# Talk:Cusp (singularity)

WikiProject Mathematics (Rated Start-class, Mid-importance)
This article is within the scope of WikiProject Mathematics, a collaborative effort to improve the coverage of Mathematics on Wikipedia. If you would like to participate, please visit the project page, where you can join the discussion and see a list of open tasks.
Mathematics rating:
 Start Class
 Mid Importance
Field: Geometry

The term cusp in the title should be ordinary cusp. The ordinary cusp is given by the equation x2 + y3 = 0. There are many different kinds of cusp, e.g. a rhamphoid cusp (coming from the Greek meaning "Beak-like") given by the equation x2 + y5 = 0. Notice that the ordinary cusp and the rhamphoid cusp are not diffeomorphic to each other, and so from a singularity point of view are quiet different. I shall start making some changes now. ~~ Dr Dec (Talk) ~~ 17:25, 28 July 2009 (UTC)

• The original claim that the Jacobian and the Hessian must be singular is not enough. The cusps are given as zero-level-sets of the A2k-singularities, i.e. with normal forms x2 + y2k+1 where k is a natural number. The A2k+1-singularities, i.e. with normal forms x2 + y2k where k is a natural number are clearly not cusps. ~~ Dr Dec (Talk) ~~ 18:01, 28 July 2009 (UTC)
• I've made some pretty wholesale changes. Let me know what you think... ~~ Dr Dec (Talk) ~~ 19:16, 28 July 2009 (UTC)
• Dear Dr. Dec, Well, I am not happy with your changes: I cannot understand your definition. I suggest that you keep the original one, fix it, if necessary and then introduce in another section your. TomyDuby (talk) 20:43, 28 July 2009 (UTC)

Hi TomyDuby, I hope you're well. The old definition was wrong! It was trying to say that f needed to have a degenerate singularity, i.e. both the Jacobian matrix and the Hessian matrix need to be singular. This is a necessary, but not sufficient condition. Take f(x,y) = x4 + y4, for example. Here we see that f(0,0) = 0 and both the Jacobian and Hessian matrices are singular at (0,0), i.e. the origin is a degenerate critical point of f. The zero-level-set of f consists of a single point: the origin. A single point is not a cusp! The cusps are given, and trust me on this one, by the family of zero-level-sets x2 - y2k+1 = 0, where k ≥ 1 is an integer.

Now, this doesn't mean that every curve with a cusp is given by one of these equations. Every cusp is locally diffeomorphic to one of these forms. So there will be a diffeomorphism taking a neighbourhood of any old cusp point onto a neighbourhood of one of our standard pictures (x2 - y2k+1 = 0) such that the curve is carried to the curve.

I think that the old article didn't cover enough ground. The ordinary cusp is just one member in a countably infinite family of cusps. To focus on the ordinary cusp (as the old article did) is, in my opinion, a little narrow sighted.

If you didn't understand the definition then that's not a problem with the article per se, maybe we need to give more explanation. You obviously know about the ordinary cusp. What didn't you understand about the definition? Did you understand the diffeomorphism article? (This article is key). We can't talk about singularities of plane curves without talking about diffeomorphisms, they form the basic equivalences, just like homeomorphisms do in topology, and isomorphisms do in group theory.

I know it might seem like over complication, but if you want to get things right then you need to refine things, and the only way to refine things is to use more sophisticated ideas.

As a refernce, see J. W. Bruce and P. J. Giblin's book giving in the main article. The latter was my PhD supervisor and he taught me all about cusps. RSVP ~~ Dr Dec (Talk) ~~ 23:21, 29 July 2009 (UTC)