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• CommentRowNumber1.
• CommentAuthorMike Shulman
• CommentTimeMay 14th 2014

In talking and thinking about second derivatives recently, I’ve been led towards a different point of view on higher differentials. We (or Toby, really) invented cogerm differential forms (including cojet ones) in order to make sense of an equation like

$\mathrm{d}^2f(x) = f''(x) \mathrm{d}x^2 + f'(x) \mathrm{d}^2x$

with $\mathrm{d}x$ an object that’s actually being squared in a commutative algebra. But it seems to be difficult to put these objects together with $k$-forms for $k\gt 1$ in a sensible way. Moreover, they don’t seem to work so well for defining differentiability: the definition of the cogerm differential in terms of curves is an extension of directional derivatives, but it’s not so clear how to ask that a function (or form) be “differentiable” in a sense stronger than having directional derivatives.

The other point of view that I’m proposing is to use the iterated tangent bundles instead of the cojet bundles: define a $k$-form to be a function $T^k X \to \mathbb{R}$. A point of $T^k X$ includes the data of $k$ tangent vectors, so an exterior $k$-form in the usual sense can be regarded as a $k$-form in this sense. And for each $k$, such functions are still an algebra and can be composed with functions on $\mathbb{R}$, so that we still have things like $\sqrt{\mathrm{d}x^2 + \mathrm{d}y^2}$ and ${|\mathrm{d}x|}$ as 1-forms.

However, the difference is in the differentials. Now the most natural differential is one that takes a $k$-form to a $(k+1)$-form: since $T^k X$ is a manifold (at least if $X$ is smooth), we can ask that a function $\omega:T^k X \to \mathbb{R}$ be differentiable in the sense of having a linear approximation, and if so its differential is a map $T^{k+1} X \to \mathbb{R}$ that is fiberwise linear.

I think it would make the most sense to denote this differential by $\mathrm{d}\otimes \omega$, because in general it is neither symmetric nor antisymmetric. For instance, if $\omega = x \, \mathrm{d}y$ is a 1-form, then $\mathrm{d}\otimes \omega = \mathrm{d}x \otimes \mathrm{d}y + x\, \mathrm{d}^2 y$, where $\mathrm{d}x \otimes \mathrm{d}y$ is the 2-form that multiplies the $x$-component of the first underlying vector of its argument by the $y$-component of the second underlying vector, and $\mathrm{d}^2 x$ is the 2-form that takes the $x$-component of the “second-order” part of its argument (which looks in coordinates like another tangent vector, but doesn’t transform like one). However, if $f$ is a twice differentiable function, then $\mathrm{d}\otimes \mathrm{d}f$ is symmetric, e.g. we have

$\mathrm{d}\otimes \mathrm{d}f(x) = f''(x) \mathrm{d}x \otimes \mathrm{d}x + f'(x) \mathrm{d}^2x.$

So this amounts to regarding the second differential as a bilinear form (plus the $\mathrm{d}^2x$ bit) rather than a quadratic form, which appears to be necessary in order to characterize second-differentiability.

In general, we can symmetrize or antisymmetrize $\mathrm{d}\otimes \omega$, by adding or subtracting its images under the action of the symmetric group $S_{k+1}$ on $T^{k+1} X$. The first gives a result that we might denote $\mathrm{d}\omega$; while the second restricts to the exterior derivative on exterior forms, so we may denote it $\mathrm{d}\wedge \omega$. So in this way, the symmetric and the antisymmetric differential fit into the same picture.

I think this is basically the same approach taken in the reference cited by the answer here.

• CommentRowNumber2.
• CommentAuthorTobyBartels
• CommentTimeMay 21st 2014

Wow, this might be just what I've been looking for all along! I was early on very suspicious about having just a symmetric notion of higher differential, but it was working, so I decided to accept it. But I did once hope for a non-symmetric version whose antisymmetric part would give exterior forms.

• CommentRowNumber3.
• CommentAuthorMike Shulman
• CommentTimeMay 21st 2014

• What are these forms called? I want to call them “co-X differential forms” where “X” is the name of a point of an iterated tangent bundle. But what is that called? The only thing I can think of is than in SDG they are “microcubes”, i.e. maps out of $D^k$ where $D$ is the space of nilsquare infinitesimals; but do they have a name in classical differential geometry?

• How to notate the nonsymmetric differential of a 1-form (for example)? I think I can say “given a 1-form like $x^2\,\mathrm{d}x$, we take its differential by treating $x$ and $\mathrm{d}x$ as separate variables, just like $\mathrm{d}(x^2\,y) = 2x y\,\mathrm{d}x + x^2 \,\mathrm{d}y$, only now $y$ is $\mathrm{d}x$.” But in this setup, the differential of $x$ that appears when we take the second differential is different from the previous variable $\mathrm{d}x$. We could call them $\mathrm{d}_1 x$ and $\mathrm{d}_2 x$, or $\mathrm{d}x$ and $\hat{\mathrm{d}}x$, but nothing I’ve thought of looks very pleasant to me.

Perhaps the most consistent thing would be to say that $\mathrm{d}_k$ is the differential from $k$-forms to $(k+1)$-forms, but that we omit the subscript $k$ when it is zero. So we would have $\mathrm{d}_1(x^2\,\mathrm{d}x) = 2x\,\mathrm{d}_1x \, \mathrm{d}x + x^2 \,\mathrm{d}_1\mathrm{d}x$, and we could abbreviate $\mathrm{d}_1\mathrm{d}x$ by $\mathrm{d}^2x$ and observe that $\mathrm{d}_1x \; \mathrm{d}x = \mathrm{d}x \; \mathrm{d}_1x = \mathrm{d}x\otimes \mathrm{d}x$ (where $\otimes$ on a pair of 1-forms means the 2-form that applies them to the two underlying vectors and multiplies the result). But this notation is still heavier than I’d prefer.

