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1. • CommentRowNumber1.
   • CommentAuthorTobyBartels
   • CommentTimeApr 21st 2012
   • (edited Apr 21st 2012)
   • PermaLink
   Author: TobyBartels
   Format: MarkdownItexAfter a few days' editing, I'm announcing [[absolute differential form]] (using my neologism). This is a notion of differential form that can be integrated on a completely *unoriented* submanifold. Examples from classical differential geometry include the arclength element on a Riemannian manifold and ${|\mathrm{d}z|}$ on the complex plane. Since there are classical examples, people must have thought about these before me, but I have never heard of them. Absolute differential forms are not linear (although they must satisfy a restricted linearity condition), and many typical examples are not smooth (although they are still continuous), so they don't show up in the usual classification theorems. Has anybody heard of them before?

   After a few days’ editing, I’m announcing absolute differential form (using my neologism). This is a notion of differential form that can be integrated on a completely unoriented submanifold. Examples from classical differential geometry include the arclength element on a Riemannian manifold and ${|\mathrm{d}z|}$ on the complex plane.

   Since there are classical examples, people must have thought about these before me, but I have never heard of them. Absolute differential forms are not linear (although they must satisfy a restricted linearity condition), and many typical examples are not smooth (although they are still continuous), so they don’t show up in the usual classification theorems. Has anybody heard of them before?

2. • CommentRowNumber2.
   • CommentAuthorEric
   • CommentTimeApr 22nd 2012
   • PermaLink
   Author: Eric
   Format: MarkdownItexHi Toby, As I watched you compose that page, you had something like "Definition: Well that is the hard part." A thought came to my mind about integration without integration (which I still do not completely understand). Is it true that an absolute 1-form on $M$ is an an absolute 0-form on path space $PM$? I wonder if there is some way to define absolute forms in terms of transgression or something?

   Hi Toby,

   As I watched you compose that page, you had something like “Definition: Well that is the hard part.”

   A thought came to my mind about integration without integration (which I still do not completely understand).

   Is it true that an absolute 1-form on $M$ is an an absolute 0-form on path space $PM$?

   I wonder if there is some way to define absolute forms in terms of transgression or something?

3. • CommentRowNumber3.
   • CommentAuthorTobyBartels
   • CommentTimeApr 22nd 2012
   • (edited Mar 1st 2013)
   • PermaLink
   Author: TobyBartels
   Format: MarkdownItexAn absolute $0$-form is just a $0$-form, which is just a scalar field, that is a function of points. So to say that an absolute $1$-form on $M$ is an absolute $0$-form on $P M$ is to say that an absolute $1$-form on $M$ is a function of paths. This is certainly true *if* the paths are simply curves, functions from (say) $[0,1]$ to $M$, but not if the paths are equivalence classes of curves (say homotopy equivalence classes, or even thin homotopy equivalence classes as in the [[path groupoid]]). This is because an absolute $1$-form is determined by its integrals on all curves, but the integrals of distinct curves can never be guaranteed to be the same. This is a difference from ordinary $1$-forms, since an ordinary $1$-form *can* be thought of as a function on the path groupoid. This is because a given $1$-form has the equal integrals on thin-homotopy-equivalent curves. Pseudo-$1$-forms are even worse, since they can't be thought of as a function of paths at all; a path is an oriented curve, and an absolute $1$-form can ignore the orientation but a pseudo-$1$-form needs a pseudo-orientation, which isn't available. In any case, you can't say that a $1$-form (ordinary or absolute) is an *arbitrary* $0$-form on any sort of path space, but only one satisfying certain conditions. Edit: spelling typo

   An absolute $0$-form is just a $0$-form, which is just a scalar field, that is a function of points. So to say that an absolute $1$-form on $M$ is an absolute $0$-form on $P M$ is to say that an absolute $1$-form on $M$ is a function of paths. This is certainly true if the paths are simply curves, functions from (say) $[0,1]$ to $M$, but not if the paths are equivalence classes of curves (say homotopy equivalence classes, or even thin homotopy equivalence classes as in the path groupoid). This is because an absolute $1$-form is determined by its integrals on all curves, but the integrals of distinct curves can never be guaranteed to be the same.

