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• CommentRowNumber1.
• CommentAuthorUrs
• CommentTimeSep 27th 2010
• (edited Dec 13th 2012)

created Stokes theorem

• CommentRowNumber2.
• CommentAuthorEric
• CommentTimeSep 28th 2010
• (edited Sep 28th 2010)

The boundary of this simplex in $X$ is the the chain (formal linear combination of smooth $(k-1)$-simplices)

$\partial \Delta^k = \sum_{i = 0}^k (-1)^i \sigma \circ \partial_i$

Should that be

$\sigma\circ \partial \Delta^k = \sum_{i = 0}^k (-1)^i \sigma \circ \partial_i\Delta^k$

?

Or would it be better to write simply

$\partial\Delta^k = \sum_{i = 0}^k (-1)^i \partial_i\Delta^k$

or maybe even

$\partial = \sum_{i = 0}^k (-1)^i \partial_i$

? In any case, what is written there doesn’t seem quite right, i.e. some kind of typo.

• CommentRowNumber3.
• CommentAuthorEric
• CommentTimeSep 28th 2010

By the way, is it possible to define the de Rham differential in terms of the boundary and Stokes’ theorem?

Stokes’ theorem is so important, I’ve often thought it should be taken as the definition of $d$ somehow. Coordinates are yucky.

• CommentRowNumber4.
• CommentAuthorUrs
• CommentTimeSep 28th 2010

some kind of typo.

Yes, thanks, I fixed it: the first $\Delta^k$ should have been the simplex $\sigma : \Delta^k \to X$:

$\partial \sigma = \sum_{i = 0}^n (-1)^i \sigma \circ \partial_i \,.$
• CommentRowNumber5.
• CommentAuthorTobyBartels
• CommentTimeSep 28th 2010

is it possible to define the de Rham differential in terms of the boundary and Stokes’ theorem?

Yes, if $\omega$ is a $p$-form, then $\mathrm{d}\omega$ is the unique $(p+1)$-form such that $\int_C \mathrm{d}\omega = \int_{\partial{C}} \omega$ for every $(p+1)$-chain $C$. However, one still has to prove that such a $(p+1)$-form exists. (Proving that it’s unique is fairly easy.) I remember that Tom Apostol’s textbook Calculus defined the divergence and curl of a vector field in essentially this way.

Coordinates are yucky.

Who would write down the definition of the differential using coordinates? You don’t need Stokes’s Theorem to avoid that, just a lot of alternating signs.

• CommentRowNumber6.
• CommentAuthorEric
• CommentTimeSep 28th 2010
• (edited Sep 28th 2010)

Yes

Cool. Thanks.

Who would write down the definition of the differential using coordinates?

I’m pretty sure that some of the references I learned from (Flanders?) defined it in local coordinates via

$d(\alpha_{\mu_1,\cdots,\mu_p} d x^{\mu_1}\wedge\cdots\wedge d x^{\mu_p}) = (\partial_\mu \alpha_{\mu_1,\cdots,\mu_p}) d x^\mu\wedge d x^{\mu_1}\wedge\cdots\wedge d x^{\mu_p} .$

I think that is pretty common (?). Or maybe I’m just old :)

Wikipedia has a nice coordinate-free definition

The exterior derivative is defined to be the unique R-linear mapping from k-forms to (k+1)-forms satisfying the following properties:

1. dƒ is the differential of ƒ for smooth functions ƒ.
2. d(dƒ) = 0 for any smooth function ƒ.
3. d(α∧β) = dα∧β + (−1)p(α∧dβ) where α is a p-form. That is to say, d is a derivation of degree 1 on the exterior algebra of differential forms.

I remember one time trying to derive these three properties (1. and 2. are trivial) directly from boundary, join, and Stokes’.

• CommentRowNumber7.
• CommentAuthorTobyBartels
• CommentTimeSep 30th 2010

I think that is pretty common

Certainly. But only silly people would do it. (^_^)

Wikipedia has a nice coordinate-free definition

Essentially that characterisation is also used at differential form. (Actually, we characterise the entire algebra of smooth differential forms, complete with both $\mathrm{d}$ and $\wedge$, at once.)

Of course, you also have to prove that such a mapping exists. (This also holds for the definition through Stokes’s Theorem.) Here you might resort to coordinates, showing that the definition in those terms satisfies it. OTOH, all definitions using local coordinates also rely on a theorem (although the one theorem suffices for all applications): that every form is a sum of forms defined only on local coordinate patches, where only finitely many nonzero terms appear in the sum at any point. (This has to do with partitions of unity.)

However, you can also write down a direct definition of $\mathrm{d}$ for all forms, in elementary terms, referring directly to elementary definitions. If you don’t see this, here’s my offer: tell me your favourite elementary definition of what a form is, and if this refers to some other concept (such as tangent vectors), define those too. (You may take as given the meaning of a smooth curve near a point and a smooth scalar field near a point; you don’t have to define those.) Then I should be able to write down the definition of $\mathrm{d}$ for you.

By the way, I don’t mean to disagree with you that

defining the exterior derivative via Stokes’ theorem provides a clearer picture of what the exterior derivative really is

(from the top of page 101 of your thesis cited above).

