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• CommentRowNumber1.
• CommentAuthorTobyBartels
• CommentTimeFeb 24th 2014

I added some fine print, examples, and counterexamples to cogerm differential form.

• CommentRowNumber2.
• CommentAuthorTobyBartels
• CommentTimeFeb 24th 2014

Note to newbies: there is previous discussion of these at What is a variable? and related discussion at differentials. (Both of these began as discussions about teaching Calculus but got much more abstract near the end.)

• CommentRowNumber3.
• CommentAuthorMike Shulman
• CommentTimeFeb 24th 2014

Thanks!

• CommentRowNumber4.
• CommentAuthorMike Shulman
• CommentTimeMar 6th 2014

I added a bit more about the disappointing “exterior differentials” of cogerm forms.

• CommentRowNumber5.
• CommentAuthorTobyBartels
• CommentTimeMar 12th 2014

Here’s a way in which we want to be able to take exterior (wedge) products of cogerm forms. Recall that a Riemannian metric, while usually thought of as a symmetric bilinear covector form, can be equivalently thought of as a quadratic covector form, hence in particular a cogerm 1-form. I argued years ago, in the great forms thread on s.p.r with Eric Forgy, that the volume pseudo-form on an (unoriented) $n$-dimensional Riemannian manifold is the principle square root of the $n$-fold wedge product of the metric with itself. I'm not sure why I said that, exactly; you also have to divide by $n!$. At the time, I considered that this calculation took place in the second symmetric power of the exterior algebra, which is correct enough; but I'd also like it to take place in the exterior cogerm algebra. Even better, since the arclength element is the principal square root of the metric, I'd like the volume pseudo-form to simply be the $n$-fold wedge product of the arclength element (divided by $\sqrt{n!}$).

Here is an example, to help you follow me. Given $g = \mathrm{d}x^2 + \mathrm{d}y^2$ on $\mathbb{R}^2$,

$\vol = \sqrt{\frac{g \wedge g}{2!}} = \sqrt{\frac{(\mathrm{d}x^2 + \mathrm{d}y^2) \wedge (\mathrm{d}x^2 + \mathrm{d}y^2)}{2}} = \sqrt{\frac{\mathrm{d}x^2 \wedge \mathrm{d}x^2 + \mathrm{d}x^2 \wedge \mathrm{d}y^2 + \mathrm{d}y^2 \wedge \mathrm{d}x^2 + \mathrm{d}y^2 \wedge \mathrm{d}y^2}{2}} = \sqrt{\frac{(\mathrm{d}x \wedge \mathrm{d}x)^2 + (\mathrm{d}x \wedge \mathrm{d}y)^2 + (\mathrm{d}y \wedge \mathrm{d}x)^2 + (\mathrm{d}y \wedge \mathrm{d}y)^2}{2}} = \sqrt{\frac{(0)^2 + (\mathrm{d}x \wedge \mathrm{d}y)^2 + (-\mathrm{d}x \wedge \mathrm{d}y)^2 + (0)^2}{2}} = \sqrt{\frac{2 (\mathrm{d}x \wedge \mathrm{d}y)^2}{2}} = \sqrt{(\mathrm{d}x \wedge \mathrm{d}y)^2} = {|\mathrm{d}x \wedge \mathrm{d}y|} = {|\mathrm{d}x|} \wedge {|\mathrm{d}y|} .$

(This last expression is traditionally called ‘$\mathrm{d}x \,\mathrm{d}y$’, but the absolute value is really there, as you can see because reversing the sign of one variable leaves the area unchanged.)

That's how I did it with Eric (except for the last step which I would not have allowed then); here's how I'd also like to do it:

$\vol = \frac{đ{s} \wedge đ{s}}{\sqrt{2!}} = \frac{\sqrt{\mathrm{d}x^2 + \mathrm{d}y^2} \wedge \sqrt{\mathrm{d}x^2 + \mathrm{d}y^2}}{\sqrt{2}} = \sqrt{\frac{(\mathrm{d}x^2 + \mathrm{d}y^2) \wedge (\mathrm{d}x^2 + \mathrm{d}y^2)}{2}} = \cdots = {|\mathrm{d}x|} \wedge {|\mathrm{d}y|} .$

In all of these calculations, operations that normally distribute over multiplication (square, principal square root, and their composite, absolute value) distribute over the wedge product, and it works.

• CommentRowNumber6.
• CommentAuthorMike Shulman
• CommentTimeMar 12th 2014

It’s nice to have an example! Now let’s see whether we can make it an example of anything. I wonder whether germs of parametrized surfaces are too general?

• CommentRowNumber7.
• CommentAuthorMike Shulman
• CommentTimeMar 13th 2014

Here’s one idea I had had for the exterior product of cogerm forms. Let $c$ be a parametrized surface; then for any angle $\theta$ we have a curve defined by $c_\theta(t) = c(t\cos\theta,t\sin\theta)$. Now given cogerm 1-forms $\eta$ and $\omega$, we evaluate them on the “perpendicular” curves $c_\theta$ and $c_{\theta+\pi/2}$ respectively, multiply the results, and average over $\theta$:

$\frac{1}{2\pi}\, \int_{\theta=0}^{2\pi} \langle \eta {|} c_{\theta} \rangle \cdot \langle \omega {|} c_{\theta+\pi/2} \rangle \, \mathrm{d}\theta.$

I think for exterior 1-forms this gives the right answer, working it out in coordinates and using the facts that $\int_0^{2\pi} \cos^2(\theta) \mathrm{d}\theta = \int_0^{2\pi} \sin^2(\theta) \mathrm{d}\theta = \pi$ and $\int_0^{2\pi} \sin(\theta) \cos(\theta) \mathrm{d}\theta = 0$. Does it give anything sensible for the metric or the arclength element?

• CommentRowNumber8.
• CommentAuthorMike Shulman
• CommentTimeMar 19th 2014

Suppose we had a wedge product of cogerm forms which, when restricted to exterior 1-forms $\eta$ and $\omega$, yields a cogerm 2-form which acts like the ordinary wedge product of exterior forms:

$\langle\eta\wedge\omega{|}c\rangle = \eta(v) \omega(w) - \eta(w)\omega(v)$

where $v,w$ are the canonical tangent vectors to $c$ at $0$, i.e. $v = D c(e_0)$ and $w = D c(e_1)$. Then we would have

$\langle(\eta\wedge\omega)^2{|}c\rangle = \langle\eta\wedge\omega{|}c\rangle^2 = (\eta(v) \omega(w) - \eta(w)\omega(v))^2 = \eta^2(v) \omega^2(w) + \eta^2(w) \omega^2(v) - 2 \eta(v) \omega(w)\eta(w)\omega(v)$

which seems to say that the cogerm 2-form $(\eta\wedge\omega)^2$ doesn’t depend only on $\eta^2$ and $\omega^2$, but also on $\eta$ and $\omega$. This makes it seem unlikely that it could be equal to $\eta^2 \wedge \omega^2$. Is this a wrong conclusion to draw?

• CommentRowNumber9.
• CommentAuthorMike Shulman
• CommentTimeMar 19th 2014

By the way, I just had occasion to look at the book A geometric approach to differential forms by David Bachman, and noticed that although he talks mostly about ordinary exterior forms, he’s very explicit about there being other sorts which one may want to integrate:

…a more general integral [over a domain parametrized by $\phi$] would be $\int f(\phi(a))\, \omega(\frac{d\phi}{d a})\, d a$, where $f$ is a function of points and $\omega$ is a function of vectors. It is not the purpose of the present work to undertake a study of integrating with respect to all possible functions, $\omega$. However, as with the study of functions of real variables, a natural place to start is with linear functions. This is the study of differential forms… The strength of differential forms lies in the fact that their integrals do not depend on a choice of parametrization. (p31)

In the appendix, he considers the arc length element and the unoriented area form as particular examples of non-linear $\omega$ that one may want to integrate, but he doesn’t give any general theory.

• CommentRowNumber10.
• CommentAuthorTobyBartels
• CommentTimeMar 20th 2014

Those two examples share the strength of differential forms —that they don't depend on parametrization— and improve it: they don't depend on orientation either. (This is because they are absolute differential forms.)

I need to go and think about $(\omega \wedge \eta)^2$ right now.

• CommentRowNumber11.
• CommentAuthorMike Shulman
• CommentTimeMar 20th 2014

Yes, I was surprised that he didn’t point that out. In the appendix he seems to change his mind a bit and writes

The thing that makes (linear) differential forms so useful is the generalized Stokes’ Theorem. We do not have anything like this for non-linear forms…

• CommentRowNumber12.
• CommentAuthorMike Shulman
• CommentTimeMar 24th 2014

Today I decided that I don’t like the definition of integration which I proposed in the other thread (and wrote on the page). Here’s why.

Right now I’m teaching my students to apply integrals to calculate volumes, arc lengths, etc., and I gave them the following method: slice up the quantity you want to compute into small pieces, find a differential form that approximates the value of each piece to first order, then integrate that differential form. For instance, in the basic case when finding an area, we slice it into pieces which are rectangles to first-order, hence have area $f(x) dx$ (where $f(x)$ is the height and $dx$ the width), then we integrate that. But suppose we instead used a differential form that is a better approximation. For instance, if we instead approximate that piece by a trapezoid, then its area would be $f(x) dx + \frac{1}{2}(f(x+dx)-f(x)) dx$, which is about $f(x) dx + \frac{1}{2} f'(x) dx^2$. I’d like to say that if we integrate this differential form, we should get the same area; but with the definition of integration that I proposed before, we don’t.

