Jacobian Conjecture for unit triangular matrices

Question

This question is about the Jacobian conjecture for a special case. I will first explain the Jacobian conjecture (since it is something every mathematician should know about).

Let $k$ be an algebraically closed field.

Consider a map $$F: k^n \rightarrow k^n,$$ defined by $$F(x_1,\ldots,x_n)=(f_1(x_1,\ldots,x_n),\ldots,f_n(x_1,\ldots,x_n)),$$ where $f_1,\ldots,f_n$ are polynomials.

The Jacobian of $F$, which I denote $J$, is the determinant of the matrix $dF$ where the $(i,j)$-th entry of $dF$ is $\partial f_i/\partial x_j$. (The matrix $dF$ gives the induced map on the tangent bundle, or maybe it's the cotangent bundle; it doesn't matter for this question.)

Since $F$ is given by polynomials, the entries of $dF$ are polynomials and $J$ is a polynomial. Hence $F$ is a nonsingular map if and only if $J$ is a constant.

The inverse function theorem tells us that $F$ has a smooth inverse map if and only if $J$ is constant. The Jacobian conjecture says that this smooth map is in fact also given by polynomials (in the case where the original map $F$ is given by polynomials).

Question: I would like to know if the Jacobian conjecture is known (or trivial) for the special case where the matrix $dF$ is a triangular matrix with $1$'s on the diagonal. If it is not known, I would like to know if the full Jacobian conjecture is known to be equivalent to this special case.

Motivation: This is a possible strategy for proving that a particular family of maps I have constructed for a particular purpose is in fact invertible within the category of affine algebraic varieties.

EDIT: Clarifying in light of Tom's remark. The inverse function theorem just says that $F$ has a local inverse. The Jacobian conjecture is that $F$ has a global inverse which is given by polynomials.

Answer 1

More simply:

(I'll write this down for the case $n=3$ because writing and reading subscripts makes me tired.)

Let $(u,v,w)=F(x,y,z)$. By hypothesis $u-x$ has derivative $0$ with respect to $x$, so $$u=x+P(y,z)$$ for some $P$. And $v-y$ has derivative $0$ with respect to $x$ and $y$, so $$v=y+Q(z)$$ for some $Q$. And $w-z$ has derivative $0$ with respect to all three variables, so $$w=z+R$$ for some constant $R$.

Now just write down the inverse: $$z=w-R$$ $$y=v-Q(z)=v-Q(w-R)$$ $$x=u-P(y,z)=u-P(v-Q(w-R),w-R).$$

Answer 2

The Jacobian conjecture is trivial in this case.

Write $M$ for $dF$ and $N$ for its inverse, which is also triangular with 1's on the diagonal. (I assume upper triangular.)

Now just look at the consequences of $MN=I$.

First, for each $j$, we have $M_{j,j+1}+N_{j,j+1}=0$. This shows that every entry of the form $N_{j,j+1}$ is a polynomial.

Next, for each $j,t$ with $t>0$ we have $\sum_p M_{j,j+p}N_{j+p,j+t}=0$. This plus induction shows that $N_{j,j+t}$ is a polynomial. (The induction is on the difference between the two indices on the entry of $N$.)

Edit: As I said in comments, this never actually uses the upper triangularity and doesn't establish the existence of a global inverse.

Here, I think, is a much better answer:

Write $F=(F_1,\ldots,F_n)$. We want to know whether $F$ is globally injective.

Suppose $F(s_1,\ldots,s_n)=F(t_1,\ldots,t_n)$. We know that $F_n$ is an affine function of the last variable only, so $s_n=t_n$. We know that $F_{n-j}$ is of the form $Ax_{n-j}+G(x_{n-j+1},\ldots,x_n)$, so by backward induction on $n$ we know that $s_{n-j}=t_{n-j}$ for all $j$. So $F$ is globally injective, which is to say that the Jacobian conjecture holds.