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Say $P$ is a convex Euclidean polytope, where the origin is not contained in any bounding hyperplane containing a facet of $P$, with $n$ facets given by $\langle f_i , x\rangle = 1$ and $m$ vertices $v_j$ with $1\leq i\leq n$ and $1\leq j\leq m$. (That is, each facet equation has been multiplied through as needed to obtain $1$ on the right side.) Is it true that the polytope $P’$ with $m$ facets given by $\langle v_j , x\rangle = 1$ and $n$ vertices $f_i$ is combinatorially equivalent to the polar dual $P^*$ of $P$?

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Yes. Your polytope $P'$ is exactly the polar $P^*$ of $P$ with respect to the standard unit sphere. Choosing another conic instead of the unit sphere will in general give another polar. See this wikpedia page for the two-dimensional case of polarity with respect to other conics. So your "Cartesian" dual corresponds to the unit sphere.

Polarity/duality of polytopes or more generally, convex bodies, is treated in many books on convexity. Unfortunately, many of them leave a lot of detail to the reader. I recommend Webster's Convexity, Brønsted's An Introduction to Convex Polytopes or Matousek's Lectures on Discrete Geometry.

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Thanks! I will check out these references. I omitted from the question whether P is fully dimensional - i.e., a d-dimensional polytope in R^d. If P is k-dimensional (k < d) in R^d, and P is contained in a k-dimensional hyperplane, then the facet equations for P are not uniquely determined - so the Cartesian dual would depend on the choice of these equations as well. (I'm ditching the scare quotes, and the pun is intended.) –  Dan Moore Sep 10 '10 at 13:53

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