Following along with this example: Understanding Ramification Points makes me question some things about ramification, and in the context of this curve:

$$c(x,y,e) = 4\,x\,y^4 + 2\,y^2 - e\,y - 1$$ (e is a real, positive parameter)

I understand this curve to have two simply ramified points above $x=-1/4$, i.e. two sheets tend towards $y=1$ and the two other sheets tend towards $y=-1$. I am trying to see if this expected behavior is preserved as $e \to 0$ from some positive value, or alternatively, if I can separate these two ramification points away from both occurring at $x=-1/4$ as $e$ grows from $0$.

Miranda's book, Algebraic Curves and Riemann Surfaces, and the link above, seem to say that a map $\pi: X \to \mathbb{C}$ is ramified at $p \in X$ iff $\partial\,f / \partial\,y \;(p)=0$. And I do follow along with the implicit function theorem.

  • $$e = 0$$

spts: solve([ c(x,y,0)=0, diff(c(x,y,0),y)=0],[x,y]) gives $$ (x,y)=(-1/4,\pm 1). $$ So by the 2 references, I assume this means these are possible ramification points. The Taylor series of each are $$(\textrm{about }y=-1)\;\; a_0 + a_1\,(y+1) + a_2\,(y+1)^2 + a_3\,(y+1)^3 + \cdots,$$ and $$(\textrm{about }y=1)\;\; b_0 + b_1\,(y-1) + b_2\,(y-1)^2 + b_3\,(y-1)^3 + \cdots.$$ This would lead me to believe that these points have a ramification index of 1. Correct? So the polynomial $$c(x,\pm 1, e) \sim \prod_i (y \mp y_i)^{v_i} \to \prod_i (y \mp y_i)^1$$ has no multiple roots?

But the Puiseux series (I use a CAS to compute Taylor and Puiseux series) are $$puiseux( c(x,y,0), x,y, -1/4, \pm 1, 4) \to \left[ {{\pm 35\,t^4}\over{8}}-{{5\,t^3}\over{2}} \pm {{3\,t^2 }\over{2}}-t \pm 1 , t^2=x+{{1}\over{4}} \right], $$ where only even powers of $t$ (a parameterization?) are altered. Why?

Doesn't the 'parameterization' imply a double cover - 2 sheets, except at $t=0$?

Isn't this in contradiction to what the Taylor series says? (I'm assuming I use both correctly in the software).

If I replace the 'parameterization' in the Puiseux series result by $\pm$ absolute value of square roots, I get (just for $(x,y) = (-1/4,-1)$,

$$ -{{35\,\left(x+{{1}\over{4}}\right)^2}\over{8}}-{{5\,\left(x+{{1 }\over{4}}\right)^{{{3}\over{2}}}}\over{2}}-\sqrt{x+{{1}\over{4}}}- {{3\,\left(x+{{1}\over{4}}\right)}\over{2}}-1 $$ and $$ -{{35\,\left(x+{{1}\over{4}}\right)^2}\over{8}}+{{5\,\left(x+{{1 }\over{4}}\right)^{{{3}\over{2}}}}\over{2}}+\sqrt{x+{{1}\over{4}}}- {{3\,\left(x+{{1}\over{4}}\right)}\over{2}}-1 $$ So I see ascending orders of half-powers (again with the non-integer powers alternating sign - why?) - isn't this another indication that there are two sheets for this point? ($(x,y)=(-1/4,-1)$)

  • $$e \neq 0$$

I notice that if I solve for these points for general $e$, $$ s1: solve([c(x,y,e)=0, diff(c(x,y,e),y)=0],[x,y]), $$ I get $$\left[ x=-{{27\,e^4+\sqrt{9\,e^2+64}\,\left(9\,e^3+64\,e \right)+288\,e^2+512}\over{2048}} , y=-{{\sqrt{9\,e^2+64}-3\,e }\over{8}} \right] , \left[ x={{-27\,e^4+\sqrt{9\,e^2+64}\,\left(9 \,e^3+64\,e\right)-288\,e^2-512}\over{2048}} , y={{\sqrt{9\,e^2+64}+ 3\,e}\over{8}} \right],$$ and taking the limit $e \to 0$ yields the same points as above. So I don't expect any complications from this 'deformation'.

Hope my question(s) fit the allowable format. Please feel free to provide reasons or critiques on improving it.


The Taylor series computation is the issue. The curve is $$c(x, y, e) = 4 x y^4 + 2 y^2 - e y - 1 = 0,$$ $x$ is simply $$x = -\frac {2 y^2 - e y - 1} {4y^4},$$ and, for $e = 0$, the expansions around $y = \pm 1$ are $$x = -\frac 1 4 + (y - 1)^2 + \dots\,, \\ x = -\frac 1 4 + (y + 1)^2 + \dots\,.$$ This agrees with the Puiseux expansions $$y = 1 + \sqrt {x + \frac 1 4} + \dots\,, \\ y = -1 + \sqrt {x + \frac 1 4} + \dots\,.$$ This gives two pairs of glued sheets, or two cycles $(12)(34)$. For a small positive $e$, there are two nearby branch points $x_i$. At each of those points, $c(x_i, y, e)$ has one double zero. The structure is similar in the sense that there is the cycle $(12)$ over $x_1$ and the cycle $(34)$ over $x_2$. In particular, the genus is the same as for $e = 0$.

  • $\begingroup$ Hhmmm my Taylor series calculation must have been off then. I knew, of course, x(y) for the whole curve... These "cycles"... are these like a deck transformation or some homology thing? $\endgroup$ – nate Apr 15 '18 at 23:49
  • $\begingroup$ Cycles in the sense of the monodromy group, describing how we move from one sheet to another when circling around a branch point: sheets $1 \to 2 \to 1$ and sheets $3 \to 4 \to 3$. $\endgroup$ – Maxim Apr 16 '18 at 1:58

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