A computation with derivations and skew-derivations
$begingroup$
On a smooth manifold $M$ a derivation $D$ of degree $k$ is an endomorphism of the algebra of forms defined on $M$, which is denoted by $Omega(M) = bigoplus_{k geq 0} Omega^k(M)$, such that
$Dcolon Omega^r(M) to Omega^{r+k}(M)$ for all $r$.
$D$ satisfies $D(alpha wedge beta) = Dalpha wedge beta + alpha wedge Dbeta.$
It is called skew-derivation if 2. is replaced by $D(alpha wedge beta) = Dalpha wedge beta + (-1)^{l}alpha wedge Dbeta$, where $l$ is the degree of $alpha$, and means that $alpha in Omega^l(M)$.
Prove that if $D$ is a derivation of degree $k$ and $D'$ is a skew-derivation of degree $k'$, then $[D,D']:= DD'-D'D$ is a skew-derivation of degree $k+k'$.
I cannot see where my computations go wrong:
begin{align}
[D,D'](alpha wedge beta) & = DD'(alpha wedge beta) - D'D(alpha wedge beta) \
& = D(D'alpha wedge beta + (-1)^{deg alpha}alpha wedge D'beta)-D'(Dalpha wedge beta + alpha wedge Dbeta) \
& = DD'alpha wedge beta + D'alpha wedge Dbeta + (-1)^{deg alpha}Dalpha wedge D'beta + (-1)^{deg alpha}alpha wedge DD'beta \
& quad -D'Dalpha wedge beta - (-1)^{deg alpha+k}Dalpha wedge D'beta-D'alpha wedge Dbeta-(-1)^{deg alpha}alpha wedge D'Dbeta \
& = [D,D']alpha wedge beta + (-1)^{deg alpha}alpha wedge [D,D']beta+(-1)^{deg alpha}(1-(-1)^k)Dalpha wedge D'beta.
end{align}
So the statement seems to hold only when $k$ is even. Where is the problem?
differential-geometry smooth-manifolds differential-forms
asked Jan 22 at 15:32
GibbsGibbs
5,3373827
$endgroup$
add a comment |
$begingroup$
On a smooth manifold $M$ a derivation $D$ of degree $k$ is an endomorphism of the algebra of forms defined on $M$, which is denoted by $Omega(M) = bigoplus_{k geq 0} Omega^k(M)$, such that
$Dcolon Omega^r(M) to Omega^{r+k}(M)$ for all $r$.
$D$ satisfies $D(alpha wedge beta) = Dalpha wedge beta + alpha wedge Dbeta.$
It is called skew-derivation if 2. is replaced by $D(alpha wedge beta) = Dalpha wedge beta + (-1)^{l}alpha wedge Dbeta$, where $l$ is the degree of $alpha$, and means that $alpha in Omega^l(M)$.
Prove that if $D$ is a derivation of degree $k$ and $D'$ is a skew-derivation of degree $k'$, then $[D,D']:= DD'-D'D$ is a skew-derivation of degree $k+k'$.
I cannot see where my computations go wrong:
begin{align}
[D,D'](alpha wedge beta) & = DD'(alpha wedge beta) - D'D(alpha wedge beta) \
& = D(D'alpha wedge beta + (-1)^{deg alpha}alpha wedge D'beta)-D'(Dalpha wedge beta + alpha wedge Dbeta) \
& = DD'alpha wedge beta + D'alpha wedge Dbeta + (-1)^{deg alpha}Dalpha wedge D'beta + (-1)^{deg alpha}alpha wedge DD'beta \
& quad -D'Dalpha wedge beta - (-1)^{deg alpha+k}Dalpha wedge D'beta-D'alpha wedge Dbeta-(-1)^{deg alpha}alpha wedge D'Dbeta \
& = [D,D']alpha wedge beta + (-1)^{deg alpha}alpha wedge [D,D']beta+(-1)^{deg alpha}(1-(-1)^k)Dalpha wedge D'beta.
end{align}
So the statement seems to hold only when $k$ is even. Where is the problem?
differential-geometry smooth-manifolds differential-forms
asked Jan 22 at 15:32
GibbsGibbs
5,3373827
$endgroup$
add a comment |
$begingroup$
On a smooth manifold $M$ a derivation $D$ of degree $k$ is an endomorphism of the algebra of forms defined on $M$, which is denoted by $Omega(M) = bigoplus_{k geq 0} Omega^k(M)$, such that
$Dcolon Omega^r(M) to Omega^{r+k}(M)$ for all $r$.
