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For a general vector bundle, I there is no "$E$-valued integration" as you put it. You are trying to add up elements in the fibres of $E$, but since the fibres over different points are not the same vector space you can't add their elements.

For the trivial bundle $M \times \mathbb{R}^k$ - and with a fixed choice of trivialisation! - you can carry out the integral component by component, But if you change the trivialisation you will get a different answer. Moreover, you can change the trivialisation in a way that varies over the manifold, so there is no hope that the integral will just change by a linear map of $\mathbb{R}^k$. You can see this behaviour even with ordinary functions. A function is a section of the trivialised rank 1 bundle. If you change the trivialisation, but insist on regarding ordinray $p$-forms as bundle valued, you multiply all your forms by a fixed nowhere vanishing function. This can change the integral over a $p$-cycle in a more-or-less arbitrary way.

What you can integrate $E$-valued forms against is $E^*$-valued ones. Given $a \in \Omega^p(E)$ and $b \in \Omega^q(E^*)$ their wedge product is an ordinary $(p+q)$-form which you can then integrate over a $(p+q)$-cycle. Now you have a version of Stokes theorem. If you have a connection $A$ in $E$ then you can check that $$ d(a \wedge b) = d_A(a) \wedge b \pm a \wedge d_A(b). $$ So Stokes theorem gives $$ \int_{M}d_A(a) \wedge b \pm a \wedge d_A(b) = \int_{\partial M} a \wedge b. $$ In the case when $b$ is a covariant constant section of $E$ and $M$ has dimension one more than the degree of $a$, we get $$\int_M \langle d_A(a) , b \rangle=\int_{\partial M} \langle a, b \rangle$$ This is just the usual Stokes theorem, for the component of $a$ in the direction $b$. Since $b$ is covariant-constant, $\langle d_A(a), b \rangle = d \langle a, b \rangle$.

For a general vector bundle, I there is no "$E$-valued integration" as you put it. You are trying to add up elements in the fibres of $E$, but since the fibres over different points are not the same vector space you can't add their elements.

For the trivial bundle $M \times \mathbb{R}^k$ - and with a fixed choice of trivialisation! - you can carry out the integral component by component, But if you change the trivialisation you will get a different answer. Moreover, you can change the trivialisation in a way that varies over the manifold, so there is no hope that the integral will just change by a linear map of $\mathbb{R}^k$. You can see this behaviour even with ordinary functions. A function is a section of the trivialised rank 1 bundle. If you change the trivialisation, but insist on regarding ordinary $p$-forms as bundle valued, you multiply all your forms by a fixed nowhere vanishing function. This can change the integral over a $p$-cycle in a more-or-less arbitrary way.

What you can integrate $E$-valued forms against is $E^*$-valued ones. Given $a \in \Omega^p(E)$ and $b \in \Omega^q(E^*)$ their wedge product is an ordinary $(p+q)$-form which you can then integrate over a $(p+q)$-cycle. Now you have a version of Stokes theorem. If you have a connection $A$ in $E$ then you can check that $$ d(a \wedge b) = d_A(a) \wedge b \pm a \wedge d_A(b). $$ So Stokes theorem gives $$ \int_{M}d_A(a) \wedge b \pm a \wedge d_A(b) = \int_{\partial M} a \wedge b. $$ In the case when $b$ is a covariant constant section of $E$ and $M$ has dimension one more than the degree of $a$, we get $$\int_M \langle d_A(a) , b \rangle=\int_{\partial M} \langle a, b \rangle$$ This is just the usual Stokes theorem, for the component of $a$ in the direction $b$. Since $b$ is covariant-constant, $\langle d_A(a), b \rangle = d \langle a, b \rangle$.

For a general vector bundle, I there is no "$E$-valued integration" as you put it. You are trying to add up elements in the fibres of $E$, but since the fibres over different points are not the same vector space you can't add their elements.

For the trivial bundle $M \times \mathbb{R^k}$ - and with a fixed choice of trivialisation! - you can carry out the integral component by component, But if you change the trivialisation you will get a different answer. Moreover, you can change the trivialisation in a way that varies over the manifold, so there is no hope that the integral will just change by a linear map of $\mathbb{R^k}$. You can see this behaviour even with ordinary functions. A function is a section of the trivialised rank 1 bundle. If you change the trivialisation, but insist on regarding ordinray $p$-forms as bundle valued, you multiply all your forms by a fixed nowhere vanishing function. This can change the integral over a $p$-cycle in a more-or-less arbitrary way.

What you can integrate $E$-valued forms against is $E^*$-valued ones. Given $a \in \Omega^p(E)$ and $b \in \Omega^q(E^*)$ their wedge product is an ordinary $(p+q)$-form which you can then integrate over a $(p+q)$-cycle. Now you have a version of Stokes theorem. If you have a connection $A$ in $E$ then you can check that $$ d(a \wedge b) = d_A(a) \wedge b \pm a \wedge d_A(b). $$ So Stokes theorem gives $$ \int_{M}d_A(a) \wedge b \pm a \wedge d_A(b) = \int_{\partial M} a \wedge b. $$ In the case when $b$ is a covariant constant section of $E$ and $M$ has dimension one more than the degree of $a$, we get $$\int_M \langle d_A(a) , b \rangle=\int_{\partial M} \langle a, b \rangle$$ This is just the usual Stokes theorem, for the component of $a$ in the direction $b$. Since $b$ is covariant-constant, $\langle d_A(a), b \rangle = d \langle a, b \rangle$.

