# Measuring contact between algebraic varieties

I have two regular surfaces in three space, both of which are given by an equation. I would like to measure the contact between the two surfaces using only their equations. Usually, one would find a local parametrisation for one of the surfaces, and then substitute this into the other surface's equation. This would give a function in two variables, and the singularity type of this map would give the contact between the two surfaces. However, as I have mentioned: I only want to use the equations.

Is there a way to do this? For example, by looking at the dimension of some suitable ideal?

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What do you mean by contact? Do you already know the surfaces are tangent? – Igor Rivin Aug 22 '12 at 17:54
The definition of contact is given above: it's the singularity type of the contact function, i.e. the composite of a parametrisation with a defining question. Consider for example, the contact between the plane $z=0$ and the surface $f(x,y,z)=0$. The contact function is $f(x,y,0)$. If the resulting contact function is non-zero then the surfaces don't intersect. If it does have a zero then the type of singularity at that zero gives the type of contact. – Fly by Night Aug 22 '12 at 18:40
For example: a smooth surface has contact of type $A_1^+$ with its tangent plane at ordinary elliptic points and type $A_1^-$ at hyperbolic points. This means the contact function is $\mathscr{A}$-equivalent to $x^2 \pm y^2$. At ordinary parabolic points the contact is of type $A_2$, i.e. equivalent to $x^2 + y^3$. At umbilic points, the surface and its tangent plane have type $D_4^{\pm}$ meaning the contact function is $\mathscr{A}$-equivalent to $x^3 \pm xy^2$. – Fly by Night Aug 22 '12 at 18:50
Ah, thanks for the explanation! – Igor Rivin Aug 22 '12 at 20:01
Your claim about the 'type' of umbilic points does not appear to be correct. For example, the surface $z = x^2+y^2$ has an umbilic point at the origin (in the usual sense of differential geometry), but, by your definition of contact type, its contact with the tangent plane is of type $A^+_1$. In general, notions such as 'umbilic' and 'contact with tangent plane' depend on geometric things that are not visible once you localize, and, in particular, are not well-defined under local (analytic) changes of coordinates. Maybe you are regarding these surfaces as lying in projective 3-space? – Robert Bryant Aug 23 '12 at 12:40

While I suspect that you are looking for some kind of homological algebra answer, here's a naive algorithm to get what you want:

Suppose that one is trying to determine the 'contact $k$-type' of a pair of algebraic surfaces at a point $p\in\mathbb{R}^3$. One may as well assume that $p$ is the origin and let the surfaces be defined by polynomial equations $f(x,y,z)=0$ and $g(x,y,z)=0$.

Of course, one must have $f(0,0,0)=g(0,0,0)$ or else the surfaces don't both pass through $p$.

Also, you are assuming that the 'surfaces are regular', by which, I am guessing that you want that $\nabla f$ and $\nabla g$ don't vanish at $p$, so I'll assume that. If $\nabla f\wedge\nabla g$ does not vanish at $p$, then the surfaces aren't tangent at $p$, so assume that $\nabla f\wedge\nabla g$ vanishes at $p$. Under these assumptions, you can, by a linear change of coordinates, assume that $f$ has the form $$f(x,y,z) = z - f_2(x,y,z),$$ where $f_2$ vanishes to order $2$ at $p=(0,0,0)$. Then, of course, one has $$g(x,y,z) = az + g_2(x,y,z)$$ for some $a\not=0$ and some polynomial $g_2$ that vanishes to order $2$ at $p$.

Now define a sequence of polynomials $h_i(x,y,z)$ as follows: $$h_2(x,y,z) = g\bigl(x,y,f_2(x,y,z)\bigr)$$ and, for $k\ge 2$, $$h_{k+1}(x,y,z) = h_k\bigl(x,y,f_2(x,y,z)\bigr).$$ One can now prove, by induction, that, when one writes, for $k\ge 1$, $$h_{k+1}(x,y,z) = p_k(x,y,z) + R_{k+1}(x,y,z),$$ where $p_k$ has degree at most $k$ and $R_{k+1}$ vanishes to order $k{+}1$ at $p$, then $p_k$ is a polynomial in $x$ and $y$ only, and it defines the $k$-th order contact type between the two surfaces at $p$. (Of course, $p_1=0$.)

There is still the task of determining when two $p_k$'s are equivalent under change of variable in $x$ and $y$, but that's another issue.

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Thanks Robert, but how does this relate to the $\mathscr{A}$-equivalence of the contact function? (Usually defined as the composition of a parametrisation and a defining function). – Fly by Night Aug 24 '12 at 17:11
I don't really know, since I don't have a precise definition of '$\mathscr{A}$-equivalence'. I assume that this means 'equivalence up to formal changes of variables that fix the origin', but you should let me know if otherwise. – Robert Bryant Aug 24 '12 at 17:23
If you type "A equivalence" into Google then the first hit is a pretty good definition. As I mentioned: usually, we have two surfaces, one given by a parametrisation and one given by a defining equation. We substitute the parametrisation into the equation to give a function in two variables. $\mathscr{A}$-equivalence means that we allow diffeomorphic changes of variable in both the source and the target. – Fly by Night Aug 24 '12 at 23:29
To bring things back into focus: I would like to know if there is a way to gain information about the singularity type of the contact function (the above composition) from then equations alone. I suspect that some algebraic consideration might suffice. For example, the dimension of a well-chosen ideal could give the Milnor number of the singularity type of the contact function. (The Milnor number of a function germ $f$ being the dimension of the local ring of function germs $\mathscr{O}_{\mathbb{R}^2,0}$, quotient the Jacobian ideal $\mathbb{R}\langle f_x,f_y\rangle$.) – Fly by Night Aug 24 '12 at 23:39
In order to make sense of '$\mathscr{A}$-equivalence' according to the definition on the Wikipedia page, you need to specify a domain and a range. Are you just taking the sequence of polynomials $p_k$, regarding them as pointed maps $p_k:\bigl(\mathbb{R}^2,0\bigr)\to\bigl(\mathbb{R},0\bigr)$, and asking for $\mathscr{A}$-equivalence in that sense? For example, you might want to know that there is a $K>0$ such that all of the $p_k$ with $k\ge K$ are $\mathscr{A}$-equivalent. (This would happen for $K=2$ if $p_2$ were a nondegenerate quadratic polynomial.) – Robert Bryant Aug 25 '12 at 14:22