MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top


I have the following question: Let $U\subset \mathbb{R}^{n}$ be an open subset. Furthermore, let $\nabla$ be a flat connection on $U$ (not necessary Levi-Civita). How can one show that the space of covariant constant one forms on $U$ is finite dimensional, i.e. $n$-dimensional real vector space ? is it true ? or no ?


share|cite|improve this question
anyone has an idea ? – hapchiu Mar 3 '13 at 7:51

The covariant constant 1-forms are parallel. The value of a covariant constant 1-form is determined throughout each connected component by its value at any one point: just take that value and parallel transport along any path to any other point. This recipe will succeed in producing a 1-form just when the 1-form you start with is invariant under the holonomy group of the connection; otherwise there is no covariant constant 1-form with the given initial value. However, if $U$ has infinitely many connected components, and there is a nonzero covariant constant 1-form in infinitely many of those path components, then the space of covariant constant 1-forms will have infinite dimension. If $U$ has a finite number of path components then the dimension of the space of covariant constant 1-forms is at most the product of the dimension of $U$ and the number of its path components. In case the connection is flat, the holonomy group is discrete, but might still be very complicated, so you can't guarantee that there are any nonzero covariant constant 1-forms.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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