# By Sheaves!

1. Oct 23, 2013

### Mandelbroth

cf. Musicals that, unfortunately, would not work. :tongue:

I recently started to go a little further in depth into studying sheaves.

I have two questions. Firstly, is the presheaf of exact forms on a smooth manifold necessarily a sheaf? I was under the impression that it wasn't. Additionally, is the space of once differentiable (but not necessarily continuously differentiable) functions together with their domains on a smooth manifold (with restriction maps being restrictions of domains) a sheaf? I'm having trouble looking at these two.

Additionally, can anyone point me to any good internet resources for learning more about using sheaves? I've got a couple of sources from a quick google search, but I'm intending to go a little further in depth this afternoon.

Thank you.

2. Oct 23, 2013

### R136a1

Yes to both, because they are local conditions. Proving that something is a sheaf is essentially the same as proving that some definition is local, that is: the condition holds for each point iff for each point there is a neighborhood base for which the condition holds.

A standard example of a non-local condition is being constant. A locally constant function is not a constant function, for example take: $X=(0,1)\cup (2,3)$ and set elements of $(0,1)$ to 1$and elements of$(2,3)$to$2$. Thus the presheaf of constant functions is not a sheaf, but the presheaf of locally constant functions is a sheaf, called the constant sheaf. If you want to learn about sheaf, then I highly recommend the book by Tennison. If you want online notes, then this is very good: http://math.stanford.edu/~vakil/216blog/ 3. Oct 23, 2013 ### Mandelbroth This is wonderful. Thank you! 4. Nov 24, 2013 ### mathwonk this is not the usual meaning of "exact" which is a global condition. i.e. locally exact is usually called "closed" for forms. so no, in my terminology, exact forms are not a sheaf. 5. Nov 25, 2013 ### Mandelbroth Can you give an example where they are not a sheaf? I thought, intuitively, that they weren't, but I couldn't really think why. I consider a form$\alpha$to be exact if(f) there exists a form$\beta$such that$\alpha=\mathrm{d}\beta$. That is, an exact form is a form in the image of the exterior derivative. I'm thinking that, if there was a problem with it being a sheaf, it would be from the locality condition, but I can't come up with a good counterexample. Or, maybe I'm just going about this the wrong way. I don't know. 6. Nov 27, 2013 ### R136a1 Yes, I'm so sorry. They don't form a sheaf. It's a theorem that closed 1-forms are always locally exact. So if closed implies exact, then you should have a sheaf. One place where this is true is in simply connected situations. I leave the verification of this to you (and you have seen that I can be easily wrong, so don't trust my word for it :D). Anyway, a locally exact 1-form that is not exact is a very standard counterexample: Take on$\mathbb{R}^2\setminus\{(0,0)\}$the 1-form given by $$\frac{xdy - ydx}{x^2 + y^2}$$ This is closed (and thus locally exact), but not exact. So exactness is not a local condition. To derive as consequence that the exact forms don't yield a sheaf, I leave to you. Let$M$be a manifold, and let$\mathcal{O}_n(M)$be the$n##-forms, then we can form the following sequence
$$\mathcal{O}_0(M)\rightarrow \mathcal{O}_1(M)\rightarrow ...\rightarrow \mathcal{O}_n(M)\rightarrow ...$$
The arrows are given by the exterior derivative. Now, closed forms are simply the sheaf kernels of this exterior derivative. Exact forms, however, are presheaf images of the exterior derivative. The annoying part is that presheaf images are not sheafs in general (presheaf kernels always are sheafs however, which yields that closed forms do form a sheaf). In order to form the "sheaf image", one needs to proceed by taking sheafifications. However, the sheaf image are no longer the exact forms.

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