$D$ satisfies $D(alpha wedge beta) = Dalpha wedge beta + alpha wedge Dbeta.$
It is called skew-derivation if 2. is replaced by $D(alpha wedge beta) = Dalpha wedge beta + (-1)^{l}alpha wedge Dbeta$, where $l$ is the degree of $alpha$, and means that $alpha in Omega^l(M)$.
Prove that if $D$ is a derivation of degree $k$ and $D'$ is a skew-derivation of degree $k'$, then $[D,D']:= DD'-D'D$ is a skew-derivation of degree $k+k'$.
I cannot see where my computations go wrong:
begin{align}
[D,D'](alpha wedge beta) & = DD'(alpha wedge beta) - D'D(alpha wedge beta) \
& = D(D'alpha wedge beta + (-1)^{deg alpha}alpha wedge D'beta)-D'(Dalpha wedge beta + alpha wedge Dbeta) \
& = DD'alpha wedge beta + D'alpha wedge Dbeta + (-1)^{deg alpha}Dalpha wedge D'beta + (-1)^{deg alpha}alpha wedge DD'beta \
& quad -D'Dalpha wedge beta - (-1)^{deg alpha+k}Dalpha wedge D'beta-D'alpha wedge Dbeta-(-1)^{deg alpha}alpha wedge D'Dbeta \
& = [D,D']alpha wedge beta + (-1)^{deg alpha}alpha wedge [D,D']beta+(-1)^{deg alpha}(1-(-1)^k)Dalpha wedge D'beta.
end{align}
So the statement seems to hold only when $k$ is even. Where is the problem?
differential-geometry smooth-manifolds differential-forms
asked Jan 22 at 15:32
GibbsGibbs
5,3373827
$endgroup$
On a smooth manifold $M$ a derivation $D$ of degree $k$ is an endomorphism of the algebra of forms defined on $M$, which is denoted by $Omega(M) = bigoplus_{k geq 0} Omega^k(M)$, such that
$Dcolon Omega^r(M) to Omega^{r+k}(M)$ for all $r$.
$D$ satisfies $D(alpha wedge beta) = Dalpha wedge beta + alpha wedge Dbeta.$
It is called skew-derivation if 2. is replaced by $D(alpha wedge beta) = Dalpha wedge beta + (-1)^{l}alpha wedge Dbeta$, where $l$ is the degree of $alpha$, and means that $alpha in Omega^l(M)$.
Prove that if $D$ is a derivation of degree $k$ and $D'$ is a skew-derivation of degree $k'$, then $[D,D']:= DD'-D'D$ is a skew-derivation of degree $k+k'$.
I cannot see where my computations go wrong:
begin{align}
[D,D'](alpha wedge beta) & = DD'(alpha wedge beta) - D'D(alpha wedge beta) \
& = D(D'alpha wedge beta + (-1)^{deg alpha}alpha wedge D'beta)-D'(Dalpha wedge beta + alpha wedge Dbeta) \
& = DD'alpha wedge beta + D'alpha wedge Dbeta + (-1)^{deg alpha}Dalpha wedge D'beta + (-1)^{deg alpha}alpha wedge DD'beta \
& quad -D'Dalpha wedge beta - (-1)^{deg alpha+k}Dalpha wedge D'beta-D'alpha wedge Dbeta-(-1)^{deg alpha}alpha wedge D'Dbeta \
& = [D,D']alpha wedge beta + (-1)^{deg alpha}alpha wedge [D,D']beta+(-1)^{deg alpha}(1-(-1)^k)Dalpha wedge D'beta.
end{align}
So the statement seems to hold only when $k$ is even. Where is the problem?