This argument suggests that integrating a cogerm 1-form of “degree $\gt 1$” should always result in zero. Symmetrically, one might argue that integrating a cogerm 1-form of “degree $\lt 1$” should always result in divergence.

Here’s a definition of integration which does have that property, and at least looks vaguely reasonable. As before, if $c$ is a curve, denote by $c_h$ the curve with $c_h(t) = c(t+h)$. Also, denote by $h \cdot c$ the curve with $(h \cdot c)(t) = c(h t)$. Then we can say that a cogerm 1-form $\omega$ is “degree-$n$ oriented-homogeneous” if $\langle \omega {|} h \cdot c\rangle = h^n \langle \omega{|}c\rangle$, and “degree-$n$ absolute-homogeneous” if $\langle \omega {|} h \cdot c\rangle = {|h|}^n \langle \omega{|}c\rangle$.

Suppose $\omega$ is a cogerm 1-form and $c$ a curve on $[a,b]$, and say we have a tagged partition $a = x_0 \lt x_1 \lt \cdots \lt x_N = b$, with tags $t_i \in [x_{i-1},x_i]$. Write $\Delta x_i = x_i - x_{i-1}$, and define the Riemann sum of $\omega$ over this partition to be

$\sum_{i=1}^n \langle \omega {|} \Delta x_i \cdot c_{t_i} \rangle$

Now the integral $\oint_c \omega$ is just the limit of these Riemann sums as the mesh goes to zero, if it exists. (We can take this limit either in the Riemann way or the Henstock way.)

If $\omega$ is degree-1 homogeneous of either sort, then this should agree with the other definition of the integral: just pull the $\Delta x_i$ out by homogeneity and you have the ordinary integral over $[a,b]$. But this definition should also give zero if $\omega$ is degree-$n$ homogeneous with $n\gt 1$, and diverge if it is degree-$n$ homogeneous with $n\lt 1$, at least if $\omega$ is otherwise well-behaved (e.g. continuous).

I believe this integral also has another nice property: if $\oint_c \omega$ exists for some $\omega$, then it is invariant under orientation-preserving affine reparametrization of $c$.

I don’t know whether this would help to solve any of the other problems. It does reinforce the idea that the exterior differential only makes sense on degree-1 forms (and not even all of those): if we define $d\wedge \omega$ in the “Arnold” way as the limit of integrals of $\omega$ over shrinking loops, then it will automatically vanish if $\omega$ has degree $\gt 1$ and diverge if it has degree $\lt 1$, because the integral does so.

• CommentRowNumber13.
• CommentAuthorMike Shulman
• CommentTimeMar 26th 2014

My new proposed definition of integration in #12 would mean that only linear 1-forms satisfy FTC. But perhaps that’s not too bothersome in light of the fact that FTC is a special case of Stokes’ theorem, and the general form of Stokes’ theorem requires the exterior differential.

I also think there’s the germ of a nice proof of the 1-variable FTC somewhere in this. Consider the fact that for any function $f$, the difference $f(x+dx) - f(x)$ is itself a cogerm form, and differentiability of $f$ says that this form differs from the differential $d (f(x)) = f'(x) dx$ by something of second order. Thus, given a general theorem like “the integral of a form depends only on its first-order part”, we could say that $\int_a^b f'(x) dx = \int_a^b (f(x+dx) - f(x))$, and by choosing a suitable Riemann sum the latter simply telescopes to $f(b) -f(a)$. That’s not very precise, of course, but I like that way of thinking about it.

• CommentRowNumber14.
• CommentAuthorTobyBartels
• CommentTimeMar 29th 2014

This argument suggests that integrating a cogerm 1-form of “degree $\gt 1$” should always result in zero. Symmetrically, one might argue that integrating a cogerm 1-form of “degree $\lt 1$” should always result in divergence.

OK, I'll buy this, for integrals along curves. But then a cogerm $p$-form of homogeneous degree $p$ should still be integrable over a $p$-dimensional submanifold.

• CommentRowNumber15.
• CommentAuthorMike Shulman
• CommentTimeMar 29th 2014

Yes, I agree.

• CommentRowNumber16.
• CommentAuthorMike Shulman
• CommentTimeApr 6th 2014

Re #5, can you say anything intuitive about why you expect this to be the case, other than the fact that the formal calculations work if you make certain assumptions? Having thought about it a bit more, my intuition doesn’t match. For instance, normally the exterior square of anything is zero; why would you expect the nonlinear case to be different? Also, the main intuition that I’ve been able to think of for why the area element should be the square of the arc length element is that area is length times width — but that’s only true for rectangles, not for arbitrary parallelograms, and a 2-form acts on all pairs of vectors (or cojets or cogerms), not just orthogonal ones.

• CommentRowNumber17.
• CommentAuthorTobyBartels
• CommentTimeApr 6th 2014

normally the exterior square of anything is zero

That's not true for, for example, scalar fields (or more generally exterior forms with even rank). And scalar fields are, besides being exterior $0$-forms, also cogerm $1$-forms. So, we already know that the square of a cogerm $1$-form need not be $0$, and it shouldn't be surprising if $g$ and $đ{s}$ are further counterexamples.

area is length times width — but that’s only true for rectangles, not for arbitrary parallelograms

That's just the sort of thing that we'd expect the wedge product to handle. Indeed, the exterior square is zero (when it is zero) because it gives us a flat parallelogram.

But you're mainly right, that my main reason for believing this is that the calculation seems to work. I mean, the original calculation definitely works, in the second symmetric power of the exterior calculus. But to do it in an exterior cogerm calculus, well, I don't know that calculus, so I can only say that it seems to work.

In particular, I don't know how to justify the factorial, other than that it has to be there to make the answer come out correctly.

• CommentRowNumber18.
• CommentAuthorTobyBartels
• CommentTimeApr 6th 2014

Here I try to give a geometric justification of the factorial, for a low-level sense of ‘geometric’ and a very hand-waving sense of ‘justification’.

Start with the usual picture of $d{s}$ as the hypotenuse of a right triangle with legs $d{x}$ and $d{y}$. Now double this right triangle to a rectangle, with diagonal $d{s}$ and area $d{A} = d{x} \,d{y}$. In this picture, how do we express $d{A}$ in terms of $d{s}$?

In one sense, we can't, of course. If the rectangle is a square, then $d{A} = d{s}^2/\sqrt{2}$, just as I want. But if the rectangle is flat, then $d{A} = 0$ even though $d{s} \ne 0$.

But my claim is not that $đA = đs^2/\sqrt{2}$ but that $đA = đs \wedge đs/\sqrt{2}$. And the exterior product only counts the part that's perpendicular. (That's why the exterior square is zero, after all, at least in the simplest case.) But how can any part of $đs$ be perpendicular to itself?

We have a rectangle with a diagonal. But this has another diagonal, which is also $d{s}$. When the rectangle is a square, these diagonals are perpendicular, and we get $d{A} = d{s}^2/\sqrt{2}$. But when the rectangle is flat, these are parallel, and we get $d{A} = 0$. In general, $d{A} = d{s}^2 \sin \theta/\sqrt{2}$, where $\theta$ is the angle1 that the two diagonals make with each other. And that's exactly the perpendicular part; it's just what the wedge product ought to give us.

Somehow, this should reflect something about the exterior cogerm calculus. But I don't know how. (Particularly when all of this talk of angles and orthogonality refers to the metric, or at least to the conformal structure, which we have when we talk about $đs$ but which should not be available in the exterior cogerm calculus itself.)

1. Of course, they cross and so make four angles, which are only equal in pairs. But these angles' sines are all equal.

• CommentRowNumber19.
• CommentAuthorTobyBartels
• CommentTimeApr 6th 2014
• (edited Apr 6th 2014)

Well, that's extremely hand-wavy when you realize that every time that I wrote $\sqrt{2}$ in that last comment, I was really only justified in writing $2$. And if I generalize to $n$ dimensions, it becomes an argument that $vol = đs \wedge \cdots \wedge đs/n$ instead of $vol = đs \wedge \cdots \wedge đs/\sqrt{n!}$ as my algebraic calculation gave. So I don't know what's up with that!

• CommentRowNumber20.
• CommentAuthorMike Shulman
• CommentTimeApr 6th 2014

I haven’t digested #18 yet, but re #17, you’re right that I should have said the exterior square of any 1-form is zero. however, if we are to embed the cogerm 1-forms in some context that also contains cogerm $k$-forms for other $k$, then it might be that we will have to regard scalar functions as cogerm 0-forms separately from their incarnation as cogerm 1-forms. It would then be perfectly consistent for the exterior square of a scalar function qua cogerm 0-form to be nonzero (being another cogerm 0-form), while its square qua cogerm 1-form is zero (specifically, the zero cogerm 2-form). For instance, that’s what happens for the “cojet differential forms” which I proposed in the other thread.

A notational question: is there a deep reason that you usually write $đ{s}$ for the arc length element? Is it just that you want a symbol that looks like the traditional notation $\mathrm{d}s$ but doesn’t suggest that it is actually the differential of something?

• CommentRowNumber21.
• CommentAuthorTobyBartels
• CommentTimeApr 7th 2014

Yes, that's it. It's a notation that's used in thermodynamics; for example in the first law of thermodynamics

$\mathrm{d}U = đQ + đW$

(the total change in the energy of a subsystem is the sum of the heat transferred into that subsystem and the mechanical work done on that subsystem). It's an important point that (contrary to the caloric theory of heat) there is no such thing as $Q$, only $đQ$.

• CommentRowNumber22.
• CommentAuthorMike Shulman
• CommentTimeApr 7th 2014

Nice, thanks.

