer


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

Not the answer you're looking for? Browse other questions tagged or ask your own question.