   This is a difference from ordinary $1$-forms, since an ordinary $1$-form can be thought of as a function on the path groupoid. This is because a given $1$-form has the equal integrals on thin-homotopy-equivalent curves. Pseudo-$1$-forms are even worse, since they can’t be thought of as a function of paths at all; a path is an oriented curve, and an absolute $1$-form can ignore the orientation but a pseudo-$1$-form needs a pseudo-orientation, which isn’t available.

   In any case, you can’t say that a $1$-form (ordinary or absolute) is an arbitrary $0$-form on any sort of path space, but only one satisfying certain conditions.

   Edit: spelling typo

4. • CommentRowNumber4.
   • CommentAuthorTobyBartels
   • CommentTimeApr 22nd 2012
   • PermaLink
   Author: TobyBartels
   Format: MarkdownItexThere's no doubt that $1$-forms are better behaved than either pseudo-$1$-forms or absolute $1$-forms. My main reason for thinking about absolute forms is just that they appear in classical examples, so one ought to understand them.

   There’s no doubt that $1$-forms are better behaved than either pseudo-$1$-forms or absolute $1$-forms. My main reason for thinking about absolute forms is just that they appear in classical examples, so one ought to understand them.

5. • CommentRowNumber5.
   • CommentAuthorEric
   • CommentTimeApr 22nd 2012
   • PermaLink
   Author: Eric
   Format: MarkdownItexThanks Toby! Ok. So an ordinary 1-form is a 0-form on the corresponding path groupoid. That makes sense. My intuition tells me there should be some kind of absolute path space that would allow to make the analogous statement: An absolute 1-form is a 0-form on the corresponding absolute path space. Is there no hope in making sense out of this? I am vaguely imagining some kind of transgression to absolute path space.

   Thanks Toby!

   Ok. So an ordinary 1-form is a 0-form on the corresponding path groupoid. That makes sense.

   My intuition tells me there should be some kind of absolute path space that would allow to make the analogous statement:

   An absolute 1-form is a 0-form on the corresponding absolute path space.

   Is there no hope in making sense out of this? I am vaguely imagining some kind of transgression to absolute path space.

6. • CommentRowNumber6.
   • CommentAuthorTobyBartels
   • CommentTimeApr 22nd 2012
   • PermaLink
   Author: TobyBartels
   Format: MarkdownItexYou can just use the space of paths, rather than the path groupoid (in which the paths are modded out by thin homotopy) as the absolute path space. For that matter, you could use that space for the ordinary $1$-forms; it's just that thin-homotopic paths must have the same value (integral) for a $1$-form but not for an absolute $1$-form.

   You can just use the space of paths, rather than the path groupoid (in which the paths are modded out by thin homotopy) as the absolute path space. For that matter, you could use that space for the ordinary $1$-forms; it’s just that thin-homotopic paths must have the same value (integral) for a $1$-form but not for an absolute $1$-form.

7. • CommentRowNumber7.
   • CommentAuthorEric
   • CommentTimeApr 22nd 2012
   • PermaLink
   Author: Eric
   Format: MarkdownItexOk. So an absolute 1-form can be thought of as a 0-form on the space of paths. Can it be iterated? An absolute 2-form can be thought of as a 0-form on the space of paths on the space of paths? In other words, an absolute 2-form can be thought of as an absolute 1-form on the space of paths? Some kind of iterative definition would be nice if possible given that 0-forms are so non-mysterious. PS: I like the page. Thanks for writing it.

   Ok. So an absolute 1-form can be thought of as a 0-form on the space of paths.

   Can it be iterated?

   An absolute 2-form can be thought of as a 0-form on the space of paths on the space of paths?

   In other words, an absolute 2-form can be thought of as an absolute 1-form on the space of paths?

   Some kind of iterative definition would be nice if possible given that 0-forms are so non-mysterious.

   PS: I like the page. Thanks for writing it.

8. • CommentRowNumber8.
   • CommentAuthorTobyBartels
   • CommentTimeApr 22nd 2012
   • PermaLink
   Author: TobyBartels
   Format: MarkdownItexRemember not *every* $0$-form on path space (in any sense) is a $1$-form (in any sense). So we haven't even got the first definition done in this style. I *think* that it should be possible. PS: Thank you.

   Remember not every $0$-form on path space (in any sense) is a $1$-form (in any sense). So we haven’t even got the first definition done in this style. I think that it should be possible.

   PS: Thank you.

