2.1 Exterior algebra

For the sake of completeness and clarity, all needed definitions from exterior algebra are given here in a brief and noncanonical way. We assume that the equivalence of our definitions to the usual ones is quite obvious. As a proper introduction to multilinear algebra, we refer to Greub’s book [Reference Greub9].

Let H be a finite-dimensional vector space with inner product $\left \langle \cdot , \cdot \right \rangle .$ We define the vector space $\Lambda ^{\ell } (H )$ as the space of the multilinear skew-symmetric functions on $\ell $ vectors of H with the natural linear structure. The vectors of $\Lambda ^{\ell } (H )$ are called $\ell $ -forms. As we consider only linear spaces with inner products, we assume that a space H and its dual $H^{*}$ coincide. This allows us to simplify the following definitions.

For a sequence of $\ell $ vectors $\{x_1, \dots , x_{\ell } \} \subset H,$ we define an $\ell $ -form $x_1 \wedge \dots \wedge x_{\ell }$ by its evaluation on vectors $\{y_1, \dots , y_{\ell }\} \subset H$ given by

$$ \begin{align*} x_1 \wedge \dots \wedge x_{\ell} (y_1, \dots, y_{\ell}) = \det M, \qquad \text{where} \qquad M_{[ij]} = \left\langle x_i,y_j\right\rangle, \quad i,j \in [\ell]. \end{align*} $$

By the properties of the determinant, $x_1 \wedge \dots \wedge x_{\ell }$ is a multilinear skew-symmetric function of $\ell $ vectors of H. For the sake of convenience, for given n vectors $\{x_i\}_1^{n}$ and $\ell $ -tuple $L =\{i_1, \dots , i_{\ell }\} \in \binom {[n]}{\ell },$ we denote by $x_L$ the $\ell $ -form $x_{i_1} \wedge \dots \wedge x_{i_{\ell }}$ .

As usual, we use the set of $\ell $ -forms $e_{i_1} \wedge \dots \wedge e_{i_{\ell }},$ where $\{i_1, \dots , i_{\ell }\} \in \binom {[n]}{\ell }$ , as the standard basis of $\Lambda ^{\ell } \left ( \mathbb {R}^n \right ).$ We use the lexicographical order on $\ell $ -tuples $\{i_1, \dots , i_{\ell }\} \in \binom {[n]}{\ell }$ to assign a number to $e_{i_1} \wedge \dots \wedge e_{i_{\ell }}$ in this basis.

Recall that for a vector space $H,$ an $\ell $ -form $w \in \Lambda ^{\ell } (H)$ is said to be decomposable if it can be represented in the form $x_1 \wedge \dots \wedge x_{\ell }$ for $\{x_i\}_1^{\ell } \subset H.$ Every $\ell $ -form is a linear combination of some decomposable $\ell $ -forms (e.g., of the forms from the standard basis), but not all $\ell $ -forms are decomposable, e.g., $e_1 \wedge e_2 + e_3 \wedge e_4 \in \Lambda ^2 (\mathbb {R}^4)$ is not decomposable. A line $l \subset \Lambda ^{\ell } \left ( \mathbb {R}^n \right )$ has a decomposable directional vector iff l is “generated” by some $\ell $ -dimensional subspace $H_{\ell } \subset \mathbb {R}^n.$

By linearity, it suffices to define a linear operator (or an inner product) on $\Lambda ^{\ell }(\mathbb {R}^n)$ only for decomposable $\ell $ -forms, since the decomposable forms span $\Lambda ^{\ell } (\mathbb {R}^n).$

Recall that the exterior $\ell $ -power of an operator A on $\mathbb {R}^n$ is a linear operator on $\Lambda ^{\ell } (\mathbb {R}^n),$ which is defined on decomposable forms by

$$ \begin{align*} \wedge^{\ell} A (x_1 \wedge \dots \wedge x_{\ell}) = Ax_1 \wedge \dots \wedge Ax_{\ell}. \end{align*} $$

Recall that the inner product of two decomposable $\ell $ -forms $a = a_1 \wedge \dots \wedge a_{\ell }$ and $b = b_1 \wedge \dots \wedge b_{\ell }$ is defined by

$$ \begin{align*} \left\langle a,b\right\rangle = \left\langle a_1 \wedge \dots \wedge a_{\ell},b_1 \wedge \cdots \wedge b_{\ell}\right\rangle = a_1 \wedge \dots \wedge a_{\ell} (b_1 , \dots, b_{\ell}). \end{align*} $$

In this way, once we fix an inner product on $\mathbb {R}^n$ , we fix the inner product on $\Lambda ^{\ell } (\mathbb {R}^n).$ Then, for $\ell \in [n]$ , we can define a special isometry $ \star : \Lambda ^{\ell } (\mathbb {R}^n) \to \Lambda ^{n - \ell } (\mathbb {R}^n),$ the so-called Hodge star operator, by the following equation:

$$ \begin{align*} a \wedge \star(b) = \left\langle a,b\right\rangle e_1 \wedge \dots \wedge e_n, \end{align*} $$

where $a,b$ are $\ell $ -forms.

Using this, one can show the following. For given $a=a_1 \wedge \dots \wedge a_{\ell } \in \Lambda ^{\ell } (\mathbb {R}^n)$ and $b = b_1 \wedge \dots \wedge b_{n - \ell } \in \Lambda ^{n -\ell } (\mathbb {R}^n),$ we have

(2.2) $$ \begin{align} \left\langle a,\star (b)\right\rangle = (-1)^{\ell (n - \ell)}\det \{a_1, \dots, a_{\ell}, b_1, \dots,b_{n - \ell}\}. \end{align} $$

For a k-dimensional subspace $H_k \subset \mathbb {R}^n,$ we use $\wedge ^{\ell } H_k$ to denote the linear hull of forms $x_1 \wedge \dots \wedge x_{\ell }$ in $\Lambda ^{\ell } \left ( \mathbb {R}^n \right )$ such that $\{x_1, \dots , x_{\ell }\} \subset H_k.$ Since $H_k$ inherits the inner product of $\mathbb {R}^n$ , the space $\wedge ^{\ell } H_k$ inherits the inner product of $\Lambda ^{\ell } (\mathbb {R}^n)$ and thus can be identified with the space $\Lambda ^{\ell } (H_k)$ in a tautological way. This allows us to use the Hodge star operator for the spaces $\Lambda ^{\ell } (H_k)$ and $\Lambda ^{(k -\ell )} (H_k)$ for $\ell \in [k].$ Also, identity (2.2) can be rewritten in the following form. For given $a=a_1 \wedge \dots \wedge a_{\ell } \in \Lambda ^{\ell } (H_k)$ and $b = b_1 \wedge \dots \wedge b_{k - \ell } \in \Lambda ^{k -\ell } (H_k),$ we get

(2.3) $$ \begin{align} \left\langle a,\star (b)\right\rangle = (-1)^{\ell (k - \ell)}\det \{a_1, \dots, a_{\ell}, b_1, \dots,b_{k - \ell}\}, \end{align} $$

where we understand the determinant as the determinant of k-vectors in a k-dimensional space $H_k$ .

