The Maximum Number of Faces of the Minkowski Sum of Two Convex Polytopes

Abstract

We derive tight bounds for the maximum number of k-faces, $0\le k\le d-1$, of the Minkowski sum, $P_1+P_2$, of two d-dimensional convex polytopes $P_1$ and $P_2$, as a function of the number of vertices of the polytopes. For even dimensions $d\ge 2$, the maximum values are attained when $P_1$ and $P_2$ are cyclic d-polytopes with disjoint vertex sets. For odd dimensions $d\ge 3$, the maximum values are attained when $P_1$ and $P_2$ are $\lfloor \frac{d}{2}\rfloor$-neighborly d-polytopes, whose vertex sets are chosen appropriately from two distinct d-dimensional moment-like curves.

Appendix

1.1 Proof of Lemma 3

Proof

McMullen [20] in his original proof of the Upper Bound Theorem for polytopes proved that for any d-polytope P the following relation holds:

$$\begin{aligned} (k+1)h_{k+1}(\partial {P})+(d-k)h_k(\partial {P}) =\sum _{v\in \,{\mathrm{vert}}(P)}h_k({\partial {P}}/{v}),\qquad 0\le k\le d-1. \end{aligned}$$

(59)

Multiplying both sides of (59) by $t^{d-k-1}$ and summing over all $0 \le k \le d$, we get

$$\begin{aligned}&\sum _{k=0}^{d} (k+1)h_{k+1}(\partial {P}) t^{d-k-1} + \sum _{k=0}^{d} (d-k)h_k(\partial {P})t^{d-k-1} \nonumber \\&\quad = \sum _{k=0}^{d} \sum _{v\in \,{\mathrm{vert}}(P)}h_k({\partial {P}}/{v})t^{d-k-1}. \end{aligned}$$

(60)

For the right-hand side of (60), we have

$$\begin{aligned} \sum _{k=0}^{d} \sum _{v\in \,{\mathrm{vert}}(P)}h_k({\partial {P}}/{v})t^{d-k-1}= & {} \sum _{v\in \,{\mathrm{vert}}(P)}\sum _{k=0}^{d}h_k({\partial {P}}/{v}) t^{d-1-k}\nonumber \\= & {} \sum _{v\in \,{\mathrm{vert}}(P)} {\mathtt {h}}({{\partial {P}}/{v}};t), \end{aligned}$$

(61)

whereas for the left-hand side of (60), we get

$$\begin{aligned}&\sum _{k=0}^{d} (k+1)h_{k+1}(\partial {P}) t^{d-k-1} + \sum _{k=0}^{d} (d-k)h_k(\partial {P})t^{d-k-1} \nonumber \\&\quad =\sum _{k=0}^{d} k h_{k}(\partial {P}) t^{d-k} + \sum _{k=0}^{d} (d-k)h_k(\partial {P})t^{d-k-1}\nonumber \\&\quad = d \sum _{k=0}^{d} h_{k}(\partial {P}) t^{d-k} +(1-t)\sum _{k=0}^{d} (d-k) h_{k}(\partial {P}) t^{d-k-1}\nonumber \\&\quad =d\, {\mathtt {h}}({\partial {P}};t)+(1- t)\, {\mathtt {h'}}({\partial {P}};t). \end{aligned}$$

(62)

Substituting in (60), from (61) and (62), we recover the relation in the statement of the lemma. $\square$

1.2 Proof of an Identity

In this section we prove the following identity, used in Sect. 4, to prove the upper bound for $f_{k-1}({\mathcal {F}})$ (see relations (50) and (51)).

Lemma 23

For any $d\ge 2$ and any sequence of numbers $\alpha _i$, where $0\le i\le \lfloor \frac{d+1}{2}\rfloor$, we have

$$\begin{aligned}&\sum _{i=0}^{\lfloor \frac{d+1}{2}\rfloor }\Big (\begin{array}{cc}{d+1-i}\\ {k-i} \end{array}\Big )\alpha _i +\sum _{i=0}^{\lfloor \frac{d}{2}\rfloor }\Big (\begin{array}{cc}{i}\\ {k-d-1+i} \end{array}\Big )\alpha _i\\&\quad = \sum \limits _{i=0}^{\frac{d+1}{2}}{^{^{*}}} \Big (\Big (\begin{array}{cc}{d+1-i}\\ {k-i} \end{array}\Big ) +\Big (\begin{array}{cc}{i}\\ {k-d-1+i} \end{array}\Big )\Big )\alpha _i. \end{aligned}$$

