How is this not a proof of the Jacobian conjecture in the complex case?An explanation for undergraduated students about why the Jacobian conjecture is hardInverse Function Theorem/ PolynomialDerivative of a determinantIs the first variation of a Jacobian determinant always zero?If $f$ is holomorphic and $left| f right|$ is constant then $f$ is constantDerivate of the cofactor and the determinantExpressing the determinant of a restricted map in terms of its cofactor matrixDeterminant of the Jacobian of a short mapProof for $A^-1=fracadj(A_j)det(A)$Differentiability of a determinant and its inverseEquation containing cofactor of derivative and Kronecker-delta
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How is this not a proof of the Jacobian conjecture in the complex case?
An explanation for undergraduated students about why the Jacobian conjecture is hardInverse Function Theorem/ PolynomialDerivative of a determinantIs the first variation of a Jacobian determinant always zero?If $f$ is holomorphic and $left| f right|$ is constant then $f$ is constantDerivate of the cofactor and the determinantExpressing the determinant of a restricted map in terms of its cofactor matrixDeterminant of the Jacobian of a short mapProof for $A^-1=fracadj(A_j)det(A)$Differentiability of a determinant and its inverseEquation containing cofactor of derivative and Kronecker-delta
$begingroup$
I've just been reading the Wikipedia entry regarding the Jacobian conjecture, and it said that either the conjecture is true for all fields of characteristic zero, or it is false for all such fields.
Hence, I wonder, shouldn't this be an easy problem that yields to methods from real or complex analysis? After all, it involves only simple terms like determinant, inverse, constant, polynomial etc.
Specifically, the determinant condition gives a relation between the derivatives, which one may then be able to integrate in order to possibly obtain polynomials.
To make this more specific, say that we have a polynomial function $f: mathbb K^n to mathbb K^n$, where $mathbb K = mathbb R$ or $mathbb C$. Then $det J_f$ is a polynomial in the derivatives of the components and hence itself a polynomial. By the inverse rule and Cramer's rule, the derivative of the (local) inverse has the form
$$
frac1det(J_f) operatornameCof(J_f),
$$
where by assumption $det(J_f)$ is constant. Also, the cofactor matrix is a polynomial matrix. Thus, we integrate any of its entries for each component to obtain a local polynomial inverse, which is also global due to the identity theorem (at least in the complex case).
What makes this approach fail?
(This main part of my question makes it unique among other questions regarding the Jacobian conjecture, which have been completely falsely suggested to be a duplicate of this one.)
real-analysis linear-algebra complex-analysis proof-verification open-problem
asked 2 days ago
AlgebraicsAnonymousAlgebraicsAnonymous
1,434113
$endgroup$
|
show 4 more comments
$begingroup$
I've just been reading the Wikipedia entry regarding the Jacobian conjecture, and it said that either the conjecture is true for all fields of characteristic zero, or it is false for all such fields.
Hence, I wonder, shouldn't this be an easy problem that yields to methods from real or complex analysis? After all, it involves only simple terms like determinant, inverse, constant, polynomial etc.
Specifically, the determinant condition gives a relation between the derivatives, which one may then be able to integrate in order to possibly obtain polynomials.
To make this more specific, say that we have a polynomial function $f: mathbb K^n to mathbb K^n$, where $mathbb K = mathbb R$ or $mathbb C$. Then $det J_f$ is a polynomial in the derivatives of the components and hence itself a polynomial. By the inverse rule and Cramer's rule, the derivative of the (local) inverse has the form
$$
frac1det(J_f) operatornameCof(J_f),
$$
where by assumption $det(J_f)$ is constant. Also, the cofactor matrix is a polynomial matrix. Thus, we integrate any of its entries for each component to obtain a local polynomial inverse, which is also global due to the identity theorem (at least in the complex case).
What makes this approach fail?
(This main part of my question makes it unique among other questions regarding the Jacobian conjecture, which have been completely falsely suggested to be a duplicate of this one.)
real-analysis linear-algebra complex-analysis proof-verification open-problem
asked 2 days ago
AlgebraicsAnonymousAlgebraicsAnonymous
1,434113
$endgroup$
2
$begingroup$
Possible duplicate of An explanation for undergraduated students about why the Jacobian conjecture is hard
$endgroup$
– Brevan Ellefsen
2 days ago
$begingroup$
As far as I can tell, the given question does not suggest a specific approach, and neither does its answer address one (except for attempting to find a counter-example). So could you please un-duplicate this question?