• CommentRowNumber23.
• CommentAuthorMike Shulman
• CommentTimeApr 12th 2014

Here’s an improvement of my vague remarks in #12. Let’s say that a cogerm 1-form $\omega$ is “$o(dx^n)$” if for any curve $c$, we have

$\lim_{h\to 0} \frac{\langle \omega {|} h\cdot c\rangle}{h^n} = 0$

A degree-$n$ homogeneous form is clearly $o(dx^k)$ for any $k\lt n$. I claim that if $\omega$ is $o(dx)$, then it is Henstock integrable over any curve $c$ (defined on a closed interval $[a,b]$), and $\oint_c \omega = 0$.

Suppose $\epsilon\gt 0$; then for any $t\in [a,b]$ there is a $\delta(t)$ such that ${\langle \omega {|} h \cdot c_t\rangle} \lt \frac{\epsilon}{b-a} h$ for any $0\lt h\lt \delta(t)$. This defines a gauge $\delta$ on $[a,b]$. Now suppose we have a $\delta$-fine tagged partition $a = x_0 \lt \cdots \lt x_n = b$ with tags $t_i \in [x_{i-1},x_i]$, so that $\Delta x_i = x_i - x_{i-1} \lt \delta(t_i)$. Then the corresponding Riemann sum is, by definition, $\sum_{i} \langle \omega {|} \Delta x_i \cdot c_{t_i}\rangle$. Since $\Delta x_i \lt \delta(t_i)$, each $\langle \omega {|} \Delta x_i \cdot c_{t_i}\rangle \lt \frac{\epsilon}{b-a} \Delta x_i$. Thus, when we sum them up, we get something less than $\frac{\epsilon}{b-a} \sum_i \Delta x_i = \frac{\epsilon}{b-a} (b-a) = \epsilon$. Thus, for any $\epsilon$ there is a gauge $\delta$ such that the Riemann sum over any $\delta$-fine tagged partition is $\lt\epsilon$; so the Henstock integral is zero.

It follows that if any such $\omega$ is Riemann integrable, then its Riemann integral is zero. I suspect that there are probably forms that are $o(dx)$ but not Riemann integrable, but I haven’t come up with any yet.

I think my suggested proof of FTC in #13 can also be made to work. Say we assume that $df$ is continuous, hence integrable (I think the usual proof that continuous functions are integrable will show that any continuous linear 1-form, such as the differential of a function, is integrable). Then by definition, $df$ differs from $f(x+dx)-f(x)$ by something $o(dx)$, and the sum of integrable forms is integrable; thus $f(x+dx)-f(x)$ is integrable. But any left Riemann sum of $f(x+dx)-f(x)$ telescopes to $f(b) - f(a)$; hence that must be its integral, and thus also the integral of $df$.

• CommentRowNumber24.
• CommentAuthorMike Shulman
• CommentTimeApr 13th 2014

Here’s another way to approach the question of #5: suppose we know how to measure lengths (with some 1-form $đl$); how can we measure (say) areas? The first thing we might do is use the polarization identity to recover the inner product:

$v\cdot w = \textstyle\frac{1}{4}(đl(v+w)^2 - đl(v-w)^2)$

Note that if we replace $đl$ by a linear form $\eta$, then this would give $\eta(v)\,\eta(w)$.

Now $v\cdot w = \Vert v\Vert \,\Vert w\Vert \,\cos \theta$, whereas the area spanned by $v$ and $w$ should be $\Vert v\Vert \,\Vert w\Vert \,\sin \theta$. Thus, we can compute the area as

$\sqrt{đl(v)^2\, đl(w)^2 - (v\cdot w)^2}.$

And if we replace $đl$ by a linear form $\eta$ here, we would get zero. So at least we have some operation on “nonlinear covector” 1-forms that yields a “nonlinear covector 2-form”, which when applied to $đl$ gives the area form, and when applied to a linear 1-form gives zero, so we might hope to call it the “exterior square”.

But I can’t think of a way to tease an operation on pairs of 1-forms out of this. I’m also ambivalent about it anyway because it doesn’t seem canonical; e.g. there are several forms of polarization, which are the same for $đl$ or a linear form, but it seems that they might be different on other forms.

• CommentRowNumber25.
• CommentAuthorMike Shulman
• CommentTimeApr 13th 2014

Re: #18, it doesn’t seem right to me to say “the exterior product only counts the part that’s perpendicular”. There is no notion of “perpendicular” until we have a metric, and the exterior product is defined without any metric. I feel like $dx\wedge \dy$ is measuring area in one way (sort of declaring $dx$ and $dy$ to “be perpendicular” for the purposes of area), while a “product” $đA = đl \wedge đl$ would be measuring area in a different way (obtaining a notion of “perpendicular” from the factors $đl$ rather than declaring them to be perpendicular to each other).

Here’s a question: suppose a manifold has two metrics $g_1$ and $g_2$, giving rise to two length elements $đl_1$ and $đl_2$. What would you expect $đl_1 \wedge đl_2$ to measure?

• CommentRowNumber26.
• CommentAuthorMike Shulman
• CommentTimeApr 13th 2014
• (edited Apr 17th 2014)

Re #24, here’s something. Maybe I was half asleep last night, because now I think this is the “obvious” thing to try. Given two covector 1-forms $\alpha$ and $\beta$, consider the covector 2-form defined by

$(\alpha\cdot\beta)(v,w) = \textstyle\frac{1}{4}(\alpha(v+w)\beta(v+w) - \alpha(v-w)\beta(v-w)).$

If $\alpha=\beta=đl$, this gives the inner product as in #24. But if $\alpha$ and $\beta$ are linear, then this gives the symmetrized product

$(\alpha\cdot\beta)(v,w) = \textstyle\frac{1}{2}(\alpha(v)\beta(w) + \beta(v)\alpha(w)).$

Now consider the covector 2-form

$(\alpha\barwedge\beta)(v,w) = \sqrt{|\alpha(v)\beta(v)\alpha(w)\beta(w) - (\alpha\cdot\beta)(v,w)^2|}.$

If $\alpha=\beta=đl$, this gives the area element as in as in #24. But if $\alpha$ and $\beta$ are linear, then it gives the absolute value of their exterior product,

${|\alpha\wedge\beta|}(v,w) = \textstyle\frac{1}{2} | \alpha(v)\beta(w) - \beta(v)\alpha(w)|.$

(with the unpopular convention $\frac{1}{2!}$ rather than $\frac{1}{1!1!}$ for the factorial factors). This raises a bunch of questions:

1. Can it be extended to higher dimensions to get volume forms?
2. Can it be extended to cojet or cogerm forms, since we can’t add and subtract jets or germs?
3. What’s up with the absolute value? Is there any way to get the non-absolute-valued exterior product?
4. What’s up with the absolute value inside the square root? I put it there because if I’m not mistaken, the relative signs of the terms being subtracted are different in the metric case and in the linear case. Is there some sense in which the two are “imaginary rotations” of each other?
• CommentRowNumber27.
• CommentAuthorTobyBartels
• CommentTimeApr 17th 2014

there are several forms of polarization, which are the same for $đl$ or a linear form, but it seems that they might be different on other forms

They are the same if and only if the parallelogram identity holds. Which, of course, it typically won't. (This result is for any operation on an abelian group in which $2$ is invertible; the operation doesn't have to have to have any properties of the square of a norm. And while I'm at it, $2$ doesn't even have to be invertible if you take the polarization identities to define $4$ times the inner product.)

Your version of the polarization identity is perhaps the most even-handed. But in general, it's not even-handed enough; besides $v + w$ and $v - w$, you also ought to include $-v + w$ and $-v - w$.

$(\alpha\barwedge\beta)(v,w) = \sqrt{|\alpha(v)\beta(v)\alpha(w)\beta(w) - (\alpha\cdot\beta)(v,w)|}.$

You mean

$(\alpha\barwedge\beta)(v,w) = \sqrt{|\alpha(v)\beta(v)\alpha(w)\beta(w) - (\alpha\cdot\beta)(v,w)^2|}$

(squaring the dot product).

What’s up with the absolute value? Is there any way to get the non-absolute-valued exterior product?

It looks like getting the non-absolute-valued (or rather non-normed) cross product. This is fixed (up to orientation, when even that can be done) by also requiring the cross product of two vectors to also be orthogonal to both vectors. But I don't see an analogue of that condition here.

• CommentRowNumber28.
• CommentAuthorTobyBartels
• CommentTimeApr 17th 2014

it doesn’t seem right to me to say “the exterior product only counts the part that’s perpendicular”

Not in general, but in this particular case we do have a metric; you talked this way in #16 too. But I'm pretty much convinced that #18 is nonsense anyway; it gives the wrong answer, after all.

Here’s a question: suppose a manifold has two metrics $g_1$ and $g_2$, giving rise to two length elements $đl_1$ and $đl_2$. What would you expect $đl_1 \wedge đl_2$ to measure?

Good question! I don't know, but I think that I can calculate it.

For definiteness, let the manifold be $\mathbb{R}^2$, so we have

$g_1 = A \,\mathrm{d}x^2 + 2 B \,\mathrm{d}x \,\mathrm{d}y + C \,\mathrm{d}y^2 ,$ $g_2 = D \,\mathrm{d}x^2 + 2 E \,\mathrm{d}x \,\mathrm{d}y + F \,\mathrm{d}y^2 ,$

with $A, C, D, F \gt 0$, $A C \gt B^2$, and $D F \gt E^2$.