Proof

We start by recalling the definition of the symbol $\sum \nolimits _{i=0}^{\frac{\delta }{2}}{^{^{*}}}T_i$. This symbol denotes the sum of the elements $T_0,T_1,\ldots ,T_{\lfloor \frac{\delta }{2}\rfloor }$, where the last term is halved if $\delta$ is even. More precisely:

$$\begin{aligned} \sum \limits _{i=0}^{\frac{\delta }{2}}{^{^{*}}}T_i= {\left\{ \begin{array}{ll} T_0+T_1+\cdots +T_{\lfloor \frac{\delta }{2}\rfloor -1}+\frac{1}{2}T_{\lfloor \frac{\delta }{2}\rfloor } &{}\text {if }\delta \text { is even},\\ T_0+T_1+\cdots +T_{\lfloor \frac{\delta }{2}\rfloor -1}+T_{\lfloor \frac{\delta }{2}\rfloor } &{}\text {if }\delta \text { is odd}. \end{array}\right. } \end{aligned}$$

Let us now first consider the case d odd. In this case $d+1$ is even, and we have

$$\begin{aligned}&\sum _{i=0}^{\lfloor \frac{d+1}{2}\rfloor }\Big (\begin{array}{cc}{d+1-i}\\ {k-i}\end{array}\Big )\alpha _i +\sum _{i=0}^{\lfloor \frac{d}{2}\rfloor }\Big (\begin{array}{cc}{i}\\ {k-d-1+i}\end{array}\Big )\alpha _i\\&\quad =\sum _{i=0}^{\lfloor \frac{d+1}{2}\rfloor }\Big (\begin{array}{cc}{d+1-i}\\ {k-i}\end{array}\Big )\alpha _i +\sum _{i=0}^{\lfloor \frac{d+1}{2}\rfloor -1}\Big (\begin{array}{cc}{i}\\ {k-d-1+i}\end{array}\Big )\alpha _i\\&\quad =\sum _{i=0}^{\lfloor \frac{d+1}{2}\rfloor -1}\Bigg (\left( {\begin{array}{c}d+1-i\\ k-i\end{array}}\right) +\left( {\begin{array}{c}i\\ k-d-1+i\end{array}}\right) \Bigg )\alpha _i +\left( {\begin{array}{c}d+1-\lfloor \frac{d+1}{2}\rfloor \\ k-\lfloor \frac{d+1}{2}\rfloor \end{array}}\right) \alpha _{\lfloor \frac{d+1}{2}\rfloor }\\&\quad = \sum _{i=0}^{\lfloor \frac{d+1}{2}\rfloor -1}\Bigg (\left( {\begin{array}{c}d+1-i\\ k-i\end{array}}\right) +\left( {\begin{array}{c}i\\ k-d-1+i\end{array}}\right) \Bigg )\alpha _i\\&\qquad \,\, +{\textstyle \frac{1}{2}}\Bigg (\left( {\begin{array}{c}d+1-\lfloor \frac{d+1}{2}\rfloor \\ k-\lfloor \frac{d+1}{2}\rfloor \end{array}}\right) +\left( {\begin{array}{c}\lfloor \frac{d+1}{2}\rfloor \\ k-d-1+\lfloor \frac{d+1}{2}\rfloor \end{array}}\right) \Bigg )\alpha _{\lfloor \frac{d+1}{2}\rfloor }\\&\quad =\sum \limits _{i=0}^{\frac{d+1}{2}}{^{^{*}}} \Bigg (\left( {\begin{array}{c}d+1-i\\ k-i\end{array}}\right) +\left( {\begin{array}{c}i\\ k-d-1+i\end{array}}\right) \Bigg )\alpha _i. \end{aligned}$$

The case d even is even simpler to prove. In this case $d+1$ is odd; hence

$$\begin{aligned}&\sum _{i=0}^{\lfloor \frac{d+1}{2}\rfloor }\Big (\begin{array}{cc}{d+1-i}\\ {k-i}\end{array}\Big )\alpha _i +\sum _{i=0}^{\lfloor \frac{d}{2}\rfloor }\Big (\begin{array}{cc}{i}\\ {k-d-1+i}\end{array}\Big )\alpha _i\\&\quad =\sum _{i=0}^{\lfloor \frac{d+1}{2}\rfloor }\Big (\begin{array}{cc}{d+1-i}\\ {k-i}\end{array}\Big )\alpha _i +\sum _{i=0}^{\lfloor \frac{d+1}{2}\rfloor }\Big (\begin{array}{cc}{i}\\ {k-d-1+i}\end{array}\Big )\alpha _i\\&\quad =\sum _{i=0}^{\lfloor \frac{d+1}{2}\rfloor }\Big (\Big (\begin{array}{cc}{d+1-i}\\ {k-i}\end{array}\Big ) +\Big (\begin{array}{cc}{i}\\ {k-d-1+i}\end{array}\Big )\Big )\alpha _i\\&\quad =\sum \limits _{i=0}^{\frac{d+1}{2}}{^{^{*}}} \Big (\Big (\begin{array}{cc}{d+1-i}\\ {k-i}\end{array}\Big )+\Big (\begin{array}{cc}{i}\\ {k-d-1+i}\end{array}\Big )\Big )\alpha _i \end{aligned}$$

