User fly by night - MathOverflow

Answer by Fly by Night for $A_{\infty}$ singularity

2012-08-01T00:52:58Z

For $\mu \ge 0,$ the $A_{\mu}$ series of map germs $f : (\mathbb{R}^n,0) \to (\mathbb{R},0)$ is, for $\varepsilon_i = \pm 1$, given by $f(x) = \varepsilon_1x_1^2 + \cdots + \varepsilon_{n-1}x_{n-1}^2 \pm x_n^{\mu +1}$. These all have algebraically isolated singularities at $0 \in \mathbb{R}^n$. I this setting, finite Milnor number is equivalent to the singularity being isolated. For $\mu = \infty$, the $A_{\infty}$ singularity is non-isolated, and is given by $f(x) = \varepsilon_1x_1^2 + \cdots + \varepsilon_{n-1}x_{n-1}^2$. Notice the lack of $x_n$. If you're working over $\mathbb{C}$ then drop all of the $\pm$s.

Measuring contact between algebraic varieties

2012-08-22T17:16:12Z

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?

Equivalence of Level Sets

2012-07-31T23:42:19Z

Consider the zero level set of $f : \mathbb{R}^3 \to \mathbb{R}$, where $0$ is a regular value. Consider also the space of planes passing through the origin, i.e. $\mathbb{RP}^2$. For a fixed plane $P \in \mathbb{RP}^2$, consider the orthogonal projection of $f^{-1}(0)$ onto $P$. The apparent contour, say $A_P \subset P$, is the critical value set of the projection of $f^{-1}(0)$ onto $P$.

Now consider a fixed set of morphisms of $\mathbb{R}^2$, say $M$, e.g. diffeomorphisms, homeomorphisms, isotopy equivalence, homotopy equivalence. I'm interested in the following space:

$\mathbb{RP}^2 / \sim$, where $P \sim P'$ if and only if there exists $\mu \in M$ such that $\mu(A_P) = A_{P'}$.

I'm thinking of all of the apparent contours as "living" in the same plane, and getting a two-parameter family of plane curves parametrised by the the choice of $P \in \mathbb{RP}^2$.

What is known, and what references can be given regarding this set-up. Especially when $M$ is the set of diffeomorphisms and homeomorphisms?

Comment by Fly by Night

2012-08-30T22:47:53Z

Who said it was surprising? I said it was interesting, because I thought we might be getting somewhere. Obviously I already knew that the contact function has a degenerate singularity at a vertex -- that's why I chose the example! I was interested because I didn't expect the dimensions of the vector spaces to reflect this. Again, I think you misunderstand my standing.

Comment by Fly by Night

2012-08-27T18:15:33Z

Let's try an example. Consider the parabola $2y−x^2=0$ and the circle $x^2+(y−r)^2−r^2=0$. When $r=1$, the circle is the oscillating circle to the parabola at the origin; if $r\neq 1$ then it is ordinarily tangent. Let's try two cases: $r=1,2$. In the first case we have $\mathcal{A}_{\mathbb{R}^2,0} / (2y−x^2,x2+(y−1)^2−1)\cong \mathbb{R}\langle 1, x, x^2, x^3\rangle$. In the second case we have $\mathcal{A}_{\mathbb{R}^2,0} / (2y−x^2,x2+(y−2)^2−4)\cong \mathbb{R}\langle 1, x, \rangle$ Interestingly, we get different quotient vector spaces! But the difference in dimension is two, not one...

Comment by Fly by Night

2012-08-25T15:36:11Z

I'm not considering a sequence of polynomials. Your idea of contact seems to be different to mine. I'm already able to calculate the contact type using local parametrisations, e.g. formal power series, and then bring everything into a normal form. I was just looking for a different way to do it. I appreciate you taking the time to help, but I don't think we're going to make any progress here.

Comment by Fly by Night

2012-08-25T01:08:42Z

@Michael: The short answer is no. We parametrise a manifold in a neighbourhood of the contact point and compose this with the equation of the other manifold. The singularity type of the resulting function germ $(\mathbb{R}^n,0) \to (\mathbb{R},0)$ denotes the contact. The equivalence relation is $\mathscr{A}$-equivalence, i.e. diffeomorphic changes of independent and dependent variables. The simple contact types are $A_{\ge 1},$ $D_{\ge 4}$ and $E_6$, $E_7$ and $E_8$. Although there is a functional moduli zoo of many more, non-simple, orbits under the the action of $\mathscr{A}$-equivalence.

Comment by Fly by Night

2012-08-24T23:39:06Z

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 $\mathbb{R}\langle f_x,f_y\rangle$.)

Comment by Fly by Night

2012-08-24T23:29:08Z

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.

Comment by Fly by Night

2012-08-24T17:11:43Z

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).

Comment by Fly by Night

2012-08-24T17:09:12Z

@Robert: Yes, you're right! The umbilics are given by $D_4^{\pm}$ of the contact function between the surface and an osculating sphere. The $D_4$ contact between a surface and a tangent plane is a flat umbilic, i.e. both parabolic and umbilic.

Comment by Fly by Night

2012-08-22T18:50:43Z

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 $x^3 \pm xy^2$.

Comment by Fly by Night

2012-08-22T18:40:46Z

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.

Comment by Fly by Night

2012-08-01T15:13:33Z

@David: They're the same notation. In Francesco's link $n = 2$ and we get the $A_{\mu}$ series given by $f(x) = x_1^2 + x_2^{\mu+1}$, with $A_{\infty}$ being $f(x) = x_1^2$.

Comment by Fly by Night

2012-08-01T15:03:36Z

The $A_P$ should be in brackets, i.e. $\mu(A_P)$. It's the image of an apparent contour. Two planes, say $P and P'$, are equivalent if and only if the apparent contour, i.e. the critical value set, of the orthogonal projection of $f^{-1}(0)$ onto $P$ is diffeo/homeo-morphic to the apparent contour of the orthogonal projection of $f^{-1}(0)$ onto $P'$.