• CommentRowNumber4.
• CommentAuthorTobyBartels
• CommentTimeMay 21st 2014

Yeah, it's all still a bit crazy. Sorry, I don't have any answers. Although it's good to see that $d_1 x \ne d x$; that was tripping me up when I tried to do this before.

• CommentRowNumber5.
• CommentAuthorMike Shulman
• CommentTimeMay 21st 2014

I’m thinking about ways to finesse (2). Suppose we restrict attention to “differential polynomials” of the sort that one gets by repeated differentiation of a function, and then just say that multiplication of differentials doesn’t commute. So for instance we have

$\mathrm{d}(x^2\, \mathrm{d}y) = 2x\,\mathrm{d}x\,\mathrm{d}y + x^2 \mathrm{d}^2y$

in which $\mathrm{d}x \,\mathrm{d}y \neq \mathrm{d}y \,\mathrm{d}x$ (being secretly a $\otimes$). This already breaks for general 3-forms, however: in

$\mathrm{d}(2x \, \mathrm{d}x\, \mathrm{d}y +x^2 \,\mathrm{d}^2y) = 2 \,\mathrm{d}x^2\,\mathrm{d}y + 2x\,\mathrm{d}^2x\,\mathrm{d}y + 2x\,\mathrm{d}x\,\mathrm{d}^2y + 2x\,\mathrm{d}x\,\mathrm{d}^2 y + x^2\,\mathrm{d}^3 y$

the two “$\mathrm{d}x\,\mathrm{d}^2y$” terms are actually different. If we coordinatize $T^3X$ by $(v^1,v^2,v^3,v^{12},v^{13},v^{23},v^{123})$, then one of them gives $v^2_x v^{13}_y$ while the other gives $v^1_x v^{23}_y$.

However, if we symmetrize, then these two terms become equal, and the same ought to be true more generally. So for taking symmetric differentials, maybe we don’t need to worry about heavy notation. And possibly the same is true for antisymmetric ones, once we have a Leibniz rule for the wedge product. I’m not entirely positive, but now I have to run off and give a final…

• CommentRowNumber6.
• CommentAuthorMike Shulman
• CommentTimeMay 21st 2014

For (1), so far my best invention is “(co)flare” — like a jet, but less unidirectional.

• CommentRowNumber7.
• CommentAuthorMike Shulman
• CommentTimeMay 21st 2014

I think it does basically work.

The symmetric and antisymmetric $k$-forms are vector subspaces of all $k$-forms, and symmetrization and antisymmetrization are linear retractions onto them. There is a tensor product from $k$-forms $\otimes$ $\ell$-forms to $(k+\ell)$-forms, obtained by restricting along two projections $T^{k+\ell}X \to T^k X$ and $T^{k+\ell}X \to T^{\ell} X$ and then multiplying. Symmetrizing and antisymmetrizing this makes the symmetric and antisymmetric forms into algebras as well.

In a coordinate system $(x^a)$, the coordinates of $T^k X$ are $(\mathrm{d}_I x^a)$, where $I\subseteq \{1,2,\dots,k\}$ (the case $I=\emptyset$ giving the coordinate functions $x^a$). We may write $\mathrm{d}^k x^a$ for $\mathrm{d}_{\{1,\dots,k\}} x^a$. Say that a (degree-$k$ homogeneous) differential polynomial is a $k$-form that is a sum of terms of the form

$f(x)\; \mathrm{d}_{I_1} x^{a_1} \cdots \mathrm{d}_{I_m} x^{a_m}$

where $I_1,\dots,I_m$ is a partition of $\{1,2,\dots,k\}$. The $k^{\mathrm{th}}$ differential of a function is a symmetric differential polynomial.

Each $I\subseteq \{1,2,\dots,k\}$ determines ${|I|}!$ maps $T^{k} X \to T^{|I|}X$, allowing us to unambiguously regard a symmetric degree-${|I|}$ differential polynomial as a degree-$k$ differential polynomial. In particular, for any change of coordinates we can substitute the ${|I|}^{\mathrm{th}}$ differential of $x^1(u)$ for each $\mathrm{d}_{I} x^a$ in a differential polynomial in $x$, obtaining a differential polynomial in $u$. Since this is how $k$-forms transform, the notion of “differential polynomial” is well-defined independent of coordinates.

Now the differential of a differential polynomial is again a differential polynomial, as is the symmetrization or antisymmetrization. The symmetrization of $\mathrm{d}_{I_1} x^{a_1} \cdots \mathrm{d}_{I_m} x^{a_m}$ is

$\frac{1}{k!} \sum_{\sigma \in S_k} \mathrm{d}_{\sigma(I_1)} x^{a_{\sigma(1)}} \cdots \mathrm{d}_{\sigma(I_m)} x^{a_{\sigma(m)}}$

and terms like this are a basis for the symmetric degree-$k$ differential polynomials (with functions as coefficients). However, this symmetrized term is also the product

$\mathrm{d}^{|I_1|}x^{a_1}\cdots \mathrm{d}^{|I_m|}x^{a_m}$

in the algebra of symmetric forms, and if we use this notation then the symmetric differential polynomials look much cleaner.

The antisymmetric case is even easier: since the antisymmetrization of $\mathrm{d}_I x^a$ is $0$ if ${|I|}\gt 1$, the antisymmetric differential polynonmials are just the ordinary exterior $k$-forms, and the antisymmetric differential is the usual exterior differential.

I don’t think we’ve run into any situations yet where we want to take differentials of forms that aren’t differential polynomials. We want to integrate forms like $|\mathrm{d}x|$ and $\sqrt{\mathrm{d}x^2+\mathrm{d}y^2}$, but we haven’t found a good meaning/use for their differentials yet.

