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A barycentric coordinate system assigns to a point in the affine span of the vertices of a nondegenerate simplex its unique vertex weights ${\lambda }_0,\dots ,{\lambda }_d$, constrained to sum to one. Uniqueness follows because $v_1-v_0,\dots ,v_d-v_0$ form a basis of the translation space: expand $x-v_0$ in that basis and set ${\lambda }_0=1-{\sum }_{i=1}^d{\lambda }_i$. A point lies in the closed simplex exactly when every weight is nonnegative. An affine function satisfies $f(x)={\sum }_i{\lambda }_{i}f(v_i)$, which makes these coordinates useful for interpolation and the integral of an affine function over a simplex.

For a nondegenerate $d$-dimensional simplex $S$ with vertices $v_0,\dots ,v_d$ and an affine function $f$, the integral equals the volume times the value at the centroid $c=(v_0+\cdots +v_d)/(d+1)$. To prove this, write $x={\sum }_i{\lambda }_i(x)v_i$ in a barycentric coordinate system. On the standard simplex, symmetry permuting the vertices gives equal integrals of all ${\lambda }_i$; since ${\sum }_i{\lambda }_i=1$, each is $\mathrm{vol}(S)/(d+1)$. An invertible affine map transfers this result to any simplex. Affineness gives $f(x)={\sum }_i{\lambda }_i(x)f(v_i)$, so linearity of integration proves the formula. For a tetrahedron the factor is one quarter. The rule is exact for affine integrands; a nonlinear integrand generally needs additional moments.

For positive coordinate intercepts $A=(a,0,0)$, $B=(0,b,0)$ and $C=(0,0,c)$, their plane equation is $\mathbf{n}\cdot \mathbf{r}=1$. The closest point on a plane to the origin is $\mathbf{d}=\mathbf{n}/\Vert \mathbf{n}{\Vert }^2$, with distance $1/\Vert \mathbf{n}\Vert$. Its barycentric coordinate system coefficients are proportional to $a^{-2},b^{-2},c^{-2}$ and sum to one.

The projection is also the orthocenter of triangle $ABC$. Indeed $\mathbf{a}\cdot (\mathbf{b}-\mathbf{c})=0$, and $\mathbf{d}\cdot \mathbf{b}=\mathbf{d}\cdot \mathbf{c}$ because each equals $1/\Vert \mathbf{n}{\Vert }^2$. Hence $(\mathbf{a}-\mathbf{d})\cdot (\mathbf{b}-\mathbf{c})=0$. Cycling the vertices gives all three altitude directions. This connects an ambient-space orthogonal projection with intrinsic triangle geometry.

Label the polygon vertices of the nondegenerate triangle $K$ as $a_1,a_2,a_3$, and put $m_{ij}=(a_i+a_j)/2$ for its three edge midpoints. The six monomials $1,x_1,x_2,x_1^2,x_1{x}_2,x_2^2$ form a basis of ${\mathcal{P}}_2(K)$, so its dimension is six. Nondegeneracy matters: $K$ has nonempty interior, and a polynomial vanishing on that interior is the zero polynomial.

Let ${\lambda }_1,{\lambda }_2,{\lambda }_3$ be the affine functions of the barycentric coordinate system, characterized by ${\lambda }_i(a_j)={\delta }_{ij}$ and ${\sum }_i{\lambda }_i=1$. Define six quadratic polynomialsAt $a_k$, $N_i(a_k)={\delta }_{ik}$ and all $N_{ij}(a_k)$ vanish. At $m_{kl}$, exactly ${\lambda }_k$ and ${\lambda }_l$ equal $1/2$, while the third is zero. Thus every $N_i$ vanishes at all three edge midpoints, and $N_{ij}(m_{kl})$ is one for the matching unordered pair and zero otherwise.

These six polynomials have the cardinal property for the six point evaluation functionals. Evaluating any vanishing linear combination at the nodes forces each of its coefficients to vanish. They are therefore six linearly independent members of a six-dimensional vector space, hence a basis.