I need to establish a lemma. I have already surmised that

$\alpha^2 \wedge \beta^2 = (\alpha \wedge \beta)^2 ,$

but what about $\alpha \beta$? (which is also a symmetric bilinear form and so may appear in a metric). Polarization lets me rewrite this using squares:

$(2 \alpha \beta) \wedge (2 \gamma \delta) = ((\alpha + \beta)^2 - \alpha^2 - \beta^2) \wedge ((\gamma + \delta)^2 - \gamma^2 - \delta^2) = (\alpha + \beta)^2 \wedge (\gamma + \delta)^2 - \alpha^2 \wedge (\gamma + \delta)^2 - \beta^2 \wedge (\gamma + \delta)^2 - (\alpha + \beta)^2 \wedge \gamma^2 + \alpha^2 \wedge \gamma^2 + \beta^2 \wedge \gamma^2 - (\alpha + \beta)^2 \wedge \delta^2 + \alpha^2 \wedge \delta^2 + \beta^2 \wedge \delta^2 = ((\alpha + \beta) \wedge (\gamma + \delta))^2 - (\alpha \wedge (\gamma + \delta))^2 - (\beta \wedge (\gamma + \delta))^2 - ((\alpha + \beta) \wedge \gamma)^2 + (\alpha \wedge \gamma)^2 + (\beta \wedge \gamma)^2 - ((\alpha + \beta) \wedge \delta)^2 + (\alpha \wedge \delta)^2 + (\beta \wedge \delta)^2 = (\alpha \wedge \gamma + \alpha \wedge \delta + \beta \wedge \gamma + \beta \wedge \delta)^2 - (\alpha \wedge \gamma + \alpha \wedge \delta)^2 - (\beta \wedge \gamma + \beta \wedge \delta)^2 - (\alpha \wedge \gamma + \beta \wedge \gamma)^2 + (\alpha \wedge \gamma)^2 + (\beta \wedge \gamma)^2 - (\alpha \wedge \delta + \beta \wedge \delta)^2 + (\alpha \wedge \delta)^2 + (\beta \wedge \delta)^2 = (\alpha \wedge \gamma)^2 + 2 (\alpha \wedge \gamma) (\alpha \wedge \delta) + 2 (\alpha \wedge \gamma) (\beta \wedge \gamma) + 2 (\alpha \wedge \gamma) (\beta \wedge \delta) + (\alpha \wedge \delta)^2 + 2 (\alpha \wedge \delta) (\beta \wedge \gamma) + 2 (\alpha \wedge \delta) (\beta \wedge \delta) + (\beta \wedge \gamma)^2 + 2 (\beta \wedge \gamma) (\beta \wedge \delta) + (\beta \wedge \delta)^2 - (\alpha \wedge \gamma)^2 - 2 (\alpha \wedge \gamma) (\alpha \wedge \delta) - (\alpha \wedge \delta)^2 - (\beta \wedge \gamma)^2 - 2 (\beta \wedge \gamma) (\beta \wedge \delta) - (\beta \wedge \delta)^2 - (\alpha \wedge \gamma)^2 - 2 (\alpha \wedge \gamma) (\beta \wedge \gamma) - (\beta \wedge \gamma)^2 + (\alpha \wedge \gamma)^2 + (\beta \wedge \gamma)^2 - (\alpha \wedge \delta)^2 - 2 (\alpha \wedge \delta) (\beta \wedge \delta) - (\beta \wedge \delta)^2 + (\alpha \wedge \delta)^2 + (\beta \wedge \delta)^2 = 2 (\alpha \wedge \gamma) (\beta \wedge \delta) + 2 (\alpha \wedge \delta) (\beta \wedge \gamma) ;$

dividing by $4$, $\alpha \beta \wedge \gamma \delta = \frac{1}{2} (\alpha \wedge \gamma) (\beta \wedge \delta) + \frac{1}{2} (\alpha \wedge \delta) (\beta \wedge \gamma)$. Note the special case $\alpha \beta \wedge \gamma^2 = (\alpha \wedge \gamma) (\beta \wedge \gamma)$ (and similarly with the square on the other side.)

Then I get

$g_1 \wedge g_2 = (A \,\mathrm{d}x^2 + 2 B \,\mathrm{d}x \,\mathrm{d}y + C \,\mathrm{d}y^2) \wedge (D \,\mathrm{d}x^2 + 2 E \,\mathrm{d}x \,\mathrm{d}y + F \,\mathrm{d}y^2) = (A D \,\mathrm{d}x^2 \wedge \,\mathrm{d}x^2 + 2 A E \,\mathrm{d}x^2 \wedge \mathrm{d}x \,\mathrm{d}y + A F \,\mathrm{d}x^2 \wedge \mathrm{d}y^2 + 2 B D \,\mathrm{d}x \,\mathrm{d}y \wedge \mathrm{d}x^2 + 4 B E \,\mathrm{d}x \,\mathrm{d}y \wedge \mathrm{d}x \,\mathrm{d}y + 2 B F \,\mathrm{d}x \,\mathrm{d}y \wedge \mathrm{d}y^2 + C D \,\mathrm{d}y^2 \wedge \mathrm{d}x^2 + 2 C E \,\mathrm{d}y^2 \wedge \mathrm{d}x \,\mathrm{d}y + C F \,\mathrm{d}y^2 \wedge \mathrm{d}y^2 = A D (\mathrm{d}x \wedge \mathrm{d}x)^2 + 2 A E (\mathrm{d}x \wedge \mathrm{d}x) (\mathrm{d}x \wedge \mathrm{d}y) + A F (\mathrm{d}x \wedge \mathrm{d}y)^2 + 2 B D (\mathrm{d}x \wedge \mathrm{d}x) (\mathrm{d}y \wedge \mathrm{d}x) + 2 B E (\mathrm{d}x \wedge \mathrm{d}x) (\mathrm{d}y \wedge \mathrm{d}y) + 2 B E (\mathrm{d}x \wedge \mathrm{d}y) (\mathrm{d}y \wedge \mathrm{d}x) + 2 B F (\mathrm{d}x \wedge \mathrm{d}y) (\mathrm{d}y \wedge \mathrm{d}y) + C D (\mathrm{d}y \wedge \mathrm{d}x)^2 + 2 C E (\mathrm{d}y \wedge \mathrm{d}x) (\mathrm{d}y \wedge \mathrm{d}y) + C F (\mathrm{d}y \wedge \mathrm{d}y)^2 = 0 + 0 + A F (\mathrm{d}x \wedge \mathrm{d}y)^2 + 0 + 0 - 2 B E (\mathrm{d}x \wedge \mathrm{d}y)^2 + 0 + C D (\mathrm{d}x \wedge \mathrm{d}y)^2 + 0 + 0 = (A F - 2 B E + C D) (\mathrm{d}x \wedge \mathrm{d}y)^2 .$

Taking square roots, $đl_1 \wedge đl_2 = \sqrt{A F - 2 B E + C D} \,{|\mathrm{d}x \wedge \mathrm{d}y|}$.

• CommentRowNumber29.
• CommentAuthorTobyBartels
• CommentTimeApr 17th 2014

If $g_1 = g_2$, then the expression under the square root is twice the determinant of the matrix of the coefficients of $g$, so we have $\sqrt{2}$ times the area element (which is the same answer that I've gotten before in that case). But I'm not sure what to make of $A F - 2 B E + C D$ itself; at least it is positive, since

$A F - 2 B E + C D \geq A F - 2 {|B|} {|E|} + C D \gt A F - 2 \sqrt{A C} \sqrt{D F} + C D = (\sqrt{A F} - \sqrt{C D})^2 \geq 0 .$

(And strictly so; that is, the wedge product of two positive-definite length elements is also positive-definite; we never get zero.)

• CommentRowNumber30.
• CommentAuthorMike Shulman
• CommentTimeApr 17th 2014

Thanks for the missing square; I’ve fixed it in the original comment.

in general, it's not even-handed enough; besides $v + w$ and $v - w$, you also ought to include $-v + w$ and $-v - w$.

You’re absolutely right. And it seems to me that even that might not be even-handed enough, when our forms may not be either linear or quadratic. It feels like maybe we ought to include everything in the lattice generated by $v$ and $w$.

• CommentRowNumber31.
• CommentAuthorMike Shulman
• CommentTimeApr 17th 2014

Re: #28 and 29, that’s intriguing. I don’t know what to make of $A F - 2 B E + C D$ either. And I guess I don’t really know how to think about a manifold that has two different metrics.

• CommentRowNumber32.
• CommentAuthorMike Shulman
• CommentTimeApr 23rd 2014

I’ve added #12 and #23 to the page cogerm differential form. One thing that is still missing is a good theorem about which cogerm forms are integrable (with either definition).

• CommentRowNumber33.
• CommentAuthorMike Shulman
• CommentTimeApr 23rd 2014

I added a theorem about existence and parametrization-invariance of integration. I’m not completely satisfied with it, but it’s the best I’ve been able to come up with.

• CommentRowNumber34.
• CommentAuthorMike Shulman
• CommentTimeApr 27th 2014

Here’s an interesting example: $\sqrt[3]{\mathrm{d}x^3 +\mathrm{d}y^3}$. It is degree-1 homogeneous with sign, so its integral reverses with orientation. Its integral around any closed circle or parallelogram is zero, but around other closed curves (like triangles) it has a nonzero integral.

In general, its integral along a line of angle $\theta$ to the $x$-axis is $\sqrt[3]{(\sin\theta)^3+(\cos\theta)^3}$ times the length of that line. Thus, along horizontal and vertical lines it just measures (signed) length, while along $45^\circ$ lines it measures something less than length, and along $135^\circ$ lines it measures zero. (This is similar to $\mathrm{d}x+\mathrm{d}y$, except that the latter measures more than length along $45^\circ$ lines, in such a way that its integral around any closed curve works out to zero — as it must, since the form is exact.)