The next result is a straightforward consequence of the definitions. We show that the outer product and the tensor product commute for the exterior powers of operators in our case. Usually, definitions of the outer product involve some kind of (anti)symmetrization, which can give an additional factor when we exchange the $\wedge $ -product and $\otimes $ -product. To avoid misunderstandings, we prove the following statement.

Proof We can assume that $\ell \leq \mathop {\mathrm {dim}} \mathop {\mathrm {Lin}} \{v_1, \dots , v_t\} \leq t,$ otherwise $\wedge ^{\ell } A = 0 = \sum \limits _{L \in \binom {[n]}{\ell }} v_{L} \otimes v_L.$ It suffices to prove the identity only for the decomposable $\ell $ -forms. Fix $\{x_1, \dots , x_{\ell }\} \subset \mathbb {R}^n.$ We have

$$ \begin{align*} \wedge^{\ell} A (x_{1} \wedge \dots \wedge x_{\ell}) = Ax_{1} \wedge \dots \wedge Ax_{\ell} = \sum\limits_{i=1}^t \left\langle v_i,x_1\right\rangle v_i \wedge \dots \wedge \sum\limits_{i=1}^t \left\langle v_i,x_{\ell}\right\rangle v_i. \end{align*} $$

By linearity and by skew symmetry, we expand the last identity, and the coefficient at $v_L = v_{i_1} \wedge \dots \wedge v_{i_{\ell }}$ is

$$ \begin{align*} \sum\limits_{\sigma} \operatorname{sgn}{\sigma} \prod\limits_{j = 1}^{\ell} \left\langle v_{i_j}, x_{\sigma(j)}\right\rangle, \end{align*} $$

where the summation is taken over all permutations of $[\ell ].$ This is the determinant of the matrix $M_{ij} = \left \langle v_{i_j}, x_i\right \rangle ,$ where $i,j \in [\ell ].$ By the definition of the inner product, this determinant is just $\left \langle v_L,(x_{1} \wedge \dots \wedge x_{\ell })\right \rangle\! ,$ that is,

$$ \begin{align*} \wedge^{\ell} A (x_{i_1} \wedge \dots \wedge x_{i_{\ell}}) = \sum\limits_{L \in \binom{[t]}{\ell}} \left\langle v_{L}, (x_{1} \wedge \dots \wedge x_{\ell})\right\rangle v_L = \sum\limits_{L \in \binom{[t]}{\ell}} v_{L} \otimes v_L (x_{1} \wedge \dots \wedge x_{\ell}). \end{align*} $$

This completes the proof. ▪

Identifying $\wedge ^{k-1} H_k$ with $\Lambda ^{k-1}(H_k)$ , we identify $H_k$ with $\wedge ^{k-1}{H_k}$ using the Hodge star operator. Now, we can define the cross product of $k-1$ vectors $\{x_1, \dots , x_{k-1}\}$ by

$$ \begin{align*} [x_1, \dots, x_{k-1}] = \star (x_1 \wedge \dots \wedge x_{k-1}). \end{align*} $$

Or, in other words, by linearity of the determinant, the cross product $x = [x_1, \dots , x_{k-1}]$ of $k-1$ vectors $\{x_1, \dots , x_{k-1}\}$ in a k-dimensional space $H_k$ with the fixed inner product $\left \langle \cdot ,\cdot \right \rangle $ is the vector defined by

$$ \begin{align*} \left\langle x,y\right\rangle = \det ( x_1, \dots, x_{k-1}, y) \quad \text{for all} \quad y \in H_k. \end{align*} $$

3.1 Lagrange’s identity and McMullen’s formula

McMullen’s formula (1.3) follows from Shepard’s formula [Reference Shephard17] for the volume of a $\sum _{i=1}^{n} [0, b_i] \subset \mathbb {R}^k$

(3.3) $$ \begin{align} \operatorname{vol}_{k} \left(\sum\limits_{i=1}^{n} [0, b_i]\right) = \sum\limits_{\{i_1, \dots, i_k\} \in \binom{[n] }{k}} \left|\det \{b_{i_1}, \dots, b_{i_k}\} \right|, \end{align} $$

and identity $\left |d_S (\{L\})\right | = \left |d_S (\{[n] \backslash L\})\right |$ for a tight frame S and a subset L of $[n].$ By assertion (4) of Lemma 3.1, this identity is equivalent to identity $|U_L| = |U_{[n] \backslash L}|$ for an orthogonal matrix U of rank n and any $L \subset [n],$ which is known as Lagrange’s identity. However, it is natural to prove it in terms of properties of a projection operator.

Lemma 3.6 Let $H_{n-k}$ be the orthogonal complement of $H_k$ in $\mathbb {R}^n.$ Denote by P and $P^\perp $ the projections onto $H_k$ and $H_{n-k},$ respectively. Then,

$$ \begin{align*} \star \left(\wedge^k P [x] \right) = \left( \wedge^{(n- k)} P^\perp \right) [\star x] \end{align*} $$

