advanced35 min studyLesson 615 in Mathematics

Minkowski sums and mixed volumes

Expand the volume of scaled convex sums and interpret mixed coefficients without confusing them with ordinary intersections.

On this page

What you’ll understand

  • Compute support functions and simple shapes under Minkowski addition.
  • Read mixed-volume normalization from the volume polynomial.
  • Distinguish full-dimensional body assumptions from extensions to lower-dimensional sets.

Adding two shapes means more than placing their drawings on top of each other. Minkowski addition forms every sum of one point from each shape, so it describes combined displacement, uncertainty, and geometric offsets. The volume of a scaled sum varies polynomially, and mixed volumes measure the interaction coefficients. These coefficients are neither the volume of an intersection nor an arbitrary correction term. Their normalization matters: even in the plane, a factor of two separates the cross coefficient from the mixed area.

Sum points, not membership indicators

For sets K and L, their Minkowski sum contains all x + y with x in K and y in L. It is different from the union, which contains the original points, and from the intersection, which contains shared points. Minkowski addition preserves convexity and compactness for nonempty compact convex sets. A positive scaling multiplies every point, and translation of one summand simply translates the entire sum, leaving its Euclidean volume unchanged.

Support functions linearize this addition. Maximizing a linear functional over x + y separates into independent maxima over x and y, so the support of K + L is the sum of their supports. This equality provides both a proof technique and a practical way to recognize resulting shapes. For example, adding two coordinate rectangles adds their corresponding side lengths, while adding two nonparallel line segments creates a parallelogram rather than another segment.

K+L={x+y:x∈K, y∈L},hλK+μL(u)=λhK(u)+μhL(u)(λ,μ≥0)K+L=\{x+y:x\in K,\ y\in L\},\qquad h_{\lambda K+\mu L}(u)=\lambda h_K(u)+\mu h_L(u)\quad(\lambda,\mu\ge0)
The support identity holds for nonempty compact convex sets, and zero scaling is interpreted as the singleton containing the origin.↔ Scroll wide equations horizontally.

Normalize the polynomial coefficients

For convex bodies in real d-space, the volume of a nonnegative linear combination is a homogeneous polynomial of degree d in the scaling parameters. We use the standard convention that the mixed volume of d identical copies of K equals its ordinary d-dimensional volume. For two bodies the binomial coefficient accompanies each mixed coefficient. In the plane this gives area(K + tL) equal to area(K), plus twice t times the mixed area, plus t squared times area(L).

Mixed volume is symmetric in its d arguments and linear in each under Minkowski addition and nonnegative scaling. It is nonnegative, translation invariant in each argument, and continuous under suitable convex-body convergence. These are structural theorems, not properties obtained by integrating pointwise products of the two shapes. For example, two bodies can be disjoint after translation yet retain exactly the same mixed volume, proving it cannot measure ordinary overlap.

Vol⁡d(K+tL)=∑j=0d(dj)V(K[d−j],L[j])tj,d=2:∣K+tL∣=∣K∣+2tV(K,L)+t2∣L∣\operatorname{Vol}_d(K+tL)=\sum_{j=0}^{d}\binom dj V(K[d-j],L[j])t^j,\qquad d=2:\quad |K+tL|=|K|+2tV(K,L)+t^2|L|
Here t ≥ 0, K[d−j] means d−j repeated arguments, and V(K[d]) = Vol_d(K). In dimension two, bars denote Euclidean area.↔ Scroll wide equations horizontally.

State when degeneracy changes the interpretation

We use convex body to mean compact, convex, and with nonempty interior in the ambient space, so its volume is positive. The volume polynomial and mixed volumes extend to nonempty compact convex sets of smaller dimension by continuity, but some coefficients or leading terms may vanish. One must not infer positive volume for each summand merely because their sum has positive volume: two transverse planar segments already give a counterexample.

