Measuring contact between algebraic varieties - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T13:23:39Z http://mathoverflow.net/feeds/question/105250 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/105250/measuring-contact-between-algebraic-varieties Measuring contact between algebraic varieties Fly by Night 2012-08-22T17:16:12Z 2012-08-24T12:39:12Z <p>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.</p> <p>Is there a way to do this? For example, by looking at the dimension of some suitable ideal?</p> http://mathoverflow.net/questions/105250/measuring-contact-between-algebraic-varieties/105385#105385 Answer by Robert Bryant for Measuring contact between algebraic varieties Robert Bryant 2012-08-24T12:39:12Z 2012-08-24T12:39:12Z <p>While I suspect that you are looking for some kind of homological algebra answer, here's a naive algorithm to get what you want: </p> <p>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 $g(x,y,z)=0$. </p> <p>Of course, one must have $f(0,0,0)=g(0,0,0)$ or else the surfaces don't both pass through $p$. </p> <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 $p$. </p> <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, $p_1=0$.)</p> <p>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.</p>