Eric Forgy
Modified Wedge Product

Warning: Speculative material under construction.

The following material began as a post on the nCafe…

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Initial Thought

Given a manifold $M$ and differential forms $\Omega (M)$ on $M$, we can construct a simplicial complex $S$ with cochains $C^{*}(S)$. The de Rham map $R$ takes forms on $M$ and turns them into cochains on $S$

$$R:\Omega (M)\to C^{*}(S).$$R:\Omega(M)\to C^*(S).

The Whitney map ($W$) takes cochains on $S$ and turns them into (Whitney) forms on $M$

$$W:C^{*}(S)\to \Omega (M).$$W:C^*(S)\to\Omega(M).

We have

$$R\circ W={\mathrm{Id}}_{C^{*}(S)}$$R\circ W=Id_{C^*(S)}

and

$$W\circ R\sim {\mathrm{Id}}_{\Omega (M)}.$$W\circ R \sim Id_{\Omega(M)}.

The wedge product in $\Omega (M)$ is graded commutative and the cup product in $C^{*}(S)$ is not graded commutative “on the nose” so all these maps do not quite fit together perfectly, e.g.

$$W(a⌣b)\ne W(a)\wedge W(b)$$W(a\smile b)\ne W(a)\wedge W(b)

and

$$R(\alpha \wedge \beta )\ne R(\alpha )⌣R(\beta ).$$R(\alpha\wedge\beta)\ne R(\alpha)\smile R(\beta).

Some people have proposed a modified cup product for computational physics via

$$a\widetilde{⌣}b≔R(W(a)\wedge W(b)).$$a\tilde\smile b \coloneqq R(W(a)\wedge W(b)).

I don’t think this helps much with the “cochain problem”. In particular, this modified cup product is not even associative!

I have proposed an alternative, which at first will seem radical but I actually think might help with things like the “cochain problem”. That is to introduce a modified wedge product

$$\alpha \tilde{\wedge }\beta ≔W(R(\alpha )⌣R(\beta )).$$\alpha\tilde\wedge\beta\coloneqq W(R(\alpha)\smile R(\beta)).

This modified wedge product has the (uber) nice property that it is an algebra homomorphism “on the nose”, i.e.

$$R(\alpha \tilde{\wedge }\beta )=R(\alpha )⌣R(\beta ).$$R(\alpha\tilde\wedge\beta)=R(\alpha)\smile R(\beta).

The undesirable property of this modified wedge product is that it will depend on triangulation $S$, but that dependence disappears when you pass to cohomology, homotopy, etc, which is what the mathematicians really care about.

My gut tells me that having a true algebra morphism like this will give you true functors and the category theoretic analysis will be much cleaner. But that is just a hunch.

Also note that in a suitable limit of refinements of S, i.e. a kind of “continuum limit” we have

$$\alpha \wedge \beta =\underset{\mathrm{continuum}}{\lim }\alpha \tilde{\wedge }\beta .$$\alpha\wedge\beta = \lim_{continuum} \alpha\tilde\wedge\beta.

Unit Elements

Both the Whitney map and de Rham map preserve unit elements, i.e.

$$W(1_{C^{*}(S)})=1_{\Omega (M)}$$W(1_{C^*(S)}) = 1_{\Omega(M)}

and

$$R(1_{\Omega (M)})=1_{C^{*}(S)}.$$R(1_{\Omega(M)}) = 1_{C^*(S)}.

Projection from Smooth Forms to Whitney Forms

Given a differential form $\alpha$, let $\tilde{\alpha }$ denote the corresponding Whitney form given by

$$\tilde{\alpha }=W\circ R(\alpha ).$$\tilde\alpha = W\circ R(\alpha).

The Whitney form $\tilde{\alpha }$ may be thought of as a piecewise linear approximation to the smooth differential form $\alpha$.

The map

$$\pi ≔W\circ R$$\pi\coloneqq W\circ R

is a projection from smooth forms to Whitney forms, which can be seen by expanding

$$W\circ R(\tilde{\alpha })=W\circ \stackrel{=1}{(R\circ W)}\circ R(\alpha )=\tilde{\alpha }$$W\circ R(\tilde\alpha) = W\circ \stackrel{=1}{(R\circ W)}\circ R(\alpha) = \tilde\alpha

and noting that $R\circ W=1$ so that

$$\pi \circ \pi =\pi .$$\pi\circ\pi = \pi.

Algebra of Whitney Forms

Now, I have proposed above using this modified wedge product on smooth forms, but that might not be the right thing to do. For example, consider the modified wedge product of a smooth form $\alpha$ and the unit 0-form 1

$$\alpha \tilde{\wedge }1=W(R(\alpha )⌣R(1))=\pi (\alpha )=\tilde{\alpha }.$$\alpha\tilde\wedge 1 = W(R(\alpha)\smile R(1)) = \pi(\alpha) = \tilde\alpha.

However,

$$\tilde{\alpha }\tilde{\wedge }1=\pi (\tilde{\alpha })=\tilde{\alpha }.$$\tilde\alpha\tilde\wedge 1 = \pi(\tilde\alpha) = \tilde\alpha.

Therefore, what I have really done is define an algebra of Whitney forms.