This completes the proof. $\square$

1.3 Proof of Lemma 19

Before proceeding with the proof of Lemma 19, we need to introduce Vandermonde and generalized Vandermonde determinants. Given a vector of $n\ge 2$ real numbers ${\varvec{x}}=(x_1,x_2,\ldots ,x_n)$, the Vandermonde determinant $\text {VD}({\varvec{x}})$ of ${\varvec{x}}$ is the $n\times n$ determinant

$$\begin{aligned} \text {VD}({\varvec{x}})=\left| \begin{array}{ccccc} 1&{}1&{}\cdots &{}1\\ x_1&{}x_2&{}\cdots &{}x_n\\ x_1^2&{}x_2^2&{}\cdots &{}x_n^2\\ \vdots &{}\vdots &{}&{}\vdots \\ x_1^{n-1}&{}x_2^{n-1}&{}\cdots &{}x_n^{n-1}\\ \end{array} \right| =\prod _{1\le i<j\le n}(x_j-x_i). \end{aligned}$$

From the above expression, it is readily seen that if the elements of ${\varvec{x}}$ are in strictly increasing order, then $\text {VD}({\varvec{x}})>0$. A generalization of the Vandermonde determinant is the generalized Vandermonde determinant: if, in addition to ${\varvec{x}}$, we specify a vector of exponents ${\varvec{\mu }}=(\mu _1,\mu _2,\ldots ,\mu _n)$, where we require that $0\le \mu _1<\mu _2<\cdots <\mu _n$, we can define the generalized Vandermonde determinant $\text {GVD}({\varvec{x}};{\varvec{\mu }})$ as the $n\times n$ determinant:

$$\begin{aligned} \text {GVD}({\varvec{x}};{\varvec{\mu }})=\left| \begin{array}{ccccc} x_1^{\mu _1}&{}\quad x_2^{\mu _1}&{}\cdots &{}x_n^{\mu _1}\\ x_1^{\mu _2}&{}\quad x_2^{\mu _2}&{}\cdots &{}x_n^{\mu _2}\\ x_1^{\mu _3}&{}\quad x_2^{\mu _3}&{}\cdots &{}x_n^{\mu _3}\\ \vdots &{}\vdots &{}&{}\vdots \\ x_1^{\mu _n}&{}\quad x_2^{\mu _n}&{}\cdots &{}x_n^{\mu _n}\\ \end{array} \right| . \end{aligned}$$

It is a well-known fact that, if the elements of ${\varvec{x}}$ are in strictly increasing order, then $\text {GVD}({\varvec{x}};{\varvec{\mu }})>0$ (for example, see [8] for a proof of this fact).

To prove Lemma 19 we exploit the Cauchy–Binet formula (cf. [3]). Let M be a $n\times n$ square matrix factorized into a product LR of an $n \times m$ and an $m \times n$ matrix L and R respectively, with $m \ge n$. If J is a subset of $\{1,2,\ldots , m \}$ of size n, we denote by $L_{[n],J}$ the $n \times n$ matrix whose columns are the columns of L at indices from J and by $R_{J,[n]}$ the $n \times n$ matrix whose rows are the rows of R at indices from J. The Cauchy–Binet theorem states that

$$\begin{aligned} \det (M) = \det (L R) = \sum \limits _{J\in \left( {\begin{array}{c}[m]\\ n\end{array}}\right) } \det (L_{[n],J}) \det (R_{J,[n]}), \end{aligned}$$

(63)

where $\left( {\begin{array}{c}[m]\\ n\end{array}}\right)$ denotes the set of subsets of [m] of size n.

Proof

The determinant $D_{k,l}(\tau )$ is clearly a polynomial function of $\tau$. To prove our lemma, it suffices to show that the coefficient of the minimum exponent of $\tau$ in $D_{k,l}(\tau )$ is strictly positive.