• CommentRowNumber8.
• CommentAuthorTobyBartels
• CommentTimeMay 22nd 2014
• (edited May 22nd 2014)

So for taking symmetric differentials, maybe we don't need to worry about heavy notation. And possibly the same is true for antisymmetric ones, once we have a Leibniz rule for the wedge product.

Yes, that's the conclusion that I came to. So they live in different systems. But you are saying that maybe you can make them work together after all.

• CommentRowNumber9.
• CommentAuthorTobyBartels
• CommentTimeMay 22nd 2014
• (edited May 22nd 2014)

It's interesting that exterior differentials are already known to be the antisymmetric part of a more general algebra. But that algebra (consisting of what physicists call covariant tensors) acts on $(T X)^k$, not on $T^k X$. And there is no natural notion of differential of a general form in this algebra; but if you pick a notion (using an affine connection), then the antisymmetrization of the differential of an exterior form always matches the exterior differential. I was always afraid of recreating that, which would have to be wrong.

But you are saying that you have a natural notion of differential of the general form; it's a different kind of form but the antisymmetric ones are still precisely the exterior forms.

• CommentRowNumber10.
• CommentAuthorMike Shulman
• CommentTimeMay 22nd 2014

Yes, I agree with all of that. Your observation about covariant tensors and connections in #9 is interesting, and makes me wonder how the general “coflare differential” might be related to affine connections. Can a connection be equivalently formulated as some way of “restricting the coflare differential to act on tensors”?

• CommentRowNumber11.
• CommentAuthorMike Shulman
• CommentTimeMay 22nd 2014

Oh, of course: a connection is just a fiberwise-bilinear section of the projection $T^2 X\to (T X)^2$, so we can obtain the covariant derivative of a 1-form by taking the coflare differential and precomposing with the connection. Presumably something analogous works for other covariant derivatives.

• CommentRowNumber12.
• CommentAuthorTobyBartels
• CommentTimeMay 22nd 2014

Brilliant!

• CommentRowNumber13.
• CommentAuthorTobyBartels
• CommentTimeMay 5th 2015
• (edited May 20th 2015)

[Edit for people reading through the discussion history: Note the dates; the conversation has been resumed after a pause of nearly a year, and in fact I'd kind of forgotten about everything in this thread during that break!]

Wow, apparently I never clicked through to the M.O answer that you mention in the top comment, where they quote somebody quoting anonymous ‘classical calculus books’ as writing

$d^2f = f'_x \,d^2x + f'_y \,d^2y + f''_{x^2} \,d{x}^2 + 2f''_{x y} \,d{x} \,d{y} + f''_{y^2} \,d{y}^2 .$

This is our $\mathrm{d}^2f$!

• CommentRowNumber14.
• CommentAuthorMike Shulman
• CommentTimeMay 5th 2015

Yep!

• CommentRowNumber15.
• CommentAuthorMike Shulman
• CommentTimeMay 6th 2015

So how do we integrate these “coflare” forms? For coflare forms the order is bounded above by the rank (terminology from here), so a rank-1 coflare form (a function on $T X$) is roughly the same as a 1-cojet form, and can be integrated in either the “genuine” or “affine” way as proposed on the page cogerm differential form. For a rank-2 coflare form, a function on $T T X$, my first thoughts in the “affine” direction are to cover the parametrizing surface by a grid of points $t_{i,j}$, regard our form $\omega$ as a function on a point and a triple of vectors (though the third one doesn’t transform as a vector), and sum up the values of

$\omega(c(t^*_{i,j}), \Delta^1 c_{i,j}, \Delta^2 c_{i,j}, \Delta^{12} c_{i,j})$

where

$\Delta^1 c_{i,j} = c(t_{i,j}) - c(t_{i,j-1})$ $\Delta^2 c_{i,j} = c(t_{i,j}) - c(t_{i-1,j})$ $\Delta^{12} c_{i,j} = c(t_{i,j}) - (c(t_{i-1,j-1}) + \Delta^1 c_{i,j} + \Delta^2 c_{i,j}).$

I suspect that there won’t be many forms other than exterior ones that will have convergent integrals. Are there any that we should expect?

• CommentRowNumber16.
• CommentAuthorTobyBartels
• CommentTimeMay 6th 2015

Well, I'd really want the absolute forms to converge, and to the correct value. Anything that doesn't recreate integration of exterior and absolute forms isn't doing its job.

• CommentRowNumber17.
• CommentAuthorMike Shulman
• CommentTimeMay 6th 2015

Right, I agree, thanks. I think it’s reasonable to hope the absolute forms to work too, although it’ll probably be some work to check. Maybe there is a Lipschitz condition like in the rank-1 case.

• CommentRowNumber18.
• CommentAuthorMike Shulman
• CommentTimeMay 8th 2015

Here’s a proposal for how to integrate coflare 2-forms, which I conjecture should suffice to integrate exterior and absolute forms, and also be invariant under orientation-preserving twice-differentiable reparametrization for Lipschitz forms. Let a “partition” be an $N\times M$ grid of points $t_{i,j}$ such that the quadrilaterals with vertices $t_{i,j}, t_{i+1,j}, t_{i+1,j+1}, t_{i,j+1}$ partition the region disjointly and have the correct orientation. Tag it with points $t^*_{i,j}$ in each such quadrilateral. If $\delta$ is a gauge ($\mathbb{R}$-valued function) on the region, say the tagged partition is $\delta$-fine if for all $i,j$:

1. $t_{i+1,j}- t_{i,j}$ and $t_{i,j+1}-t_{i,j}$ are both less that $\delta(t^*_{i,j})$, and also
2. $t_{i+1,j+1} - t_{i+1,j} - t_{i,j+1} + t_{i,j}$ is less than $\delta(t^*_{i,j})^2$.

The Riemann sum of a 2-form $\omega$ over such a tagged partition is defined as in #15, and the integral is the limit of these over gauges as usual.

