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.