differential-geometry smooth-manifolds differential-forms
differential-geometry smooth-manifolds differential-forms
asked Jan 22 at 15:32
GibbsGibbs
5,3373827
asked Jan 22 at 15:32
GibbsGibbs
5,3373827
asked Jan 22 at 15:32
GibbsGibbs
5,3373827
asked Jan 22 at 15:32
GibbsGibbs
5,3373827
asked Jan 22 at 15:32
GibbsGibbs
5,3373827
5,3373827
add a comment |
add a comment |
1 Answer
1
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oldest
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$begingroup$
Assume $alpha, beta$ are differential forms and let $D$ be a derivation of order $k$. We have two relevant cases:
$alpha$ of even degree and $beta$ of odd degree.- Both $alpha,beta$ of odd degree.
In the first case we have
begin{align}
D(alpha wedge beta) & = Dalpha wedge beta + alpha wedge Dbeta,\
D(beta wedge alpha) & = Dbeta wedge alpha + beta wedge Dalpha = alpha wedge Dbeta +(-1)^kDalpha wedge beta.
end{align}
First and second line must now agree, which implies that either $k$ is even or $D$ (of odd order) vanishes on even degree forms.
Second case:
begin{align}
D(alpha wedge beta) & = Dalpha wedge beta + alpha wedge Dbeta,\
-D(beta wedge alpha) & = -Dbeta wedge alpha - beta wedge Dalpha = (-1)^k(alpha wedge Dbeta +Dalpha wedge beta).
end{align}
Again, the two lines agree. Now, either $k$ is even, or it is odd and then $D$ vanishes on odd degree forms.
Summing up, either $k$ is even, or $D$ vanishes. This should prove that derivations of odd order vanish and that non-trivial derivations have even order.
Thus, there is no loss of generality in assuming $k$ even. In this case the term giving problems vanishes and the statement holds true.
answered Jan 22 at 18:42
GibbsGibbs
5,3373827
$endgroup$
$begingroup$
This doesn't look correct. For example, when $M$ is one-dimensional, the exterior derivative is both a derivation of degree one and a skew-derivation of degree one which is not trivial...
$endgroup$
– levap
Jan 22 at 19:18
$begingroup$
@levap You are right, I agree. Thanks for giving your feedback, I spent a big part of the afternoon on this problem and ended up with a wrong answer. Good... Then I do not see how the statement I posted could be correct. It was taken by "Kobayashi and Nomizu - Foundations of Differential Geometry", vol. 1. Do you know how to prove it?
$endgroup$
– Gibbs
Jan 22 at 21:54
add a comment |
Your Answer
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$begingroup$
Assume $alpha, beta$ are differential forms and let $D$ be a derivation of order $k$. We have two relevant cases:
$alpha$ of even degree and $beta$ of odd degree.- Both $alpha,beta$ of odd degree.
In the first case we have
begin{align}
D(alpha wedge beta) & = Dalpha wedge beta + alpha wedge Dbeta,\
D(beta wedge alpha) & = Dbeta wedge alpha + beta wedge Dalpha = alpha wedge Dbeta +(-1)^kDalpha wedge beta.
end{align}
First and second line must now agree, which implies that either $k$ is even or $D$ (of odd order) vanishes on even degree forms.
Second case:
begin{align}
D(alpha wedge beta) & = Dalpha wedge beta + alpha wedge Dbeta,\
-D(beta wedge alpha) & = -Dbeta wedge alpha - beta wedge Dalpha = (-1)^k(alpha wedge Dbeta +Dalpha wedge beta).
end{align}
Again, the two lines agree. Now, either $k$ is even, or it is odd and then $D$ vanishes on odd degree forms.
Summing up, either $k$ is even, or $D$ vanishes. This should prove that derivations of odd order vanish and that non-trivial derivations have even order.
Thus, there is no loss of generality in assuming $k$ even. In this case the term giving problems vanishes and the statement holds true.
answered Jan 22 at 18:42
GibbsGibbs
5,3373827
$endgroup$
$begingroup$
This doesn't look correct. For example, when $M$ is one-dimensional, the exterior derivative is both a derivation of degree one and a skew-derivation of degree one which is not trivial...
