Hi, mathoverflow, I am currently working on finite element exterior calculus, and right now my concern is the construction of some certain kind of "conforming" basis functions for a finite element space. Guess most of you guys are working on pure math, hence before I asked my questions I would like to briefly introduce what this is.
For simplicity consider a 3D simply-connected polyhedral domain $\Omega$ which has a triangulation consisting of tetrahedra, we have the following deRham sequence: $$ H^1_0(\Omega) \xrightarrow{\nabla} H_0(\mathbf{curl}) \xrightarrow{\nabla\times} H_0(\mathrm{div})\xrightarrow{\nabla\cdot} L^2(\Omega) $$ And its discrete counterpart reads: $$ P^1 \xrightarrow{\nabla} \mathcal{Nd}^0 \xrightarrow{\nabla\times} \mathcal{RT}^0 \xrightarrow{\nabla\cdot} P^0_{-1} $$ in which each space is the conforming finite element space of its continuous counterpart, $P^1$ is the continuous piecewise degree one polynomial on this triangulation, within each tetrahedron, it consists four basis functions in the form of $\lambda_i$ of which the degrees of freedom associated with the four vertices $V_j$ of this tetrahedron, $\lambda_i(V_j) = \delta_{ij}$ is the barycentric coordinate of each vertices. $\mathcal{Nd}^0$ is the lowest order Nédélec elements, it basis functions are in the form of $\lambda_i\nabla\lambda_j - \lambda_j\nabla\lambda_i$ for each of the six edges $\overrightarrow{V_i V_j}$. $\mathcal{RT}^0$ is the lowest order Raviart-Thomas element, the basis functions are $\lambda_i \nabla\lambda_j\times \nabla \lambda_k + \lambda_j \nabla\lambda_k\times \nabla \lambda_i + \lambda_j \nabla\lambda_i\times \nabla \lambda_j$ for the face $V_i V_j V_k$. Lastly $P^1_{-1}$ is the constant on each tetrahedra with no continuity across the inter-element faces.
Use the construction of $\mathcal{Nd}^0$ as an example, the basis functions are obtained by Whitney 1-form interpolation, we would like to construct 1-form such that: $$ \int_{V_k V_l} \boldsymbol{\varphi}_{ij} d\boldsymbol{l}= \begin{cases} 1 & \text{if $i=k$, $j=l$}\newline -1 & \text{if $i=l$, $j=k$}\newline 0 & \text{otherwise} \end{cases} $$
and what we get would be $\boldsymbol{\varphi}_{ij} =\lambda_i\nabla\lambda_j - \lambda_j\nabla\lambda_i $ which is the lowest order Whitney form.
Now my question is: Instead of using the standard interpolation technique to construction these element, can we directly use the deRham sequence to construct the finite element basis functions? Say we know that $R(\nabla) = \nabla H^1_0= \ker(\mathbf{curl})$, hence we have $$ H_0(\mathbf{curl}) = R(\nabla) \oplus R(\nabla)^{\perp} $$ where the polar set $R(\nabla)^{\perp}$ is isomorphic to $R(\mathbf{curl})$. And this decomposition also holds for discrete spaces: $$ \mathcal{Nd}^0 = \nabla P^1 \oplus W $$ where $W$ is isomorphic to the range of $\mathbf{curl}$-operator into the space $\mathcal{RT}^0$, since $R(\mathbf{curl}) = \ker(\mathrm{div})$, now if we cook up a finite element space for divergence free functions, we could know what $W$ is.
However, if following this route, we wouldn't have something nice like Whitney 1-form: $\lambda_i\nabla\lambda_j - \lambda_j\nabla\lambda_i$, so where does the construction of discrete 1-form on a tetrahedron falls into above deRham sequence framework? or rather to say, how does this deRham cohomology framework affect or guide our construction of the discrete k-forms basis on a simplex?
springerlink.com
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