9. • CommentRowNumber9.
   • CommentAuthorEric
   • CommentTimeApr 22nd 2012
   • PermaLink
   Author: Eric
   Format: MarkdownItexTo refresh my memory, are two different parameterization of the same underlying point set making up a curve considered different points in path space? If so, we might want to consider path space modded out by parameterizations. Does that space have a name? The full path space may not be the most natural thing to consider.

   To refresh my memory, are two different parameterization of the same underlying point set making up a curve considered different points in path space?

   If so, we might want to consider path space modded out by parameterizations. Does that space have a name? The full path space may not be the most natural thing to consider.

10. • CommentRowNumber10.
    • CommentAuthorMike Shulman
    • CommentTimeApr 22nd 2012
    • PermaLink
    Author: Mike Shulman
    Format: MarkdownItex> There's no doubt that 1-forms are better behaved than either pseudo-1-forms or absolute 1-forms. Perhaps algebraically. But geometrically, I would be inclined to say that pseudo 1-forms, at least, are just as natural and well-behaved as their duals. Also: this is a very interesting page, thanks!

    There’s no doubt that 1-forms are better behaved than either pseudo-1-forms or absolute 1-forms.

    Perhaps algebraically. But geometrically, I would be inclined to say that pseudo 1-forms, at least, are just as natural and well-behaved as their duals.

    Also: this is a very interesting page, thanks!

11. • CommentRowNumber11.
    • CommentAuthorTobyBartels
    • CommentTimeApr 22nd 2012
    • PermaLink
    Author: TobyBartels
    Format: MarkdownItex@Mike - Yes, I meant algebraically. Geometrically (at least at the level of naïve geometric intuition), all three are quite natural. @Eric - There are several different notions of [[path space]], so it's up to you what to mod out by. But there's more ... as a function of paths, a $1$-form must satisfy $\int_{C ; D} \omega = \int_C \omega + \int_D \omega$ and $\int_{-C} \omega = -\int_C \omega$; that is, it must be a functor from the [[path groupoid]] to the additive monoid of real numbers. For an absolute $1$-form, the former requirement stays, but the latter becomes $\int_{-C} \omega = \int_C \omega$. And I don't think that these are the only requirements, especially for the ordinary $1$-forms.

    @Mike - Yes, I meant algebraically. Geometrically (at least at the level of naïve geometric intuition), all three are quite natural.

    @Eric - There are several different notions of path space, so it’s up to you what to mod out by. But there’s more … as a function of paths, a $1$-form must satisfy $\int_{C ; D} \omega = \int_C \omega + \int_D \omega$ and $\int_{-C} \omega = -\int_C \omega$; that is, it must be a functor from the path groupoid to the additive monoid of real numbers. For an absolute $1$-form, the former requirement stays, but the latter becomes $\int_{-C} \omega = \int_C \omega$. And I don’t think that these are the only requirements, especially for the ordinary $1$-forms.

12. • CommentRowNumber12.
    • CommentAuthorEric
    • CommentTimeApr 22nd 2012
    • (edited Apr 22nd 2012)
    • PermaLink
    Author: Eric
    Format: MarkdownItexWith absolute forms, it seems almost like you want absolute chains and we are not even allowed to discuss orientation so something like $\int_{-C} | \alpha |$ might not even parse and we can only write things like $\int_{|C|} | \alpha |.$ Something like "chai", i.e. "chains without negatives" :) Edit: Instead of path groupoid, maybe we need a path rigoid :)

    With absolute forms, it seems almost like you want absolute chains and we are not even allowed to discuss orientation so something like

    $\int_{-C} | \alpha |$

    might not even parse and we can only write things like

    $\int_{|C|} | \alpha |.$

    Something like “chai”, i.e. “chains without negatives” :)

    Edit: Instead of path groupoid, maybe we need a path rigoid :)

13. • CommentRowNumber13.
    • CommentAuthorTobyBartels
    • CommentTimeApr 23rd 2012
    • (edited Mar 1st 2013)
    • PermaLink
    Author: TobyBartels
    Format: MarkdownItexI think that $-C$ should parse, but we might well write $-C = C$, that is, we may identify a curve with its reversal in the "absolute path space". (Similarly, in the path groupoid for ordinary forms, one identifies thin-homotopic curves.) In other words, ${|{-C}|} = {|C|}$. A "groupoid without negatives" is a category! We would indeed need some sort of "unoriented path category". Edit: formatting.