Note that because grids can be rotated arbitrarily, there’s no way that a form like $\mathrm{d}x \otimes \mathrm{d}y$ is going to be integrable with this definition. We essentially need to perform some antisymmetrization to make the pieces rotationally invariant, but we should be free to insert norms or absolute values afterwards to get absolute forms. I don’t know whether this integral will satisfy any sort of Fubini theorem, though.

• CommentRowNumber19.
• CommentAuthorTobyBartels
• CommentTimeMay 12th 2015

I’m not finished looking at this integral, but I want to record here a fact about second-order differentials for when we write an article on them:

+-- {: .num_theorem #2ndDTest}
###### The second-differential test


Given a twice-differentiable space $X$, a twice-differentiable quantity $u$ defined on $X$, and a point $P$ in $X$. If $u$ has a local minimum at $P$, then $\mathrm{d}u|_P$ is zero and $\mathrm{d}^2u|_P$ is positive semidefinite. Conversely (mostly), if $\mathrm{d}u|_P$ is zero and $\mathrm{d}^2u|_P$ is positive definite, then $u$ has a local minimum at $P$.

=--


Here we may view $\mathrm{d}^2u$ as either a cojet $1$-form (the question is whether $\langle{\mathrm{d}^2u|C}\rangle$ is always (semi)-positive for any curve $C$ through $P$) or a coflare $2$-form (the question is whether $\mathrm{d}^2u|_{R = P, \mathrm{d}R = \mathbf{v}, \mathrm{d}_1R = \mathbf{v}, \mathrm{d}^2R = \mathbf{w}}$ is always (semi)-positive for any vector $\mathbf{v}$ (note the repetition) and second-order vector $\mathbf{w}$).

Of course, this is well-known in the case of the second derivative (which does not act on $\mathbf{w}$, but of course the action on $\mathbf{w}$ is zero when $\mathrm{d}u = 0$).

• CommentRowNumber20.
• CommentAuthorTobyBartels
• CommentTimeMay 12th 2015

Actually, this is still more complicated than it needs to be! The theorem is simply this:

+-- {: .num_theorem #2ndDTest}
###### The second-differential test


Given a twice-differentiable space $X$, a twice-differentiable quantity $u$ defined on $X$, and a point $P$ in $X$. If $u$ has a local minimum at $P$, then $\mathrm{d}^2u|_P$ is positive semidefinite. Conversely (mostly), if $\mathrm{d}^2u|_P$ is positive definite, then $u$ has a local minimum at $P$.

=--


In terms of the cojet version, $\langle{\mathrm{d}^2u|C}\rangle = u''(P) \cdot C'(0) \cdot C'(0) + u'(P) \cdot C''(0)$ is (semi)-positive1 for all twice-differentiable $C$ iff $u''(P)$ is positive (semi)-definite and $u'(P)$ is zero. (Just look at examples where $C'$ or $C''$ is zero.) In terms of the coflare version, $\mathrm{d}^2u|_{R = P, \mathrm{d}R = \mathbf{v}, \mathrm{d}_1R = \mathbf{v}, \mathrm{d}^2R = \mathbf{w}} = u''(P) \cdot \mathbf{v} \cdot \mathbf{v} + u'(P) \cdot \mathbf{w}$ is (semi)-positive for all $\mathbf{v}$ and all $\mathbf{w}$ iff the same condition holds. (Inasmuch as the cojet version of $\mathrm{d}^2u$ is the symmetrization of the coflare version, then properties of values of the cojet form correspond precisely to properties of values of the coflare form when $\mathrm{d}R$ and $\mathrm{d}_1R$ are set to the same value.)

So one never needs to mention $\mathrm{d}u$ as such at all!

1. where ‘semipositive’ means $\geq 0$, which hopefully was obvious in context

• CommentRowNumber21.
• CommentAuthorMike Shulman
• CommentTimeMay 12th 2015

Nice! The improved version of the theorem gives another argument for why the $\mathrm{d}^2 x$ terms should be included in $\mathrm{d}^2 u$.

I like the word “semipositive”; maybe I will start using it in other contexts instead of the unlovely “nonnegative”.

• CommentRowNumber22.
• CommentAuthorTobyBartels
• CommentTimeMay 13th 2015

It also works better when the context consists of more than just the real numbers. For example, $2 + 3\mathrm{i}$ is nonnegative, but it's not semipositive. (Another term that I've used is ‘weakly positive’, which goes with the French-derived ‘strictly positive’ rather than the operator-derived ‘semidefinite’.)

• CommentRowNumber23.
• CommentAuthorMike Shulman
• CommentTimeMay 13th 2015

Maybe it’s because I’m more familiar with operators than with French, but “semipositive” is more intuitive for me than “weakly positive”. The word I’ve most often heard used for $\le$ (in opposition to “strictly” for $\lt$) is the equally unlovely “non-strictly”.

• CommentRowNumber24.
• CommentAuthorTobyBartels
• CommentTimeMay 13th 2015

Wow, ‘non-strictly’!? As far as I know, ‘weakly’ here doesn't come from French; in French, they say simply ‘positif’ for the weak concept (which is why they must say ‘strictement positif’ for the strict one). But it seems to me that ‘weakly’ is the established antonym of ‘strictly’.

• CommentRowNumber25.
• CommentAuthorMike Shulman
• CommentTimeMay 13th 2015

Unless it’s “strongly” or “pseudo”…

• CommentRowNumber26.
• CommentAuthorDavid_Corfield
• CommentTimeMay 14th 2015

In that funny way that sometimes happens on the $n$ forum/Lab/Café where separate conversations turn out to be related, is there anything you guys are finding out in these threads which makes contact with differential cohesion?