For polytopes, Minkowski addition combines their supporting directions and normal information, making the operation useful in motion planning and sparse-polynomial geometry. The volume polynomial does not by itself enumerate faces or construct the sum efficiently in high dimension. Distinguish a structural identity from an algorithm, and specify whether a subsequent inequality requires positive full-dimensional volume. The next lesson uses that positive-volume setting to relate mixed geometry to concavity of a volume root.

Extract mixed area from rectangles

Let K = [0, 2] × [0, 1] and L = [0, 1] × [0, 3]. Find the area polynomial for K + tL and the mixed area V(K, L).

  1. For t nonnegative, the summed rectangle has side lengths 2 + t and 1 + 3t. Its area is their product, which expands to 2 + 7t + 3t².

  2. Compare with the standard planar formula |K| + 2tV(K, L) + t²|L|. The constant area is two, the area of L is three, and twice the mixed area is seven.

  3. Therefore V(K, L) = 7/2. At t = 2, the direct side lengths are four and seven, giving area 28; the polynomial gives 2 + 14 + 12 = 28 as a check.

The result

The area polynomial is 2 + 7t + 3t², and the mixed area is 3.5, not the cross coefficient 7.

Explore the model

Interactive lab

Interactive lab · matrix

A parallelogram as a sum of segments

Pause and Reset, then set diagonal scale a = 1.5 and shear b = 0.5. Predict that the selected unit-square image is the Minkowski sum of the segments from zero to (1.5, 0) and from zero to (0.5, 1), with area 1.5. Play to inspect the resulting parallelogram. Readouts describe the final matrix; this family does not calculate sums or mixed volumes of arbitrary bodies.

A parallelogram as a sum of segmentsThe selected matrix is A = [[a,b],[0,1]]. Playback interpolates geometry between the identity and A; readouts always describe A. Its determinant is a and its singular values come from AᵀA. Positive a makes every supported matrix invertible. This two-dimensional family is not a general eigensolver or a learned network. Current readings: Area scale · det A: 1.4; Singular values · σmax, σmin: 1.597, 0.876; Condition number · κ₂(A): 1.823. All parameters are adjustable with the labeled controls below.Matrix A with first row 1.4, 0.6 and second row zero, one; display interpolates between identity and A while final determinant remains 1.4.−4−2024−2−1012x · equal geometric scaleyA = [ 1.4 0.6 ; 0 1 ]Display blend: 0% · final values below
Interpolated geometryOriginal unit squareSelected final transform
Area scale · det A
1.4
Singular values · σmax, σmin
1.597, 0.876
Condition number · κ₂(A)
1.823
t = 0 s

The selected matrix is A = [[a,b],[0,1]]. Playback interpolates geometry between the identity and A; readouts always describe A. Its determinant is a and its singular values come from AᵀA. Positive a makes every supported matrix invertible. This two-dimensional family is not a general eigensolver or a learned network. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forA Minkowski sum can gain dimension even when each summand has zero area.

Check your understanding

If area(K + tL) = 4 + 12t + 9t², what is the standard mixed area V(K, L)?

Try it yourself

Let K be the horizontal segment from (0, 0) to (2, 0), and L the vertical segment from (0, 0) to (0, 3). Compute |K + tL| for t ≥ 0 and the extended mixed area. Explain why positive area of the sum does not imply either segment is full dimensional.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

The summed set is a rectangle whose horizontal and vertical side lengths come from different summands.

Compare with the solution

The sum K + tL is [0, 2] × [0, 3t], with area 6t. Both original planar areas are zero, so the extended mixed-area formula becomes 6t = 2tV(K, L), giving V = 3. For every t > 0 the sum has positive area because the segment directions span the plane. Each summand remains one dimensional with empty planar interior. Thus the polynomial extends naturally to these compact convex sets, but hypotheses requiring full-dimensional bodies cannot be checked from the sum alone.

Take it with you

  • Minkowski addition combines point displacements and adds support functions.
  • Mixed-volume coefficients have a fixed binomial normalization.
  • Lower-dimensional summands can produce a full-dimensional sum, so ambient-volume hypotheses must be checked separately.

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