    I think that $-C$ should parse, but we might well write $-C = C$, that is, we may identify a curve with its reversal in the “absolute path space”. (Similarly, in the path groupoid for ordinary forms, one identifies thin-homotopic curves.) In other words, ${|{-C}|} = {|C|}$.

    A “groupoid without negatives” is a category! We would indeed need some sort of “unoriented path category”.

    Edit: formatting.

14. • CommentRowNumber14.
    • CommentAuthorTim_Porter
    • CommentTimeApr 23rd 2012
    • (edited Apr 23rd 2012)
    • PermaLink
    Author: Tim_Porter
    Format: MarkdownItexYou probably know this, but do look at some of the directed homotopy stuff. [[Martin Raussen]] and Uli Fahrenberg wrote a JHRS paper a few years ago on reparametrisation and Martin followed this up with [this](http://www.math.aau.dk/fileadmin/user_upload/www.math.aau.dk/Forskning/Rapportserien/R-2011-11.pdf). (I have also done something similar.) (Edit: On a silly note for Eric: _chai or cha is the generic word for "tea" in many European and Asian languages. The widespread form chai comes from Persian چای chay. This derives from Mandarin Chinese chá._)

    You probably know this, but do look at some of the directed homotopy stuff. Martin Raussen and Uli Fahrenberg wrote a JHRS paper a few years ago on reparametrisation and Martin followed this up with this. (I have also done something similar.)

    (Edit: On a silly note for Eric:

    chai or cha is the generic word for “tea” in many European and Asian languages. The widespread form chai comes from Persian چای chay. This derives from Mandarin Chinese chá.)

15. • CommentRowNumber15.
    • CommentAuthorTobyBartels
    • CommentTimeApr 26th 2012
    • PermaLink
    Author: TobyBartels
    Format: MarkdownItexI really got around to creating [[absolute differential form]] because I was writing [[line integral]], and that now exists.

    I really got around to creating absolute differential form because I was writing line integral, and that now exists.

16. • CommentRowNumber16.
    • CommentAuthorTobyBartels
    • CommentTimeMar 2nd 2013
    • (edited Mar 2nd 2013)
    • PermaLink
    Author: TobyBartels
    Format: MarkdownItexThese seem to be related to (if not the same as, or meant to be the same as) certain densities in a generalised sense due to Gelfand & Gindikin; see [Math](http://mathoverflow.net/questions/99488/the-ds-which-appears-in-an-integral-with-respect-to-arclength-is-not-a-1-form/99496#99496) [Overflow](http://mathoverflow.net/questions/90455/why-do-i-need-densities-in-order-to-integrate-on-a-non-orientable-manifold/90714#90714).

    These seem to be related to (if not the same as, or meant to be the same as) certain densities in a generalised sense due to Gelfand & Gindikin; see Math Overflow.

17. • CommentRowNumber17.
    • CommentAuthorMirco Richter
    • CommentTimeMar 3rd 2013
    • (edited Mar 3rd 2013)
    • PermaLink
    Author: Mirco Richter
    Format: MarkdownItexAt least for absolute forms of maximal degree this is intrinsically defined in [Wikipedia](https://en.wikipedia.org/wiki/Density_on_a_manifold) [Mathoverflow](http://mathoverflow.net/questions/121076/pseudo-differentialforms) However the question, how to define absolute forms of lower degree still remains. My first thought was to just take the absolute value of the representations of the general linear group that gives the $k$-form bundles. The appropriate associated vector bundles, then *maybe* qualify as bundles of absolute $k$-forms. Someone should calculate an example integration of the 2-dim volume of a particular Moebius-strip. There is a lot of confusion about how its done out there.

    At least for absolute forms of maximal degree this is intrinsically defined in

    Wikipedia

    Mathoverflow

    However the question, how to define absolute forms of lower degree still remains. My first thought was to just take the absolute value of the representations of the general linear group that gives the $k$-form bundles. The appropriate associated vector bundles, then maybe qualify as bundles of absolute $k$-forms.

    Someone should calculate an example integration of the 2-dim volume of a particular Moebius-strip. There is a lot of confusion about how its done out there.