In order to apply the Cauchy–Binet formula in our case, we factorize the matrix $\Delta _{k,l}(\tau )$, corresponding to the determinant $D_{k,l}(\tau )$, into the product of an $(m+3)\times 2(m+1)$ and a $2(m+1) \times (k+l)$ matrix L and R, respectively, as follows (recall that $m+3=k+l$):

The numbers over and sideways of L indicate the column and row indices, respectively, with $\hat{k} := k+m+1$. We partition the index set J into $J_1 \cup J_2$ where $J_1 \subseteq \{1,\ldots ,m+1\}$ and $J_2\subseteq \{\widehat{1}, \widehat{2},\ldots ,\widehat{m+1} \}.$ Notice that a term $\det (L_{[m+3],J}) \det (R_{J,[k+l]})$ in the Cauchy–Binet expansion of $D_{k,l}(\tau )$ vanishes in the following two cases:

• (i) $i \in J_1$ and $\hat{\imath }\in J_2$ for some $3 \le i \le m+1$; in this case the i-th and $\hat{\imath }$-th columns of $L_{[m+3],J}$ are identical, and thus $\det (L_{[m+3],J})=0$.

• (ii) $|J_1| \ne k$ or $|J_2| \ne l$; in this case $R_{J,[k+l]}$ is a block-diagonal square matrix with non-square non-zero blocks. The determinant of such a matrix is always zero.⁶

Furthermore, notice that for any non-vanishing term in the Cauchy–Binet expansion of $D_{k,l}(\tau )$, we have

$$\begin{aligned} \det (R_{J,[k+l]})= & {} \text {GVD}(\tau {\varvec{x}};{\varvec{\mu }}_1)\,\text {GVD}({\varvec{y}};{\varvec{\mu }}_2),\\ \tau {{\varvec{x}}}= & {} (\tau x_1,\ldots ,\tau x_k),\\ {\varvec{y}}= & {} (y_1,\ldots ,y_l), \end{aligned}$$

where ${\varvec{\mu }}_1$ (resp., ${\varvec{\mu }}_2$) is the vector consisting of the elements in $\{i-1\mid i\in J_1\}$ (resp., $\{i-(m+1)-1\mid i\in J_2\}$) ordered increasingly. The parameter $\tau$ appears only in $\text {GVD}(\tau {\varvec{x}};{\varvec{\mu }}_1)$ and can be factored out (see 64). We thus have

$$\begin{aligned} \det (R_{J,[k+l]} ) =\tau ^{M(J)}\,\text {GVD}({\varvec{x}};{\varvec{\mu }}_1)\,\text {GVD}({\varvec{y}};{\varvec{\mu }}_2)>0, \qquad M(J) = \sum _{i \in J_1}(i-1), \end{aligned}$$

since $\text {GVD}({\varvec{x}};{\varvec{\mu }}_1)$ and $\text {GVD}({\varvec{y}};{\varvec{\mu }}_2)$ are positive due to the way we have chosen ${\varvec{x}}$ and ${\varvec{y}}$.

Among all possible index sets $J=J_1\cup J_2$ for which the product $\det (L_{[m+3],J})\det (R_{J,[k+l]})$ does not vanish, we have to find the index set that gives the minimum exponent for $\tau$. Recall that, in view of condition (ii), the size of $J_1$ is k and the size of $J_2$ is l. The minimum exponent M(J) is then attained when $J_1=J_1^\star :=\{1,\ldots ,k \}$. In view of condition (i), we have $J_2 \subseteq \{\widehat{1}, \widehat{2},\ldots ,\widehat{m+1}\}\setminus \{\widehat{3},\ldots ,\widehat{k} \},$ which leaves no other choice but $J_2=J_2^\star :=\{ \widehat{1}, \widehat{2}, \widehat{k+1},\ldots ,\widehat{m+1} \}$.

For $J^\star =J_1^\star \cup J_2^\star$, the matrix $L_{[m+3],J}$ is

$$\begin{aligned} L_{[m+3],J^\star } = \left( \begin{array}{cccc} \mathrm{I}_2&{}\quad 0&{}\quad 0&{}\quad 0\\ 0&{}\quad 0&{}\quad \mathrm{I}_2&{}\quad 0\\ 0&{}\quad \mathrm{I}_k&{}\quad 0&{}\quad 0\\ 0&{}\quad 0&{}\quad 0&{}\quad \mathrm{I}_l \end{array}\right) . \end{aligned}$$

Clearly, $L_{[m+3],J^\star }$ becomes the identity matrix after 2k row swaps. Hence $\det (L_{[m+3],J^\star })=1$, which further implies that

$$\begin{aligned} \text {sign}(D_{k,l}(\tau ))=\text {sign}(R_{J^\star ,[k+l]}) > 0. \end{aligned}$$

$\square$