This is significant because Whitney forms are not closed under the ordinary wedge product, i.e.

$$\tilde{\alpha }\wedge \tilde{\beta }$$\tilde\alpha\wedge\tilde\beta

is not a Whitney form (which always bugged me). However

$$\tilde{\alpha }\tilde{\wedge }\tilde{\beta }$$\tilde\alpha\tilde\wedge\tilde\beta

is a Whitney form.

Exterior Derivative and Cohomology

The exterior derivative behaves nicely with respect to both the Whitney and de Rham maps, i.e.

$$d\circ W=W\circ d$$d\circ W = W\circ d

and

$$d\circ R=R\circ d.$$d\circ R = R\circ d.

In particular, this means that

$$d\tilde{\alpha }=d[W\circ R(\alpha )]=W\circ R(d\alpha )=\widetilde{d\alpha }$$d\tilde\alpha = d\left[W\circ R(\alpha)\right] = W\circ R(d\alpha) = \widetilde{d\alpha}

so the exterior derivative of a Whitney form is a Whitney form.

For smooth forms $\alpha$ and $\beta$, we have the Leibniz rule

$$d(\alpha \wedge \beta )=(d\alpha )\wedge \beta +(-1{)}^{\mid \alpha \mid }\alpha \wedge (d\beta ).$$d(\alpha\wedge\beta) = (d\alpha)\wedge\beta + (-1)^{|\alpha|} \alpha\wedge(d\beta).

For cochains $a$ and $b$, we have the Leibniz rule

$$d(a⌣b)=(da)⌣b+(-1{)}^{\mid a\mid }a⌣(db).$$d(a\smile b) = (d a)\smile b + (-1)^{|a|} a\smile(d b).

The Leibniz rule is important for many reasons, not least of them being that it preserves cohomological properties.

Therefore, it is important to note that Whitney forms together with the modified wedge product also satisfy the Leibniz rule

$$\begin{array}{rl}d(\tilde{\alpha }\tilde{\wedge }\tilde{\beta })& =d[W(R(\tilde{\alpha })⌣R(\tilde{\beta })]\\ & =W[R(d\tilde{\alpha })⌣R(\tilde{\beta })+(-1{)}^{\mid \tilde{\alpha }\mid }R(\tilde{\alpha })⌣R(d\tilde{\beta })]\\ & =(d\tilde{\alpha })\tilde{\wedge }\tilde{\beta }+(-1{)}^{\mid \tilde{\alpha }\mid }\tilde{\alpha }\tilde{\wedge }(d\tilde{\beta }).\end{array}$$\begin{aligned} d(\tilde\alpha\tilde\wedge\tilde\beta) &= d\left[W(R(\tilde\alpha)\smile R(\tilde\beta)\right] \\ &= W\left[R(d\tilde\alpha) \smile R(\tilde\beta)+(-1)^{|\tilde\alpha|} R(\tilde\alpha)\smile R(d\tilde\beta)\right] \\ &= (d\tilde\alpha)\tilde\wedge\tilde\beta + (-1)^{|\tilde\alpha|} \tilde\alpha\tilde\wedge (d\tilde\beta). \end{aligned}

Differential Graded Non-Commutative Algebra

All this taken together means that on a manifold $M$ we have a full fledged graded differential algebra of Whitney forms

$$\tilde{\Omega }(M)=\underset{r=0}{\overset{n}{⨁}}{\tilde{\Omega }}^r(M)$$\tilde\Omega(M) = \bigoplus_{r=0}^n \tilde\Omega^r(M)

where ${\tilde{\Omega }}^r(M)$ is the space of Whitney $r$-forms on $M$. We have an exterior derivative

$$d:{\tilde{\Omega }}^r(M)\to {\tilde{\Omega }}^{r+1}(M)$$d:\tilde\Omega^r(M)\to\tilde\Omega^{r+1}(M)

that is nilpotent ($d^2=0$) and satisfies the graded Leibniz rule

$$d(\tilde{\alpha }\tilde{\wedge }\tilde{\beta })=(d\tilde{\alpha })\tilde{\wedge }\tilde{\beta }+(-1{)}^{\mid \tilde{\alpha }\mid }\tilde{\alpha }\tilde{\wedge }(d\tilde{\beta }).$$d(\tilde\alpha\tilde\wedge\tilde\beta) = (d\tilde\alpha)\tilde\wedge\tilde\beta + (-1)^{|\tilde\alpha|} \tilde\alpha\tilde\wedge(d\tilde\beta).

However, the modified wedge product of Whitney forms is not graded commutative, i.e.

$$\tilde{\alpha }\tilde{\wedge }\beta \ne (-1{)}^{\mid \tilde{\alpha }\mid \mid \tilde{\beta }\mid }\tilde{\beta }\tilde{\wedge }\tilde{\alpha }.$$\tilde\alpha\tilde\wedge\beta \ne (-1)^{|\tilde\alpha||\tilde\beta|} \tilde\beta\tilde\wedge\tilde\alpha.

This failure to be graded commutative is a feature (and not a bug that) leads to all kinds of fascinating concepts related to, among other things, quantum mechanics and noncommutative geometry.

Cochain Problem

To be continued…

References

• The search for discrete differential geometry (String Coffee Table)
• Gugenheim, On the multiplication structure of the de Rham theory
• Forgy & Schreiber, Discrete differential geometry on causal graphs
• Baez, This Week’s Finds in Mathematical Physics (Week 288)