Kemperman's theorem

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Statement

For two subsets

Suppose G is a compact connected topological group with a Haar measure \mu of total volume 1 (so that the Haar measure is a probability measure). Suppose A,B are compact subsets of G. Then, the product of subsetsAB is also a compact subset and:

\! \mu(AB) \ge \min \{ \mu(A) + \mu(B), 1 \}

For finitely many subsets

Suppose G is a compact connected topological group with a Haar measure \mu of total volume 1 (so that the Haar measure is a probability measure). Suppose A_1,A_2,\dots,A_n are compact subsets of G. Then, the product of subsetsA_1A_2\dots A_n is also a compact subset and:

\! \mu(A_1A_2 \dots A_n) \ge \min \{ \mu(A_1) + \mu(A_2) + \dots + \mu(A_n), 1 \}

Note for abelian groups

When, G satisfies the additional condition of being an abelian group, and the group operations are denoted additively, the product of subsets is termed the Minkowski sum. Thus, this provides a lower bound on the size of the Minkowski sum.

Caveats

Importance of connectedness

Connectedness is important because if G has a proper open subgroup, then by compactness it must have finite index, and then the measure of the subgroup is the reciprocal of that index. Taking A,B to both be that subgroup violates the theorem.

Sharpness of estimate

We can take G to be the circle group and A,B to be arcs starting at the identity element to show that the estimates cannot be improved.

Related facts

External links