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To make the problem simple, consider an undirected graph $G$ with vertices $V$ and edges $E$.

A discrete "scalar function" (0-cochain) can be defined on the vertices, taking values $f(v)$ on each vertex $v$.

A discrete "1-form" (1-cochain) can be defined as an anti-symmetric function on pairs of vertices $u(v_1,v_2) = -u(v_2,v_1)$ whose value is 0 if $v_1$ and $v_2$ are not connected by an edge. We can naturally think of the non-zero values of $u$ as being encoded on the edges of the graph supplemented with an orientation of each bond, i.e. $u$ is essentially a weighted, directed version of $G$. For example, by choosing an orientation of each bond one could define $u$ to be the finite difference gradient of $f$.

For concreteness, I am considering $f$ and $u$ to be real or complex valued.

A lot of intuitive ideas from differential geometry can be brought over (and generalized further if we consider for example simplicial complexes instead of graphs), like boundary/coboundary etc. My question is, is there a natural analog of $C^\infty$ multiplication in the discrete case? I don't see an obvious way to achieve this because the scalar functions are defined on vertices while the discrete 1-cochains are defined on the edges, so we can't just multiply them pointwise like we can on a manifold.

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This somewhat depends on what features of the module structure $\Omega^0M\times\Omega^1M\to\Omega^1M$ you're interested in replicating. One common choice is the cup product, which replicates the properties of the wedge product of differential forms in simplicial cohomology. It is defined in various ways in various contexts, and oftentimes one is only interested in the induced product on cohomology. Using your definition of simplicial cochains, we could, for instance, define $$ (a^p\smile b^q)(v_0,\cdots,v_{p+q}) \\ =\frac{1}{(1+p+q)!}\sum_{\pi\in\mathcal{P}(0,\cdots,p+q)}\operatorname{sgn}(\pi)a^p(v_{\pi(0)},\cdots,v_{\pi(p)})b^q(v_{\pi(p)},\cdots,v_{\pi(p+q)}) $$ where $a^p$ is a $p$-cochain, $b^q$ is a $q$-cochain, $v_0,\cdots,v_{p+q}$ are vertices of of a $p+q$-simplex, $\mathcal{P}$ denotes the set of permutations of a sequence, and $\operatorname{sgn}$ denotes the sign of a permutation. This product has a lot of the properties of the wedge product, for instance being graded-commutative, and making the coboundary operator an antiderivation.

For $0$ and $1$-cochains, for instance, this would give $$ (a^0\smile b^1)(v_0,v_1)=\frac{1}{2}(a^0(v_0)+a^0(v_1))b^1(v_0,v_1) $$

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  • $\begingroup$ Thanks! This makes a lot of sense. I was playing around with this earlier looking at how a "left"/"right" product could be used, which I think one would need to consider to emulate the Leibniz rule for finite-difference gradients. Of course this version basically averages the two. I've been meaning to learn more about the cup product, so this is quite useful. $\endgroup$
    – Kai
    Feb 23, 2021 at 0:30

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