I believe we have orientation forms (nowhere vanishing n-forms) and orientable manifolds (manifolds that admit an orientation form). What is your definition of volume form?

math.stackexchange.com/questions/903440/…

@l4teLearner I define a volume form as nowhere vanishing n-form. An orientation form is an n-form that defines the manifold's orientation. An orientation form is a volume form I believe. Edit I've seen that link, hence why I asked to prove what's been said in the top voted answer

@Kajelad I just modified the question, because there was clearly a contradiction.

In my reference book (Loring Tu, Introduction to Manifolds) a n-form needs compact support to be integrable. Please consider that a closed subset of a compact set is compact: math.stackexchange.com/questions/212181/…

@l4teLearner I edited the question so that M is compact, so yes the support of the n-form is automatically compact.

According to the answer to the related question I provided in the comment, the integral of a volume form is necessarily non zero. What is less clear to me is your second question, i.e. what do you mean by an "oriented" form.

@l4teLearner an orientation form is a volume form that is positively oriented, meaning it's positively oriented in each tangent space. I'm still not convinced why the integral of a volume form (not necessarily an orientation form) is nonzero

@l4teLearner sorry if my question wasn't well worded, I obviously don't just want to know if it's true or not, but I want to be convinced why in either case

I may be mistaken but I believe you cannot say if a volume form is positively or negativey oriented in general, that would depend on the atlas you choose. Just think of $\mathbb{R}^2$ and flip the sign of one coordinate. You can classify orientation forms in two equivalence classes but the sign that you will apply to each equivalence class would be conventional... as for the rigorous proof of that answer, it would be too long to be put in a comment, but it is basically sections 23.3 and 23.4 of Tu's Introduction to manifolds.

@l4teLearner I've seen your book but nothing new I've seen there. Check John Lee's book with the same title for a more detailed construction. I just want us to agree on the definitions. A volume form is a nowhere vanishing n-form (nothing to do with orientation). Check chapter 15 for the definition of a manifold orientation (it's defined independently of any n-form), and most importantly prop 15.5 that gives the rise of a precious volume form called "orientation form" that is in some way compatible with the orientation of a manifold

@l4teLearner Let M be an oriented manifold. Prop 15.5 gives us the existence of an orientation form that is compatible with the orientation. A positively oriented form is basically a positive function times the orientation form, a negatively oriented form is defined similarly. My question is about a volume form picked arbitrarily. You don't have the right to choose that orientation form. That would be too easy

I am sorry but I don't have that book at hand. At the risk of being repetitive, but hoping that this helps you, I would also have a look at section 21.4 of Tu, where a 1 to 1 correspondence of orientations and equivalence classes of nowhere vanishing n-forms is established. Let's wait for a more authoritative answer, then.

@l4teLearner update: I'm now convinced that if M is connected, then the answer is yes. Prop 15.9 (John Lee) says that if M is connected, then there are exactly two opposite orientations on M. So a volume form is either positively oriented (therefore its integral is positive) or negatively oriented (negative integral). So for now, I will be satisfied if you give me a counterexample (volume form with integral=0) in the case of a nonconnected manifold, or if you prove that the integral is nonzero in general

@l4teLearner I see the reasoning. It might work. Sorry, can I ask another question? it's related to a comment you made saying that a volume form doesn't need to be positively or negatively. My question is this: take two volume forms on a compact oriented manifold, is it true that one of them is a smooth function times the other? And why? I hope you see how my question relates to your comment

more thanwelcome to ask, keep in mind I am mostly self taught and as my avatar says... still learning:) I did not mean that a volume form does not need to be positive or neg. You split them in two equivalence classes, but the label you give to the two classes is arbitrary... as you will need to specify an orientation upfront to say if a form agrees with it or not. Once you define an orientation, then ok, the volume form will either agree everywhere or disagree everywhere with it. Also, keep in mind that for the integration of a form to be meaningful, you have to choose an orientation first.

@l4teLearner I hope you understand that's quite a big claim (for a beginner like me) that's worthwhile proving. So, let's formulate it. If M is a compact manifold with an orientation, every volume form must be positively oriented or negatively oriented. Can you prove it rigorously?

hi, I think that it is better to move this discussion in the chat. as I am a bit lazy today, but I think this snapshot could be useful for you, here is an extract from my book:

postimg.cc/D8ygmg1z

postimg.cc/9r6rsXXj this is better, the other got cropped

so if $\omega$ determines an orientation, all the $f \omega$ forms with $f>0$ will determine the samd orientation. while the others, $f<0$ will disagree. these are all and only the possibilies.

I guess there are only two orientations IF M is connected. If M is not connected, let's say there are two connected components. we can define 4 orientations: let's pick an arbitrary orientation ++ on M (imagine it like a magnet). We can define a second orientation +-(positive on the first connected component, negative on the second), -+ and -- defined similarly.

Your dichotomy falls flat if M is not connected.

yep sure. all this holds if M is connected!

But with 4 orientations, it obviously doesn't hold

You can see it this way : the function f can have two different signs on the two connected components.

yeah, it will be nicely different from zero. i take it as the manifold equivalent of the zero theorem

If you're talking about Intermediate value theorem, it does not hold on non connected manifols. That's why connectedness is so important. If M is has two connected components for example, you can define a nowhere vanishing function on M that has 2 different signs on each component

Take the function x->x^3 on the manifold M=]-2,-1[U]1,2[. f is nowhere nonvanishing but has two different signs.

postimg.cc/phrV1r7x indeed.

Do you see why it's so crucial to have connectedness? It's such a life saver because it allows to use the Intermediate value theorem

And about the last snap, that's what I've been saying

yeah I think that we agree on this. hope this is a bit more clear now.

So, let's rewind. If M is nonconnected, the integral of a volume form is nonzero. If not, there are volume forms whose integral vanishes. So the answer you cited in the beginning math.stackexchange.com/questions/903440/… is false sadly.

yes you are right, the answer I cited only holds for connected manifolds. this is also what Moishe said yesterday

i am not sure who closed your answer. i didnt.

Hold on a minute, I think I understand everything now. The orientation is just a matter of convention chosen arbitrarily. So if I have a volume form, I can choose THE orientation such as my volume form is positively oriented. And I define the integral using this orientation that I've chosen. I have the right to choose after all. Well thanks for your time. I'm satisfied now :)

you are welcome. everyone needs to figure out things so that they click for them :) ( by the way the answer mentioned that the manifold has to be connected, so maybe it is my fault to have cited it a bit out of context). cheers!

You all provoked me into writing a blog post about how to define orientation. deaneyang.github.io/blog/blog/math/differential-geometry/…

Thanks buddy