• CommentRowNumber27.
• CommentAuthorUrs
• CommentTimeMay 14th 2015
• (edited May 14th 2015)

To the extent that this is about higher differentials, it might be (I haven’t followed closely) that the concepts here overlap with Kochan-Severa’s “differential gorms and differential worms” (arXiv:math/0307303). To the extent that this is so, then this is seamlessly axiomatized in solid cohesion.

• CommentRowNumber28.
• CommentAuthorDavid_Corfield
• CommentTimeMay 14th 2015

Well that paper

From a more conceptual point of view, differential forms are functions on the superspace of maps from the odd line to M

seems to relate to our discussion here, where I was wondering about how the shift to the super gave rise to differential forms.

• CommentRowNumber29.
• CommentAuthorUrs
• CommentTimeMay 14th 2015
• (edited May 14th 2015)

I may have misunderstood your aim there. It seemed to me that you were wondering there about the role of “$(0\vert 1)$-Euclidean field theory” in this business, to which I replied that I find it a red herring in the present context.

That the $\mathbb{Z}/2\mathbb{Z}$-graded algebra of differential forms on a (super-)manifold $X$ is $C^\infty([\mathbb{R}^{0\vert 1},X])$ is a standard fact of supergeometry, and the observation that the natural $Aut(\mathbb{R}^{0\vert 1})$-action on this gives the refinement to $\mathbb{Z}$-grading as well as the differential originates in Kontsevich 97, and was long amplified by Severa, see also at odd line – the automorphism supergroup.

In the article on gorms and worms they generalize this by replacing $\mathbb{R}^{0|1}$ by $\mathbb{R}^{0|q}$.

• CommentRowNumber30.
• CommentAuthorDavid_Corfield
• CommentTimeMay 14th 2015

Re #29, I returned the discussion to the original thread.

• CommentRowNumber31.
• CommentAuthorMike Shulman
• CommentTimeMay 14th 2015

I think we just had a drive-by categorification. (-:

The “higher” in this thread refers essentially to roughly the same number that indexes the usual 1-forms, 2-forms, $n$-forms; the generalization is rather to eliminate the requirement of antisymmetry. It sounds from the abstract like Kochan-Severa instead keep the “antisymmetry” (encoded using super-ness) and generalize in a different direction.

On the other hand, re #26 as stated, coflare differential forms and their differentials make perfect sense in SDG, whereas integration is always tricky in such contexts.

• CommentRowNumber32.
• CommentAuthorUrs
• CommentTimeMay 14th 2015
• (edited May 14th 2015)

It sounds from the abstract like Kochan-Severa instead keep the “antisymmetry”

They also get commuting differentials: the algebra $C^\infty(\mathbb{R}^{0\vert 2})$ is spanned over $\mathbb{R}$ by the two odd-graded coordinate functions $\theta_1$ and $\theta_2$ as well as their even graded product $\theta_1 \theta_2$. Under the identification of $C^\infty([\mathbb{R}^{0\vert 2}, X])$ with second order differential forms, the first two generators give anticommuting forms, but the second gives commuting second order forms.

This seems close to what you start with in #1, but I haven’t really thought through it.

• CommentRowNumber33.
• CommentAuthorTobyBartels
• CommentTimeMay 18th 2015
• (edited Dec 13th 2019)

And another issue that I've been thinking of:

I've occasionally seen the claim (and now I wish that I knew where, and it may have only ever been oral) that the higher differentials are the terms in the Taylor series; that is (for $f$ analytic at $c$ with a radius of convergence greater than $h$),

$f(c + h) = \sum_{i=0}^\infty \frac{1}{i!} f^{(i)}(c) h^i = \sum_{i=0}^\infty \frac{1}{i!} \,\mathrm{d}^{i}f(x)|_{x\coloneqq{c}\atop\mathrm{d}x\coloneqq{h}} = \mathrm{e}^{\mathrm{d}}f(x)|_{x\coloneqq{c}\atop\mathrm{d}x\coloneqq{h}}.$

This works if $\mathrm{d}^{i}f(x) = f^{(i)}(x) \,\mathrm{d}x^i$, but our claim is that $\mathrm{d}^{i}f(x)$ is much more complicated. However, the extra terms all involve higher differentials of $x$, so if we evaluate $\mathrm{e}^{\mathrm{d}}f(x)$ with not only $x \coloneqq c$ and $\mathrm{d}x \coloneqq h$ but also $\mathrm{d}^{i}x \coloneqq 0$ for $i \geq 2$, then it still works.

• CommentRowNumber34.
• CommentAuthorMike Shulman
• CommentTimeMay 18th 2015

@Toby: Yes, I agree. In general, the sum of the higher differentials seems like some generalization of a power series in which you have not only a “first-order-small variable” but also a separate $n$th-order-small variable for all $n$.

• CommentRowNumber35.
• CommentAuthorUrs
• CommentTimeMay 19th 2015
• (edited May 19th 2015)

Please help me: to which extent is the following definition (first full paragraph on p. 4 of arXiv:math/0307303) what you are after, or to which extent is it not what you are after:

for $X = \mathbb{R}^n$ a Cartesian space, say that its $\mathbb{Z}/2\mathbb{Z}$-bigraded commutative differential algebra of second order differential forms is the smooth algebra freely generated over the smooth functions on $X$ in even degree from two differentials $d_1$ and $d_2$. In particular for $f \colon \mathbb{R}^n \to \mathbb{R}$ then

$d_1 d_2 f = \frac{\partial ^2 f}{\partial x^i \partial x^j} d_1 x^1 d_2 x^2 + \frac{\partial f}{\partial x^i} d_1 d_2 x^i \,,$

where $d_1 x^i$ is in bidegree $(1,0)$, $d_2 x^i$ is in bidegree $(0,1)$ and $d_1 d_2 x^i$ is in bidegree $(1,1)$.

