The de Rham theorem equates simplicial cohomology with de Rham cohomology. The correspondence is explicit. Start with a smooth differential k-form f, it defines a discrete differential form by assigning the value to each k-simplex x. This produces so a discrete differential form. A discrete differential form on the other hand can be seen as a de Rham current. One can apply the heat flow using the Hodge Laplacian to get from this again a smooth differential form on the manifold. To get explicit maps between the cohomologies (= harmonic forms), just do the same construction for harmonic forms but apply the heat flow afterwards in each direction to get harmonic forms in the image. The composition of these two maps defines a linear map on the harmonic forms (a finite dimensional space) and is so given by a matrix, once a basis for the harmonic k-forms are chosen. This map is close to the identity if the triangulation is good enough.
Quadratic cohomology does not have a continuum analog yet but it appears as if one can build a continuum de Rham theory. Just build the exterior differential forms on and restrict to the diagonal. If that works, then this could explain why we have for orientable discrete manifolds that the quadratic Betti vector is equal to , where is the Betti vector for $M$. This relation is not true in the non-orientable case. The projective plane is the simplest example, where it fails. There is a small projective plane with f-vector (12,38,27) and Betti vector (1,0,0) for which the quadratic cohomology with quadratic f-vector (15, 168, 602, 784, 336) has the quadratic Betti vector (0,0,0,0,1). For a 2-sphere, one has the Betti vector (1,0,1) and the quadratic Betti vector (0,0,1,0,1). The picture suggests that the quadratic cohomology for orientable q-manifolds is just the cohomology of the delta set obtained by taking the product of a q-ball with M. Note that the product of two finite simplicial complexes is not a simplicial complex a priory but is equivalent. Of course, quadratic cohomology is much more interesting than that. It allows to distinguish homotopic but not homeomorphic spaces or distinguish spaces that are homotopic to 1 but not collapsible. The dunce hat for example has a non-trivial quadratic cohomology.
During our visit in Switzerland, we stayed a few days in wallis. (See “Tracing Jean Piaget and George de Rham” from 2014). Here is a recent 360 degree panorama showing the region. And here is a page about George de Rham , the mountaineer. One story tells about some climbing of de Rham with Milnor and two others in the summer of 1966. My parents had bought the alp hat in the spring of 1962. My personal time line never crossed with de Rham, but we spent maybe 12 summers in Lausanne as my parents guided “Sprach kolonien” (language summer camps there). Here are some photos. We would later be in Chexbres (where my brother and I once in 1981 also joined as guides. Chexbres is between Lausanne and Roche, a village between Montreux and Martigni). I have very fond memories from our vacation times in the Lac Lemon region as well as the mountain region in Visp. Both places are associated to de Rham. By the way, in the panorama below, just made 2 weeks ago, one can see the settlement Ranft above Visp, where Jean Piaget lived. Here is a photo of him from there.
