# determining a convex set by mixed volumes

For a convex set $K \subset \mathbb{R}^2$ let $\phi_K:$ convexsets in $\mathbb{R}^2 \rightarrow [0,\infty), A \mapsto MV(A,K)$. Where by $MV(A,K)$ I mean the mixed volume of $A$ and $K$ in the dimension two.

1. Is the map $K \mapsto \phi_{K}$ a bijection?

2. What about higher dimensions? i.e Could we identify a convex set in $\mathbb{R}^n$ by the knowledge of it's mixed volume with other convex sets?

3. What if we only consider convex polytopes?

Excuse my naïveté on the subject. Some references would be appreciated.

-
I believe $MV(A,K) = \int_{S^1} h_A(\theta) dS_K(\theta)$, where $h_A$ is the support height function of $A$ and $S_K$ is the surface area measure (perimeter measure in this case) of $K$. Clearly if $MV(\cdot,K)=MV(\cdot,K')$ then $S_K=S_{K'}$. The fact that $S_K$ uniquely determines $K$ is a well-studied problem, which if I remember correctly is due to Aleksandrov. I will have to look for the appropriate references, but everything should be in the Handbook of Convex Geometry. –  Yoav Kallus Feb 9 '13 at 18:46
Yes, this is the convex-geometric proof. $K$ needs to be centrally-symmetric if one wants uniqueness. –  alvarezpaiva Feb 9 '13 at 18:58

Dear Karl, if by $K$ convex you mean centrally-symmetric convex body, the answer is yes, the map is injective. There must be a neat proof somewhere, but on the spur of the moment I came up with this one which works in $n$-dimensions if you use the mixed volume $$MV(A,K) := \lim_{h \rightarrow +0} (V(A + hK) - V(A))/h .$$ Define a norm $\psi_K$ in the space of $(n-1)$-vectors in $\mathbb{R^n}$ by setting $$\psi(v_1 \wedge \cdots \wedge v_{n-1}) = MV([v_1,\cdots,v_{n-1}],K),$$ where $[v_1,\cdots,v_{n-1}]$ is the parallelotope defined by the vectors $v_1$,$\ldots$,$v_{n-1}$.
Assume you have two centrally symmetric bodies $K$ and $K'$ giving rise to the same norm in $\Lambda^{n-1} \mathbb{R}^n$, then they must be the same because both bodies can be reconstructed from this norm (up to a constant multiplcative factor depending on dimensions, which I've been lazy to specify) by the Wulff construction: Let $B \subset \Lambda^{n-1} \mathbb{R}^n$ be the unit ball of the norm, and let $B^* \subset \Lambda^{n-1} \mathbb{R}^{n*}$ be its polar body. Note that the volume form $\Omega$ in $\mathbb{R}^n$ defines an isomorphism between $\mathbb{R^n}$ and $\Lambda^{n-1} \mathbb{R}^{n*}$ by the contraction map $$v \mapsto i_v\Omega = \Omega(v\wedge \cdot).$$ Up to a dilation that depends only on the dimension of the space, your convex body $K$ (and $K'$) are the images of $B^*$ under this isomorphism.