Typo in a section title fixed

]]>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.)

]]>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.

]]>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.

]]>[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.]

]]>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.)

]]>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 $\mathbb{R}^p$ …?)

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

]]>**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

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.

]]>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 $\sqrt{(\dx\wedge \dy)^2 + (dy\wedge \dz)^2 + (\dz\wedge \dx)^2}$.

]]>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 $d^2x$.

]]>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.

]]>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.

]]>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.

]]>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.

]]>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

- Construct $\sqrt{\frac{g\wedge g}{2}}$,
- Transpose its arguments, and
- 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 $g\wedge g\wedge g$.

]]>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?

]]>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.

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

]]>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.

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

]]>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.

]]>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.

Hmm, you may be right.

]]>