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

Consider a polynomial $p \in \mathbb C[x_0,\cdots, x_n]$ that is homogeneous under a weighted $\mathbb C^*$ action (we can assume for simplicity that the weights are coprime). I would ideally like to know

What is/are necessary and sufficient conditions on the monomials and coefficients of a polynomial for its singular locus to be the origin.

By singular locus, I mean the variety defined as the common vanishing locus of $p$ and its derivatives. I think this probably requires some sort of sufficiently generic choice of terms (both monomials and coefficients), but I don't know how to properly characterize it.

For example, in $\mathbb C[x,y]$, setting $p = x^3$ is homogeneous under any weighted action, but its singular locus is $\mathbb V(x)$. Similarly, $p = (x-y)^3$ is homogeneous under the diagonal action, but its singular locus is $\mathbb V(x-y)$. If I take $p = x^2 + y^2$, however, the singular locus is $\mathbb V(x,y)$.

From playing around with various polynomials, I suspect I am looking for something like "$p$ must have 'sufficiently many monomials with generic coefficents'."

Edit: Sorry I was not clear; I'm not beginning with a fixed polynomial. I want to characterize polynomials whose singular locus is the origin. I suspect there should be a condition that says "if you pick exponents with such and such properties, then a polynomial with those monomials and generic complex coefficients will have singular locus at the origin."

If I pick any non-zero coefficent $a$, the polynomial $p=a x^3$, the singular locus will always be $\mathbb V(x)$, but for arbitrary non-zero coefficents $a,b$, the polynomial $p=a x^2 + b y^2$, the singular locus will always be $\mathbb V(x,y)$. So a condition might be "if the newton polytope is codimension 1 or less, for generic coefficients the singular locus is the origin".

share|improve this question
    
I don't think your question is too clear at the moment to get a decent answer, as you answer the question yourself - a neccessary and sufficient condition is given by looking at where the partial derivatives vanish. –  Daniel Loughran Sep 30 '10 at 15:19
2  
However a "general" hypersurface in projective space will always be non-singular, in that the smooth hypersurfaces form a Zariski open subset of corresponding linear system. I would guess the same is true for weighted projective space but unfortunately I am not an expert on it, and perhaps the singularities of weighted projective space might mess things up (but I would guess not as these occur in codimension one). Note that a singularity of your variety which is not the origin is exactly a singularity of the corresponding variety in weighted projective space. –  Daniel Loughran Sep 30 '10 at 15:20
    
On the weighted hypersurface the quotient singularities can be in any codimension. It is also possible to construct hypersurfaces for which there are nontrivial intersections between the quotient singularities and the singularities that arise when the defining polynomial is not transverse. –  Laie Sep 30 '10 at 15:51
    
If your weights are pairwise coprime and divide the degree d, then a general degree d hypersurface in this weighted projective space will be smooth. If your weights don't divide the degree, then the hypersurface will always be singular (because it will pass through one of the singular points of the weighted projective space) –  mdeland Sep 30 '10 at 18:07
    
@Daniel: I don't know the definition of general, and googling doesn't reveal anything enlightening. Would you mind pointing me to a reference? –  James Davidoff Sep 30 '10 at 19:22
show 7 more comments

Your Answer

 
discard

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

Browse other questions tagged or ask your own question.