$endgroup$
– AlgebraicsAnonymous
2 days ago
$begingroup$
Run through the hypothetical argument outlined in one of those answers. While not a perfect duplicate, the fact you propose solving this problem as if it were a simple bit of ODE/Linear-Algebra implies to me you don't actually understand why the problem is nuanced and complicated. I agree with you it's not a perfect duplicate, but it's better than my immediate answer to your question, which is that the Jacobian Conjecture involves multiple variables. You can't just integrate to cancel out the differentiation immediately and get polynomials.
$endgroup$
– Brevan Ellefsen
2 days ago
$begingroup$
"you don't actually understand why the problem is nuanced and complicated" - Yes, that's why I'm asking. Since I've only just met the conjecture, this doesn't imply that I'm stupid though.
$endgroup$
– AlgebraicsAnonymous
2 days ago
1
$begingroup$
I never stated anything about your intelligence, so please don't accuse me of doing so. It's perfectly understandable that you wouldn't get the nuance looking at the conjecture for the first time - I surely didn't when I first saw it. That is why I: 1) connected a question I think answers most aspects of this question decently well enough to be considered a duplicate, and 2) answered the one part of your question that isn't a duplicate via my above comment.
$endgroup$
– Brevan Ellefsen
2 days ago
|
show 4 more comments
$begingroup$
I've just been reading the Wikipedia entry regarding the Jacobian conjecture, and it said that either the conjecture is true for all fields of characteristic zero, or it is false for all such fields.
Hence, I wonder, shouldn't this be an easy problem that yields to methods from real or complex analysis? After all, it involves only simple terms like determinant, inverse, constant, polynomial etc.
Specifically, the determinant condition gives a relation between the derivatives, which one may then be able to integrate in order to possibly obtain polynomials.
To make this more specific, say that we have a polynomial function $f: mathbb K^n to mathbb K^n$, where $mathbb K = mathbb R$ or $mathbb C$. Then $det J_f$ is a polynomial in the derivatives of the components and hence itself a polynomial. By the inverse rule and Cramer's rule, the derivative of the (local) inverse has the form
$$
frac1det(J_f) operatornameCof(J_f),
$$
where by assumption $det(J_f)$ is constant. Also, the cofactor matrix is a polynomial matrix. Thus, we integrate any of its entries for each component to obtain a local polynomial inverse, which is also global due to the identity theorem (at least in the complex case).
What makes this approach fail?
(This main part of my question makes it unique among other questions regarding the Jacobian conjecture, which have been completely falsely suggested to be a duplicate of this one.)
real-analysis linear-algebra complex-analysis proof-verification open-problem
asked 2 days ago
AlgebraicsAnonymousAlgebraicsAnonymous
1,434113
$endgroup$
I've just been reading the Wikipedia entry regarding the Jacobian conjecture, and it said that either the conjecture is true for all fields of characteristic zero, or it is false for all such fields.
Hence, I wonder, shouldn't this be an easy problem that yields to methods from real or complex analysis? After all, it involves only simple terms like determinant, inverse, constant, polynomial etc.
Specifically, the determinant condition gives a relation between the derivatives, which one may then be able to integrate in order to possibly obtain polynomials.
To make this more specific, say that we have a polynomial function $f: mathbb K^n to mathbb K^n$, where $mathbb K = mathbb R$ or $mathbb C$. Then $det J_f$ is a polynomial in the derivatives of the components and hence itself a polynomial. By the inverse rule and Cramer's rule, the derivative of the (local) inverse has the form
$$
frac1det(J_f) operatornameCof(J_f),
$$
where by assumption $det(J_f)$ is constant. Also, the cofactor matrix is a polynomial matrix. Thus, we integrate any of its entries for each component to obtain a local polynomial inverse, which is also global due to the identity theorem (at least in the complex case).
What makes this approach fail?