18. • CommentRowNumber18.
    • CommentAuthorMirco Richter
    • CommentTimeMar 3rd 2013
    • (edited Mar 3rd 2013)
    • PermaLink
    Author: Mirco Richter
    Format: MarkdownItexMoreover, in local coordinates I would vote for the "$\cdot$" as a replacement for the $\wedge$, i.e. absolute forms look like $f(x) dx^{i_1}\cdot \ldots \cdot dx^{i_k}$ .This is consistent with the Lebesque-Measure description and reflects the fact that they are symmetric, but different from 'symmetric forms' like $dx^{i_1} \vee \ldots \vee dx^{i_k}$.

    Moreover, in local coordinates I would vote for the “$\cdot$” as a replacement for the $\wedge$, i.e. absolute forms look like

    $f(x) dx^{i_1}\cdot \ldots \cdot dx^{i_k}$

    .This is consistent with the Lebesque-Measure description and reflects the fact that they are symmetric, but different from ’symmetric forms’ like $dx^{i_1} \vee \ldots \vee dx^{i_k}$.

19. • CommentRowNumber19.
    • CommentAuthorDmitri Pavlov
    • CommentTimeMar 3rd 2013
    • PermaLink
    Author: Dmitri Pavlov
    Format: Markdown* Isn't it true that the notion of absolute p-forms makes sense for any real vector space, and absolute p-forms discussed in the article are merely absolute p-forms associated to the tangent bundle of X? * It seems to me that one can simplify Definition 1 by saying that an absolute p-form on a real vector space V is an even positively homogeneous function on the variety of nonzero decomposable p-vectors on V. Am I missing something? This makes sense in particular because a morphism of vector spaces V→W induces a linear map between their p-fold exterior powers, which preserves decomposable p-vectors, hence one immediately obtains pullbacks of absolute p-forms.

    • Isn't it true that the notion of absolute p-forms makes sense for any real vector space, and absolute p-forms discussed in the article are merely absolute p-forms associated to the tangent bundle of X?

    • It seems to me that one can simplify Definition 1 by saying that an absolute p-form on a real vector space V is an even positively homogeneous function on the variety of nonzero decomposable p-vectors on V. Am I missing something? This makes sense in particular because a morphism of vector spaces V→W induces a linear map between their p-fold exterior powers, which preserves decomposable p-vectors, hence one immediately obtains pullbacks of absolute p-forms.

20. • CommentRowNumber20.
    • CommentAuthorTobyBartels
    • CommentTimeMar 4th 2013
    • PermaLink
    Author: TobyBartels
    Format: MarkdownItex@Dmitri: The answer to both questions is Yes. The definition given at the [second MO link in my comment #16 above](http://mathoverflow.net/questions/90455/why-do-i-need-densities-in-order-to-integrate-on-a-non-orientable-manifold/90714#90714) does just that. (My comment #16 was more hesitant, but I now believe that the even densities defined at that MO link are exactly the same as continuous absolute forms.)

    @Dmitri: The answer to both questions is Yes. The definition given at the second MO link in my comment #16 above does just that. (My comment #16 was more hesitant, but I now believe that the even densities defined at that MO link are exactly the same as continuous absolute forms.)

21. • CommentRowNumber21.
    • CommentAuthorTobyBartels
    • CommentTimeMar 4th 2013
    • PermaLink
    Author: TobyBartels
    Format: MarkdownItex@Marco: I would go farther and remove the dot entirely; this agrees with the usual notation for volume integrals etc. However, there is still an ambiguity between the exterior form $\mathrm{d}x$ and the absolute form $\mathrm{d}x$; the latter is the absolute value of the former. So it would be more proper to write ${|\mathrm{d}x^{i_1}|} \cdots {|\mathrm{d}x^{i_k}|}$, which is equivalent to ${|\mathrm{d}x^{i_1} \wedge \cdots \wedge \mathrm{d}x^{i_k}|}$, but I find it hard to imagine that this will catch on. But note that most absolute forms cannot be expressed in local coordinates as linear combinations of such expressions. For example, the arclength element $\sqrt{(\mathrm{d}x)^2 + (\mathrm{d}y)^2}$ is not a linear combination of ${|\mathrm{d}x|}$ and ${|\mathrm{d}y|}$.

    @Marco: I would go farther and remove the dot entirely; this agrees with the usual notation for volume integrals etc. However, there is still an ambiguity between the exterior form $\mathrm{d}x$ and the absolute form $\mathrm{d}x$; the latter is the absolute value of the former. So it would be more proper to write ${|\mathrm{d}x^{i_1}|} \cdots {|\mathrm{d}x^{i_k}|}$, which is equivalent to ${|\mathrm{d}x^{i_1} \wedge \cdots \wedge \mathrm{d}x^{i_k}|}$, but I find it hard to imagine that this will catch on.

