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.

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    $\begingroup$ Just out of curiosity, is your family of maps so twisted that proving the Jacobian conjecture is a feasible alternative to proving directly what you want? $\endgroup$ – Gunnar Þór Magnússon Dec 14 '11 at 6:47
  • $\begingroup$ @Gunnar: The answer is no; I am hoping this special case of the Jacobian conjecture is easy. Also, the problem is more that the map is actually hard to write down in specific coordinates, but it should be fairly easy to show that the derivative matrix is unit upper triangular (for a given ordering of coordinates) because I don't have to care what the entries strictly above the diagonal are. $\endgroup$ – Alexander Woo Dec 14 '11 at 7:32
  • $\begingroup$ The inverse function theorem only gives a local inverse. The Jacobian Conjecture doesn't just say that the inverse is polynomial; it says that a global inverse exists. $\endgroup$ – Tom Goodwillie Dec 14 '11 at 14:16

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: $$z=w-R$$ $$y=v-Q(z)=v-Q(w-R)$$ $$x=u-P(y,z)=u-P(v-Q(w-R),w-R).$$


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.

  • $\begingroup$ To complete this answer, $N$ is the Jacobian of the reciprocal map $F^{-1}$. The latter, having polynomial derivatives, is polynomial. $\endgroup$ – Denis Serre Dec 14 '11 at 9:51
  • $\begingroup$ Denis Serre: Yes, I ought to have said that explicitly. Thanks for filling it in. $\endgroup$ – Steven Landsburg Dec 14 '11 at 15:33
  • $\begingroup$ I've just realized that this argument doesn't actually use the upper triangularity. $M=dF$ is a matrix over the ring of polynomials with unit determinant, so of course it has a polynomial inverse. If the map $F$ has a global inverse, this proves that the global inverse is polynomial --- but it doesn't prove the existence of that global inverse. $\endgroup$ – Steven Landsburg Dec 14 '11 at 15:48

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