$endgroup$
– levap
Jan 22 at 19:18
$begingroup$
@levap You are right, I agree. Thanks for giving your feedback, I spent a big part of the afternoon on this problem and ended up with a wrong answer. Good... Then I do not see how the statement I posted could be correct. It was taken by "Kobayashi and Nomizu - Foundations of Differential Geometry", vol. 1. Do you know how to prove it?
$endgroup$
– Gibbs
Jan 22 at 21:54
add a comment |
$begingroup$
Assume $alpha, beta$ are differential forms and let $D$ be a derivation of order $k$. We have two relevant cases:
$alpha$ of even degree and $beta$ of odd degree.- Both $alpha,beta$ of odd degree.
In the first case we have
begin{align}
D(alpha wedge beta) & = Dalpha wedge beta + alpha wedge Dbeta,\
D(beta wedge alpha) & = Dbeta wedge alpha + beta wedge Dalpha = alpha wedge Dbeta +(-1)^kDalpha wedge beta.
end{align}
First and second line must now agree, which implies that either $k$ is even or $D$ (of odd order) vanishes on even degree forms.
Second case:
begin{align}
D(alpha wedge beta) & = Dalpha wedge beta + alpha wedge Dbeta,\
-D(beta wedge alpha) & = -Dbeta wedge alpha - beta wedge Dalpha = (-1)^k(alpha wedge Dbeta +Dalpha wedge beta).
end{align}
Again, the two lines agree. Now, either $k$ is even, or it is odd and then $D$ vanishes on odd degree forms.
Summing up, either $k$ is even, or $D$ vanishes. This should prove that derivations of odd order vanish and that non-trivial derivations have even order.
Thus, there is no loss of generality in assuming $k$ even. In this case the term giving problems vanishes and the statement holds true.
answered Jan 22 at 18:42
GibbsGibbs
5,3373827
$endgroup$
$begingroup$
This doesn't look correct. For example, when $M$ is one-dimensional, the exterior derivative is both a derivation of degree one and a skew-derivation of degree one which is not trivial...
$endgroup$
– levap
Jan 22 at 19:18
$begingroup$
@levap You are right, I agree. Thanks for giving your feedback, I spent a big part of the afternoon on this problem and ended up with a wrong answer. Good... Then I do not see how the statement I posted could be correct. It was taken by "Kobayashi and Nomizu - Foundations of Differential Geometry", vol. 1. Do you know how to prove it?
$endgroup$
– Gibbs
Jan 22 at 21:54
add a comment |
$begingroup$
Assume $alpha, beta$ are differential forms and let $D$ be a derivation of order $k$. We have two relevant cases:
$alpha$ of even degree and $beta$ of odd degree.- Both $alpha,beta$ of odd degree.
In the first case we have
begin{align}
D(alpha wedge beta) & = Dalpha wedge beta + alpha wedge Dbeta,\
D(beta wedge alpha) & = Dbeta wedge alpha + beta wedge Dalpha = alpha wedge Dbeta +(-1)^kDalpha wedge beta.
end{align}
First and second line must now agree, which implies that either $k$ is even or $D$ (of odd order) vanishes on even degree forms.
Second case:
begin{align}
D(alpha wedge beta) & = Dalpha wedge beta + alpha wedge Dbeta,\
-D(beta wedge alpha) & = -Dbeta wedge alpha - beta wedge Dalpha = (-1)^k(alpha wedge Dbeta +Dalpha wedge beta).
end{align}
Again, the two lines agree. Now, either $k$ is even, or it is odd and then $D$ vanishes on odd degree forms.
Summing up, either $k$ is even, or $D$ vanishes. This should prove that derivations of odd order vanish and that non-trivial derivations have even order.
Thus, there is no loss of generality in assuming $k$ even. In this case the term giving problems vanishes and the statement holds true.
answered Jan 22 at 18:42
GibbsGibbs
5,3373827
$endgroup$
Assume $alpha, beta$ are differential forms and let $D$ be a derivation of order $k$. We have two relevant cases:
$alpha$ of even degree and $beta$ of odd degree.- Both $alpha,beta$ of odd degree.
