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Codifferential of k-th exterior power endmorphism applied to differential form $delta(wedge^p Aalpha)$?

2

1

$begingroup$

Let $(M,g)$ be an $n$-dimensional smooth Riemannian manifold without boundary. Given an endomorphism $A(x) : sf T_x^*M to sf T_x^*M$ on the cotangent bundle, we can extend $A$ to an endomorphism on the $p$-th exterior power acting on a $p$-form $alpha$ by
begin{align*}
wedge^p A : wedge^p sf T_x^*M &to wedge^p sf T_x^*M\
wedge^p A(alpha_1 wedge ... wedge alpha_p) &:= Aalpha_1 wedge ... wedge Aalpha_p.
end{align*}
How can we make sense of the codifferential
$$delta(wedge^p Aalpha)?$$
The problem is that $dim wedge^p sf T_x^*M = {n choose p}$, unlike the case $dim wedge^n sf T_x^*M = {n choose n} = 1$, where $wedge^n A$ must act as a multiplication operator, i.e. the determinant.

But $delta$ is not a derivation of the algebra, i.e. there is no nice version of the Leibniz rule. Either using the definition of $delta = ast , mathbf d , ast $ directly or using the definition that it is the formal adjoint of the exterior derivative $mathbf d$, the question boils down to: What is the exterior derivative of this $k$-th power endomorphism $wedge^p A$?

differential-geometry differential-forms

share|cite|improve this question

edited Mar 29 at 0:05

wueb

asked Mar 22 at 18:49

wuebwueb

34129

$endgroup$

add a comment | 

2

1

$begingroup$

Let $(M,g)$ be an $n$-dimensional smooth Riemannian manifold without boundary. Given an endomorphism $A(x) : sf T_x^*M to sf T_x^*M$ on the cotangent bundle, we can extend $A$ to an endomorphism on the $p$-th exterior power acting on a $p$-form $alpha$ by
begin{align*}
wedge^p A : wedge^p sf T_x^*M &to wedge^p sf T_x^*M\
wedge^p A(alpha_1 wedge ... wedge alpha_p) &:= Aalpha_1 wedge ... wedge Aalpha_p.
end{align*}
How can we make sense of the codifferential
$$delta(wedge^p Aalpha)?$$
The problem is that $dim wedge^p sf T_x^*M = {n choose p}$, unlike the case $dim wedge^n sf T_x^*M = {n choose n} = 1$, where $wedge^n A$ must act as a multiplication operator, i.e. the determinant.

But $delta$ is not a derivation of the algebra, i.e. there is no nice version of the Leibniz rule. Either using the definition of $delta = ast , mathbf d , ast $ directly or using the definition that it is the formal adjoint of the exterior derivative $mathbf d$, the question boils down to: What is the exterior derivative of this $k$-th power endomorphism $wedge^p A$?

differential-geometry differential-forms

share|cite|improve this question

edited Mar 29 at 0:05

wueb

asked Mar 22 at 18:49

wuebwueb

34129

$endgroup$

add a comment | 

$begingroup$

Let $(M,g)$ be an $n$-dimensional smooth Riemannian manifold without boundary. Given an endomorphism $A(x) : sf T_x^*M to sf T_x^*M$ on the cotangent bundle, we can extend $A$ to an endomorphism on the $p$-th exterior power acting on a $p$-form $alpha$ by
begin{align*}
wedge^p A : wedge^p sf T_x^*M &to wedge^p sf T_x^*M\
wedge^p A(alpha_1 wedge ... wedge alpha_p) &:= Aalpha_1 wedge ... wedge Aalpha_p.
end{align*}
How can we make sense of the codifferential
$$delta(wedge^p Aalpha)?$$
The problem is that $dim wedge^p sf T_x^*M = {n choose p}$, unlike the case $dim wedge^n sf T_x^*M = {n choose n} = 1$, where $wedge^n A$ must act as a multiplication operator, i.e. the determinant.

But $delta$ is not a derivation of the algebra, i.e. there is no nice version of the Leibniz rule. Either using the definition of $delta = ast , mathbf d , ast $ directly or using the definition that it is the formal adjoint of the exterior derivative $mathbf d$, the question boils down to: What is the exterior derivative of this $k$-th power endomorphism $wedge^p A$?

differential-geometry differential-forms

share|cite|improve this question

edited Mar 29 at 0:05

wueb

asked Mar 22 at 18:49

wuebwueb

34129

$endgroup$

share|cite|improve this question

edited Mar 29 at 0:05

wueb

asked Mar 22 at 18:49

wuebwueb

34129

share|cite|improve this question

edited Mar 29 at 0:05

wueb

asked Mar 22 at 18:49

wuebwueb

34129

asked Mar 22 at 18:49

wuebwueb

34129

asked Mar 22 at 18:49

wuebwueb

34129