    But note that most absolute forms cannot be expressed in local coordinates as linear combinations of such expressions. For example, the arclength element $\sqrt{(\mathrm{d}x)^2 + (\mathrm{d}y)^2}$ is not a linear combination of ${|\mathrm{d}x|}$ and ${|\mathrm{d}y|}$.

22. • CommentRowNumber22.
    • CommentAuthorMirco Richter
    • CommentTimeMar 5th 2013
    • PermaLink
    Author: Mirco Richter
    Format: MarkdownItexSo absolute p-forms do not arise as sections of some kind of fiber bundle?

    So absolute p-forms do not arise as sections of some kind of fiber bundle?

23. • CommentRowNumber23.
    • CommentAuthorDavidRoberts
    • CommentTimeMar 6th 2013
    • (edited Mar 6th 2013)
    • PermaLink
    Author: DavidRoberts
    Format: MarkdownItexThere is such a thing as a sheaf of densities, which I guess is constructed from the orientation bundle and exterior powers of the cotangent bundle...

    There is such a thing as a sheaf of densities, which I guess is constructed from the orientation bundle and exterior powers of the cotangent bundle…

24. • CommentRowNumber24.
    • CommentAuthorTobyBartels
    • CommentTimeMar 6th 2013
    • PermaLink
    Author: TobyBartels
    Format: MarkdownItexSure, they\'re sections of a fibre bundle, even of a vector bundle; the absolute forms at a given point form a vector space. (I added a brief remark on this to the article in a relevant place.) It\'s just that the products of the absolute values of the differentials of the coordinate functions don\'t form a basis of that vector space.

    Sure, they're sections of a fibre bundle, even of a vector bundle; the absolute forms at a given point form a vector space. (I added a brief remark on this to the article in a relevant place.) It's just that the products of the absolute values of the differentials of the coordinate functions don't form a basis of that vector space.

25. • CommentRowNumber25.
    • CommentAuthorMirco Richter
    • CommentTimeMar 6th 2013
    • (edited Mar 6th 2013)
    • PermaLink
    Author: Mirco Richter
    Format: MarkdownItexOk. But under coordinate changes they transform like ordinary anti-symmetric form, just without any sign changes? If this is true, then the bundle should be the one I proposed in #17. Not?

    Ok. But under coordinate changes they transform like ordinary anti-symmetric form, just without any sign changes?

    If this is true, then the bundle should be the one I proposed in #17. Not?

26. • CommentRowNumber26.
    • CommentAuthorDmitri Pavlov
    • CommentTimeMar 6th 2013
    • PermaLink
    Author: Dmitri Pavlov
    Format: Markdown@Mirco Richter: What you claim in #25 cannot be true because the bundle that you are proposing is finite-dimensional, whereas the bundle of absolute p-forms is infinite-dimensional, as you can see from the description in #19.

    @Mirco Richter: What you claim in #25 cannot be true because the bundle that you are proposing is finite-dimensional, whereas the bundle of absolute p-forms is infinite-dimensional, as you can see from the description in #19.

27. • CommentRowNumber27.
    • CommentAuthorMirco Richter
    • CommentTimeMar 6th 2013
    • (edited Mar 6th 2013)
    • PermaLink
    Author: Mirco Richter
    Format: MarkdownItexCan you write that bundle down explicitly? I can't see any natural definition. What irritates me, is that absolute forms of maximum degree are sections in a finite dim bundle. The bundle related to the frame bundle by the absolute value of the determinant representation of $Gl(n)$ on $\mathbb{R}$.

    Can you write that bundle down explicitly? I can’t see any natural definition. What irritates me, is that absolute forms of maximum degree are sections in a finite dim bundle. The bundle related to the frame bundle by the absolute value of the determinant representation of $Gl(n)$ on $\mathbb{R}$.

