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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.

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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 $e = 0$.

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  • $\begingroup$ 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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