for any $x \in \Lambda ^k \left (\mathbb {R}^n \right ).$

Proof We use $\pi $ to denote a unit directional vector of the line ${\wedge ^k} H_k$ generated by $H_k$ in $\Lambda ^k \left (\mathbb {R}^n \right ).$ Clearly, $\star \pi $ is the unit directional vector (with a proper orientation) of the line $\wedge ^{(n-k)} H_{n-k}$ generated by $H_{n-k}$ in $\Lambda ^{n-k} \left (\mathbb {R}^n \right ).$ For an arbitrary $x \in \Lambda ^k \left (\mathbb {R}^n \right )$ and $y \in \Lambda ^{n-k} \left (\mathbb {R}^n \right ),$ we have ${\wedge ^k}P [x] = c_1 \pi \in {\wedge ^k} H_k$ and $\left ({\wedge ^{(n - k)}} P^\perp \right ) [ y ] = c_2 (\star \pi ) \in {\wedge ^{(n - k)}}H_{n-k}$ for some scalars $c_1, c_2.$ From this and by linearity of operators $\star \left ({\wedge ^k} P\right )$ and $\left (\wedge ^{(n - k)} P^\perp \right ) \star ,$ we obtain that $\star \left ({\wedge ^k} P\right ) [x] = c \left (\wedge ^{(n - k)} P^\perp \right ) [\star x]$ for some scalar c. From the chain

$$ \begin{align*} \star \left({\wedge^k} P\right) \pi = \star \pi = \left(\wedge^{(n - k)} P^\perp \right) [\star \pi], \end{align*} $$

we conclude that $c=1.$ This completes the proof. ▪

So, Lagrange’s identity takes the following form with a simple geometric meaning.

Corollary 3.7 In the notation of Lemma 3.6 , we have

(3.4) $$ \begin{align} P_L = P^\perp_{[n]\backslash L} \end{align} $$

for $L \in \binom {[n]}{k}.$

4.1 Equivalent formulation

Tight frame $\{P^H e_1, \dots , P^H e_n \},$ where H is a k-dimensional subspace of $\mathbb {R}^n$ and $P^H$ is the orthogonal projection onto $H,$ can be considered as the ordered set of n vectors of $\mathbb {R}^k.$ That allows us to avoid working in Grassmannian $\operatorname {Gr}\!\left ( n, k \right )\!.$ However, there is an ambiguity in correspondence $H \to \{P^H e_1, \dots , P^H e_n \}.$ Any choice of orthonormal basis of H gives its own tight frame in $\mathbb {R}^k,$ all of them are isometric but different from each other. Let us formalize our technique and explain how one can deal with this ambiguity.

We endow $\Omega (n,k)$ with the metric

$$ \begin{align*} \operatorname{dist}\!\left( \{v_1, \dots, v_n\}, \{w_1, \dots, w_n\} \right) = \sqrt{\sum\limits_1^n |v_i - w_i|^2}. \end{align*} $$

The orthogonal group $\operatorname {O}\left ( k \right )$ acts on $\Omega (n,k)$ in a usual way

$$ \begin{align*} U \{v_1, \dots, v_n\} = \{U v_1, \dots, U v_n\} \quad \text{for} \quad U \in \operatorname{O}\left( k \right). \end{align*} $$

The equivalence classes of this action are called O-classes, and the O-class of a tight frame S is denoted as $\left [ S \right ].$ We say that a function $\Psi : \Omega (n, k) \to \mathbb {R}$ is O-invariant if it is constant on each O-class.

There is a one-to-one correspondence between $\operatorname {Gr}\!\left ( n, k \right )$ and $\frac {\Omega (n,k)}{\operatorname {O}\left ( k \right )}:$

• • the map $\alpha : \operatorname {Gr}\!\left ( n, k \right ) \to \frac {\Omega (n,k)}{\operatorname {O}\left ( k \right )}$ is defined by

  $$ \begin{align*} \alpha(H) = \left[ P^H e_1, \dots, P^H e_n \right] \quad \text{for} \quad H \in \operatorname{Gr}\!\left( n, k \right). \end{align*} $$

• • the map $\beta : \frac {\Omega (n,k)}{\operatorname {O}\left ( k \right )} \to \operatorname {Gr}\!\left ( n, k \right )$ is defined as follows. Let $\{v_1, \dots , v_n\} \in \Omega (n,k).$ By Lemma 3.1, the Gram matrix P of $\{v_1, \dots , v_n\}$ is the matrix of projection onto a k-dimensional subspace $H.$ Then, $\beta \left ( \left [\{v_1, \dots , v_n\} \right ] \right ) = H.$ Clearly, an orthogonal transformation does not change the Gram matrix. Therefore, $\beta $ is defined correctly.

By assertion (4) of Lemma 3.1, the sets of vectors $\{v_1, \dots , v_n\}$ and $\{P e_1, \dots , P e_n\}$ are isometric. Thus, $\alpha $ and $\beta $ are the inverse functions of each other.

Consequently, there is a one-to-one correspondence between the space of functions $\operatorname {Gr}\!\left ( n, k \right ) \to \mathbb {R}$ and the space of O-invariant functions $\Omega (n,k) \to \mathbb {R}.$ For any function $\Psi : \operatorname {Gr}\!\left ( n, k \right ) \to \mathbb {R},$ we define its frame-function $\gamma (\Psi ): \Omega (k, n) \to \mathbb {R}$ by

$$ \begin{align*} \gamma(\Psi)\left( S \right) = \Psi(\beta([S])). \end{align*} $$

Clearly, the frame-function is O-invariant. Thus, by the above argument, $\Psi $ and $\gamma (\Psi )$ have the same global extrema. Let us show that the local extrema of any continuous function $\operatorname {Gr}\!\left ( n, k \right ) \to \mathbb {R}$ and its frame-functions are the same up to factorization by $\operatorname {O}\left ( k \right )\!.$ Of course, we need to specify a topology or metric on $\operatorname {Gr}\!\left ( n, k \right )\!.$ We consider the natural one as a topology of a homogeneous space $\left (\operatorname {Gr}\!\left ( n, k \right ) = \operatorname {O}\left ( n \right )/ (\operatorname {O}\left ( k \right ) \times \operatorname {O}\left ( n-k \right )) \right )$ , which is the metric topology generated by metric

$$ \begin{align*} \operatorname{Dist} (H, H') = \left\| P^H - P^{H'} \right\|, \end{align*} $$

where $\left \| \cdot \right \|$ denotes the operator norm.

Proof The equivalence of the first two assertions is trivial as $\gamma (\Psi )$ is O-invariant. Let us show that they are equivalent to the third one.