(This main part of my question makes it unique among other questions regarding the Jacobian conjecture, which have been completely falsely suggested to be a duplicate of this one.)
real-analysis linear-algebra complex-analysis proof-verification open-problem
real-analysis linear-algebra complex-analysis proof-verification open-problem
asked 2 days ago
AlgebraicsAnonymousAlgebraicsAnonymous
1,434113
asked 2 days ago
AlgebraicsAnonymousAlgebraicsAnonymous
1,434113
edited 2 days ago
AlgebraicsAnonymous
asked 2 days ago
AlgebraicsAnonymousAlgebraicsAnonymous
1,434113
asked 2 days ago
AlgebraicsAnonymousAlgebraicsAnonymous
1,434113
asked 2 days ago
AlgebraicsAnonymousAlgebraicsAnonymous
1,434113
1,434113
2
$begingroup$
Possible duplicate of An explanation for undergraduated students about why the Jacobian conjecture is hard
$endgroup$
– Brevan Ellefsen
2 days ago
$begingroup$
As far as I can tell, the given question does not suggest a specific approach, and neither does its answer address one (except for attempting to find a counter-example). So could you please un-duplicate this question?
$endgroup$
– AlgebraicsAnonymous
2 days ago
$begingroup$
Run through the hypothetical argument outlined in one of those answers. While not a perfect duplicate, the fact you propose solving this problem as if it were a simple bit of ODE/Linear-Algebra implies to me you don't actually understand why the problem is nuanced and complicated. I agree with you it's not a perfect duplicate, but it's better than my immediate answer to your question, which is that the Jacobian Conjecture involves multiple variables. You can't just integrate to cancel out the differentiation immediately and get polynomials.
$endgroup$
– Brevan Ellefsen
2 days ago
$begingroup$
"you don't actually understand why the problem is nuanced and complicated" - Yes, that's why I'm asking. Since I've only just met the conjecture, this doesn't imply that I'm stupid though.
$endgroup$
– AlgebraicsAnonymous
2 days ago
1
$begingroup$
I never stated anything about your intelligence, so please don't accuse me of doing so. It's perfectly understandable that you wouldn't get the nuance looking at the conjecture for the first time - I surely didn't when I first saw it. That is why I: 1) connected a question I think answers most aspects of this question decently well enough to be considered a duplicate, and 2) answered the one part of your question that isn't a duplicate via my above comment.
$endgroup$
– Brevan Ellefsen
2 days ago
|
show 4 more comments
2
$begingroup$
Possible duplicate of An explanation for undergraduated students about why the Jacobian conjecture is hard
$endgroup$
– Brevan Ellefsen
2 days ago
$begingroup$
As far as I can tell, the given question does not suggest a specific approach, and neither does its answer address one (except for attempting to find a counter-example). So could you please un-duplicate this question?
$endgroup$
– AlgebraicsAnonymous
2 days ago
$begingroup$
Run through the hypothetical argument outlined in one of those answers. While not a perfect duplicate, the fact you propose solving this problem as if it were a simple bit of ODE/Linear-Algebra implies to me you don't actually understand why the problem is nuanced and complicated. I agree with you it's not a perfect duplicate, but it's better than my immediate answer to your question, which is that the Jacobian Conjecture involves multiple variables. You can't just integrate to cancel out the differentiation immediately and get polynomials.
$endgroup$
– Brevan Ellefsen
2 days ago
$begingroup$
"you don't actually understand why the problem is nuanced and complicated" - Yes, that's why I'm asking. Since I've only just met the conjecture, this doesn't imply that I'm stupid though.
$endgroup$
– AlgebraicsAnonymous
2 days ago
1
$begingroup$
I never stated anything about your intelligence, so please don't accuse me of doing so. It's perfectly understandable that you wouldn't get the nuance looking at the conjecture for the first time - I surely didn't when I first saw it. That is why I: 1) connected a question I think answers most aspects of this question decently well enough to be considered a duplicate, and 2) answered the one part of your question that isn't a duplicate via my above comment.
$endgroup$
– Brevan Ellefsen
2 days ago
2
2
$begingroup$
Possible duplicate of An explanation for undergraduated students about why the Jacobian conjecture is hard
$endgroup$
– Brevan Ellefsen
2 days ago
$begingroup$
Possible duplicate of An explanation for undergraduated students about why the Jacobian conjecture is hard
$endgroup$
– Brevan Ellefsen
2 days ago
$begingroup$
As far as I can tell, the given question does not suggest a specific approach, and neither does its answer address one (except for attempting to find a counter-example). So could you please un-duplicate this question?