1. Re #35: here bidifferential means that the two differentials anti-commute $d_1 d_2=-d_2 d_1$ right?

Another approach to higher differentials, which I don’t know if it had been mentioned already, is in Laurent Schwartz “Geometrie differentielle du 2ème ordre, semi-martingales et equations differentielles stochastiques sur une variete differentielle” p.3. There he defines $d^k f$, for a scalar valued function $f$ on some manifold $M$, to be the k-th jet of f “modulo 0-th order information”. More precisely: if $f$ vanishes at $p\in M$, then $d^k f(p)$ is the equivalence class of $f$ modulo all functions that vanish to $k-th$ order at $p$. If $f$ does not vanish at $p$, then $d^k f(p)$ is the equivalence class of $f-f(p)$ modulo functions vanishing to order $k$ at $p$. In coordinates this should amount to the same as considering the k-th Taylor polynomials of $f$, but forgetting the constant term.

In this approach it is not immediately clear (to me), what the product of two higher differentials is and in which sense say $d^2=dd$ etc. But I thought I’ll add it to the mix.

• CommentRowNumber37.
• CommentAuthorMike Shulman
• CommentTimeMay 19th 2015

Urs, I need you to back up a moment as I’m not used to thinking of differential forms in this way. Even for ordinary differential forms, how do you get the partial derivatives of $f$ appearing from only algebraic assumptions? I can see it if $f$ is a polynomial, but in general, $f$ might not even be analytic!

• CommentRowNumber38.
• CommentAuthorTobyBartels
• CommentTimeMay 19th 2015

I found Schwartz’s paper here: http://www.numdam.org/item?id=SPS_1982__S16__1_0.

It seems to me that Schwartz's differentials contain the same information as mine (if I may lay claim to them even after seeing citations to them in 19th-century books). In particular, the space of order-$2$ forms at a given point in a manifold of dimension $n$ is $n + n(n+1)/2$, that is $n$ for the order-$2$ differentials of the local coordinates and $n(n+1)/2$ for the products of two order-$1$ differentials. (These are the symmetrized products, producing my cojet forms, not Mike's coflare forms.)

• CommentRowNumber39.
• CommentAuthorMike Shulman
• CommentTimeMay 19th 2015

Perhaps Schwartz is one of the “classical calculus books” mentioned in the answer linked in #1 and referred to in #13.

• CommentRowNumber40.
• CommentAuthorTobyBartels
• CommentTimeMay 19th 2015

Schwartz has the good point that the space of order-$1$ forms (at a given point) is a quotient of the space of order-$2$ forms; the kernel consists of those forms which appear in the order-$2$ terms of a Taylor series.

• CommentRowNumber41.
• CommentAuthorMike Shulman
• CommentTimeMay 19th 2015

Re #40: is that something special about 1 and 2, or is it true more generally?

• CommentRowNumber42.
• CommentAuthorUrs
• CommentTimeMay 19th 2015

Mike, the reasoning with the graded infinitesimals here is the same as for the non-graded ones familiar from SDG, e.g. here.

• CommentRowNumber43.
• CommentAuthorTobyBartels
• CommentTimeMay 19th 2015

In local coordinates, the quotient map is

$\sum_{i\leq{j}} A_{i,j} \mathrm{d}x_i \,\mathrm{d}x_j + \sum_i B_i \,\mathrm{d}^2x_i \mapsto \sum_i B_i \,\mathrm{d}x_i .$

In $1$ dimension, an analogous quotient map from order-$3$ forms to order-$2$ forms would be

$A \,\mathrm{d}x^3 + B \,\mathrm{d}^2x \,\mathrm{d}x + C \,\mathrm{d}^3 x \mapsto B \,\mathrm{d}x^2 + C \,\mathrm{d}^2x ,$

but that is not coordinate-invariant (most simply, if $A, B = 0$, $C = 1$, and $x = t^2$). And I don't even know how I would write down a map from order $4$ to order $3$ in $2$ dimensions (in particular, for $\mathrm{d}^2x \,\mathrm{d}^2y$).

However, it seems to me that we can make a quotient map from order-$p$ forms to order-$1$ forms for any $p$, so it's just special about $1$. But in this case, I don't see anything especially special about the kernel.

• CommentRowNumber44.
• CommentAuthorTobyBartels
• CommentTimeMay 19th 2015

H'm, but this map from order $3$ to order $2$ is invariant:

$A \,\mathrm{d}x^3 + B \,\mathrm{d}^2x \,\mathrm{d}x + C \,\mathrm{d}^3 x \mapsto B \,\mathrm{d}x^2 + 3 C \,\mathrm{d}^2x$

(note the factor of $3$). I'm not sure what to make of that!

• CommentRowNumber45.
• CommentAuthorMike Shulman
• CommentTimeMay 20th 2015

Thanks Urs, that was the reminder I needed.

The formula for the second differential does indeed look just like the one Toby and I are working with. But is Michael also right that the differentials anticommute? $d_1d_2 = -d_2d_1$ and/or $d_1x \, d_2x = - d_2 x\, d_1x$? And does their setup include nonlinear differential forms such as $\sqrt{d x^2 + d y^2}$?

• CommentRowNumber46.
• CommentAuthorMike Shulman
• CommentTimeMay 20th 2015

From the coflare perspective, the general linear order-2 form is

$A \mathrm{d}_{2\cdot 1\cdot 0}x + B \mathrm{d}_{21\cdot 0}x + C\mathrm{d}_{2\cdot10}x + D \mathrm{d}_{1\cdot 20}x + E\mathrm{d}_{210} x$

where $\mathrm{d}_{2\cdot 1 0}x$ is shorthand for $\mathrm{d}_2x \, \mathrm{d}_{1}\mathrm{d}_0 x$, and so on. From this perspective the factor of 3 arises because your map to order-2 forms would have to give something like

$(B+C+D) \mathrm{d}_{1\cdot 0} x + E \mathrm{d}_{10} x$

whereas under a coordinate change $B$, $C$, and $D$ all get their own correction term involving $E$.

