# Is there a variant of the implicit function theorem covering a branch of a curve around a singular point?

I am looking for a variant of the Implicit function theorem such that

• it guarantees the analyticity of the implicitly defined function, and
• works for a single branch of a curve around a singular point?

What kind of such variants of the IFT are out there?

Hopefully clarifying background.

I was teaching a short course on series and their applications to sophomores last fall. In their last set of homework problems I had a question about the Folium of Descartes $$x^3+y^3=3xy.$$ I asked the students to find two first non-zero terms of the Taylor series of the branch $y=y(x)$ that obviously has a local minimum at $x=0$ (shown in thick red in the image below).

My goals with this exercise were modest. Basically I wanted the students to do a minimum amount of work with the arithmetic of formal power series. I didn't ask them anything about convergence this time. We had not covered any tools for that (they were scheduled to take a first course in complex analysis only later).

My crude thinking. The folium is a genus zero algebraic curve, and has the rational parametrization (coming from blowing up the singularity with $y=tx$) $$x=\frac{3t}{1+t^3},\qquad y=\frac{3t^2}{1+t^3}.$$ The branch of interest is gotten by restricting $t$ to a suitable neigborhood of $t=0$ (when producing that image I used $t\in[-1/2,1/2]$). When viewed as complex variables, we can invert the $x=x(t)$ to a function $t=t(x)$ that is holomorphic in a neighborhood of $x=0$. Therefore it is represented by a convergent Taylor series. Consequently so is $y=tx$.

But.

• This depends heavily on our ability to blow up the singularity allowing us to isolate the chosen branch. I am too ignorant to know for sure, but I suspect that we may need some structures from algebraic geometry to be able to even define a branch around the singular point in such a way that the IFT can be applied.
• My solution (if you can call it that) also depended on the genus. Having a rational parametrization made it easier. I'm aware of the existence of analytic parametrizations of some curves (like the formal group law parameter of an elliptic curve), but I'm ignorant about the more general results.

So the main question may have evolved to. (feel free to discuss your favorite variants)

• Do we need the singularity to come from an algebraic equation to be able to discuss the branches, and study the analyticity of an individual branch?

• Is there a variant of the IFT that guarantees the analyticity of a single branch?

Extra credit for students. (in case some students read my question up to this point, and want to work on something related)

Show that in a solution $y=\sum_{n=0}^\infty a_n x^n$ to the equation $x^3+y^3=3xy$ only every third term can be non-zero. More precisely, show that if $a_n\neq0$ then $n\equiv2\pmod3$.

Edit: Thinking about this topic more (particularly in light of Cactus314's answer) it dawned on me that type of algebraic singularity makes a huge difference.

• When we work around a multiple point, with distinct tangents for each branch, we can, like in the case of the Folium, work with any branch as long as the corresponding tangent isn't vertical.
• But, when we have a cusp things can go wrong. Consider the textbook cusp that the curve $y^2=x^3$ has at the origin. No amount of resolving of the singularity is going to make either $y=x^{3/2}$ or $y=-x^{3/2}$ analytic!
• So my updated guess is that for this to work we need to able to isolate branch such that $x-x_0$ becomes a local parameter along that branch.
• Can we formulate this in a non-algebraic way?
• I realize that there are many questions in this post. I welcome answers covering a single aspect only. This may be a bit orthogonal to how things are usually done in Math.SE, but I think this could work. Imagine a local equivalent of a bunch of musicians coming to jam together riffing on a theme :-) Commented Jan 14, 2018 at 13:18
• what statements of implicit function theorem have you been able to find so far... and why do they come up short? Wikipedia does mention an "analytic" implicit function theorem. I think you want a more reliable source. The Folium of Descartes has two tangents at $(x,y) = (0,0)$ so the Taylor expansion depends on which direction you are moving. Commented Jan 14, 2018 at 15:42
• @cactus314 Thanks for the comment. I had a vague recollection of having seen analytic versions (at least in the complex domain where holomorphic => analytic). But they all assumed a non-singular point. I was hoping the example would make it clear that I'm looking for a version that allows singularities. Obviously we then need to select a branch, and exercise a modicum of care in the selection. For example, the other branch of the folium does not give $y$ as a function of $x$ in any neighborhood of $x=0$. Sorry about not making all this clear. Commented Jan 14, 2018 at 16:29
• The reason for including the algebraic-geometry tag is that we may need such machinery to have a working definition of a branch. The process of resolving a singularity may make the question a bit uninteresting in those cases. I guess I may also be interested in seeing non-algebraic ways of isolating a branch and dealing with those. Commented Jan 14, 2018 at 16:36
• Possibly useful: given a harmonic function $u(x,y),$ the level set passing through a critical point meets at equal angles, like your example, or the set where the real part of $(x+iy)^3$ is zero. Commented Jan 15, 2018 at 22:37

