Correction to: Intersection bodies of polytopes

The proof of the original article, Theorem2.6 contains an error. However all statements in the article remain correct. To be precise, the mistake lies in the equation that in the paper is wrongly given as ρI P (u) = ∑ j∈J 1 d! ∣det ( Mj (u) )∣∣ . Instead of the multiplying factor 1 d! , the summands of ρI P (u) should be multiplied by 1 (d−1)! . Therefore the correct version is ρI P (u) = ∑ j∈J 1 (d − 1)! ∣det ( Mj (u) )∣∣ .

Instead of the multiplying factor 1 d! , the summands of ρ I P (u) should be multiplied by 1 (d−1)! . Therefore the correct version is The same happens in the proofs of Theorem 3.2 and Proposition 5.5. Similarly, in line 13 of the original article, Algorithm 1 and line 5 of the original article, Algorithm 2, formula As a consequence, the examples in the paper are also affected by this mistake. We list here the examples and provide the correct expressions. We note that the figures remain correct.

Example 2.3
If the intersection is a square, then the radial function in a neighborhood of that point will be a constant term over a coordinate variable, e.g. 4 z . On the other hand, when the intersection is a hexagon, the radial function is a degree two polynomial over x yz.

Example 2.8
In six regions the radial function has the following shape (up to permutation of the coordinates and sign): There are then 18 = 6 + 12 regions in which the radial function looks like In the remaining six regions we have Example 5.3 The facet exposed by the vector (1, 0, 0) is the intersection of z = 4 with the convex cone In other words, the variety V(z − 4) is one of the irreducible components of ∂ a I P.

Example 5.8
The polynomial that defines the boundary of I P in the region C 1 is a quartic, namely The polynomial that defines the boundary in the region C 2 is a cubic Example 5.10 In the 12 regions which are spanned by five rays, the polynomial that defines the boundary of I P has degree 5 and it looks like In the other 20 regions spanned by three rays, ∂ I P is the zero set of a sextic polynomial with the following shape Example 6.2 The linear face exposed by (1, 0, 0, 0) is cut out by the hyperplane w = 8. The second family of chambers is made of cones with 5 extreme rays, where the boundary is defined by a cubic equation with shape Finally there are 64 cones spanned by 4 rays such that the boundary of the intersection body is a quartic, such as Open Access This article is licensed under a Creative Commons Attribution 4.0 International License, which permits use, sharing, adaptation, distribution and reproduction in any medium or format, as long as you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons licence, and indicate if changes were made. The images or other third party material in this article are included in the article's Creative Commons licence, unless indicated otherwise in a credit line to the material. If material is not included in the article's Creative Commons licence and your intended use is not permitted by statutory regulation or exceeds the permitted use, you will need to obtain permission directly from the copyright holder. To view a copy of this licence, visit http://creativecommons.org/licenses/by/4.0/.
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