In the first case we have
begin{align}
D(alpha wedge beta) & = Dalpha wedge beta + alpha wedge Dbeta,\
D(beta wedge alpha) & = Dbeta wedge alpha + beta wedge Dalpha = alpha wedge Dbeta +(-1)^kDalpha wedge beta.
end{align}
First and second line must now agree, which implies that either $k$ is even or $D$ (of odd order) vanishes on even degree forms.
Second case:
begin{align}
D(alpha wedge beta) & = Dalpha wedge beta + alpha wedge Dbeta,\
-D(beta wedge alpha) & = -Dbeta wedge alpha - beta wedge Dalpha = (-1)^k(alpha wedge Dbeta +Dalpha wedge beta).
end{align}
Again, the two lines agree. Now, either $k$ is even, or it is odd and then $D$ vanishes on odd degree forms.
Summing up, either $k$ is even, or $D$ vanishes. This should prove that derivations of odd order vanish and that non-trivial derivations have even order.
Thus, there is no loss of generality in assuming $k$ even. In this case the term giving problems vanishes and the statement holds true.
answered Jan 22 at 18:42
GibbsGibbs
5,3373827
answered Jan 22 at 18:42
GibbsGibbs
5,3373827
answered Jan 22 at 18:42
GibbsGibbs
5,3373827
answered Jan 22 at 18:42
GibbsGibbs
5,3373827
5,3373827
$begingroup$
This doesn't look correct. For example, when $M$ is one-dimensional, the exterior derivative is both a derivation of degree one and a skew-derivation of degree one which is not trivial...
$endgroup$
– levap
Jan 22 at 19:18
$begingroup$
@levap You are right, I agree. Thanks for giving your feedback, I spent a big part of the afternoon on this problem and ended up with a wrong answer. Good... Then I do not see how the statement I posted could be correct. It was taken by "Kobayashi and Nomizu - Foundations of Differential Geometry", vol. 1. Do you know how to prove it?
$endgroup$
– Gibbs
Jan 22 at 21:54
add a comment |
$begingroup$
This doesn't look correct. For example, when $M$ is one-dimensional, the exterior derivative is both a derivation of degree one and a skew-derivation of degree one which is not trivial...
$endgroup$
– levap
Jan 22 at 19:18
$begingroup$
@levap You are right, I agree. Thanks for giving your feedback, I spent a big part of the afternoon on this problem and ended up with a wrong answer. Good... Then I do not see how the statement I posted could be correct. It was taken by "Kobayashi and Nomizu - Foundations of Differential Geometry", vol. 1. Do you know how to prove it?
$endgroup$
– Gibbs
Jan 22 at 21:54
$begingroup$
This doesn't look correct. For example, when $M$ is one-dimensional, the exterior derivative is both a derivation of degree one and a skew-derivation of degree one which is not trivial...
$endgroup$
– levap
Jan 22 at 19:18
$begingroup$
This doesn't look correct. For example, when $M$ is one-dimensional, the exterior derivative is both a derivation of degree one and a skew-derivation of degree one which is not trivial...
$endgroup$
– levap
Jan 22 at 19:18
$begingroup$
@levap You are right, I agree. Thanks for giving your feedback, I spent a big part of the afternoon on this problem and ended up with a wrong answer. Good... Then I do not see how the statement I posted could be correct. It was taken by "Kobayashi and Nomizu - Foundations of Differential Geometry", vol. 1. Do you know how to prove it?
$endgroup$
– Gibbs
Jan 22 at 21:54
$begingroup$
@levap You are right, I agree. Thanks for giving your feedback, I spent a big part of the afternoon on this problem and ended up with a wrong answer. Good... Then I do not see how the statement I posted could be correct. It was taken by "Kobayashi and Nomizu - Foundations of Differential Geometry", vol. 1. Do you know how to prove it?
$endgroup$
– Gibbs
Jan 22 at 21:54
add a comment |
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