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A mesh connecting a surface's control points according to the representation's indexing topology. A tensor-product surface basis gives a rectangular index grid and quadrilateral cells, even when its points in space are neither planar nor rectangular. The drawn mesh is a geometric skeleton, not generally the exact limit surface or an enclosing solid. Positive partition of unity coefficients give a convex hull bound on the relevant controls.

A solid body transform is a rigid transformation: it preserves distances within the object. In three-dimensional coordinates it has the form $T(x)=Rx+b$, with translation $b\in {\mathbb{R}}^3$ and an orthogonal matrix $R$. For a proper physical rotation, the coefficient constraints areEquivalently the columns of $R$ are an orthonormal basis with positive orientation. A reflection also preserves distance but has $\det R=-1$ and is excluded when “solid-body motion” means a rotation and translation. There are three rotational and three translational degrees of freedom, not twelve independent matrix coefficients.

Computationally use the homogeneous coordinates representationThe last row is $(0,0,0,1)$; matrix multiplication composes transforms. Alternatively store a unit quaternion together with $b$, maintaining its unit norm. A transpose gives the rotational inverse: $T^{-1}(y)=R^T(y-b)$.

Write the cubic Bézier curve as $P(t)={\sum }_{i=0}^3{B}_i^3(t)p_i$, where $B_i^3(t)=\left(\genfrac{}{}{0ex}{}{3}{i}\right)(1-t{)}^{3-i}{t}^i$. The binomial theorem gives ${\sum }_i{B}_i^3(t)=1$. Transforming the controls producesThis proves pointwise equivariance, including the original parameter values; it is stronger than a statement that the two curves merely have the same shape.

Orthogonality was not used in the calculation. The identical proof applies to every affine map $T(x)=Ax+b$, including scales, shears and even singular affine maps. It applies to every representation $P(u)={\sum }_i{\varphi }_i(u)p_i$ with scalar, control-independent partition of unity basis functions: Bézier curves of every degree, B-splines, surfaces built on a tensor-product surface basis, triangular Bézier patches, and more general linear control-point representations. Positivity is not needed for affine equivariance of a geometric basis. For linear stationary subdivision matrices with row sums one, each refinement step commutes with the affine transformation; taking a convergent limit proves the corresponding assertion for subdivision curves and subdivision surfaces. Rules with geometry-dependent nonlinear weights require separate equivariance tests and are not covered automatically.

The affine class is the largest class that preserves every such Euclidean-control construction with unchanged basis weights, under the usual continuity assumption. In fact the degree-one case would forcefor all $x,y$ and $0\le t\le 1$. This segment identity, followed by continuity, forces $T$ to be affine. Nonlinear transforms such as perspective division generally fail it.

There is a further extension when the representation itself is enlarged: apply a projective transformation linearly to homogeneous controls $(w_i{p}_i,w_i)$, then dehomogenize the weighted sum. The transformed objects are rational Bézier curves or non-uniform rational B-splines, with transformed weights as well as transformed Euclidean controls. This works wherever the new denominator is nonzero. It does not assert projective invariance of ordinary polynomial Bézier curves with their old weights.

A tensor-product surface basis is formed from two independent univariate basis families ${\varphi }_i(u)$ and ${\psi }_j(v)$ on separate parameter intervals. Given a rectangular control net $p_{ij}$, its parametric surface isThe parameters and indices vary independently over a product domain. The representation can be evaluated first in either parameter, giving a convenient computational advantage. Assume a finite basis, or locally finite spline bases, so differentiation and summation below are justified locally.

For positivity, if each factor basis is nonnegative, then ${\varphi }_i(u){\psi }_j(v)\ge 0$. If both factors are strictly positive at a point, their product is strictly positive there. Nonnegative weights are the usual geometric meaning of positivity; zero boundary weights are allowed. With partition of unity, this gives the local convex hull enclosure of the active control points.

For continuity, suppose the first family is $C^r$ and the second $C^s$, including matching derivatives across their respective spline knots. On every polynomial piece, and by matching at the knot lines,Every factor on the right is continuous in its own variable, so its product is jointly continuous. Thus all these mixed derivatives are continuous, in particular the surface is jointly $C^{\min (r,s)}$. The separate coordinate smoothness can be better: $r$ orders in $u$ and $s$ in $v$. For piecewise polynomials of degrees $d_u,d_v$, an interior knot of multiplicity $k$ in the first factor normally gives $C^{d_u-k}$ matching in $u$, inherited along that knot line; similarly for $v$. The statement is an inherited lower bound. Special control points can cancel derivative jumps, so a particular surface may be smoother than a generic member of its basis.