Being degree-1 homogeneous with sign means in particular that if we take a region $R$ and divide it into two regions $R_1$ and $R_2$, then the integral around the boundary of $R$ is equal to the sum of the integrals around the boundaries of $R_1$ and $R_2$. This suggests that there might be a way to define the exterior differential of this 1-form in such a way that Stokes’ theorem would hold. (Note that since the analogous subdivision property for boundary integrals fails for $đl=\sqrt{\mathrm{d}x^2+\mathrm{d}y^2}$, it can’t have an exterior differential satisfying Stokes’ theorem.) However, I haven’t been able to think of such a definition.

• CommentRowNumber35.
• CommentAuthorTobyBartels
• CommentTimeApr 29th 2014

Working in the plane, we ought to be able to calculate this exterior derivative as an ordinary $2$-form. If every sufficiently regular oriented region $R$ is given a measure by integrating $\root{3}{\mathrm{d}x^3 + \mathrm{d}y^3}$ around it, and if this is additive and reasonably continuous, then this integral must be given by an absolutely continuous Radon pseudo-measure, hence an exterior $2$-form.

But this can't be right, because (as you say) the measure of any rectangle is zero, while the measure of some triangles is not. So it must violate continuity. Indeed, if we approximate a triangle from inside and out by rectangles, then we approximate its measure as zero, which it is not, so we don't have a Radon measure.

Well, this is just reporting a negative result, but those don't get published as much as they should, so here it is.

• CommentRowNumber36.
• CommentAuthorMike Shulman
• CommentTimeApr 29th 2014

That’s a good observation, thanks. So maybe we should ask, is integration of cogerm 2-forms on the plane necessarily continuous? Because if so, that means we can’t define such an exterior derivative there either.

While I’ve got you here, I have an unrelated question. When teaching multiple integration, what notation do you use for iterated integrals as in Fubini’s theorem? Always insisting on writing

$\int \Big(\int f(x,y)\, \mathrm{d}x \Big) \mathrm{d}y$

seems tedious and cumbersome, but omitting the parentheses and writing

$\int \int f(x,y)\, \mathrm{d}x \, \mathrm{d}y$

looks like we are integrating with respect to a product $\mathrm{d}x \, \mathrm{d}y$, which we are not (in particular, because $\int \int f(x,y)\, \mathrm{d}x \, \mathrm{d}y = \int \int f(x,y)\, \mathrm{d}y \, \mathrm{d}x$, both being equal to $\iint f(x,y) \,đA$).

• CommentRowNumber37.
• CommentAuthorTobyBartels
• CommentTimeApr 30th 2014

Well, I wouldn't write either of those, at least not until very late in the course when we all know what it means and I'm abbreviating. You're not specifying the region of integration, and that's key.

$\int_{y=a}^b \Big(\int_{x=g(y)}^{h(y)} f(x,y) \,\mathrm{d}x\Big) \,\mathrm{d}y$

but abbreviate this fairly early on as

$\int_a^b \int_{g(y)}^{h(y)} f(x,y) \,\mathrm{d}x \,\mathrm{d}y$

(which is what they write in the textbook). This is an iterated integral.

Then I introduce the separate concept

$\iint_{(x,y) \in R} f(x,y) \,\mathrm{d}x \,\mathrm{d}y ,$

which is defined (for bounded $R$) as the limit of Riemann sums, each given by a tagged rectangular mesh. This is pretty obviously equal to

$\iint_{(x,y) \in R} f(x,y) \,\mathrm{d}y \,\mathrm{d}x$

(with an obvious bijective correspondence between the relevant tagged meshes), and these can be abbreviated as

$\iint_R f(x,y) \,đA .$

This is a double integral. (I am not yet discussing what sort of thing $đA$ actually is, or even that the $\mathrm{d}x$ and $\mathrm{d}y$ in this expression are really ${|\mathrm{d}x|}$ and ${|\mathrm{d}y|}$. In fact, at this point I'm still writing ‘$\mathrm{d}A$’, like they do in the book. I don't really examine what kind of a thing this is until I get to change of variables. It's no secret that I'm going to discuss this, but I don't discuss it right away.)

Then I state (without proof, we don't have that kind of time) the Fubini Theorem as the statement that the iterated integral is equal to the obvious corresponding double integral (when $f$, $g$, and $h$ are continuous). In particular, if a region can be written as an iterated integral in both ways, then these iterated integrals are equal. Then further immediate corollaries, not explicitly stated in detail, about regions that can be divided into subregions amenable to iterated integrals. So the lesson is that we really care about double integrals, while iterated integrals are the method used to evaluate them analytically.

• CommentRowNumber38.
• CommentAuthorMike Shulman
• CommentTimeApr 30th 2014

Ok, but you do write $\mathrm{d}x \,\mathrm{d}y$ in both the double integral and the iterated integral.

• CommentRowNumber39.
• CommentAuthorMike Shulman
• CommentTimeApr 30th 2014
• CommentRowNumber40.
• CommentAuthorTobyBartels
• CommentTimeApr 30th 2014

Ok, but you do write $\mathrm{d}x \,\mathrm{d}y$ in both the double integral and the iterated integral.

Yes, but I make it clear (or try to) that the iterated integral is fundamentally the expression with parentheses, so that $\mathrm{d}x \,\mathrm{d}y$ is not a thing in it. It is a thing in the double integral, but I don't examine closely what that thing is until we have a little practical experience.

By the way, there is still a slight issue in the inner integral in the iterated integral. I do line integrals before area integrals (which the book also does but in the case of arclengths only), originally so that they'll have had some experience manipulating differential forms before I spring ${|\mathrm{d}x \wedge \mathrm{d}y|}$ on them when we get to change of variables. But another benefit is that I can now point out that

$\int_{g(y)}^{h(y)} f(x,y) \,\mathrm{d}x$

is really a line integral along a line of constant $y$-value.

• CommentRowNumber41.
• CommentAuthorMike Shulman
• CommentTimeMay 8th 2014

Have you read anything by Solomon Leader? I’m just having a look at his book The Kurzweil-Henstock integral and its differentials in which he has an interesting definition of differentials. He defines a summant to be a function on intervals tagged with an endpoint, which is basically to say tangent vectors, and then defines the integral of a summant basically just as I did for cogerm forms. Then he defines a differential to be an equivalence class of summants, where $S\sim T$ if $\int |S-T| = 0$. Then the differential of a function $f$ is the equivalence class $\mathrm{d}f$ of the summant $\Delta f$ which is defined by $\Delta f([a,b]) = f(b) - f(a)$; we could write that as $f(x+\mathrm{d}x)-f(x)$. If $f$ is differentiable, then $\mathrm{d}f = f'(x) \,\mathrm{d}x$ as differentials, i.e. their summants are equivalent – this is what contains the content of the fundamental theorem of calculus. But $\mathrm{d}f$ is defined for any function $f$, and its integrals are Riemann–Stieltjes integrals.

• CommentRowNumber42.
• CommentAuthorMike Shulman
• CommentTimeMay 13th 2014

I added some remarks about delta functions and Riemann-Stieltjes integrals to cogerm differential form.

• CommentRowNumber43.
• CommentAuthorTobyBartels
• CommentTimeMay 21st 2014

No, I've never heard of Solomon Leader. It looks like I should read him!

• CommentRowNumber44.
• CommentAuthorTobyBartels
• CommentTimeMay 4th 2015

I’m coming back to the idea of parametrization invariance, such as the parametrization invariance (or lack thereof) of the integration of higher-order1 forms such as $\sqrt{\mathrm{d}^{2}f}$. (Recall from the page that $\int_C \sqrt{\mathrm{d}^{2}f} = \int_{t \in \dom C} \sqrt{(f \circ C)''(t)} \,{|\mathrm{d}t|}$, at least when $f$ and $C$ are each twice continuously differentiable, for the so-called ‘genuine’ integral.)

It seems to me that parametrization invariance is a form of the Chain Rule, and we know that the Chain Rule can be tricky to apply to higher-order derivatives. I believe that the formula above for the integral of $\sqrt{\mathrm{d}^2f}$ is wrong in the same way that $f''(x) = \mathrm{d}^2f(x)/\mathrm{d}x^2$ is wrong. This formula seems to work under affine change of variables, just as the ‘genuine’ integral seems to work under affine reparametrization, but otherwise we can see that it fails.

In fact, knowing that $\mathrm{d}^2f(x) = f''(x) \,\mathrm{d}x^2 + f'(x) \,\mathrm{d}^2x$, I would calculate as follows:

$\int_C \sqrt{\mathrm{d}^2f} = \int_{t \in \dom C} \sqrt{\mathrm{d}^2f(C(t))} = \int_t \sqrt{\mathrm{d}^2(f \circ C)(t)} = \int_t \sqrt{(f \circ C)''(t) \,\mathrm{d}t^2 + (f \circ C)'(t) \,\mathrm{d}^2t} .$

If $\mathrm{d}^2t = 0$, then this simplifies to the result for ‘genuine’ integral, but not otherwise.

I'm convinced that this formula is correct, but unfortunately I don't know how to interpret it!