We define a metric on $\frac {\Omega (n,k)}{\operatorname {O}\left ( k \right )}$ as follows:

(4.1) $$ \begin{align} \rho\left(O_1, O_2 \right) = \min\left\{\operatorname{dist}\!\left( S_1, S_2 \right): \;\; S_1 \in O_1, \; S_2 \in O_2 \right\}. \end{align} $$

Obviously, the minimum is attained. Let us check the triangle inequality for arbitrary O-classes $O_1, O_2$ , and $O_3.$ Since $\operatorname {O}\left ( k \right )$ acts transitively on each O-class, there is $S_2 \in O_2$ such that $\rho \left (O_1, O_2 \right ) = \operatorname {dist}\!\left ( S_1, S_2 \right )$ for any $S_1 \in O_1.$ We assume that the minimum in (4.1) is attained at $S_1$ and $S_2$ for $O_1$ and $O_2,$ and at $S_2$ and $S_3$ for $O_2$ and $O_3.$ Then,

$$ \begin{align*} \rho\left(O_1, O_2 \right) + \rho\left(O_2, O_3 \right) = \operatorname{dist}\!\left( S_1, S_2 \right) + \operatorname{dist}\!\left( S_2, S_3 \right) \geq \operatorname{dist}\!\left( S_1, S_3 \right) \geq \rho\left(O_1, O_3 \right). \end{align*} $$

Clearly, the tight frames of $[S]$ are the local extrema of $\gamma (\Psi )$ iff $[S]$ is a local extremum of the function $\Psi _O : \frac {\Omega (n,k)}{\operatorname {O}\left ( k \right )} \to \mathbb {R}$ defined by $\Psi _O\left (\left [S\right ]\right ) = \gamma (\Psi (S)).$ To complete the proof, it suffices to show that $\operatorname {Gr}\!\left ( n, k \right )$ and $\frac {\Omega (n,k)}{\operatorname {O}\left ( k \right )}$ are homeomorphic. This follows from the observation that the operator norm is equivalent to the Hilbert–Schmidt operator norm $\left (\left \| P^H \right \|_{HS} = \sqrt {\sum \limits _1^n |P^H e_i|^2} \right ),$ and the latter is equivalent to metric $\rho $ on $\frac {\Omega (n,k)}{\operatorname {O}\left ( k \right )}.$  ▪

4.2 Perturbation of frames

Our main idea of finding local extrema of an O-invariant function $\Psi $ is to transform a given tight frame S to a new one $S'.$ However, it is not convenient to restrict ourselves to the set of tight frames. For example, consider function $\Psi _1 : \Omega (n,k) \to \mathbb {R}_+$ given by $\Psi _1 \left (\{v_1, \dots , v_n\} \right ) = \operatorname {vol}_{k} \mathop {\mathrm {co}} \{\pm v_1, \dots , \pm v_n\}$ (that is, the frame-function of the volume of a projection of the cross-polytope). It is clear what happens with the convex hull when we move one or several vectors of the tight frame, but we may destroy the tight frame condition with such a transformation. It is not a problem as we can easily extend the domain.

For a frame $S = \{v_1, \dots , v_n\}$ and a linear transformation $L,$ we denote $\{Lv_1, \dots , Lv_n\}$ by $LS.$ The operator $B_S$ is well-defined as the condition $\mathop {\mathrm {Lin}} S = \mathbb {R}^k$ implies that $A_S$ is a positive definite operator. Clearly, for any frame $S,$ the frame $B_S S$ is a tight frame:

$$ \begin{align*} \sum_{i=1}^{n} B_S v_i \otimes B_S v_i = B_S \left( \sum_{i=1}^{n} v_i \otimes v_i \right) B_S^T = B_S A_S B_S = I_k. \end{align*} $$

It is easy to see that the metric $\operatorname {dist}\!\left ( \cdot , \cdot \right )$ on $\Omega (n,k)$ extends to the set of all frames. Also, a small enough perturbation of a frame S yields a small perturbation of $B_S.$

Typically, O-invariant functions can be extended to the set of all frames in a tautological way, for example, the extensions of O-invariant functions for the objects listed above might be:

1. (1) $\Psi _1 \left ( \{v_1, \dots , v_n\} \right ) = \operatorname {vol}_{k} \mathop {\mathrm {co}}\{ \pm v_1, \dots , \pm v_n\};$

2. (2) $\Psi _2 \left ( \{v_1, \dots , v_n\} \right ) = \operatorname {vol}_{k} \left (\bigcap \limits _1^n \left \{x \in \mathbb {R}^k : |\left \langle x,v_i\right \rangle | \leq 1 \right \} \right );$

3. (3) $\Psi _3 \left ( \{v_1, \dots , v_n\} \right ) = \operatorname {vol}_{k}\left ( \sum \limits _{1}^{n} [-v_i, v_i]\right ).$

We prefer to extend the functions as follows:

(4.2) $$ \begin{align} \Psi \!\! \left( \tilde{S}\right) = \frac{\Psi\!\! \left(B_{\tilde{S}} \tilde{S} \right)}{\det B_{\tilde{S}}} \quad \text{for a frame}\ \tilde{S}. \end{align} $$

One can see that $\Psi _1$ and $\Psi _3$ satisfy this condition, and $\overline {\Psi }_2 = 1 / \Psi _2$ satisfies it as well. It looks as a natural extension at least for problems related to the volume of projections or sections of bodies because of the positive homogeneity of the volume $\operatorname {vol}_{k} L(K) = |\det L| \operatorname {vol}_{k} K,$ where K is a convex set and L is a linear transformation.

In order to obtain properties of extremizers, we consider a composition of two operations:

(4.3) $$ \begin{align} S \quad \stackrel{\mathbf{T}}{\longrightarrow} \quad \tilde{S} \quad \stackrel{B_{\tilde{S}}}{\longrightarrow} \quad S', \end{align} $$

where $\mathbf {T}$ is a map from a subset of $\Omega (n, k)$ to the set of frames in $\mathbb {R}^k$ and $B_{\tilde {S}}$ just maps $\tilde {S} = T (S)$ to a new tight frame $S'.$

Using our setting, it is easy to write a necessary and sufficient condition for a local extremum of an O-invariant function.

