Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Dear all,

in some of my work I am considering the algebraic surface given implicitly by

$$ z^2\left(1-x+\frac{z^2}{4}\right) = y^2. $$

Maybe this surface is known? Does anyone recognize it and/or point me to some literature?

Thanks a lot!

share|improve this question
4  
I don't think it has a special name, but it's a singular quartic whose resolution/normalization (take the substitution $\alpha=y/z$) is a smooth quadric in $\mathbb{P}^3$. –  Charles Siegel Jun 29 '10 at 13:28

1 Answer 1

Your surface is actually a Whitney umbrella.

To see that, just perform the following substitution (which is just a translation):

$$x =1+\frac{z^2}{4} -t.$$

After this, your surface is defined by the simpler equation:

$$z^2 t =y^2.$$

This is exactly the canonical form of the Whitney umbrella. The Whitney umbrella is a singular surface in $\mathbb{C}^3$ which looks like a self-intersecting plane. It has a line of double points at $z=y=0$ and the singularity worsen to a pinch point at the origin $z=y=t=0$. You can resolve it by blowing-up the double line.

The Whitney umbrella is studied in many books of algebraic geometry as an example of a surface with a pinch point and a first tricky example of blow-up: it shows that blowing-up the worse singularity is not the best way to smooth a space (if you blow-up the pinch points you will again get the same equation). In classical singular theory, it is an example of a singular surface which does not have a regular Whitney stratification. It also appears very naturally in the study of elliptic fibration in the context of string theory (more precisely F-theory) when one consider "Sen's weak coupling limit" which gives the orientifold limit of F-theory at weak coupling.

share|improve this answer
1  
Nice! For a picture: mathematik.uni-kl.de/~zca/Reports_on_ca/29/paper_html/… . –  Joseph O'Rourke Jul 15 '10 at 14:27

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.