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Let $k$ be a field of characteristic zero.

It is well-known, see for example Corollary 10.2.21, that if $(x,y) \mapsto (p,q) \in k[x,y]^2$ is a counterexample to the two-dimensional Jacobian Conjecture, then there exists an automorphism $g$ of $k[x,y]$ such that the support of $g(p)$ is contained in some rectangular $\{ (i,j) | 0 \leq i \leq a, 0 \leq j \leq b \}$ with $(a,b)$ in the support of $g(p)$; such $g(p)$ is called subrectangular. By Lemma B or Theorem 3.4, we obtain that in this case $g(q)$ is also subrectangular (since $g(p)$ and $g(q)$ are similar).

Interestingly, the same result holds in the non-commutative case, namely, in the first Weyl algebra, see Theorem 5.12.

In both the commutative and non-commutative cases:

Is it true that we can further obtain that a possible counterexample has the form: $g(p)$ is as above and, in addition, $\{(a,v)_{0 \leq v \leq b-1}\}$ are not in the support of $g(p)$ and $\{(u,b)_{0 \leq u \leq a-1}\}$ are not in the support of $g(p)$? Something like the picture of $(P_0,Q_0)$ on page 50.

I think that I have once seen a half positive answer to my question, see the picture on page 21 (in that picture both $P$ and $\phi(P)$ are 'half good' for me). More precisely, I think that I have seen that the support of $g(p)$ is contained in the quadrangle with vertices $\{(0,0),(a,0),(a,b),(b-a)\}$, when $a \leq b$. (Perhaps this is a result of S. Abhyankar).

Edit: I see that $(P_0,Q_0)$ on page 50 is a counterexample to the two-dimensional Jacobian Conjecture of minimal degree. So my above question remains interesting only for the non-commutative case (the commutative case almost has a positive answer); does the commutative result has a non-commutative analog in the first Weyl algebra? (Corollary 7.13 has a non-commutative analog, isn't it?)

Thank you very much!

Let $k$ be a field of characteristic zero.

It is well-known, see for example Corollary 10.2.21, that if $(x,y) \mapsto (p,q) \in k[x,y]^2$ is a counterexample to the two-dimensional Jacobian Conjecture, then there exists an automorphism $g$ of $k[x,y]$ such that the support of $g(p)$ is contained in some rectangular $\{ (i,j) | 0 \leq i \leq a, 0 \leq j \leq b \}$ with $(a,b)$ in the support of $g(p)$; such $g(p)$ is called subrectangular. By Lemma B or Theorem 3.4, we obtain that in this case $g(q)$ is also subrectangular (since $g(p)$ and $g(q)$ are similar).

Interestingly, the same result holds in the non-commutative case, namely, in the first Weyl algebra, see Theorem 5.12.

In both the commutative and non-commutative cases:

Is it true that we can further obtain that a possible counterexample has the form: $g(p)$ is as above and, in addition, $\{(a,v)_{0 \leq v \leq b-1}\}$ are not in the support of $g(p)$ and $\{(u,b)_{0 \leq u \leq a-1}\}$ are not in the support of $g(p)$? Something like the picture of $(P_0,Q_0)$ on page 50.

I think that I have once seen a half positive answer to my question, see the picture on page 21 (in that picture both $P$ and $\phi(P)$ are 'half good' for me). More precisely, I think that I have seen that the support of $g(p)$ is contained in the quadrangle with vertices $\{(0,0),(a,0),(a,b),(b-a)\}$, when $a \leq b$. (Perhaps this is a result of S. Abhyankar).

Edit: I see that $(P_0,Q_0)$ on page 50 is a counterexample to the two-dimensional Jacobian Conjecture of minimal degree. So my above question remains interesting only for the non-commutative case (the commutative case almost has a positive answer); does the commutative result has a non-commutative analog in the first Weyl algebra? (Corollary 7.13 has a non-commutative analog, isn't it?)

I have asked the above question here.

Thank you very much!

Let $k$ be a field of characteristic zero.

It is well-known, see for example Corollary 10.2.21, that if $(x,y) \mapsto (p,q) \in k[x,y]^2$ is a counterexample to the two-dimensional Jacobian Conjecture, then there exists an automorphism $g$ of $k[x,y]$ such that the support of $g(p)$ is contained in some rectangular $\{ (i,j) | 0 \leq i \leq a, 0 \leq j \leq b \}$ with $(a,b)$ in the support of $g(p)$; such $g(p)$ is called subrectangular. By Lemma B or Theorem 3.4, we obtain that in this case $g(q)$ is also subrectangular (since $g(p)$ and $g(q)$ are similar).

Interestingly, the same result holds in the non-commutative case, namely, in the first Weyl algebra, see Theorem 5.12.

In both the commutative and non-commutative cases:

Is it true that we can further obtain that a possible counterexample has the form: $g(p)$ is as above and, in addition, $\{(a,v)_{0 \leq v \leq b-1}\}$ are not in the support of $g(p)$ and $\{(u,b)_{0 \leq u \leq a-1}\}$ are not in the support of $g(p)$? Something like the picture of $(p_0,q_0)$ on page 50.

I think that I have once seen a half positive answer to my question, see the picture on page 21 (in that picture both $P$ and $\phi(P)$ are 'half good' for me). More precisely, I think that I have seen that the support of $g(p)$ is contained in the quadrangle with vertices $\{(0,0),(a,0),(a,b),(b-a)\}$, when $a \leq b$. (Perhaps this is a result of S. Abhyankar).