1. I have settled on the following language for order/degree/rank: The rank of a form indicates what dimension of submanifold it is integrated on; rank is increased by using the wedge product or applying the exterior differential, the cojet differential doesn't affect the rank, and the rank of a sum must agree with all addends. The degree indicates the power to which a positive scaling factor is raised when a form is applied to a small scaled (multi)-vector/curve; the degree is increased by multiplication or by applying the cojet differential, the exterior differential doesn't affect the degree, and the degree of a sum is the minimum degree of the addends. The order of a form indicates the highest order of derivatives of a curve that affect the value of the form at that curve; the order is increased by applying the coject differential, the exterior differential doesn't affect the order, and the order of a sum or product is the maximum order of the addends. For example, $\mathrm{d} \wedge {x \,\mathrm{d}y} = \mathrm{d}x \wedge \mathrm{d}y$ has rank and degree $2$ but order $1$, $\mathrm{d}x^2$ has degree $2$ but rank and order $1$, and $\mathrm{d}^2x$ has degree and order $2$ but rank $1$. In particular, $\sqrt{\mathrm{d}^2x}$ has order $2$ but rank and degree $1$

• CommentRowNumber45.
• CommentAuthorMike Shulman
• CommentTimeMay 5th 2015

Interesting — so how do we calculate $\int_t \sqrt{\mathrm{d}^2t}$?

• CommentRowNumber46.
• CommentAuthorMike Shulman
• CommentTimeMay 5th 2015

Hmm… can we extend the “affine integral” to act on forms of order $\gt 1$? If so, maybe that is actually the right definition?

• CommentRowNumber47.
• CommentAuthorMike Shulman
• CommentTimeMay 5th 2015

Shooting from the hip, what if we consider an order-2 rank-1 form to be a function $\omega:X\times V\times V\to\mathbb{R}$ and integrate it by adding up $\omega(c(t_i^*),\Delta c_i, \Delta^2 c_i)$, where $\Delta^2 c_i=c(t_{i+1}) - 2 c(t_i) + c(t_{i-1})$?

• CommentRowNumber48.
• CommentAuthorMike Shulman
• CommentTimeMay 5th 2015

On the other hand, according to the idea here, the cojet differential would be replaced by the symmetric part of the coflare differential, whose antisymmetric part is the exterior differential — all of which increase the rank. In that world, $\sqrt{\mathrm{d}^2 x}$ isn’t something we ought to think of integrating over a curve at all.

• CommentRowNumber49.
• CommentAuthorTobyBartels
• CommentTimeMay 5th 2015

Darn it, I knew that I was missing a thread! Now I have to read and contemplate that again. (I see that I was enthusiastic about it a year ago.)

• CommentRowNumber50.
• CommentAuthorTobyBartels
• CommentTimeMay 5th 2015

I thought of #47, but I also thought that it smacked of using only right or left endpoints (or, since its so symmetric, midpoints) in an ordinary Riemann sum. Since a tagged partition is a partition within a partition, I thought of using a partition within a partition within a partition, but then where would it end?

• CommentRowNumber51.
• CommentAuthorTobyBartels
• CommentTimeMay 5th 2015

The affine integral is bad, because it only works in affine spaces, but it also suggests the Stieltjes integral, which makes sense in more general contexts. Unfortunately, I don’t know how to make it work for an arbitrary cogerm (or even cojet) form given as an operation on curves, rather than given as an expression in symbols like $\mathrm{d}x$.

• CommentRowNumber52.
• CommentAuthorMike Shulman
• CommentTimeMay 5th 2015

If we do want to integrate $\sqrt{\mathrm{d}^2f}$ along a curve, then a basic question is what the integral $\int_{x=a}^b \sqrt{\mathrm{d}^2x}$ should be, where our curve is a simple affine increasing parametrization of $[a,b]$. I can’t think of anything sensible for it to equal except $0$, with the argument that this parametrization has “no second derivative”. If that’s the case, then to make the integral parametrization-invariant with your formula, then we’d have to have $\int_t \sqrt{x''(t) \,\mathrm{d}t^2 + x'(t) \,\mathrm{d}^2t} = 0$ for any such parametrization $x(t)$, so the two terms would have to exactly cancel somehow. I can sort of imagine how a second-order Taylor formula might cause that sum to reduce to something third-degree so that its square root would be greater than first-degree and hence integrate to zero. But it seems that the same argument would suggest that the integral of $\sqrt[3]{\mathrm{d}^2x}$ should also be zero, and likewise $\sqrt[4]{\mathrm{d}^2x}$ and so on, and it seems less and less likely to me that in those cases we could make the two terms in the change-of-variables formula cancel.

• CommentRowNumber53.
• CommentAuthorTobyBartels
• CommentTimeMay 5th 2015
• (edited May 5th 2015)

Another thought that I had about affine integration is that if we really want to use it, but balk because we don't know what ‘affine’ means in general, then perhaps we can only integrate generally in the presence of an affine connection (not much there yet, so press on to Wikipedia), which tells us what is affine.

Since I first learnt differential geometry through general relativity, I learnt about the exterior differential as the antisymmetrization of the covariant derivative, which uses the affine connection and applies to any tensor. It's then a great theorem that the action of this antisymmetrization on an antisymmetric contravariant tensor (which is an exterior form) doesn't depend on which connection is used. Here, we have a notion of differential that can be applied without a connection, but maybe we still need a connection to integrate the resulting forms. But of course the result won't depend on the connection, as long as we integrate only exterior forms.

Edit: It seems that we were making the same point a year ago in the thread that I forgot.

• CommentRowNumber54.
• CommentAuthorMike Shulman
• CommentTimeMay 5th 2015

I would have thought that the affine integral could be applied on any manifold by using local charts. Do you think that’s not the case?

• CommentRowNumber55.
• CommentAuthorTobyBartels
• CommentTimeMay 6th 2015

I thought that you were suggesting on the page that it wasn't. It comes down to whether different local charts give the same results; I haven't checked.

• CommentRowNumber56.
• CommentAuthorMike Shulman
• CommentTimeMay 6th 2015

I think it does, at least in the Lipschitz case; I tried to write a proof on the page (end of the section)

• CommentRowNumber57.
• CommentAuthorTobyBartels
• CommentTimeMay 6th 2015

OK, I'll buy it in the Lipschitz case.

I used the affine integral to define integration on a curve in my Calc 3 class the other day, even though I'm not using it to prove any theorems beyond hand-waving. I defined the mesh of a Riemann sum to be the maximum distance between points in space rather the maximum difference between values of the parameter, mainly because that was easier to point to in a diagram. But now I realize that you did it the other way (implicitly in the phrase ‘as before’).

It seems to me that the slick proof that the affine integral is parametrization-dependent requires my mesh to go through, rather than yours. Of course, for Lipschitz forms, it works either way; everything is better with Lipschitz!

• CommentRowNumber58.
• CommentAuthorMike Shulman
• CommentTimeMay 6th 2015

Hmm, you may be right.

• CommentRowNumber59.
• CommentAuthorMike Shulman
• CommentTimeMay 7th 2015

Hmm $\times 2$, it seems to me that what’s needed to make the two meshes coincide is continuity of the curve, not the form.

• CommentRowNumber60.
• CommentAuthorTobyBartels
• CommentTimeMay 7th 2015

Yes, I was just thinking about that. As long as the curve is uniformly continuous, it should be all right, at least for the Riemann integral.

• CommentRowNumber61.
• CommentAuthorMike Shulman
• CommentTimeMay 21st 2015

Toby, did you ever reach a conclusion about the argument in #8 above?

• CommentRowNumber62.
• CommentAuthorTobyBartels
• CommentTimeMay 21st 2015

If $\eta$ is determined up to sign by $\eta^2$ and $\omega$ is determined up to sign by $\omega^2$, then $(\eta \wedge \omega)^2$ may still be determined by $\eta^2$ and $\omega^2$, since the signs in $\eta^2(v) \omega^2(w) + \eta^2(w) \omega^2(v) - 2 \eta(v) \omega(w) \eta(w) \omega(v)$ will cancel. However, that does seem to suggest an odd formula for the wedge product between $2$-forms.

• CommentRowNumber63.
• CommentAuthorMike Shulman
• CommentTimeMay 21st 2015

Hmm. What’s the odd formula that it suggests?

• CommentRowNumber64.
• CommentAuthorTobyBartels
• CommentTimeMay 21st 2015

Something like

$(\alpha \wedge \beta)(v,w) = \alpha(v) \beta(w) + \alpha(w) \beta(v) - 2 \sqrt{\alpha(v) \beta(w) \alpha(w) \beta(v)} = \Big(\sqrt{\alpha(v)\beta(w)} - \sqrt{\alpha(w)\beta(v)}\Big)^2,$

which doesn't entirely make sense.

But I don't really believe this anyway. Now that we're looking at coflare forms, this set-up isn't even relevant; we should be applying $(\eta \wedge \omega)^2$ to a $4$-flare, not to two tangent vectors derived from a parametrized surface. However, I'm finding it hard to think through all of that.

• CommentRowNumber65.
• CommentAuthorMike Shulman
• CommentTimeMay 22nd 2015

That seems right if by $\eta^2$ we mean $\eta\otimes \eta$, but if we go back to the motivating #5 with coflares, it seems that $\eta^2$ has to be the valuewise square (keeping the rank constant) since we want it to match with a square root?

• CommentRowNumber66.
• CommentAuthorMike Shulman
• CommentTimeMay 22nd 2015

Here’s a sort of exterior product for coflare forms that at least comes close to giving the right answer “$vol_g = \sqrt{\frac{g\wedge g}{2}}$” for a metric on a 2-manifold.