Just making stuff up, I notice that sending the above order-2 form to

$2(B+C+D) \mathrm{d}_{1\cdot 0} x + 6E \mathrm{d}_{10} x$

will also be coordinate-invariant, and there seems to be some sense in which there are two ways to “forget” from $21\cdot 0$ to $1\cdot 0$ (“send $0$ to $0$, and send $1$ to either $1$ or $2$”) and six ways to “forget” from $210$ to $10$ (the six injective maps from a 2-element set to a 3-element set).

• CommentRowNumber47.
• CommentAuthorTobyBartels
• CommentTimeMay 20th 2015

Thanks, Mike, I tried to figure out something like that, but I wasn't coming up with the right combinatorics. I like your version. It's not immediately obvious why it should work that way, but it does seem to work!

• CommentRowNumber48.
• CommentAuthorMike Shulman
• CommentTimeMay 20th 2015

Here’s an even better explanation, again from the coflare POV. At the 2-to-1 stage, there is a map $T X \to T^2 X$ that sends a tangent vector $(x;v)$ to $(x;0,0;v)$. Second-order tangent vectors don’t transform like vectors in general, but they do if the associated first-order tangent vectors are zero, so this makes sense. The quotient map from 2-forms to 1-forms is just precomposition with this map.

Now at the 3-to-2 stage, there are six natural maps $T^2 X \to T^3 X$, sending $(x;u,v;w)$ to $(x;u,0,0;0,0,v;w)$ or $(x;0,u,0;0,v,0;w)$ or $(x;0,0,u;v,0,0;w)$ or the same with $u$ and $v$ switched. The formula $2(B+C+D) \mathrm{d}_{1\cdot 0} x + 6E \mathrm{d}_{10} x$ is the sum of the precomposition with all six of these maps. (But it would probably be more natural, in a general theory, to consider all these maps separately rather than only when added up.)

• CommentRowNumber49.
• CommentAuthorUrs
• CommentTimeMay 20th 2015
• (edited May 20th 2015)

But is Michael also right that the differentials anticommute?

Yes.

And does their setup include nonlinear differential forms such as $\sqrt{d x^2 + d y^2}$?

Only for the second order differentials $d_1 d_2 x$, yes, since only these are commuting forms in their setup. In section 4.4 you see them consider Gaussians of these.

(Do you want first order differentials to not square to 0? That would sound a dubious wish to me, but I don’t really know where you are coming from here.)

• CommentRowNumber50.
• CommentAuthorTobyBartels
• CommentTimeMay 20th 2015

tl;dr: We seek an algebra that encompasses both exterior forms and equations such as the first displayed equation in the top post of this thread.

The main motivating examples for what Mike and I have been doing (beyond the exterior calculus) are these from classical Calculus:

$f''(x) = \frac{\mathrm{d}^2f(x)}{\mathrm{d}x^2}$

(which is wrong but can be fixed) and

$\mathrm{d}s = \sqrt{\mathrm{d}x^2 + \mathrm{d}y^2}$

(which is correct except for the implication that $\mathrm{d}s$ is the differential of some global $s$). Anything that treats $\mathrm{d}x^2$ as zero is not what we're looking for.

Having determined that

$f''(x) = \frac{\mathrm{d}^2f(x)}{\mathrm{d}x^2} - f'(x) \frac{\mathrm{d}^2x}{\mathrm{d}x^2}$

is correct, we're now looking more carefully at $\mathrm{d}^2$ and deciding that (if we wish to fit exterior forms into the same algebra) the two $\mathrm{d}$s are not exactly the same operator, and that furthermore it matters which of these appears in $\mathrm{d}x$.

The current working hypothesis is that

$\mathrm{d}^2f(x) = f''(x) \,\mathrm{d}x^2 + f'(x) \,\mathrm{d}^2x$

(the previous equation, rearranged) is the symmetrization (in the algebra of cojet forms) of

$\mathrm{d}_1\mathrm{d}_0f(x) = f''(x) \,\mathrm{d}_1x \,\mathrm{d}_0x + f'(x) \,\mathrm{d}_1\mathrm{d}_0x$

(in the algebra of coflare forms).

• CommentRowNumber51.
• CommentAuthorMike Shulman
• CommentTimeMay 20th 2015

And does their setup include nonlinear differential forms such as $\sqrt{d x^2 + d y^2}$?

Only for the second order differentials $d_1 d_2 x$, yes

What does that mean? Does $\sqrt{d x^2 + d y^2}$ exist or doesn’t it? Here $d x$ and $d y$ are first-order differentials, but their squares are of course second order, and then when we take a square root we get back to “first order” in a suitable sense.

• CommentRowNumber52.
• CommentAuthorUrs
• CommentTimeMay 20th 2015

By second order differentials I mean those with second derivatives, as in $d_1 d_2 x$. In $C^\infty([\mathbb{R}^{0|2},\mathbb{R}^1])$ we have $(d_1 x) (d_1 x) = 0$ but no power of $d_1 d_2 x$ vanishes.

• CommentRowNumber53.
• CommentAuthorMike Shulman
• CommentTimeMay 20th 2015

So $\sqrt{d x^2 + d y^2}=0$ in their setup, because $d x^2 = d y^2 = 0$?

I’m also confused because $d x^2=0$ seems to contradict the formula cited in #35, which ought to have a nonzero term with coefficient $\partial^2 f / (\partial x^i)^2$ when $i=j$.

• CommentRowNumber54.
• CommentAuthorMichael_Bachtold
• CommentTimeMay 21st 2015
• (edited May 21st 2015)

Wait, I’m now in doubt if indeed in Severas setting we have that differentials $d_1$, $d_2$ anti-commute. Wouldn’t the commutation rule in a $\mathbb{Z}^2$ graded commutative algebra read $a\cdot b=(-1)^{|a|\cdot |b|}b\cdot a$ where $|a|=(n_1,n_2)$ is the bi-degree of $a$ and $|a|\cdot|b|:=n_1 k_1+ n_2 k_2$, for $|b|=(k_1,k_2)$? If that’s the case then elements of bi-degree $(1,0)$ commute with things of bi-degree $(0,1)$. ?