Edit: I see that $(P_0,Q_0)$ on page 50 is a counterexample to the two-dimensional Jacobian Conjecture of minimal degree. So my above question remains interesting only for the non-commutative case (the commutative case almost has a positive answer); does the commutative result has a non-commutative analog in the first Weyl algebra?

Thank you very much!

Let $k$ be a field of characteristic zero.

It is well-known, see for example Corollary 10.2.21, that if $(x,y) \mapsto (p,q) \in k[x,y]^2$ is a counterexample to the two-dimensional Jacobian Conjecture, then there exists an automorphism $g$ of $k[x,y]$ such that the support of $g(p)$ is contained in some rectangular $\{ (i,j) | 0 \leq i \leq a, 0 \leq j \leq b \}$ with $(a,b)$ in the support of $g(p)$; such $g(p)$ is called subrectangular. By Lemma B or Theorem 3.4, we obtain that in this case $g(q)$ is also subrectangular (since $g(p)$ and $g(q)$ are similar).

Interestingly, the same result holds in the non-commutative case, namely, in the first Weyl algebra, see Theorem 5.12.

In both the commutative and non-commutative cases:

Is it true that we can further obtain that a possible counterexample has the form: $g(p)$ is as above and, in addition, $\{(a,v)_{0 \leq v \leq b-1}\}$ are not in the support of $g(p)$ and $\{(u,b)_{0 \leq u \leq a-1}\}$ are not in the support of $g(p)$? Something like the picture of $(P_0,Q_0)$ on page 50.

I think that I have once seen a half positive answer to my question, see the picture on page 21 (in that picture both $P$ and $\phi(P)$ are 'half good' for me). More precisely, I think that I have seen that the support of $g(p)$ is contained in the quadrangle with vertices $\{(0,0),(a,0),(a,b),(b-a)\}$, when $a \leq b$. (Perhaps this is a result of S. Abhyankar).

Edit: I see that $(P_0,Q_0)$ on page 50 is a counterexample to the two-dimensional Jacobian Conjecture of minimal degree. So my above question remains interesting only for the non-commutative case (the commutative case almost has a positive answer); does the commutative result has a non-commutative analog in the first Weyl algebra? (Corollary 7.13 has a non-commutative analog, isn't it?)

Thank you very much!

Let $k$ be a field of characteristic zero.

It is well-known, see for example Corollary 10.2.21, that if $(x,y) \mapsto (p,q) \in k[x,y]^2$ is a counterexample to the two-dimensional Jacobian Conjecture, then there exists an automorphism $g$ of $k[x,y]$ such that the support of $g(p)$ is contained in some rectangular $\{ (i,j) | 0 \leq i \leq a, 0 \leq j \leq b \}$ with $(a,b)$ in the support of $g(p)$; such $g(p)$ is called subrectangular. By Lemma B or Theorem 3.4, we obtain that in this case $g(q)$ is also subrectangular (since $g(p)$ and $g(q)$ are similar).

Interestingly, the same result holds in the non-commutative case, namely, in the first Weyl algebra, see Theorem 5.12.

In both the commutative and non-commutative cases:

Is it true that we can further obtain that a possible counterexample has the form: $g(p)$ is as above and, in addition, $\{(a,v)_{0 \leq v \leq b-1}\}$ are not in the support of $g(p)$ and $\{(u,b)_{0 \leq u \leq a-1}\}$ are not in the support of $g(p)$? Something like the picture of $(p_0,q_0)$ on page 50.

I think that I have once seen a half positive answer to my question, see the picture on page 21 (in that picture both $P$ and $\phi(P)$ are 'half good' for me). More precisely, I think that I have seen that
the support of $g(p)$ is contained in the quadrangle with vertices $\{(0,0),(a,0),(a,b),(b-a)\}$, when $a \leq b$. (Perhaps this is a result of S. Abhyankar).

Thank you very much!

Let $k$ be a field of characteristic zero.

It is well-known, see for example Corollary 10.2.21, that if $(x,y) \mapsto (p,q) \in k[x,y]^2$ is a counterexample to the two-dimensional Jacobian Conjecture, then there exists an automorphism $g$ of $k[x,y]$ such that the support of $g(p)$ is contained in some rectangular $\{ (i,j) | 0 \leq i \leq a, 0 \leq j \leq b \}$ with $(a,b)$ in the support of $g(p)$; such $g(p)$ is called subrectangular. By Lemma B or Theorem 3.4, we obtain that in this case $g(q)$ is also subrectangular (since $g(p)$ and $g(q)$ are similar).

Interestingly, the same result holds in the non-commutative case, namely, in the first Weyl algebra, see Theorem 5.12.

In both the commutative and non-commutative cases:

Is it true that we can further obtain that a possible counterexample has the form: $g(p)$ is as above and, in addition, $\{(a,v)_{0 \leq v \leq b-1}\}$ are not in the support of $g(p)$ and $\{(u,b)_{0 \leq u \leq a-1}\}$ are not in the support of $g(p)$? Something like the picture of $(p_0,q_0)$ on page 50.

I think that I have once seen a half positive answer to my question, see the picture on page 21 (in that picture both $P$ and $\phi(P)$ are 'half good' for me). More precisely, I think that I have seen that the support of $g(p)$ is contained in the quadrangle with vertices $\{(0,0),(a,0),(a,b),(b-a)\}$, when $a \leq b$. (Perhaps this is a result of S. Abhyankar).

Edit: I see that $(P_0,Q_0)$ on page 50 is a counterexample to the two-dimensional Jacobian Conjecture of minimal degree. So my above question remains interesting only for the non-commutative case (the commutative case almost has a positive answer); does the commutative result has a non-commutative analog in the first Weyl algebra?

Thank you very much!