Suppose $\eta$ and $\omega$ are both coflare forms of rank $n$. Then $\eta\omega$ (meaning $\eta\otimes \omega$) is a coflare form of rank $2n$. Note that the symmetric group $\Sigma_k$ acts on coflare forms of rank $k$, because it acts on $k$-flares. Let $G_{2n}$ be the subgroup of $\Sigma_{2n}$ generated by transpositions of the form $(i,i+n)$; thus for instance $G_4 = \{e, (13),(24),(13)(24) \}$. Observe that $G_{2n} \cong (\Sigma_2)^n$. Now define

$\eta\wedge\omega = \sum_{\sigma\in G_{2n}} (-1)^{sign(\sigma)} (\eta\omega)^\sigma.$

(Hmm, I suppose there should probably be a factorial coefficient in front of that.) The point is that at least if $\eta$ and $\omega$ depend only on the first-order tangent vectors in a flare, then $\sigma\in G_{2n}$ acts on $\eta\omega$ by swapping corresponding arguments of $\eta$ and $\omega$. (I find it easier to think here of $2n$ as a $2\times n$ matrix rather than $2n$ things in a row.)

When $n=1$, this clearly gives the correct wedge product for exterior 1-forms. More generally, I think it also gives the right answer when $\eta$ and $\omega$ are exterior $n$-forms: in that case $\eta\omega$ is already antisymmetric under the two actions of $\Sigma_n$, so antisymmetrizing under $G_{2n}$ forces it to be antisymmetric under all of $\Sigma_{2n}$.

But now if we have a metric regarded as a coflare 2-form:

$g = A dx^2 + B dx dy + B dy dx + C dy^2$

we can compute its exterior square as follows. By construction, $(\eta\wedge\omega)^\sigma = (-1)^{sign(\sigma)} (\eta\wedge\omega)$ for $\sigma\in G_{2n}$. In particular, $(dx^i dx^j)\wedge (dx^k dx^l) = 0$ if $i=k$ or $j=l$. Thus all the terms in $g\wedge g$ vanish except for

$A C (dx^2\wedge dy^2 + dy^2 \wedge dx^2) + B^2 (dx dy \wedge dy dx + dy dx \wedge dx dy).$

Now because $(13)(24)\in G_4$ and is even, we have $dy^2\wedge dx^2 = dx^2 \wedge dy^2$ and $dy dx \wedge dx dy = dx dy \wedge dy dx$. Finally, because $(13)\in G_4$ and is odd, we have $dx dy \wedge dy dx = - dx^2 \wedge dy^2$. Thus,

$g\wedge g = 2(A C-B^2) (dx^2 \wedge dy^2).$

Note that by definition we have

$dx^2 \wedge dy^2 = dx dx dy dy - dy dx dx dy - dx dy dy dx + dy dy dx dx.$

On the other hand, we have

$(dx\wedge dy)^2 = (dx dy - dy dx)^2 = dx dy dx dy - dx dy dy dx - dy dx dx dy + dy dx dy dx.$

These differ exactly by the action of $(23)\in \Sigma_4$, which is a “transposition” of the “$2\times 2$ matrix” representing the arguments of these coflare 4-forms. Therefore, it seems that if we

1. Construct $\sqrt{\frac{g\wedge g}{2}}$,
2. Transpose its arguments, and
3. Apply a “diagonal” to make it into a 2-form rather than a 4-form, so that e.g. the rank-2 form I’ve been writing as $dx^2$ but is really $dx\otimes dx$ becomes the rank-1 form that’s more honestly written $dx^2$.

we should get the desired $vol_g = \sqrt{det(g)}\; {| dx \wedge dy |}$.

However, right now I don’t see any way to define that diagonal in a coordinate-invariant way. (If we could solve that problem, I expect the weird-looking transposition would get incorporated into the choice of one particular “map $2\to 4$” rather than another one.)

Moreover, this only defines the wedge product of two forms of the same rank, which seems clearly unsatisfactory. I haven’t thought yet about whether it generalizes to, for instance, 3-manifolds, where we’d have to find some way to define $g\wedge g\wedge g$.

• CommentRowNumber67.
• CommentAuthorMike Shulman
• CommentTimeMay 22nd 2015

Hmm, well the diagonal (combined with transposition) does make sense coordinate-invariantly for order-1 forms, which are all that we have here. But it would be a bit disappointing to have to use in this construction an operation that doesn’t make sense in generality.

Another possibility that occurrs to me is that as we noted when I first suggested coflare forms, an affine connection is a section of the projection $T^2X \to (T X)^2$, thereby giving a way to “forget” from an arbitrary rank-2 (hence order $\le 2$) coflare form down to a rank-2 order-1 one. And we know that a metric gives rise to an affine connection. Perhaps a connection can be generalized to give a way to forget the higher-order terms in a coflare form of arbitrary rank? Performing such an operation first in order to make a “diagonal” make sense seems reasonable especially if our eventual goal is to integrate the result, since (“genuine”) integrals don’t generally notice higher-order terms.

• CommentRowNumber68.
• CommentAuthorTobyBartels
• CommentTimeMay 25th 2015

An affine connection by itself won't allow us to collapse the higher-order parts of a coflare form, which is clear from your explicit formula in the other thread for the collapsing operation:

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

As you say there, the transformation rules for Christoffel symbols ensure that this is invariant, which they can do since those transformation rules involve second derivatives. But nothing could handle $\mathrm{d}^3x$ unless its transformation rules involve third derivatives, which Christoffel symbols' rules don’t. And an affine connection is determined by its Christoffel symbols.

However, a metric should be plenty of information! For simplicity, let's take a Riemannian metric, so as not to worry about strange sign conventions in the semiRiemannian case. In that case, you simply pick (at any given point) a system of coordinates that's orthonormal relative to the metric, in which case $\Gamma^i_{j k}$ is simply $\delta^i_j \delta^i_k$, so $\mathrm{d}^2x^i \mapsto \mathrm{d}x^i \otimes \mathrm{d}x^i$ (no summation). And this generalizes to any order:

$\mathrm{d}^{n}x^i \mapsto \bigotimes_n \mathrm{d}x^i .$

Now all that we have to do is to obfuscate this by identifying what this looks like in an arbitrary system of coordinates.

• CommentRowNumber69.
• CommentAuthorTobyBartels
• CommentTimeMay 25th 2015

Actually, the previous comment is simpler than it should be, because it tacitly assumes that the system of coordinates is orthnormal not just at the point in question but on an infinitesimal neighbourhood of it, which may not be possible.

• CommentRowNumber70.
• CommentAuthorTobyBartels
• CommentTimeMay 25th 2015

Having said all of that, however, I don't think that the Levi-Civita connection can be relevant. The volume element at a given point depends only on the metric at that point, whereas the Levi-Civita connection there also involves the derivatives of the metric.

• CommentRowNumber71.
• CommentAuthorMike Shulman
• CommentTimeMay 26th 2015

An affine connection by itself won't allow us to collapse the higher-order parts of a coflare form

Right, I expected as much; I was thinking along the lines of your second paragraph, whether a metric would also give rise to a sort of “higher connection”. I wonder if this is something someone else has written down?

I don’t think that the Levi-Civita connection can be relevant. The volume element at a given point depends only on the metric at that point, whereas the Levi-Civita connection there also involves the derivatives of the metric.

The connection (or higher connection) isn’t going to arise in the volume element itself. I’m only proposing the connection as a way to make the “diagonal” operation make sense for arbitrary coflare forms. But in defining the volume element, we only need to take diagonals of order-1 forms, and in that case the diagonal already makes sense; the connection would only arise when deciding what the diagonal does to things like $d^2x$.

• CommentRowNumber72.
• CommentAuthorMike Shulman
• CommentTimeJun 8th 2015

The proposal in #66 does generalize to “$vol_g = \sqrt{\frac{1}{n!} g^{\wedge n}}$” on an $n$-manifold (BUT see my next comment for an important caveat).

In general, given $m$ coflare forms $\eta_1,\dots,\eta_m$ all of rank $k$, let $G_{m,k}$ be the subgroup of $\Sigma_{m k}$ generated by transpositions of the form $(i,i+k)$, which is isomorphic to $(\Sigma_m)^k$. Define

$\eta_1 \wedge\cdots \wedge \eta_m = \sum_{\sigma \in G_{m,k}} (-1)^{sign(\sigma)} (\eta_1 \otimes\cdots \otimes \eta_m)^\sigma.$

Now let $g = \sum_{1\le i,j\le n} g_{i,j} dx^i \otimes dx^j$ be a metric regarded as a coflare form of rank 2. Each term in $g^{\wedge n}$ looks like

$\bigwedge_{\ell =1}^{n} g_{i_\ell, j_\ell} dx^{i_\ell} \otimes dx^{j_\ell}.$

By construction, this vanishes if $i_\ell = i_{\ell'}$ or $j_\ell = j_{\ell'}$ for any $\ell\neq \ell'$. Thus, $i_1,\dots,i_n$ and $j_1,\dots,j_n$ are each permutations of $1,\dots,n$. We can permute the $i$s and $j$s together without introducing any signs, so we may as well assume that $i_\ell = \ell$ for all $\ell$. Thus we have a sum over all possible permutations $j_1,\dots,j_n$, with each permutation occurring $n!$ times:

$g^{\wedge n} = n! \, \sum_{\sigma \in \Sigma_n} \bigwedge_{\ell=1}^{n} g_{\ell, \sigma(\ell)} dx^{\ell} \otimes dx^{\sigma(\ell)}.$

Now if we permute the $j$s back to the identity, we introduce the sign of that permutation:

$g^{\wedge n} = n! \, \sum_{\sigma \in \Sigma_n} (-1)^{sign(\sigma)} \bigwedge_{\ell=1}^{n} g_{\ell, \sigma(\ell)} dx^{\ell} \otimes dx^{\ell}.$

i.e.

