Lowest-order Nedelec edge circulation unisolvence

Parametrize the oriented edge $[x_i,x_j]$ by $x(t)=(1-t)x_i+t{x}_j$, $0\le t\le 1$. Along it, ${\lambda }_i=1-t$, ${\lambda }_j=t$, so $\mathrm{\nabla }{\lambda }_j\cdot x^{\mathrm{\prime }}(t)=1$ and $\mathrm{\nabla }{\lambda }_i\cdot x^{\mathrm{\prime }}(t)=-1$. Therefore ${\varphi }_{ij}(x(t))\cdot x^{\mathrm{\prime }}(t)=(1-t)-t(-1)=1$, and ${\mathrm{\ell }}_{ij}({\varphi }_{ij})=1$.

Let $[x_r,x_s]$ be a different edge. If it shares no endpoint with $[x_i,x_j]$, then both ${\lambda }_i$ and ${\lambda }_j$ vanish identically along $[x_r,x_s]$, so its circulation is zero. If it shares $x_i$, then ${\lambda }_j$ vanishes identically along that edge and its directional derivative also vanishes; the remaining term $-{\lambda }_j\mathrm{\nabla }{\lambda }_i$ is zero. The case sharing $x_j$ is symmetric. Thus ${\mathrm{\ell }}_{rs}({\varphi }_{ij})=0$ whenever $(r,s)\mathrm{\ne }(i,j)$.

Each ${\varphi }_{ij}$ is affine in $x$. Its linear part is the skew-symmetric map $x\mapsto (\mathrm{\nabla }{\lambda }_i\cdot x)\mathrm{\nabla }{\lambda }_j-(\mathrm{\nabla }{\lambda }_j\cdot x)\mathrm{\nabla }{\lambda }_i$, hence equals $b_{ij}\times x$ for a unique $b_{ij}\in {\mathbb{R}}^3$; its constant part is a vector $a_{ij}$. The space $\{a+b\times x:a,b\in {\mathbb{R}}^3\}$ has dimension six: if $a+b\times x=0$ on the open tetrahedron, subtracting the values at two points shows $b\times y=0$ for every $y\in {\mathbb{R}}^3$, so $b=0$ and then $a=0$. The six functions are linearly independent because their circulation matrix is the identity by steps 1.1 and 2.1. They therefore form a basis of this six-dimensional space, and the six moments are unisolvent.