• CommentRowNumber55.
• CommentAuthorUrs
• CommentTimeMay 21st 2015
• (edited May 21st 2015)

Sorry, yes, I suppose you are right.

Somehow I am not fully focusing on this discussion here…

I think I am just suggesting that if in search of a generalization of anything (here: exterior calculus) it helps to have some guiding principles. What Kochan-Ševera do has the great advantage that by construction it is guaranteed to have much internal coherence, because in this approach one doesn’t posit the new rules by themselves, but derives them from some more abstract principle (of course one has to derive them correctly, I just gave an example of failing to do that :-)

• CommentRowNumber56.
• CommentAuthorMike Shulman
• CommentTimeMay 21st 2015

Which “you” of us is right, #53 or #54?

I think that what Toby and I are doing has perfectly valid guiding principles. One of those principles is, I think, that it should be as simple and concrete as possible, so that it can be taught to calculus students who don’t know what a graded commutative algebra is. (-:

• CommentRowNumber57.
• CommentAuthorTobyBartels
• CommentTimeMay 21st 2015

Those two guiding principles (a structure with internal coherence, simple calculations) should really be in concert.

• CommentRowNumber58.
• CommentAuthorMike Shulman
• CommentTimeMay 22nd 2015

Sure; I think coflare forms have plenty of internal coherence as well. We aren’t “positing” any rules, just taking seriously the idea of iterated tangent bundles and dropping any requirement of linearity on forms.

As long as I’m over here, let me record this observation here which I made in the other thread: the view of a connection in #11 gives a coordinate-invariant way (which of course depends on the chosen connection) to “forget the order-2 part” of a rank-2 coflare form, by substituting

$d^2x^i \mapsto \Gamma^i_{j k} d x^j \otimes d x^k$

wherever it appears. The transformation rule for Christoffel symbols precisely ensures that this is coordinate-invariant. Are there analogues in higher rank?

• CommentRowNumber59.
• CommentAuthorUrs
• CommentTimeMay 22nd 2015
• (edited May 22nd 2015)

Michael, re #54 wait, no, the algebra in the end is the superalgebra $C^\infty([\mathbb{R}^{0|2},X])$ and as such is $\mathbb{Z}/2\mathbb{Z}$-graded.

I’ll have a few spare minutes tomorrow morning when I am on the train. I’ll try to sort it out then and write it cleanly into the $n$Lab entry.

• CommentRowNumber60.
• CommentAuthorUrs
• CommentTimeMay 22nd 2015
• (edited May 22nd 2015)

it should be as simple and concrete as possible, so that it can be taught to calculus students who don’t know what a graded commutative algebra is.

Returning the favour, allow me to add the advise: first you need to understand it, only then should it be made simple for exposition. Not the other way around! :-)

The graded algebra is there to guide the concept. The concept in the end exists also without graded algebra.

I’ll try to find time tomorrow to extract the relevant essence in Kochan-Severa.

• CommentRowNumber61.
• CommentAuthorTobyBartels
• CommentTimeMay 22nd 2015

The simple Calculus exposition already exists, particularly if you go back to the 19th century. It is internally inconsistent, but perhaps it can be cleaned up to be consistent. (Already parts of it have been, but can the whole thing be bundled into a single system?) A unifying conceptual idea would be very valuable here, and so far the best idea that Mike and I have is coflare forms.

• CommentRowNumber62.
• CommentAuthorMichael_Bachtold
• CommentTimeMay 22nd 2015
• (edited May 22nd 2015)

Urs re #59 you are probably right, I was trying to recall from vague memory how the formula for graded commutativity works. It should follow from the definition at graded commutative algebra once we agree how a $\mathbb{Z}^2$ graded vector space is $\mathbb{Z}_2$ graded, which is probably by mapping the bi-degree $(n_1,n_2)\mapsto n_1 + n_2 mod 2$. Then I guess the right commutativity formula is $a\cdot b=(-1)^{(n_1+n_2)(k_1+k_2)}b\cdot a$.

• CommentRowNumber63.
• CommentAuthorTobyBartels
• CommentTimeApr 24th 2019
• (edited Apr 24th 2019)

I ran across a math help page that at least tries to justify the incorrect traditional formula for the second differential; it argues ‘When we calculate differentials it is important to remember that $d x$ is arbitrary and independent from $x$ number [sic]. So, when we differentiate with respect to $x$ we treat $d x$ as constant.’. Now, that's not how differentials work; you're not differentiating with respect to anything in particular when you form a differential, and so you treat nothing as constant. (If you do treat something as constant, then you get a partial differential.) But at least they realize that there is something funny going on here, and that it might be meaningful to not treat $d x$ as constant.

• CommentRowNumber64.
• CommentAuthorMichael_Bachtold
• CommentTimeApr 25th 2019
• (edited Apr 25th 2019)

The formulation on that page is indeed confusing. On the other hand, treating $d x$ constant ($d d x=0$) does lead to the incorrect traditional formula. That’s also how Leibniz, Euler etc. arrived at the (generally incorrect) equation $\frac{d}{d x}\frac{d y}{d x} = \frac{d^2 y}{d x^2}$. Nowadays that equation is true by notational convention.

• CommentRowNumber65.
• CommentAuthorTobyBartels
• CommentTimeMay 2nd 2019

That MSE.stackexchange answer that you linked is interesting, as are the comments by Francois Ziegler. I'm glad to see that the early Calculators explicitly stated that they were making an assumption, equivalent to $d^2{x}=0$, in deriving that formula.