show/hide this revision's text 2 tinkered with wording

Sorry to answer my own question, but asking this in public seems to have spurred me into thought.

As auniket suspected, the answer is "yes" in the strongest sense I'd hoped: properties 1-3 do characterize mixed volume. In fact, something slightly stronger is true: $V$ is the unique function $(\mathscr{K}_n)^n \to \mathbb{R}$ satisfying

  1. $V(A, \ldots, A) = Vol(A)$

  2. $V$ is symmetric

  3. $V(A_1 + A'_1, A_2, \ldots, A_n) = V(A_1, A_2, \ldots, A_n) + V(A'_1, A_2, \ldots, A_n)$.

In other words, we don't need multilinearity, just multiadditivity.

The proof is along the lines suggested by auniket.

Fix $n$ and $A_1, \ldots, A_n \in \mathscr{K}_n$. Write $\mathbf{n} = {1, \{1, \ldots, n}$n\}$, and for sets $R$ and $S$, write $\mathrm{Surj}(R, S)$ for the set of surjections $R \to S$.

I claim that for all subsets $S$ of $\mathbf{n}$, $$ \sum_{f \in \mathrm{Surj}(\mathbf{n}, S) S)} V(A_{f(1)}, \ldots, A_{f(n)}) $$ is uniquely determined by the properties above. The proof will be by induction on the cardinality of $S$. When $S = \mathbf{n}$, this sum is $$ n! V(A_1, \ldots, A_n), $$ so this claim will imply the characterization theorem.

To prove the claim, take $S \subseteq \mathbf{n}$. Then $$ Vol(\sum_{i \in S} A_i) = \sum_{f: \mathbf{n} \to S} V(A_{f(1)}, \ldots, A_{f(n)}) $$ by the three properties. This in turn is equal to $$ \sum_{R \subseteq S} \sum_{f \in \mathrm{Surj}(\mathbf{n}, R) R)} V(A_{f(1)}, \ldots, A_{f(n)}). $$ By the inductive assumption, every summand except all but one -of the summands in the first summation - namely, $R = S$ -- is uniquely determined. Hence the $S$-summand is uniquely determined too. This completes the induction, and so completes the proof.

The proof makes it clear that $V(A_1, \ldots, A_n)$ is some rational linear combination of ordinary volumes of Minkowski sums of some of the $A_i$s. It must be possible to unwind this proof and get an explicit expression; and that expression must be the one auniket gave (which also appears in Lemma 5.1.3 of Schneider's book Convex Bodies: The Brunn-Minkowski Theory).

This all seems rather easy, and must be well-known, though I'm a bit surprised that this characterization isn't mentioned in some of the things I've read. Incidentally, I now understand why it doesn't appear in the paper of Milman and Schneider mentioned in my question: they explicitly state that they want to avoid assuming property 1.

show/hide this revision's text 1

Sorry to answer my own question, but asking this in public seems to have spurred me into thought.

As auniket suspected, the answer is "yes" in the strongest sense I'd hoped: properties 1-3 do characterize mixed volume. In fact, something slightly stronger is true: $V$ is the unique function $(\mathscr{K}_n)^n \to \mathbb{R}$ satisfying

  1. $V(A, \ldots, A) = Vol(A)$

  2. $V$ is symmetric

  3. $V(A_1 + A'_1, A_2, \ldots, A_n) = V(A_1, A_2, \ldots, A_n) + V(A'_1, A_2, \ldots, A_n)$.

In other words, we don't need multilinearity, just multiadditivity.

The proof is along the lines suggested by auniket.

Fix $n$ and $A_1, \ldots, A_n \in \mathscr{K}_n$. Write $\mathbf{n} = {1, \ldots, n}$, and for sets $R$ and $S$, write $\mathrm{Surj}(R, S)$ for the set of surjections $R \to S$.

I claim that for all subsets $S$ of $\mathbf{n}$, $$ \sum_{f \in \mathrm{Surj}(\mathbf{n}, S) V(A_{f(1)}, \ldots, A_{f(n)}) $$ is uniquely determined by the properties above. The proof will be by induction on the cardinality of $S$. When $S = \mathbf{n}$, this sum is $$ n! V(A_1, \ldots, A_n), $$ so this claim will imply the characterization theorem.

To prove the claim, take $S \subseteq \mathbf{n}$. Then $$ Vol(\sum_{i \in S} A_i) = \sum_{f: \mathbf{n} \to S} V(A_{f(1)}, \ldots, A_{f(n)}) $$ by the three properties. This in turn is equal to $$ \sum_{R \subseteq S} \sum_{f \in \mathrm{Surj}(\mathbf{n}, R) V(A_{f(1)}, \ldots, A_{f(n)}). $$ By the inductive assumption, every summand except one -- namely, $R = S$ -- is uniquely determined. Hence the $S$-summand is uniquely determined too. This completes the induction, and so completes the proof.

The proof makes it clear that $V(A_1, \ldots, A_n)$ is some rational linear combination of ordinary volumes of Minkowski sums of some of the $A_i$s. It must be possible to unwind this proof and get an explicit expression; and that expression must be the one auniket gave (which also appears in Lemma 5.1.3 of Schneider's book Convex Bodies: The Brunn-Minkowski Theory).

This all seems rather easy, and must be well-known, though I'm a bit surprised that this characterization isn't mentioned in some of the things I've read. Incidentally, I now understand why it doesn't appear in the paper of Milman and Schneider mentioned in my question: they explicitly state that they want to avoid assuming property 1.