I think you have to make a choice of whether you want the local behavior of the paramterized curve (a circle) or to study the self-intersecting behavior of your planar curve embedded in a copy of $\mathbb{C}$ (or choice of ambient space).

Wikipedia offers two different paramterizaions:

• polar coordintes $\displaystyle r = \frac{3a \, \sin \theta \cos \theta}{\sin^3 \theta + \cos^3 \theta}$ and expand around $\theta = 0, \frac{\pi}{2}$.

• rational functions $\displaystyle x = \frac{3at}{1 + t^3}$ and $\displaystyle y = \frac{3at^2}{1 + t^3}$ and expand around $t =$.

Singular points of planar curves have been studied for a long time, in classical language (using equations) and there books on the theory of algebraic curves. Yet... the mapping between the modern and classical language is lacking in my opionion.

Here he starts to outline the different kinds of singular points that may occur. One problem is that once you have singular points of any kind these are no longer varieties, they are schemes. Even though this very basic case has been studied since Isaac Newton.

Often modern elementary discusions of algebraic geometry avoid discussions of singular points altogether. I guess in order to make the discussion smooth they rule out all the interesting parts?

She does algeraic geometry over $\mathbb{C}$, talks about Puiseux series (to handle both branches) and advocates resolution of singularities.

• Thanks! I recall seeing Kirwan's book in our library. I need to check it out. Commented Jan 14, 2018 at 18:07

Let us rewrite the equation as a level set of a multivariate polynomial $$p(x,y) = x^3+y^3-3xy=0$$

Then take gradient $$\cases{\frac{\partial p}{\partial x}=3x^2-3y\\\frac{\partial p}{\partial y}=3y^2-3x}$$

We can confirm that at $$(0,0)$$ both are zero. The curious student can probably find that this is unique for this point along the curve. So if we imagined the partial derivatives usually giving us a hint of expansion around a level set, that is not going to help us in this case. But why stop the investigation at partial first derivatives and gradients? We can verify: $$\cases{\frac{\partial p}{\partial xy}=-3\\\frac{\partial p}{\partial y^2}=6y=0}$$

Which by symmetry gives us a Hessian of:

$$\left[\begin{array}{rr}0&-3\\-3&0\end{array}\right]$$

We can verify that the eigenvalue decomposition is:

$$H=VDV^{-1} : D = \left[\begin{array}{cc}-3&0\\0&3\end{array}\right], V = \frac{1}{\sqrt{2}}\left[\begin{array}{cc}1&1\\-1&1\end{array}\right]$$

This does give us information about the local dependence of second order terms (in some sense).

Plot of exponential family map of distance to level set and the polynomial $$y=f(x)=\frac{x^2}{3}$$. Hmm, could those 3:s in the Hessian be a coincidence or how can we explain this?

If we generalize our problem to look at $$p(x,y) = x^3+y^3 -kxy = 0$$

and then consider the parabolae $$\cases{y = x^2/k\\x = y^2/k}$$

we can make an animation over increasing $$k$$:

To see that the parabolae indeed follow the curve nicely and smoothly.

• Thanks for your post. Yes, we can find the higher derivatives $y^{(n)}(x=0)$ by repeated implicit differentiation together with "branch identifying initial data" (here $y(0)=0$, $y'(0)=0$). What will this lead to? Commented Jan 14, 2018 at 13:56