4
$\begingroup$

When studying fibre bundles, connections and gauge theories it is usual to consider vector-valued differential forms, like the connection one-form, or it's pull back by a local trivialization known as the gauge potential. In those examples, the form takes value on the Lie algebra of the group, and hence, they take values on some vector space $V$.

Now, for regular $\mathbb{R}$-valued differential forms, integration is simple to define. If $c$ is a singular $k$-cube and $\omega$ a differential $k$-form, then

$$\int_{c} \omega = \int_{[0,1]^k} c^\ast \omega.$$

Now, how the integral of a $V$-valued differential form is defined? Is it simply componentwise? If so, I'm unsure if it depends on the basis chosen for the vector space.

So how one deals with this kind of integration? Is it some well-defined concept?

$\endgroup$
5
$\begingroup$

It's just componentwise. Fix a basis $e_i$ with dual basis $\theta^i$ and define $$\int \omega = \sum_i \left(\int \theta^i \circ \omega\right) e_i.$$

If you have another basis $v_i$ with dual $\pi^i$, then let $A$ be the change of basis matrix such that $v_j = \sum_i A_{ji} e_i$. A little bit of linear algebra tells us that the dual bases are then related by $\pi^j = \sum_i A^{-1}_{ij} \theta^i$, so by the linearity of integration we have

$$ \sum_i \left( \int \pi^i \circ \omega \right) v_i = \sum_{i,j} A^{-1}_{ji} \left( \int \theta^j \circ \omega \right) v_i =\sum_{i,j,k} A^{-1}_{ji} A_{ik} \left( \int \theta^j \circ \omega \right) e_k = \sum_{j} \left( \int \theta^j \circ \omega \right) e_j;$$ i.e. the integral is basis-independent.

If you want a more abstract definition then something like the Bochner integral might be more satisfying, but it's not at all necessary in finite dimension.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.