Lemma 4.2 A tight frame $S = \{v_1, \dots , v_n\}$ is a local maximum (or minimum) of a non-negative O-invariant function $\Psi $ iff the following inequality holds for any frame $\tilde {S}$ of some open neighborhood $U(S)$ of S in the set of frames:

(4.4) $$ \begin{align} \frac{\Psi(\tilde{S})}{\Psi\! \left(S\right)} \leq \sqrt{\det A_{\tilde{S}}} \quad ( or \geq). \end{align} $$

In addition, S is a global maximum (or minimum) iff inequality (4.4) holds for all frames $\tilde {S}.$

Proof As mentioned above, $B_{\tilde {S}} \tilde {S}$ is a tight frame and, by continuity, it belongs to a small enough neighborhood of S in $\Omega (n,k)$ if $\tilde {S} \in U(S).$ Using these observations and the definition of $B_S,$ we have

$$ \begin{align*} \frac{\Psi(\tilde{S})}{\Psi\! \left(S\right)} = \frac{1}{ \det B_{\tilde{S}}} \frac{\Psi(B_{\tilde{S}} \tilde{S})}{\Psi\! \left(S\right)} \leq \frac{1}{ \det B_{\tilde{S}}} = \sqrt{\det A_{\tilde{S}}}. \end{align*} $$

The equivalence for a global extremum is trivial. ▪

Choosing a proper simple operation $\mathbf {T}$ ( e.g., scaling one or several vectors, moving one vector to the origin, and mapping one vector to another), we may understand the geometric meaning of the left-hand side of (4.4). On the other hand, the determinant in the right-hand side of (4.4) can be calculated directly. In particular, the first-order approximation of the determinant is obtained in Theorem 1.2.

4.3 Computation of the determinants

Proof of Theorem 1.2

First of all, by the Cauchy–Binet formula, we have

(4.5) $$ \begin{align} \det \left( \sum\limits_1^n v_i \otimes v_i\right) = \sum\limits_{I \in \binom{[n]}{k}} P_I^S. \end{align} $$

By linearity and identity $P_I^S = ( d_S(\{I\}))^2,$ it suffices to prove the theorem for the case when only one vector of the set $\{x_i\}_1^n$ is nonzero. Thus, the theorem follows from the following lemma. ▪

Lemma 4.3 Let $S = \{v_1, \dots , v_n \}$ be a tight frame and $S'$ be a frame obtained from S by substitution $v_i \to v_i + t x,$ where $t \in \mathbb {R}, x \in \mathbb {R}^k.$ Then,

$$ \begin{align*} \sqrt{\det A_{S'}} = 1 + t \left\langle v_i,x\right\rangle + o(t). \end{align*} $$

That is, $\sqrt {\det A_{S'}}$ is a differentiable function of t at $t =0,$ and the derivative equals $\left \langle v_i,x\right \rangle .$

Proof Given a substitution $v_i \to v_i + t x,$ we modify the minor $P_I^S$ iff $i \in I.$ By the properties of minors of $P^S$ and definition of the Hodge star operator, we have that $P_I^S = (\left \langle v_i,\star (v_{I\backslash i})\right \rangle )^2$ if $i \in I.$ After the substitution, we get

$$ \begin{align*} \left(\left\langle v_i + t x, \star (v_{I\backslash i})\right\rangle\right)^2 = \left(\left\langle v_i, \star (v_{I\backslash i})\right\rangle\right)^2 + 2 t \left\langle v_i,\star (v_{I\backslash i})\right\rangle \left\langle \star (v_{I\backslash i}), x\right\rangle + o(t). \end{align*} $$

Hence,

$$ \begin{align*} \det A_{S'} = \sum\limits_{I \in {\binom{[n] \backslash i}{k}}} P_I^S + \sum\limits_{I \in \binom{[n]}{k}, i \in I} P_I^S + 2 t \sum\limits_{I \in \binom{[n]}{k}, i \in I } \left\langle v_i,\star (v_{I\backslash i})\right\rangle \left\langle\star (v_{I\backslash i}),x\right\rangle + o(t). \end{align*} $$

If $i \in J$ for $J \in \binom {[n]}{k-1},$ then we have $ \left \langle v_i,\star (v_{J})\right \rangle = 0.$ Therefore, we can rewrite the last identity

$$ \begin{align*} \det A_{S'} &= \sum\limits_{I \in \binom{[n]}{ k}} P_I^S + 2t \sum\limits_{J \in \binom{[n]}{k-1}} \left\langle v_i,\star (v_{J})\right\rangle \left\langle\star (v_{J}), x\right\rangle + o(t). \end{align*} $$

By Theorem 1.1, we know that the vectors $(\star (v_J))_{J \in \binom {[n]}{k-1} }$ form a tight frame in $H_k.$ Therefore,

$$ \begin{align*} \sum\limits_{J \in \binom{[n]}{k-1}} \left\langle v_i,\star (v_{J})\right\rangle \left\langle\star (v_{J}),x\right\rangle = \left\langle v_i,x\right\rangle. \end{align*} $$

Using this and by (4.5), we obtain

$$ \begin{align*} \sqrt{\det A_{S'}} = \sqrt{\det A_S + 2 t \left\langle v_i,x\right\rangle + o(t)} = 1 + t \left\langle v_i,x\right\rangle + o(t). \\[-35pt]\end{align*} $$

 ▪

Theorem 1.2 gives the first-order approximation of the left-hand side in inequality (4.4). It implies the first-order necessary condition for a differentiable O-invariant function $\Psi .$

Corollary 4.4 Let $\Psi $ be an O-invariant function, $S = \{v_1, \dots , v_n\}$ be a local extremum of $\Psi $ on $\Omega (n,k),$ and $\Psi $ is differentiable at $S.$ Then,

$$ \begin{align*} \Psi \!\left( \tilde{S}\right) = \Psi(S) \left(1 + \tau \sum\limits_{i = 1}^{n} \left\langle x_i,v_i\right\rangle \right) + o \left(\tau \right) \end{align*} $$

for an arbitrary $\tilde {S} = \{v_1 + \tau x_1, \dots , v_n + \tau x_n\}.$

By the dimension argument, Corollary 4.4 gives a complete system of the first-order necessary conditions. As a small enough perturbation of a frame is still a frame, we see that the set of frames looks like $\mathbb {R}^{nk}$ locally, and we can speak about local convexity of O-invariant functions. By a basic fact from subdifferential calculus, inequality (4.4) implies the differentiability of a locally convex function $\Psi $ at its local maximum.

Actually, the local convexity of $\Psi $ is a rather typical situation when we consider the volume of the projection of a polytope in $\mathbb {R}^n$ onto a k-dimensional subspace, as the latter is piecewise linear on the corresponding Grassmannian by Theorem 1 in [Reference Filliman7]. Another example is the so-called linear parameter systems (see [Reference Rogers and Shephard15]).

