# Decomposition of a k-form into projections on simplex facets

I hope the following is already been calculated somewhere, but unfortunately I have no guess where to look for it.

Suppose $s$ is a $d$-simplex in $\mathbb R^d$ with $k$-dimensional facet set $s^k$, and let $s$ contain its circumcentre. By using this circumcentre instead of the barycentre, one obtains a circumcentric subdivision $\operatorname{csd}(s)$ (geometrically different, but combinatorially equivalent to the usual subdivision complex), and the dual cell complex of $s$ is a subcomplex of $\operatorname{csd}(s)$.

Let $\alpha \in \Lambda^k(\mathbb R^d)$ be an alternating $k$-form, and for any $k$-facet $t \in s^k$, let $\pi_t \alpha$ be the restriction of $\alpha$ to the tangent space of $t$. Then I would like to prove

$\sum\limits_{t \in s^k} \pi_t \alpha |t|_k |{*}t|_{d-k} = \alpha |s|_d$,

where $|\cdot|_n$ denotes the $n$-dimensional volume (say, e.g., $n$-dimensional Hausdorff measure) and $*t$ is the dual cell of $t$, which is a $(d-k)$-cell in the dual complex.

### What I know

• There must be factors $\lambda_t$ such that $\sum \pi_t \alpha \lambda_t = \alpha$ because the sum of all facets' tangent spaces span $\mathbb R^d$. The question can only be what these $\lambda_t$ are.
• In two dimensions (where $s = ijk$ is a triangle), I was able to prove it: For $k=0,d$ the statement is always void, so all that remains is to decompose a covector $\alpha$ into parts $\alpha_{ij}$ etc. that are tangential to edges $ij$. But my computation uses that one can explicitely give $\pi_{ij}\alpha$ etc. by cosines of the angles between $\alpha^\sharp$ and the edges.
• ${d \choose k} |U(t)|_d = |t|_k \;|{*}t|_{d-k}$, where the $U(t)$ is the union of all $d$-simplices in $\operatorname{csd}(s)$ that share an interior point with $t$. The $d$-simplices in $\operatorname{csd}(s)$ all belong to exactly one of those 'subdivision neighbourhoods'.
• In two dimensions, $|U(ij)|_2 = \sin(2\phi^i) |ijk|_2$, where $\phi^i$ is the angle opposite to the edge $ij$. This is because the barycentric coordinates of the circumcentre are $\sin(2\phi^i), \sin(2\phi^j), \sin(2\phi^k)$.

### Application

The ''discrete exterior calculus'' by Hirani, Desbrun et al. calculates harmonic forms, Hodge decompositions etc. only be means of ''discrete $k$-differential forms'' that are nothing more that $k$-chains over $\mathbb R$, i.e. they map each $k$-simplex to some real number. In all examples this works remarkably well, but the convergence theorey is not yet fully established.

I believe that it would be a good idea to interpolate these ''discrete forms'' not by Whitney forms as usual (which leads to wrong derivatives and co-derivatives) but by piecewise constant forms tangent to the simplex. Of course, any form of convergence proof requires that smooth forms can be well-approximated by those piecewise constant form, and the argument then goes the usual way:

If $\alpha \in \Omega^k$ is a smooth differential $k$-form, it is constant in each simplex up to some factor depending on first derivatives of $\alpha$ (I would use covariant derivatives, but there are many possibilities). This constant form $\alpha'$ is well-approximated in $L^2$ by the $\pi_t\alpha'$ forms, which hence must also be a good approximation of $\alpha$.

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