advanced32 min studyLesson 614 in Mathematics

Polarity and dual convex bodies

Turn supporting inequalities into dual points while tracking the essential role of the origin and inverse-transpose maps.

On this page

What you’ll understand

  • Define the polar and explain why an interior origin gives a bounded dual body.
  • Derive inclusion reversal and the inverse-transpose transformation rule.
  • Compute a polar from a support function and interpret face duality.

A polytope can be described from inside by its vertices or from outside by supporting inequalities. Polarity relates these descriptions using a chosen origin: a point becomes a linear constraint, and a supporting constraint becomes a point in a dual space. This is a geometric duality, not a visual instruction to reflect a picture. Its behavior under scaling is reciprocal, its behavior under inclusion is reversed, and translating the original body can fundamentally change the construction.

Normalize support with respect to an interior origin

For a set K in real d-space, its polar consists of all vectors y whose inner product with every point of K is at most one. In this lesson K is a compact convex body with zero in its interior. The polar is then also compact, convex, full dimensional, and contains zero in its interior. Convexity follows because the definition is an intersection of halfspaces, one for each point of K.

The interior assumption does real work. A small ball around zero inside K bounds the norm of every polar vector, while boundedness of K supplies a small ball inside the polar. If zero lies only on the boundary, the polar can be unbounded: the interval [0, 1] has polar (−∞, 1] in one dimension. If the chosen origin changes, first translate K to that origin and recompute; ordinary polarity is not translation invariant.

K∘={y:⟨x,y⟩≤1 for all x∈K}={y:hK(y)≤1}K^\circ=\{y:\langle x,y\rangle\le1\text{ for all }x\in K\}=\{y:h_K(y)\le1\}
Assume K is compact and convex with 0 in its interior whenever bounded full-dimensional dual bodies are asserted.↔ Scroll wide equations horizontally.

Reverse inclusion and transform normals contragrediently

If K is contained in L, a vector satisfying every inequality from L also satisfies those from K. Therefore the polar of L is contained in the polar of K. Scaling K outward by a positive factor scales its polar inward by the reciprocal factor. More generally an invertible linear map A sends the polar to A inverse transpose applied to the old polar, because the pairing with Ax equals the pairing of x with A transpose y.

Applying polarity twice returns K under our closed-convex, origin-containing hypotheses. One inclusion follows directly from the definition; the reverse follows by separating a point outside K from K. Without convexity and closure, double polarity generally produces a closed convex hull containing zero rather than the original set. This distinction parallels optimization duality: inequalities can capture a convex envelope even when the initial collection of points was not convex.

K⊆L⟹L∘⊆K∘,(sK)∘=s−1K∘,(AK)∘=A−TK∘K\subseteq L\Longrightarrow L^\circ\subseteq K^\circ,\qquad (sK)^\circ=s^{-1}K^\circ,\qquad (AK)^\circ=A^{-T}K^\circ
Here s > 0 and A is invertible. The inverse transpose preserves the primal-dual inner product and is generally different from applying A to both objects.↔ Scroll wide equations horizontally.

Faces exchange dimension in a polar polytope

For a full-dimensional polytope containing zero in its interior, a proper nonempty face F corresponds to the dual face where the inner product with every point of F equals one. Inclusion of faces is reversed, and dimensions add to d − 1. Consequently vertices correspond to facets, while edges in three dimensions correspond to edges. The f-vector therefore reverses, but this statement concerns the polar’s combinatorics rather than arbitrary geometric dual drawings.

Support functions make calculations concrete. The support of a centered axis-aligned rectangle is the sum of its half-widths times the absolute values of the dual coordinates. Setting that expression at most one gives a diamond. A long primal direction becomes a short dual direction. This reciprocal relation is useful for norm balls and optimization constraints, but it depends on the chosen pairing and normalization; a polar is not obtained by merely swapping x and y labels.

The polar of a centered rectangle

Find the polar of K = [−2, 2] × [−1, 1], its vertices, and the product of the two areas.

  1. For dual coordinates (u, v), maximize ux + vy independently over the two coordinate intervals. The support function is 2|u| + |v|, so the polar is the diamond defined by 2|u| + |v| ≤ 1.

  2. The diamond meets its axes at (±1/2, 0) and (0, ±1). Its two diagonal lengths are one and two, so its area is one-half times 1 × 2 = 1.

  3. The original rectangle has side lengths four and two and area eight. The area product is therefore eight. Scaling K by two would multiply its area by four and divide the polar area by four, preserving that product.

The result

The polar has the four stated vertices and area 1; the original area is 8 and their product is 8.

Explore the model

Interactive lab

Interactive lab · matrix

Normals follow the inverse transpose

Pause and Reset, then set diagonal scale a = 2 and shear b = 1. Predict final edge vectors (2, 0) and (1, 1). Check that the transformed normal (0.5, −0.5) pairs with those edges as one and zero, respectively, matching A inverse transpose applied to (1, 0). The diagram transforms a unit square rather than drawing its polar; the normal pairing must be checked by hand. Readouts refer to the selected final matrix.

Normals follow the inverse transposeThe 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 forTransforming points and supporting normals by different matrices preserves the inequality pairing.

Check your understanding

Which condition ensures that the polar of a compact convex body is bounded in the stated Euclidean space?

Try it yourself

Find the polar of the Euclidean ball of radius three centered at zero. Prove both containment directions, rather than relying only on the scaling rule.

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

Give me a hint

Use the Cauchy–Schwarz inequality for one direction, and choose an original point aligned with y for the other.

Compare with the solution

If the norm of y is at most one-third, every x of norm at most three satisfies x·y ≤ ||x|| ||y|| ≤ 1, so y belongs to the polar. Conversely, for nonzero y of norm greater than one-third, choose x = 3y/||y||. That point lies in the original ball and has x·y = 3||y|| > 1, excluding y from the polar. Thus the polar is exactly the centered ball of radius one-third; y = 0 is already included.

Take it with you

  • A bounded polar body requires the origin to be interior to the original compact convex body.
  • Polarity reverses inclusion and sends linear maps to inverse-transpose maps.
  • Proper face dimensions add to d − 1 under polytope polarity.

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