$g^{\wedge n} = n! \, det(g) \bigwedge_{\ell=1}^{n} dx^{\ell} \otimes dx^{\ell}.$

Now there is a diagonal operation from rank-$2n$ forms to rank-$n$ forms (at least on order-1 forms like these, and presumably extendable with a metric to arbitrary ones in a way that reduces to the obvious one on order-1 forms) that takes this to

$n! \, det(g) \left(\bigwedge_{\ell=1}^{n} dx^{\ell}\right)^2$

If we now divide by $n!$ and take the square root, we obtain the correct volume element, as an absolute rank-$n$ coflare form.

More generally, I bet that for $k\lt n$ a similar construction $\sqrt{\frac{g^{\wedge k}}{k!}}$ will produce the standard $k$-volume element in an $n$-manifold. For $k=1$ it of course gives the line element $\sqrt{g}$, while for $k=2$, $n=3$ and the standard metric $dx^2+dy^2+dz^2$ on $\mathbb{R}^3$ I think we do get the right answer $\sqrt{(\dx\wedge \dy)^2 + (dy\wedge \dz)^2 + (\dz\wedge \dx)^2}$.

• CommentRowNumber73.
• CommentAuthorMike Shulman
• CommentTimeJun 8th 2015

BUT I was wrong in #66 that this definition of $\wedge$ (and its generalization in #72) is correct for exterior forms! I think I thought that because $G_{2n}$ together with $\Sigma_n \times\Sigma_n$ generates $\Sigma_{2n}$, but that’s not enough. E.g. for exterior 2-forms $\eta$ and $\omega$ this definition would give

$(\eta\wedge\omega) (v_1,v_2,v_3,v_4) = \eta(v_1,v_2)\omega(v_3,v_4) - \eta(v_3,v_2)\omega(v_1,v_4) - \eta(v_1,v_4)\omega(v_3,v_2) + \eta(v_3,v_4) \omega(v_1,v_2)$

whereas the correct formula is something like

$(\eta\wedge\omega) (v_1,v_2,v_3,v_4) = \eta(v_1,v_2)\omega(v_3,v_4) - \eta(v_1,v_3)\omega(v_2,v_4) + \eta(v_1,v_4)\omega(v_2,v_3) + \eta(v_2,v_3)\omega(v_1,v_4) - \eta(v_2,v_4)\omega(v_1,v_3) + \eta(v_3,v_4)\omega(v_1,v_2)$

So now I don’t know what to do.

• CommentRowNumber74.
• CommentAuthorTobyBartels
• CommentTimeMay 21st 2018

First mention of coflare forms on the Lab rather than the Forum.

• CommentRowNumber75.
• CommentAuthorTobyBartels
• CommentTimeDec 15th 2019
• (edited Dec 15th 2019)

I don't know where it's written down, but somewhere somebody (probably Mike) suggested that coflare forms of rank $p$ could be seen as functions on germs of maps from $\mathbb{R}^p$. So I put this in here, in the section near the end where coflare forms are mentioned. (This gives a notion of coflare-cogerm form that includes all cogerm forms, not just cojet forms, as well as all coflare forms, thus including exterior forms, absolute forms, and even twisted forms with enough care. And why restrict the domains to only $\mathbb{R}^p$ …?)

• CommentRowNumber76.
• CommentAuthorTobyBartels
• CommentTimeNov 4th 2020

My current opinion about how to apply $\wedge$ (and the symmetric version as well) is as follows: To multiply a $p$-form $\alpha$ and a $q$-form $\beta$, take all permutations on $p+q$ letters that keep the first $p$ in order and the last $q$ in order. So for example, if $p$ and $q$ are both $2$, then there are $6$ permitted permutations: $1234$, $1324$, $1423$, $2314$, $2413$, and $3412$. The first $p$ letters tell you which vectors to apply $\alpha$ to, while the last $q$ tell you which to apply $\beta$ too. If you're multiplying antisymmetrically, multiply by the sign of the permutation. Add up, and (if you're following this convention) divide by the number of permutations that you used. (Thus the symmetrized product is an average.)

So, if $\eta$ and $\omega$ are exterior $2$-forms, then $\eta \wedge \omega$ is a $4$-form:

$(\eta \wedge \omega)(v_1,v_2,v_3,v_4) = \frac 1 6 \eta(v_1,v_2) \omega(v_3,v_4) - \frac 1 6 \eta(v_1,v_3) \omega(v_2,v_4) + \frac 1 6 \eta(v_1,v_4) \omega(v_2,v_3) + \frac 1 6 \eta(v_2,v_3) \omega(v_1,v_4) - \frac 1 6 \eta(v_2,v_4) \omega(v_1,v_3) + \frac 1 6 \eta(v_3,v_4) \omega(v_1,v_2) ,$

which is exactly what Mike wanted in the previous comment. Notice that this is multilinear (since $\eta$ and $\omega$ are) and alternating (since $\eta$ and $\omega$ are). The usual textbook way to define the exterior product of exterior forms would give $24$ terms (and divide by $4! = 24$ or $2!\,2! = 4$, depending on convention), not just $6$, but if $\omega$ and $\eta$ are already alternating, then these come in groups of $4$ equal terms, so the result is the same. But some textbooks save on terms by defining things as I did (see this definition on Wikipedia for example, at least currently).

If $\eta$ and $\omega$ are not already alternating, then neither should $\eta \wedge \omega$ be, and the definition with $(p q)!$ terms will incorrectly make it alternating. But the definition with $(p q)!/(p!\,q!)$ terms will not. And nothing stops this from being applied to forms that aren't multilinear or always defined either.

Also note that this operation is associative. The generalized version says that to multiply a list of $m$ forms of ranks $p_1, \ldots, p_m$, you look at the $(\sum_i p_i)!/\prod_i p_i!$ permutations on $\sum_i p_i$ letters that keep the first $p_1$ letters, the next $p_2$ letters, etc through the last $p_n$ letters in order, apply the forms to the vectors given by the appropriate indexes, multiply by the sign of the permutation if you're multiplying antisymmetrically, add them up, and (optionally) divide by the number of terms.

Now to apply this to the metric $g$. There are two ways to think of $g$, as a bilinear symmetric form of rank $2$, or as a quadratic form of rank $1$. In local coordinates on a surface parametrized by $u$ and $v$ for example (Gauss's first fundamental form), the first version is $E \,\mathrm{d}u \otimes \mathrm{d}u + F \,\mathrm{d}u \otimes \mathrm{d}v + F \,\mathrm{d}v \otimes \mathrm{d}u + G \,\mathrm{d}v \otimes \mathrm{d}v$, while the second is $E \,\mathrm{d}u^2 + 2F \,\mathrm{d}u \,\mathrm{d}v + G \,\mathrm{d}v^2$.

Taking the first version, $g \wedge g$ is a $96$-term expression that simplifies (after much cancellation) to $\frac 1 3 E^2 \,\mathrm{d}u \otimes \mathrm{d}u \otimes \mathrm{d}u \otimes \mathrm{d}u + \cdots$, and I'm going to stop writing it down after (what came from) the first $6$ terms, because these are all of the ones with $E^2$, and those should have cancelled completely (not just partially as happened here). The unbalanced nature of the signs has ruined this. Taking the second version, $g \wedge g$ is an $18$-term expression that simplifies all the way to $0$; in fact, $\alpha \wedge \alpha = 0$ whenever $\alpha$ is a $1$-form, which is well-known when $\alpha$ is linear but true regardless.

So neither of these is working out! (For the record, the answer that we were looking for here is $E G - F^2$, possibly with a constant factor to worry about later.)

1. [Administrative note: I have now merged a thread entitled ’Cogerm forms’ with this one. Comments 1. - 73. and 76. are from this old ’Cogerm forms’ thread.]

• CommentRowNumber78.
• CommentAuthorTobyBartels
• CommentTimeNov 9th 2020

Richard, why does this have to be done? Although the discussion of this topic is spread out, it is all linked from the bottom of the nLab page. One of those links (to https://nforum.ncatlab.org/discussion/5700/cogerm-forms/) is now broken, and while it can simply be removed now, how do we know whether anybody has put that link anywhere else? As much as possible, reorganization should not break external links.

• CommentRowNumber79.
• CommentAuthorRichard Williamson
• CommentTimeNov 10th 2020
• (edited Nov 10th 2020)

We have been taking the Latest Changes pages as canonical for page discussion for a number of years, and doing this kind of merging of older threads into the Latest Changes threads where appropriate. Good that you noticed the link, yes, it would be good to remove it or update it. I’ll not enter into a debate about the possibility of breaking other links; it might happen, yes, but the probability of any significant negative consequences is tiny, and outweighed by the benefits of having a single, canonical page to find discussion in my opinion. E.g. it is very unlikely that I or many others would find the links to the nForum discussions when glancing at cogerm differential form, as indeed I did not in this case, but everybody can understand ’Discuss this page’ and know what to expect upon clicking upon it.

• CommentRowNumber80.
• CommentAuthorTobyBartels
• CommentTimeDec 17th 2020

There's a difference between a thread about a topic and a thread about a page (although that distinction is not made in the early comments in the merged thread). Is it possible to set up the server so that old links will redirect when merging? (That still breaks so-called PermaLinks to individual comments, but at least people will be on the correct page.)

• CommentRowNumber81.
• CommentAuthorDavidRoberts
• CommentTimeJan 28th 2021

Typo in a section title fixed