For a general vector bundle, I there is no "$E$-valued integration" as you put it. You are trying to add up elements in the fibres of $E$, but since the fibres over different points are not the same vector space you can't add their elements.

For the trivial bundle $M \times \mathbb{R}^k$ - and with a fixed choice of trivialisation! - you can carry out the integral component by component, But if you change the trivialisation you will get a different answer. Moreover, you can change the trivialisation in a way that varies over the manifold, so there is no hope that the integral will just change by a linear map of $\mathbb{R}^k$. You can see this behaviour even with ordinary functions. A function is a section of the trivialised rank 1 bundle. If you change the trivialisation, but insist on regarding ordinray $p$-forms as bundle valued, you multiply all your forms by a fixed nowhere vanishing function. This can change the integral over a $p$-cycle in a more-or-less arbitrary way.

What you can integrate $E$-valued forms against is $E^*$-valued ones. Given $a \in \Omega^p(E)$ and $b \in \Omega^q(E^*)$ their wedge product is an ordinary $(p+q)$-form which you can then integrate over a $(p+q)$-cycle. Now you have a version of Stokes theorem. If you have a connection $A$ in $E$ then you can check that $$ d(a \wedge b) = d_A(a) \wedge b \pm a \wedge d_A(b). $$ So Stokes theorem gives $$ \int_{M}d_A(a) \wedge b \pm a \wedge d_A(b) = \int_{\partial M} a \wedge b. $$ In the case when $b$ is a covariant constant section of $E$ and $M$ has dimension one more than the degree of $a$, we get $$\int_M \langle d_A(a) , b \rangle=\int_{\partial M} \langle a, b \rangle$$ This is just the usual Stokes theorem, for the component of $a$ in the direction $b$. Since $b$ is covariant-constant, $\langle d_A(a), b \rangle = d \langle a, b \rangle$.

For a general vector bundle, I there is no "$E$-valued integration" as you put it. You are trying to add up elements in the fibres of $E$, but since the fibres over different points are not the same vector space you can't add their elements.

For the trivial bundle $M \times \mathbb{R^k}$ - and with a fixed choice of trivialisation! - you can carry out the integral component by component, But if you change the trivialisation you will get a different answer. Moreover, you can change the trivialisation in a way that varies over the manifold, so there is no hope that the integral will just change by a linear map of $\mathbb{R^k}$. You can see this behaviour even with ordinary functions. A function is a section of the trivialised rank 1 bundle. If you change the trivialisation, but insist on regarding ordinray $p$-forms as bundle valued, you multiply all your forms by a fixed nowhere vanishing function. This can change the integral over a $p$-cycle in a more-or-less arbitrary way.

What you can integrate $E$-valued forms against is $E^*$-valued ones. Given $a \in \Omega^p(E)$ and $b \in \Omega^q(E^*)$ their wedge product is an ordinary $(p+q)$-form which you can then integrate over a $(p+q)$-cycle. Now you have a version of Stokes theorem. If you have a connection $A$ in $E$ then you can check that $$ d(a \wedge b) = d_A(a) \wedge b \pm a \wedge d_A(b). $$ So Stokes theorem gives $$ \int_{M}d_A(a) \wedge b \pm a \wedge d_A(b) = \int_{\partial M} a \wedge b. $$ In the case when $b$ is a covariant constant section of $E$ and $M$ has the same dimension one more than the degree of $a$, we get $$\int_M \langle d_A(a) , b \rangle=\int_{\partial M} \langle a, b \rangle$$ This is just the usual Stokes theorem, for the component of $a$ in the direction $b$. Since $b$ is covariant-constant, $\langle d_A(a), b \rangle = d \langle a, b \rangle$.

For a general vector bundle, I there is no "$E$-valued integration" as you put it. You are trying to add up elements in the fibres of $E$, but since the fibres over different points are not the same vector space you can't add their elements.

For the trivial bundle $M \times \mathbb{R^k}$ - and with a fixed choice of trivialisation! - you can carry out the integral component by component, But if you change the trivialisation you will get a different answer. Moreover, you can change the trivialisation in a way that varies over the manifold, so there is no hope that the integral will just change by a linear map of $\mathbb{R^k}$. You can see this behaviour even with ordinary functions. A function is a section of the trivialised rank 1 bundle. If you change the trivialisation, but insist on regarding ordinray $p$-forms as bundle valued, you multiply all your forms by a fixed nowhere vanishing function. This can change the integral over a $p$-cycle in a more-or-less arbitrary way.

What you can integrate $E$-valued forms against is $E^*$-valued ones. Given $a \in \Omega^p(E)$ and $b \in \Omega^q(E^*)$ their wedge product is an ordinary $(p+q)$-form which you can then integrate over a $(p+q)$-cycle. Now you have a version of Stokes theorem. If you have a connection $A$ in $E$ then you can check that $$ d(a \wedge b) = d_A(a) \wedge b \pm a \wedge d_A(b). $$ So Stokes theorem gives $$ \int_{M}d_A(a) \wedge b \pm a \wedge d_A(b) = \int_{\partial M} a \wedge b. $$ In the case when $b$ is a covariant constant section of $E$ and $M$ has dimension one more than the degree of $a$, we get $$\int_M \langle d_A(a) , b \rangle=\int_{\partial M} \langle a, b \rangle$$ This is just the usual Stokes theorem, for the component of $a$ in the direction $b$. Since $b$ is covariant-constant, $\langle d_A(a), b \rangle = d \langle a, b \rangle$.