• CommentRowNumber8.
• CommentAuthorEric
• CommentTimeSep 30th 2010

Hi Toby,

It will take a little more time than I have to spare right now to provide some info you requested (it sounds like fun though so I will try later! :)), but in the meantime, I was wondering if you or anyone else may find the motivation interesting enough to work out item 3 from boundary, join, and Stokes’ theorem? I think a clever high school student could figure it out, so it might not be “fun” enough, but I think it would be cute to see an “algebraic” derivation.

Basically, we have

$\int_{\partial(a*b)} \alpha\wedge\beta = \int_{a*b} d(\alpha\wedge\beta) = \int_{a*b} [(d\alpha)\wedge\beta + (-1)^{|\alpha|} \alpha\wedge d\beta] = \int_{(\partial a)*b + (-1)^{|a|+1} a*\partial b} \alpha\wedge\beta.$

It may be impossible, but I thought it might be cute if the integral of a join may be split into two intergals and apply Stokes to each individually. The goal was to derive the graded Leibniz rule from that of boundary and join somehow. Or something…

It’s been a while.

• CommentRowNumber9.
• CommentAuthorUrs
• CommentTimeSep 30th 2010
• (edited Sep 30th 2010)

Toby recalled

Wikipedia has a nice coordinate-free definition

whose first item is

$d f$ is the differential of $f$ for smooth functions $f$.

One can go even further and let even that come out automatic. That is the definition of the de Rham complex as Kähler differentials of $C^\infty$-rings of smooth functions.

• CommentRowNumber10.
• CommentAuthorEric
• CommentTimeSep 30th 2010
• (edited Sep 30th 2010)

Hi Urs,

I only have a vague idea of the relation between spaces and algebras, but maybe what I was trying to do is find something like a category of (noncommutative) Kähler spaces whose opposite is the category of (noncommutatve) Kähler differentials. Or something…

Kähler spaces would be something along the lines of spaces that satisfy

$\partial(a*b) = (\partial a)*b + (-1)^{|a|+1} a*\partial b$

where $a*b$ is the join of two Kähler spaces $a$ and $b$ and $|a|$ is the dimension of $a$. Or something…

I know I am not saying that very clearly, but I hope that the idea buried in there makes a little bit of sense even if the exact wording does not.

PS: Is there a such thing a “join closed” or something to describe a category of spaces closed under join operation? Or is that automatic for most spaces?

• CommentRowNumber11.
• CommentAuthorUrs
• CommentTimeSep 30th 2010
• (edited Sep 30th 2010)

The dual of $d (\lambda \wedge \omega) = (d \lambda ) \wedge \omega \pm \omega \wedge d\lambda$ is the Lie bracket

$[v,w]$

on vector fields. Specifically, the space that is the formal dual of the dg-algebra of forms on $X$ is the tangent Lie algebroid of $X$.

• CommentRowNumber12.
• CommentAuthorEric
• CommentTimeSep 30th 2010
• (edited Sep 30th 2010)

The dual of $d (\lambda \wedge \omega) = (d \lambda ) \wedge \omega \pm \omega \wedge d\lambda$ is the Lie bracket

$[v,w]$

Ok, but this is something that always seemed to trip me up. When talking about forms, there are two ways to talk about duals

1. duals in the sense of (multi)vector fields
2. duals in the sense of chains

I am talking about the second sense.

• CommentRowNumber13.
• CommentAuthorUrs
• CommentTimeSep 30th 2010

I am talking about the second sense.

Not when making statements like:

find something like a category of (noncommutative) Kähler spaces whose opposite is the category of (noncommutatve) Kähler differentials.

The opposite of the category of dg-algebras is that of $\infty$-Lie algebroids, and under this identitfication $\Omega^\bullet(X)$ maps to the tangent Lie algebroid.

• CommentRowNumber14.
• CommentAuthorzskoda
• CommentTimeSep 30th 2010
• CommentRowNumber15.
• CommentAuthorUrs
• CommentTimeSep 30th 2010

Yes.

• CommentRowNumber16.
• CommentAuthorEric
• CommentTimeSep 30th 2010

Not when making statements like:

Ok. Then forget I made a statement like that since I was on thin ice already :)

• CommentRowNumber17.
• CommentAuthorUrs
• CommentTimeSep 30th 2010

Then forget I made a statement like that

That would be too bad, because that was a good statement.

• CommentRowNumber18.
• CommentAuthorEric
• CommentTimeSep 30th 2010

That would be too bad, because that was a good statement.

Ok. Then don’t forget I made a statement like that :)

This will require some meditation and I have a headache and it is way past my bedtime.

• CommentRowNumber19.
• CommentAuthorUrs
• CommentTimeDec 13th 2012

added References on generalizations to manifolds with singular boundaries.

• CommentRowNumber20.
• CommentAuthorUrs
• CommentTimeOct 24th 2017

added the statement of Stokes’ theorem for fiber integration: here

• CommentRowNumber21.
• CommentAuthorTobyBartels
• CommentTimeSep 17th 2018

Classical forms

• CommentRowNumber22.
• CommentAuthorUrs
• CommentTimeSep 12th 2020