28. • CommentRowNumber28.
    • CommentAuthorMirco Richter
    • CommentTimeMar 6th 2013
    • (edited Mar 6th 2013)
    • PermaLink
    Author: Mirco Richter
    Format: MarkdownItexAnother definition to consider could be the following: Say $C^0(\wedge^p TM)$ is the $C^0(M)$-module of continuus $p$-multivector fields. Then a continuus absolute $p$-form is a non-negative linear map $\omega: C^0(\wedge^p TM) \to \mathbb{R}^+$. But this more like, what is called a $p$-pseudoform, right?

    Another definition to consider could be the following:

    Say $C^0(\wedge^p TM)$ is the $C^0(M)$-module of continuus $p$-multivector fields. Then a continuus absolute $p$-form is a non-negative linear map $\omega: C^0(\wedge^p TM) \to \mathbb{R}^+$.

    But this more like, what is called a $p$-pseudoform, right?

29. • CommentRowNumber29.
    • CommentAuthorDmitri Pavlov
    • CommentTimeMar 6th 2013
    • PermaLink
    Author: Dmitri Pavlov
    Format: MarkdownThe explicit description is obtained by applying the construction in the second part of #19 fiberwise to the vector bundle TM. The resulting bundle is infinite-dimensional unless p=n. >absolute $p$-form is a non-negative linear map $\omega: C^0(\wedge^p TM) \to \mathbb{R}^+$ A nonzero map to R^+ cannot be linear, it can only be positive homogeneous. Furthermore, at least in the two definitions under consideration the map is not additive with respect to the addition in Λ^p TM, and is only defined on decomposable p-vectors, which do not span a linear subspace of Λ^p TM.

    The explicit description is obtained by applying the construction in the second part of #19 fiberwise to the vector bundle TM. The resulting bundle is infinite-dimensional unless p=n.

    absolute $p$-form is a non-negative linear map $\omega: C^0(\wedge^p TM) \to \mathbb{R}^+$

    A nonzero map to R^+ cannot be linear, it can only be positive homogeneous. Furthermore, at least in the two definitions under consideration the map is not additive with respect to the addition in Λ^p TM, and is only defined on decomposable p-vectors, which do not span a linear subspace of Λ^p TM.

30. • CommentRowNumber30.
    • CommentAuthorTobyBartels
    • CommentTimeMar 6th 2013
    • (edited Mar 10th 2013)
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    Author: TobyBartels
    Format: MarkdownItexAnd also, an absolute $p$-form can take negative values. (But it\'s still not linear as a map to $\mathbb{R}$ or even defined on nonhomogenous $p$-vectors, for $0 \lt p \lt n$.) Edit: Remove spurious plus sign.

    And also, an absolute $p$-form can take negative values. (But it's still not linear as a map to $\mathbb{R}$ or even defined on nonhomogenous $p$-vectors, for $0 \lt p \lt n$.)

    Edit: Remove spurious plus sign.

31. • CommentRowNumber31.
    • CommentAuthorTobyBartels
    • CommentTimeMar 6th 2013
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    Author: TobyBartels
    Format: MarkdownItexFor $p = 0$ and $p = n$, the bundle is a line bundle ($1$-dimensional); for $p \gt n$, it\'s trivial ($0$-dimensional). For $0 \lt p \lt n$, it\'s infinite-dimensional. (I only know how to make sense of it for $p$ a natural number, so there may be other answers for generalisations to other kinds of $p$.)

    For $p = 0$ and $p = n$, the bundle is a line bundle ($1$-dimensional); for $p \gt n$, it's trivial ($0$-dimensional). For $0 \lt p \lt n$, it's infinite-dimensional. (I only know how to make sense of it for $p$ a natural number, so there may be other answers for generalisations to other kinds of $p$.)

32. • CommentRowNumber32.
    • CommentAuthorTobyBartels
    • CommentTimeMar 10th 2013
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    Author: TobyBartels
    Format: MarkdownItexOne way to see that the bundle of absolute forms is infinite-dimensional is to consider $\sqrt {\mathrm{d}x^2 + c \mathrm{d}y^2}$ for various positive $c$. These are all linearly independent. (Their squares, which are symmetric bilinear forms, are linearly dependent, in fact spanning a bundle of dimension $2$. But that's different.)

    One way to see that the bundle of absolute forms is infinite-dimensional is to consider $\sqrt {\mathrm{d}x^2 + c \mathrm{d}y^2}$ for various positive $c$. These are all linearly independent. (Their squares, which are symmetric bilinear forms, are linearly dependent, in fact spanning a bundle of dimension $2$. But that’s different.)