$endgroup$
– AlgebraicsAnonymous
2 days ago
$begingroup$
As far as I can tell, the given question does not suggest a specific approach, and neither does its answer address one (except for attempting to find a counter-example). So could you please un-duplicate this question?
$endgroup$
– AlgebraicsAnonymous
2 days ago
$begingroup$
Run through the hypothetical argument outlined in one of those answers. While not a perfect duplicate, the fact you propose solving this problem as if it were a simple bit of ODE/Linear-Algebra implies to me you don't actually understand why the problem is nuanced and complicated. I agree with you it's not a perfect duplicate, but it's better than my immediate answer to your question, which is that the Jacobian Conjecture involves multiple variables. You can't just integrate to cancel out the differentiation immediately and get polynomials.
$endgroup$
– Brevan Ellefsen
2 days ago
$begingroup$
Run through the hypothetical argument outlined in one of those answers. While not a perfect duplicate, the fact you propose solving this problem as if it were a simple bit of ODE/Linear-Algebra implies to me you don't actually understand why the problem is nuanced and complicated. I agree with you it's not a perfect duplicate, but it's better than my immediate answer to your question, which is that the Jacobian Conjecture involves multiple variables. You can't just integrate to cancel out the differentiation immediately and get polynomials.
$endgroup$
– Brevan Ellefsen
2 days ago
$begingroup$
"you don't actually understand why the problem is nuanced and complicated" - Yes, that's why I'm asking. Since I've only just met the conjecture, this doesn't imply that I'm stupid though.
$endgroup$
– AlgebraicsAnonymous
2 days ago
$begingroup$
"you don't actually understand why the problem is nuanced and complicated" - Yes, that's why I'm asking. Since I've only just met the conjecture, this doesn't imply that I'm stupid though.
$endgroup$
– AlgebraicsAnonymous
2 days ago
1
1
$begingroup$
I never stated anything about your intelligence, so please don't accuse me of doing so. It's perfectly understandable that you wouldn't get the nuance looking at the conjecture for the first time - I surely didn't when I first saw it. That is why I: 1) connected a question I think answers most aspects of this question decently well enough to be considered a duplicate, and 2) answered the one part of your question that isn't a duplicate via my above comment.
$endgroup$
– Brevan Ellefsen
2 days ago
$begingroup$
I never stated anything about your intelligence, so please don't accuse me of doing so. It's perfectly understandable that you wouldn't get the nuance looking at the conjecture for the first time - I surely didn't when I first saw it. That is why I: 1) connected a question I think answers most aspects of this question decently well enough to be considered a duplicate, and 2) answered the one part of your question that isn't a duplicate via my above comment.
$endgroup$
– Brevan Ellefsen
2 days ago
|
show 4 more comments
1 Answer
1
active
oldest
votes
$begingroup$
I now see what my mistake was. Instead of being a polynomial in $y$, the variable of the target space, the inverse is a polynomial in $f^-1(y)$. Specifically:
$$
J_f^-1(y) = J_f^-1(f^-1(y)).
$$
Thus, we only obtain a polynomial in the components of $f^-1(y)$, which is probably worthless.
The only property of the inverse we have thus shown is this: If we differentiate the function in any direction, we obtain a polynomial in the components of that function. Yet this is even true for $exp$.
answered 2 days ago
AlgebraicsAnonymousAlgebraicsAnonymous
1,434113
$endgroup$
$begingroup$
We do get a nice ODE for $f^-1$ though.
$endgroup$
– AlgebraicsAnonymous
2 days ago
add a comment |
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$begingroup$
I now see what my mistake was. Instead of being a polynomial in $y$, the variable of the target space, the inverse is a polynomial in $f^-1(y)$. Specifically:
$$
J_f^-1(y) = J_f^-1(f^-1(y)).
$$
Thus, we only obtain a polynomial in the components of $f^-1(y)$, which is probably worthless.
The only property of the inverse we have thus shown is this: If we differentiate the function in any direction, we obtain a polynomial in the components of that function. Yet this is even true for $exp$.
answered 2 days ago
AlgebraicsAnonymousAlgebraicsAnonymous
1,434113
$endgroup$
$begingroup$
We do get a nice ODE for $f^-1$ though.
$endgroup$
– AlgebraicsAnonymous
2 days ago
add a comment |
$begingroup$
I now see what my mistake was. Instead of being a polynomial in $y$, the variable of the target space, the inverse is a polynomial in $f^-1(y)$. Specifically:
$$
J_f^-1(y) = J_f^-1(f^-1(y)).
$$
Thus, we only obtain a polynomial in the components of $f^-1(y)$, which is probably worthless.