Here is another observation. One can notice that $\mathop {\mathrm {tr}} A_S = \sum \limits _1^n |v_i|^2$ for a tight frame $S = \{v_1, \dots , v_n\}.$ Hence, $\mathop {\mathrm {tr}} A_S = k$ for a tight frame $S.$ Let $\Omega ' (n,k)$ denote the class of frames $S = \{v_1, \dots , v_n \}$ such that $ \sum \limits _1^n |v_i|^2 = k( = \mathop {\mathrm {tr}} A_S).$ The same arguments as in Lemma 4.2 imply the following corollary.

Corollary 4.6 Let $\Psi $ be an O-invariant function. Then,

$$ \begin{align*} \max\limits_{S \in \Omega'(n,k)} \Psi (S) = \max\limits_{S \in \Omega(n,k)} \Psi(S). \end{align*} $$

Proof Since $\Omega (n, k) \subset \Omega '(n,k),$ it is enough to show that for any $S' \in \Omega '(n,k), S' \notin \Omega (n,k),$ there exists a tight frame S such that $ \Psi (S') < \Psi (S).$ We put $S = B_{S'} S'.$ Then, by Lemma 4.2, it is enough to prove that $\det A_{S'} < 1.$ Considering $A_{S'}$ in the basis of its eigenvectors (in which it is a diagonal operator) and using the inequality of arithmetic and geometric means, we obtain

$$ \begin{align*} \det A_{S'} \leq \left( \frac{\mathop{\mathrm{tr}} A_{S'}}{k}\right)^k = 1, \end{align*} $$

where equality is attained iff all eigenvalues of $A_{S'}$ are ones. This means, the equality is attained iff $S' \in \Omega (n,k).$ This completes the proof. ▪

4.4 Lifting of frames

Vectors $\{v_1, \dots , v_n\}$ of a tight frame $S \in \Omega (n,k)$ have a nice isometric embedding in $\wedge ^{k-1} (H^S).$ Fix $i \in [n]$ and let $d_S(i)$ be the vector in $\Lambda ^{k-1} (\mathbb {R}^n)$ such that its Lth coordinate in the standard basis of $\Lambda ^{k-1} (\mathbb {R}^n)$ is $d_S(\{i, L\}).$

Lemma 4.8 Let $S = \{v_1, \dots , v_n\}$ be an arbitrary tight frame. Then, the vectors $d_S(i), i \in [n],$ belong to $\wedge ^{k-1} (H^S) \subset \Lambda ^{k-1} (\mathbb {R}^n).$ Moreover, they form a tight frame in $\wedge ^{k-1} (H^S),$ and the following identity holds:

(4.6) $$ \begin{align} \left\langle d_S (i), d_S (j)\right\rangle = \left\langle v_i, v_j\right\rangle \end{align} $$

for all $i,j \in [n].$

Proof Let $a_j, j \in [k],$ be the rows of the $k \times n$ matrix $M^S = (v_1, \dots , v_n ).$ By the definition of $H^S,$ we know that $a_j \in H_k^{S}, j \in [k],$ and that they form an orthonormal system in $H^{S}.$ Hence, the $(k-1)$ -forms $(a_{[k] \backslash j})_{j=1}^k$ are an orthonormal basis of $\wedge ^{k-1} (H^S).$ Consider the $(k-1)$ -forms $b_i = \sum \limits _{j=1}^k (-1)^{j +1} v_i[j] \cdot a_{[k] \backslash j},$ where $i \in [n].$ Then, by the definition of the inner product in $\Lambda ^{k-1} (\mathbb {R}^n)$ and the Laplace expansion of the determinant, we have

$$ \begin{align*} b_i[L] &= \left\langle b_i,e_L\right\rangle = \sum\limits_{j=1}^k (-1)^{j+1} v_i[j] \left\langle a_{[k] \backslash j}, \right\rangle{e_L} \\ &= \sum\limits_{j=1}^k (-1)^{j+1} v_i[j] M^S_{[k] \backslash j, L} = M^S_{\{i,L\}, \{i, L\}} = d_S(i)[L] \end{align*} $$

for $L \in \binom {[n]}{k-1}, i \notin L.$ We get $b_i[L] = d_{S}(i)[L] = 0$ if $i \in L.$ Thus, $d_S(i) = b_i \in \Lambda ^{k-1} (H^{S})$ and, clearly, identity (4.6) holds. ▪

Lemma 3.1 allows us to describe all substitutions $w_i \to w^{\prime }_i, i \in [n],$ which preserve the operator $A_S$ for a frame $S = \{w_1, \dots , w_n\}.$ But we need a more suitable geometric description. So, let $S = \{w_1, \dots , w_n\}$ be a frame and $\ell \in [k], L = \{i_1, \dots , i_{\ell }\}\subset \binom {[n]}{\ell }.$ Consider a substitution $w_i \to w^{\prime }_i,$ for $i \in L,$ and $w^{\prime }_i = w_i,$ for $i \notin L,$ which preserves $A_S,$ and denote by $S' = \{w^{\prime }_1,\dots , w^{\prime }_n\}$ the new frame.

In this notation, we get the following.

Lemma 4.9 The substitution preserves $A_S$ (i.e., $A_S = A_{S'}$ ) iff there exists an orthogonal matrix U of rank $\ell $ such that

(4.7) $$ \begin{align} \begin{pmatrix} w^{\prime}_{i_1}, & \dots, & w^{\prime}_{i_{\ell}} \end{pmatrix} = \begin{pmatrix} w_{i_1}, & \dots , & w_{i_{\ell}} \end{pmatrix} U. \end{align} $$

Additionally, in case S is a tight frame, let $\{f_1, \dots , f_n\}$ be any orthonormal basis of $\mathbb {R}^n$ given by the assertion (3.1) of Lemma 3.1 . Then, the substitution preserves $A_S$ iff the vectors of $S'$ are the projection of an orthonormal basis $\{f^{\prime }_1, \dots , f^{\prime }_n\}$ of $\mathbb {R}^n,$ which is obtained from $\{f_1, \dots , f_n\}$ by an orthogonal transformation of $\{f_i\}_{i \in I}$ in their linear hull.

