1 Correction to: Beitr Algebra Geom https://doi.org/10.1007/s13366-022-00621-7

The proof of the original article, Theorem 2.6 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

$$\begin{aligned} \rho _{IP}(u) = \sum _{j\in J} \frac{1}{d!} \left| \det \left( M_j(u)\right) \right| . \end{aligned}$$

Instead of the multiplying factor \(\frac{1}{d!}\), the summands of \(\rho _{IP}(u)\) should be multiplied by \(\frac{1}{(d-1)!}\). Therefore the correct version is

$$\begin{aligned} \rho _{IP}(u) = \sum _{j\in J} \frac{1}{(d-1)!} \left| \det \left( M_j(u)\right) \right| . \end{aligned}$$

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

$$\begin{aligned} \frac{1}{\Vert x \Vert ^2}\sum _{\Delta \in {\mathcal {T}}} {{\,\mathrm{sgn}\,}}(\Delta )\det (M_\Delta ) \quad \hbox {must be replaced by}\quad \frac{1}{\Vert x \Vert ^2 (d-1)!}\sum _{\Delta \in {\mathcal {T}}} {{\,\mathrm{sgn}\,}}(\Delta )\det (M_\Delta ). \end{aligned}$$

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. \(\frac{4}{z}\). On the other hand, when the intersection is a hexagon, the radial function is a degree two polynomial over xyz.

Example 2.8

In six regions the radial function has the following shape (up to permutation of the coordinates and sign):

$$\begin{aligned} \rho (x,y,z) = \frac{4}{z}. \end{aligned}$$

There are then \(18 = 6 + 12\) regions in which the radial function looks like

$$\begin{aligned} \rho (x,y,z) = \frac{2x}{yz} \quad \hbox {or} \quad \rho (x,y,z) = \frac{2(x+2z)}{yz}. \end{aligned}$$

In the remaining six regions we have

$$\begin{aligned} \rho (x,y,z) = \frac{2(x^2+2xy+y^2+2xz+z^2)}{xyz}. \end{aligned}$$

Example 5.3

The facet exposed by the vector (1, 0, 0) is the intersection of \(z = 4\) with the convex cone

$$\begin{aligned} {\overline{C}}_1 = \text {co}\{(1, 0, 1), (-1, 0, 1), (0, 1, 1), (0, -1, 1)\}. \end{aligned}$$

In other words, the variety \({\mathcal {V}}(z-4)\) is one of the irreducible components of \(\partial _a IP\). The remaining 8 regions are spanned by 3 rays each, and the polynomial that defines the boundary of IP is a cubic, such as

$$\begin{aligned} 2 x y z - 2 x^2 - 4 x y - 2 y^2 - 4 x z + 4 y z - 2 z^2 \end{aligned}$$

in the region

$$\begin{aligned} {\overline{C}}_2 = \text {co}\{(0, 1, 1), (-1, 1, 0), (-1, 0, 1)\}. \end{aligned}$$

Example 5.8

The polynomial that defines the boundary of IP in the region \({\overline{C}}_1\) is a quartic, namely

$$\begin{aligned} q_2(x,y,z) - \frac{p_2(x,y,z)}{\Vert (x,y,z)\Vert ^2} = (x + z) (x - z) (y + z) (y - z) - 2 ( x^2 + y^2 - z^2) z. \end{aligned}$$

The polynomial that defines the boundary in the region \({\overline{C}}_2\) is a cubic

$$\begin{aligned} q_1(x,y,z) - \frac{p_1(x,y,z)}{\Vert (x,y,z)\Vert ^2} = (x - y) (x - z) (y + z) + ( x - y - z)^2. \end{aligned}$$

Example 5.10

In the 12 regions which are spanned by five rays, the polynomial that defines the boundary of IP has degree 5 and it looks like

$$\begin{aligned}&( (\sqrt{5} x + \sqrt{5} y - x + y)^2 - 4z^2) ( (\sqrt{5} x + x + 2 y)^2 - (\sqrt{5} z - z)^2) y \\&\quad + 8 \sqrt{5} x^3 y + 68 \sqrt{5} x^2 y^2 + 72 \sqrt{5} x y^3\\&\quad + 20 \sqrt{5} y^4 - 40 \sqrt{5} x y z^2 - 20 \sqrt{5} y^2 z^2 + 4 \sqrt{5} z^4 \\&\quad + 8 x^3 y + 164 x^2 y^2 + 168 x y^3 + 44 y^4 - 8 x^2 z^2 - 72 x y z^2 - 44 y^2 z^2 + 12 z^4. \end{aligned}$$

In the other 20 regions spanned by three rays, \(\partial IP\) is the zero set of a sextic polynomial with the following shape

$$\begin{aligned}&( (\sqrt{5} x + x + 2 y )^2 - ( \sqrt{5} z - z)^2 ) ( (\sqrt{5} y - 2 x - y )^2 - ( \sqrt{5} z - z)^2 ) x y + 20 \sqrt{5} x^4 y \\&\quad - 20 \sqrt{5} x^2 y^3 - 4 \sqrt{5} x y^4 + 4 \sqrt{5} y^5 - 4 \sqrt{5} x^3 z^2 - 60 \sqrt{5} x^2 y z^2\\&\quad - 12 \sqrt{5} x y^2 z^2 + 12 \sqrt{5} x z^4 + 44 x^4 y \\&\quad - 8 x^3 y^2 - 44 x^2 y^3 + 12 x y^4 + 12 y^5 - 12 x^3 z^2\\&\quad - 156 x^2 y z^2 - 60 x y^2 z^2 - 8 y^3 z^2 + 28 x z^4. \end{aligned}$$

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

$$\begin{aligned} 3 x y z - 3 w^2 - 6 x^2 - 12 x y - 6 y^2 - 12 x z + 12 y z - 6 z^2. \end{aligned}$$

Finally there are 64 cones spanned by 4 rays such that the boundary of the intersection body is a quartic, such as

$$\begin{aligned}&4 w x y z - w^3 - 3 w^2 x - 3 w x^2 - x^3 - 3 w^2 y - 6 w x y - 3 x^2 y - 3 w y^2 - 3 x y^2 \\&\quad - y^3 - 3 w^2 z - 6 w x z - 3 x^2 z + 18 w y z - 6 x y z\\&\quad - 3 y^2 z - 3 w z^2 - 3 x z^2 - 3 y z^2 - z^3. \end{aligned}$$