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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.

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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
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2 Answers 2

As Yoav says in his comment, this is essentially the Minkowski problem, which asks whether, given a measure on the unit sphere, it is the surface area measure of a convex body and whether the convex body is unique. This was originally solved by Alexandrov and independently by Fenchel and Jessen. The convex body is unique only up to translation. If you add the assumption that the center of mass of the body is at the origin, then you do get uniqueness.

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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.

As I said before, I think there is simpler than this (and I have not given thought to the non-symmetric case). But if you want more details on this see section 6 of my paper with Thompson Volumes on normed and Finsler spaces

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thank you everyone, your comments will help me a ton. –  Farhad Feb 11 '13 at 20:58
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