I've been doing some work with differential forms on manifolds. I am still new to the subject so please excuse me for any minor gaffes. Currently, every discussion of manifolds I have done is with them in some ambient space, usually a dim $k$ manifold in $R^n$. Now, (and I think this is why some books say it is often easier to work without an ambient space).

Now, when we write a $k$ differential form on this manifold, the presence of the ambient space complicates things. Specifically, if I have a form defined on some coordinate patch of a k surface in $R^n$, the form has to be written as: $$ \sum_I f_I dx_I $$ But, if we have a diffeomorphism, we can use this diffeomorphism to pull-back the form to a top form in $R^k$, giving it the neat representation $f dx_1 \wedge \cdots \wedge dx_k$. My question is: is this pulled back form "equal" to the form in the ambient space? It seems like there would be some differences, for example, where the forms are being evaluated (evaluation being the k vectors that the form takes in and spits out a real number with).


The general idea of the formalism of differential geometry is to find rules for representations of objects which actually should not depend on the representation. This includes embeddings into some ambient space or local coordinate system. A $k$- form on a manifold is a purely geometric object, so this applies to $k$-forms, too.

In older books on the topic or in phisics textbooks you'll find complicate looking equations which verify the "correct" transformation behavior for coefficients of differential geometric objects (like vector field, $k$-forms or other tensor fields) which is nothing but the assertion that these objects are, in fact, independent of the coordinate system. A similar calculation (which I will refuse to write down right now) will show that the answer to your question is 'yes'. The pulled back form is equal to the form in the ambient space in the sense that, if you plug in a set of $k$ vectors tangent to the manifold in some point and calculate the result, the result will be the same regardless of the representation (of course you have to adapt the representation of the $k$ vectors, as well).

Note though, that there are geometric quantities which do depend on the embedding (e.g. some curvatures of a curve or surfaces).


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.