In particular, every $p\in {\mathcal{P}}_2(K)$ has the unique expansionThis proves both existence and uniqueness for arbitrary six prescribed nodal values, which is precisely an unisolvent finite element. It also identifies the finite element degrees of freedom and shape functions of the quadratic Lagrange triangle explicitly.

Taking the dot product with $\mathbf{n}$ gives $\mathbf{n}\cdot \mathbf{r}=\lambda +\mu +\nu$. The necessary and sufficient condition is thereforeThese coefficients describe an affine combination of the three vertices, or their barycentric coordinate system. Nonnegativity would additionally restrict the point to the triangle; it is not required for the whole plane.

Fix a vertex of a simplicial complex $v$, and let $R$ be the set of vertices reachable from $v$ by an edge path. If a simplex contains a vertex of $R$, every vertex of that simplex lies in $R$: append its edge from the reachable vertex to the existing edge path. Thus every simplex has either all its vertices in $R$ or all outside $R$.

Partition the geometric realization as $|K|=A\bigsqcup B$, according to whether the support in the barycentric coordinate system lies in $R$ or outside $R$. On each geometric simplex, the intersection with $A$ is either the whole simplex or the empty set, and the same holds for $B$. The defining realization topology therefore makes both $A$ and $B$ closed, hence also open because they are complements.

The vertex $v$ belongs to $A$. If any vertex were unreachable, it would belong to the nonempty set $B$, giving a separation of a topological space of $|K|$. This contradicts connectedness. Therefore every vertex is reachable from $v$, and any two vertices can be joined by an edge path.

An abstract simplicial complex consists of a set $V$ of vertices of a simplicial complex and a family $K$ of finite nonempty subsets of $V$, called abstract simplexes, such that every singleton $\{v\}$ belongs to $K$ and every nonempty subset of a member of $K$ also belongs to $K$. One may include the empty set as an additional face by convention. A member with $k+1$ vertices is a $k$-simplex; its nonempty subsets are its faces of a simplex.

The geometric realization has the following explicit description using a barycentric coordinate system:Here the support is $\mathrm{supp}(t)=\{v:t(v)>0\}$. Each $\rho \in K$ gives a standard geometric simplexwith its usual Euclidean topology. Common faces of a simplex are identified using the same barycentric coordinate system.

Precisely, the topology on $|K|$ is defined by declaring $A\subseteq |K|$ closed if and only if $A\cap |\rho |$ is closed in $|\rho |$ for every $\rho \in K$. Equivalently, use the quotient topology on the disjoint union of all the standard geometric simplexes, identifying identical points of common faces of a simplex. For a finite simplicial complex, this agrees with the subspace topology in ${\mathbb{R}}^V$. For an infinite simplicial complex, the stated rule specifies the realization topology without presuming a finite-dimensional ambient space.

The quadratic Lagrange triangle uses degree-at-most-two polynomials on a nondegenerate triangle, with point evaluation at its three polygon vertices and three edge midpoints as the six finite element degrees of freedom. If ${\lambda }_i$ are the affine functions of its barycentric coordinate system, the cardinal shape functions areAt a polygon vertex, one coordinate in the barycentric coordinate system is one and the others vanish; at an edge midpoint, two are $1/2$ and the third is zero. These values prove the cardinal property. Since $\dim {\mathcal{P}}_2(K)=6$, this is an unisolvent finite element. Matching the two endpoint values and midpoint value on a shared edge of a polygon matches the whole quadratic restriction, producing a continuous global conforming finite element space.

A simplicial homeomorphism is the homeomorphism of geometric realizations induced by an isomorphism of simplicial complexes: a bijection of their vertex sets such that a finite vertex set is a simplex exactly when its image is a simplex. The map acts linearly in each barycentric coordinate system. It and its inverse are continuous by the realization topology. Between the boundaries of two triangles, any bijection of the three vertices extends linearly on edges to a simplicial homeomorphism. This ensures that the edge identifications agree at common endpoints.