The only property of the inverse we have thus shown is this: If we differentiate the function in any direction, we obtain a polynomial in the components of that function. Yet this is even true for $exp$.
answered 2 days ago
AlgebraicsAnonymousAlgebraicsAnonymous
1,434113
$endgroup$
$begingroup$
We do get a nice ODE for $f^-1$ though.
$endgroup$
– AlgebraicsAnonymous
2 days ago
add a comment |
$begingroup$
I now see what my mistake was. Instead of being a polynomial in $y$, the variable of the target space, the inverse is a polynomial in $f^-1(y)$. Specifically:
$$
J_f^-1(y) = J_f^-1(f^-1(y)).
$$
Thus, we only obtain a polynomial in the components of $f^-1(y)$, which is probably worthless.
The only property of the inverse we have thus shown is this: If we differentiate the function in any direction, we obtain a polynomial in the components of that function. Yet this is even true for $exp$.
answered 2 days ago
AlgebraicsAnonymousAlgebraicsAnonymous
1,434113
$endgroup$
I now see what my mistake was. Instead of being a polynomial in $y$, the variable of the target space, the inverse is a polynomial in $f^-1(y)$. Specifically:
$$
J_f^-1(y) = J_f^-1(f^-1(y)).
$$
Thus, we only obtain a polynomial in the components of $f^-1(y)$, which is probably worthless.
The only property of the inverse we have thus shown is this: If we differentiate the function in any direction, we obtain a polynomial in the components of that function. Yet this is even true for $exp$.
answered 2 days ago
AlgebraicsAnonymousAlgebraicsAnonymous
1,434113
edited 2 days ago
answered 2 days ago
AlgebraicsAnonymousAlgebraicsAnonymous
1,434113
answered 2 days ago
AlgebraicsAnonymousAlgebraicsAnonymous
1,434113
answered 2 days ago
AlgebraicsAnonymousAlgebraicsAnonymous
1,434113
1,434113
$begingroup$
We do get a nice ODE for $f^-1$ though.
$endgroup$
– AlgebraicsAnonymous
2 days ago
add a comment |
$begingroup$
We do get a nice ODE for $f^-1$ though.
$endgroup$
– AlgebraicsAnonymous
2 days ago
$begingroup$
We do get a nice ODE for $f^-1$ though.
$endgroup$
– AlgebraicsAnonymous
2 days ago
$begingroup$
We do get a nice ODE for $f^-1$ though.
$endgroup$
– AlgebraicsAnonymous
2 days ago
add a comment |
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2
$begingroup$
Possible duplicate of An explanation for undergraduated students about why the Jacobian conjecture is hard
$endgroup$
– Brevan Ellefsen
2 days ago
$begingroup$
As far as I can tell, the given question does not suggest a specific approach, and neither does its answer address one (except for attempting to find a counter-example). So could you please un-duplicate this question?
$endgroup$
– AlgebraicsAnonymous
2 days ago
$begingroup$
Run through the hypothetical argument outlined in one of those answers. While not a perfect duplicate, the fact you propose solving this problem as if it were a simple bit of ODE/Linear-Algebra implies to me you don't actually understand why the problem is nuanced and complicated. I agree with you it's not a perfect duplicate, but it's better than my immediate answer to your question, which is that the Jacobian Conjecture involves multiple variables. You can't just integrate to cancel out the differentiation immediately and get polynomials.
$endgroup$
– Brevan Ellefsen
2 days ago
$begingroup$
"you don't actually understand why the problem is nuanced and complicated" - Yes, that's why I'm asking. Since I've only just met the conjecture, this doesn't imply that I'm stupid though.
$endgroup$
– AlgebraicsAnonymous
2 days ago
1
$begingroup$
I never stated anything about your intelligence, so please don't accuse me of doing so. It's perfectly understandable that you wouldn't get the nuance looking at the conjecture for the first time - I surely didn't when I first saw it. That is why I: 1) connected a question I think answers most aspects of this question decently well enough to be considered a duplicate, and 2) answered the one part of your question that isn't a duplicate via my above comment.
$endgroup$
– Brevan Ellefsen
2 days ago