Question

Suppose \({{\bf{v}}_{\bf{1}}}\),….., \({{\bf{v}}_{\bf{k}}}\) are linearly independent vectors in \({\mathbb{R}^n}\left( {{\bf{1}} \le k \le n} \right)\). Then the set \({X^k} = {\bf{conv}}\;\left\{ { \pm {{\bf{v}}_{\bf{1}}},......, \pm {{\bf{v}}_k}} \right\}\) is called a k-crosspolytope.

a. Sketch \({X^{\bf{1}}}\) and \({X^{\bf{2}}}\).

b. Determine the number of k-faces of the 3-dimensional crosspolytope \({X^{\bf{3}}}\) for \(k = {\bf{0}},{\bf{1}},\,{\bf{2}}\). What is the another name of \({X^{\bf{3}}}\).

c. Determine the number of k-faces of the 4-dimensional crosspolytope \({X^{\bf{4}}}\) for \(k = {\bf{0}},{\bf{1}},\,{\bf{2}},{\bf{3}}\). Verify that your answer satisfies Euler’s formula.

d. Find a formula for \({f_k}\left( {{X^n}} \right)\), the number of k-faces of \({X^n}\) for \({\bf{0}} \le k \le n - {\bf{1}}\).

Short answer

1. 
   

b. 6, 12, 8, and another name of \({X^3}\) is an octahedron.

c. 8, 24, 32, 16

d. \({f_k}\left( {{X^n}} \right) = {2^{k + 1}}\left( {\begin{array}{*{20}{c}}n\\{k + 1}\end{array}} \right)\). Here \(\left( {\begin{array}{*{20}{c}}a\\b\end{array}} \right) = \frac{{a!}}{{b!\left( {a - b} \right)!}}\) is the binomial coefficient.

Step-by-step solution

Find the solution for part (a)

The plot of \({X^1}\) is a straight line. The figure below represents the sketch for \({X^1}\) passing through vectors 0 and \({{\bf{v}}_1}\).

The plot of \({X^2}\) is a parallelogram for \(k = 2\), and the vectors \({{\bf{v}}_1}\) and \({{\bf{v}}_2}\) that is shown below:

Find the solution for part (b)

The number offaces for \(k = 0\):

\({f_0}\left( {{X^3}} \right) = 6\)

The number of faces for \(k = 1\):

\({f_1}\left( {{X^3}} \right) = 12\)

The number of faces for \(k = 2\):

\({f_2}\left( {{X^3}} \right) = 8\)

Euler’s formulacan be verified as follows:

\(\begin{array}{c}{f_0}\left( {{X^3}} \right) - {f_1}\left( {{X^3}} \right) + {f_2}\left( {{X^3}} \right) = 6 - 12 + 8\\ = 2\end{array}\)

Another name for \({X^3}\) is Octahedron.

Find the solution for part (c)

The number facesfor \(k = 0\):

\({f_0}\left( {{X^4}} \right) = 8\)

The number of faces for \(k = 1\):

\({f_1}\left( {{X^4}} \right) = 24\)

The number of faces for \(k = 2\):

\({f_2}\left( {{X^4}} \right) = 32\)

The number of faces for \(k = 3\):

\({f_3}\left( {{X^4}} \right) = 16\)

Euler’s formula can be verified as follows:

\(\begin{array}{c}{f_0}\left( {{X^4}} \right) - {f_1}\left( {{X^4}} \right) + {f_2}\left( {{X^4}} \right) + {f_3}\left( {{X^4}} \right) = 8 - 24 + 32 - 16\\ = 0\end{array}\)

Find the solution for part (d)

The pattern for the above chart is given by the formula \({f_k}\left( {{X^n}} \right) = {2^{k + 1}}\left( {\begin{array}{*{20}{c}}n\\{k + 1}\end{array}} \right)\). Here \(\left( {\begin{array}{*{20}{c}}a\\b\end{array}} \right) = \frac{{a!}}{{b!\left( {a - b} \right)!}}\) is the binomial coefficient.