# Do most degree $d$ morphisms of $P^n$ have smooth critical locus?

Let $f=[F,G,H]:\mathbb{P}^2\to\mathbb{P}^2$ be a morphism of degree $d\ge2$. The critical locus $C_f$ of $f$ is the zero-locus of the Jacobian determinant: $$C_f = \left\{ [x,y,z]\in\mathbb{P}^2 : \det\begin{pmatrix} F_X&F_Y&F_Z\\ G_X&G_Y&G_Z\\ H_X&H_Y&H_Z\\ \end{pmatrix} = 0 \right\}.$$ In the parameter space of such $f$, it is clear that $$\{ f : \text{the critical locus C_f is smooth} \}$$ is Zariski open. Our question is whether this set is non-empty, i.e., for each $d\ge2$, does there exist at least one degree $d$ morphism $f:\mathbb{P}^2\to\mathbb{P}^2$ such that $C_f$ is smooth?

We suspect that for every $d\ge2$ the map $$f=[X^d+X^2Y^{d-2}+XZ^{d-1},Y^d+Y^2Z^{d-2}+YX^{d-1},Z^d+Z^2X^{d-2}+ZY^{d-1}]$$ has smooth $C_f$, and have verified this for all $d\le 10$ by explicitly computing resultants modulo an appropriate prime.

More generally, we would be interested in the same question for morphisms $\mathbb{P}^n\to\mathbb{P}^n$ on projective space of higher dimension.

• Is there a typo in the definition of $f$? – Kevin Ventullo Apr 11 '14 at 2:33
• @KevinVentullo Yes, thanks, I'll fix it. (The exponents aren't quite right.) – Joe Silverman Apr 11 '14 at 2:41

• Ciliberto and Flamini claim that the case $\mathbb{P}^2\to \mathbb{P}^2$ was proven by Kulikov-Kulikov in 2002. (Actually I am surprised that this result has only been proven in this century...). Both proofs are very much characteristic zero proofs. – Remke Kloosterman Apr 11 '14 at 7:16
This is addressing the second part of OP's question: could the critical locus be smooth for $n>2$? Given an $(n+1)$-tuple of degree $d$ polynomials, $(F_0,\dots,F_n)$, in $n+1$ variables, $x_0,\dots,x_n$, the Jacobian matrix $[\partial F_i/\partial x_j]$ defines local morphisms from $\mathbb{P}^n$ to the affine space of $(n+1)\times (n+1)$ - matrices. The critical locus of the morphism is the inverse image of the determinant hypersurface under this morphism.
Since the determinant hypersurface is singular in codimension $3$, i.e., codimension $4$ in the ambient space of matrices, the inverse image of the determinant hypersurface "should" be singular in codimension $3$, i.e., codimension $4$ in $\mathbb{P}^n$. In fact, using the Thom-Porteous formula, the inverse image of this codimension $4$ locus has degree $$\frac{(n+2)(n+1)^2n(d-1)^4}{12},$$ at least assuming that the inverse image has codimension $4$ (i.e., not codimension $0$, $1$, $2$ or $3$).