# Manifold Topology and Geometry

I am green in manifold and I have some geometry background. I have some question about "manifoldness" of a triangular mesh. To my understanding, charts {$\phi_{\alpha},U_{\alpha}$} and transition maps $\tau_{\alpha\beta}$ define every point p on manifold M a unique location. A mesh T is a set of space points $P\in R^3$ with index set {$V,E,F$} specifying the topology. It usually use simple bilinear Barycentric map to represent every triangular face in both charts {$\phi_{\alpha},U_{\alpha}$} and the realization $\psi_{\alpha}(\phi_{\alpha}(U_{\alpha}))\in R^3$ (Am I correct?). The transition function $\tau_{\alpha\beta}$ then is well defined both in manifold space and $R^3$ because automorphism(right?).

Then I have a question about the possibility of not using bilinear Barycentric map for $\psi_{\alpha}$ to all triangles that keeps the surface the same manifold representation. For example a cubic polynomial map for $\psi_{\alpha}$ while keeping $\phi_{\alpha}$ unchanged for some triangular faces. E.g. $\psi_{\alpha}(\phi_{\alpha}(U_{\alpha}))$ is polynomial and $\psi_{\beta}(\phi_{\beta}(U_{\beta}))$ is bilinear. Then the transition map $\tau_{\alpha\beta}$ should not be affected because it works in the domain of $\psi_{\alpha}$ and $\psi_{\beta}$.(right?) However, $R^3\ni\psi_{\alpha}\ne\psi_{\beta}\in R^3$ because they have different degree of continuity. So, the mesh now becomes topologically manifold but geometrically "half-edge" mismatched. Would it be the situation?

Thank you. I get your point. The concept of the topological structure is clearly explained by Eular-Poincare Characteristics. However, because the manifold-structure of a triangulation of space $X$ itself is topologically proven, then the problem I am think of is the incompatibility of the realization maps $\psi_\alpha$. – l l Aug 13 '12 at 2:33