Proof The identity $A_{S'} = A_S$ holds if and only if

$$ \begin{align*} \sum_{i \in I} w_i \otimes w_i = \sum_{i \in I} w^{\prime}_i \otimes w^{\prime}_i. \end{align*} $$

Writing this in a matrix form, we get another equivalent statement that the Gram matrices of the rows of the matrices $\begin {pmatrix} w^{\prime }_{i_1}, & \dots , & w^{\prime }_{i_{\ell }} \end {pmatrix} $ and $ \begin {pmatrix} w_{i_1}, & \dots , & w_{i_{\ell }} \end {pmatrix}$ are the same. The latter is equivalent to the existence of an isometry of $\mathbb {R}^{\ell }$ , which maps the rows of the first matrix to the rows of the second. This isometry defines an orthogonal matrix $U,$ which satisfies the assumptions of the lemma.

In case of a tight frame, extending U as the orthogonal transformation of $\mathop {\mathrm {Lin}} \{f_i \; : \; {i \in I}\},$ we obtain the suitable $\{f^{\prime }_1, \dots , f^{\prime }_n\}.$  ▪

Also, we can reformulate the second claim of Lemma 4.9 in the following equivalent way: given a block matrix $M = \begin {pmatrix} A & B \\ C & D \end {pmatrix}$ such that the rows of $\begin {pmatrix} A & B \end {pmatrix}$ are orthonormal and the columns of $\begin {pmatrix} A \\ C \end {pmatrix}$ are orthonormal, then D can be chosen in such a way that M will be an orthogonal matrix.

5.1 Definitions and history

A zonotope is the Minkowski sum of several segments in $\mathbb {R}^k$ . Every zonotope can be represented (up to an affine transformation) as a projection of a higher-dimensional cube $\operatorname {C}^n = [0,1]^n.$ For example, it follows from the definition of $B_S.$ To get more information about zonotopes, we refer the reader to Zong’s book [Reference Zong20, Chapter 2] and to [Reference Schneider and Weil16].

Several different bounds for the maximum in (1.2) are known. For example, G. D. Chakerian and P. Filliman [Reference Chakerian and Filliman4] prove

$$ \begin{align*} \operatorname{vol}_{k} (\operatorname{C}^n | H) \leq \sqrt{\frac{n!}{(n-k)! k!}} \quad \text{and} \quad \operatorname{vol}_{k} (\operatorname{C}^n | H) \leq \frac{\omega_{k-1}^k}{\omega_{k}^{k-1}} \left(\frac{n}{k} \right)^{k/2}, \end{align*} $$

where $w_i$ is the volume of the i-dimensional Euclidean unit ball. The right-hand side inequality is asymptotically tight, as was shown in [Reference Filliman5], and the left-hand side inequality is tight in the cases $k =1,n-1.$ The tight upper and lower bounds in the limit case for the volume of a zonotope in a specific position (even for so-called $L_p$ -zonoids) were obtained in [Reference Lutwak, Yang and Zhang12, Theorem 2].

Among different upper bounds, all maximizers of (1.2) were described in the cases $k =1,2,n-2,n-1$ and $k=3, n=6$ in [Reference Filliman5]. It appears that projections of the standard basis vectors onto $H_k$ have the same length whenever H is a maximizer of (1.2) for all cases mentioned above, which gives a rather reasonable conjecture.

We note that there are k-dimensional subspaces of $\mathbb {R}^n$ such that the projections of the vectors of the standard basis onto them have the same length. An explicit construction of such subspaces can be found in [Reference Zimmermann19]. In [Reference Ivanov10, Lemma 2.4], the author describes all possible lengths of the projections of the vectors of the standard basis.

Summarizing the observations of Section 4, we have that the problem to find and to study maximizers of

(5.1) $$ \begin{align} F_2(S) = \operatorname{vol}_{k} \left(\sum\limits_1^n [0, v_i]\right), \qquad \text{where} \qquad \{v_1, \dots, v_n\} \in \Omega(n,k), \end{align} $$

is equivalent to that of (1.2). We illustrate the developed technique and write the first-order necessary condition of local maximum for (5.1) together with some geometric consequences.

5.2 The first-order necessary condition

We start with showing that $F_2(S)$ is a differentiable function at its local maximum.

Proof Let $S = \{v_1, \dots , v_n\}$ be a local maximizer of (5.1) and $S'$ be a frame obtained from S by substitution $v_i \to v_i + t_i x_i, i \in [n].$

The function $|d_{S'}(\{L\})|$ as a function of $(t_1, \dots , t_n)$ is the absolute value of a polynomial of $(t_1, \dots , t_n).$ Hence, it is differentiable at a point $(t_1, \dots , t_n)$ whenever $d_{S'}(\{L\}) \neq 0.$ By Corollary 5.3, we have $d_S(\{L\}) \neq 0$ for all k-tuples. Therefore, the function

$$ \begin{align*} {F_2(S')} = {\sum\limits_{L \in \binom{[n]}{k}}|d_{S'}(\{L\})| } \end{align*} $$

as a function of $(t_1, \dots , t_n)$ is differentiable at the origin. ▪

Now, we show the geometric meaning of identities of Corollary 4.4 for the function $F_2.$ We define a sign function $\sigma _S(i, L)$ by

$$ \begin{align*} \sigma_S(i, L) = \begin{cases} 1/{F_2(S)} , \qquad \text{if} \qquad d(\{i, L\})>0; \\ -1/{F_2(S)} , \qquad \text{if} \qquad d(\{i, L\}) <0; \\ 0, \qquad \text{if} \qquad d(\{i, L\}) = 0, \end{cases} \end{align*} $$

for a frame $S, i \in [n]$ and $L \in \binom {[n]}{k-1}.$

In the same way as with the vectors $d_S(i), i \in [n],$ we identify $\sigma _S (i)$ with a vector in $\Lambda ^{k-1} (\mathbb {R}^n)$ such that its Lth coordinate in the standard basis of $\Lambda ^{k-1} (\mathbb {R}^n)$ is $\sigma _S({i, L}).$ As a direct consequence of Lemma 5.4, we obtain the following lemma.

Lemma 5.5 Let $S = \{v_1, \dots , v_n\}$ be a local maximizer of (5.1). Then,

(5.2) $$ \begin{align} \wedge^{k-1} P^S (\sigma_S (i)) = d_S (i) \qquad \text{and} \qquad \left\langle\sigma_S (i), d_S (j)\right\rangle = \left\langle v_i, v_j\right\rangle, \;\;\; i,j \in [n]. \end{align} $$

Proof By Lemma 4.8, we have that the vectors $(d_S(i))_{i=1}^{n}$ form a tight frame in $\wedge ^{k-1} H_k^S.$ Therefore, by Theorem 3.3, it is enough to show that the rightmost identity in (5.2) is true.