For summation to unity, independent summation factors exactly:This also explains why translations of all control points translate the surface by the same vector, as in affine equivariance of a geometric basis. These three arguments establish tensor-product inheritance of geometric basis properties directly from the factor identities.

A triangular Bézier patch is a non-tensor-product example. For barycentric parameters $u,v,w\ge 0$ with $u+v+w=1$, a quadratic patch isThe six basis functions have total degree two and sum to $(u+v+w{)}^2=1$. Their index constraint and triangular parameter domain do not arise from two independently indexed univariate factors on a rectangle. An individual triangular polynomial could be embedded in another larger representation, but this triangular basis and control net are not a tensor-product surface definition.

A tensor-product surface basis combines two independent univariate bases:The coefficients $P_{ij}$ are three-dimensional control points. Fixing one parameter gives a curve in the other univariate basis; the coefficients of that curve are the corresponding weighted combinations of rows or columns of controls.

The control net must have a rectangular two-index topology, with row/column neighbors and quadrilateral mesh cells. “Rectangular” describes the index grid, not the geometry of the points: its faces need not be flat or rectangular in space. A surface requiring a triangular domain or arbitrary extraordinary mesh connectivity is not directly described by one such tensor-product patch.

Suppose ${\varphi }_j\in C^r$ in $u$ and ${\psi }_k\in C^s$ in $v$, including their respective knot joins. The mixed derivatives of the tensor-product surface basis areThese are continuous because each factor is continuous. A finite or locally finite control sum has the same property. In the usual total-order notation the parametric surface is therefore at least $C^{\min (r,s)}$, and also has the stronger stated rectangular set of continuous mixed derivatives. Across a $u$-knot its guaranteed order is $r$, and across a $v$-knot its guaranteed order is $s$. No general improvement over a factor's smoothness is possible: setting $P_{jk}=A_j$ and using ${\sum }_k{\psi }_k=1$ recovers the univariate curve ${\sum }_j{A}_j{\varphi }_j(u)$ as a parametric surface constant in $v$.

If each univariate basis sums to one, distribute the finite sums, or use local finiteness when the basis is infinite:Thus the tensor-product surface basis is a partition of unity. In particular, translating every control point by a fixed vector translates the parametric surface by precisely that vector, and constant control data produce a constant parametric surface.

The tensor-product surface basis is $b_{jk}(u,v)={\varphi }_j(u){\psi }_k(v)$. If both univariate bases consist of nonnegative functions, thenIf both factors are strictly positive at the chosen parameters, their product is strictly positive there as well. Its support is the Cartesian product of the two factor supports. With partition of unity, this nonnegativity additionally places the parametric surface point in the convex hull of the active control points.

Let the univariate linear precision sites be ${\xi }_j,{\eta }_k$, meaningThen their Cartesian-product grid is a linear precision set for the tensor-product surface basis:Indeed, the first coordinate is $({\sum }_j{\xi }_j{\varphi }_j)({\sum }_k{\psi }_k)=u$, and the second is $({\sum }_j{\varphi }_j)({\sum }_k{\eta }_k{\psi }_k)=v$. More generally, take controls $P_{jk}=A+{\xi }_{j}B+{\eta }_{k}C$. The resulting parametric surface is exactlyThis proves linear precision of a geometric basis and affine map reproduction, including arbitrary changes of the embedding plane. In fact the products also reproduce the bilinear coordinate $uv$, since ${\sum }_{j,k}{\xi }_j{\eta }_k{\varphi }_j{\psi }_k=uv$; affine precision is the requested consequence.

Nonnegative tensor-product surface basis factors yield nonnegative product weights. Their two partition of unity sums multiply to one. If the factors are $C^r$ and $C^s$, mixed derivatives through separate orders $r,s$ are continuous, so the resulting finite or locally finite surface representation is jointly $C^{\min (r,s)}$. Control coefficients can cancel derivative jumps and yield more smoothness; inherited continuity is a guaranteed lower bound, not a claim that every surface has exactly the basis's worst regularity.