Fix $i, j \in [n].$ Let $S'$ be a frame obtained from S by the substitution $v_i \to v_i + t v_j.$ By Corollary 5.3, we have that $d_S(\{L\}) \neq 0$ for every k-tuple $L.$ Therefore, $d_{S'}(\{L\}) \neq 0$ and $d_{S'}(\{L\})$ has the same sign as $d_S(\{L\})$ for all k-tuples L and for a small enough  $t.$ Using the substitution $v_i \to v_i + t v_j,$ we only change the determinants of type $d_S(\{i, J\}),$ where $i \notin J \subset \binom {[n]}{{k-1}}.$ Thus, by the properties of absolute value (for a small enough t), we get

$$ \begin{align*} \frac{F_2(S')}{F_2(S)} = \frac{1}{F_2(S)}{\sum\limits_{L \in \binom{[n]}{k}}|d_{S'}(\{L\})| } = 1 + t \sum\limits_{J \in \binom{[n]}{k-1} , i \notin J} \sigma_S(i, J) d_S(\{j, J\}). \end{align*} $$

Since $\sigma _S(i, J) = 0$ whenever $i \in J,$ we have that the coefficient at t in the previous formula is

$$ \begin{align*} \sum\limits_{J \in \binom{[n]}{k-1}} \sigma_S(i, J) d_S(\{j, J\}). \end{align*} $$

But this is the inner product $\left \langle \sigma _S (i), d_S(j)\right \rangle $ of the $(k-1)$ -forms $\sigma _S(i)$ and $d_S(j)$ written in the standard basis of $\Lambda ^{k-1} (\mathbb {R}^n).$ By Lemma 5.4 and Corollary 4.4, we have that $ \left \langle \sigma _S (i),d_S (j)\right \rangle = \left \langle v_i, v_j\right \rangle .$  ▪

5.3 Proofs of Corollary 1.3, Theorem 1.4, and Corollary 1.5

Corollary 1.3 is a direct consequence of Lemma 5.5 with a simple geometric meaning.

Proof of Corollary 1.3

By Corollary 5.3, $|v_i| \neq 0$ for all $i \in [n].$ Let $S' = \{v^{\prime }_1, \dots , v^{\prime }_n\}$ be projections of the vectors $\{v_1, \dots , v_n\}$ onto $v_i^\perp .$ Clearly, $v^{\prime }_i = 0.$ By Shepard’s formula (3.3) and the properties of the determinant, we have

$$ \begin{align*} \operatorname{vol}_{k-1} \left(Q | v_i^\perp \right) = \sum\limits_{J \in \binom{[n]}{{k-1}}} \left| d_{S'}(\{J\}) \right| = \frac{1}{|v_i|} \sum\limits_{J \in \binom{[n]}{{k-1}}} \left| d_{S}(\{i,J\}) \right|. \end{align*} $$

By Lemma 5.5, the latter is $|v_i| \operatorname {vol}_{k} Q.$  ▪

The idea of the proof of Theorem 1.4 is the following. We use Lemma 4.9 to rotate vectors by $\pi /4$ (i.e., $v_i \to \cos (\pi /4) v_i - \sin (\pi /4) v_j, v_j \to \sin (\pi /4) v_i + \cos (\pi /4) v_j$ ), and after some simple calculations along with Corollary 4.4, we get the inequality of Theorem 1.4.

Proof of Theorem 1.4

We prove the theorem for a maximizer $S = \{ v_1, \dots , v_n\}$ of (5.1). Fix i and j in $[n].$ We assume that $|v_i|^2 < |v_j|^2,$ otherwise there is nothing to prove. Using Lemma 4.9, we have that the substitution $v_i \to \cos (\pi /4) v_i - \sin (\pi /4) v_j, v_j \to \sin (\pi /4) v_i + \cos (\pi /4) v_j$ preserves $A_S$ and the absolute value of $d_S(\{L\})$ for all $L \subset \binom {[n]}{k}$ such that $i,j \in L.$ Let $S'$ be the tight frame obtained by this substitution. From the choice of $S,$ we have $F_2(S') \leq F_2(S).$ Expanding these volumes by formula (3.3) and reducing common determinants, we get

$$ \begin{align*} \frac{\sqrt{2}}{2} \sum\limits_{J \in \binom{[n]}{{k-1}}, \; i,j \notin J} & \left( \left|d_{S}(\{i,J\}) - d_{S}(\{j,J\}) \right| + \left|d_{S}(\{i,J\}) + d_{S}(\{j,J\}) \right|\right) \\ &\quad \leq \sum\limits_{J \in \binom{[n]}{{k-1}},\; i,j \notin J} (\left|d_{S}(\{i,J\})\right| + \left| d_{S}(\{j,J\}) \right|). \end{align*} $$

By identity $|a + b| + |a-b| = 2 \max \{|a|, |b|\},$ we obtain that each summand in the left-hand side is at least $2 \max \{ \left |d_{S}(\{i,J\})\right | , \left | d_{S}(\{i,J\}) \right |\}$ and, consequently, is at least $2 \left |d_{S}(\{j,J\})\right |.$ Hence, we show that

$$ \begin{align*} (\sqrt{2} -1)\sum\limits_{J \in \binom{[n]}{k-1},\; i,j \notin J} \left| d_{S}(\{j,J\}) \right| \leq \sum\limits_{J \in \binom{[n]}{{k-1}},\; i,j \notin J} \left| d_{S}(\{i,J\}) \right|. \end{align*} $$

Consider the one-to-one correspondence between the set of all $(k-1)$ -tuples L with $i \in L$ and $j \notin L$ and the set of all $(k-1)$ -tuples J with $i \notin J$ and $j \in J$ given by $L \to (L\backslash \{i\}) \cup \{j\}.$ In this case, $|d_S(\{j, L\})| = |d_S(\{i, (L\backslash \{i\}) \cup \{j\}\})|.$ Adding all such determinants to the last inequality, we obtain

$$ \begin{align*} &\sum\limits_{J \in \binom{[n]}{{k-1}},\; i \notin J} \left| d_{S}(\{i,J\}) \right| \\ &\quad\geq (\sqrt{2} -1)\sum\limits_{J \in \binom{[n]}{{k-1}},\; i,j \notin J} \left| d_{S}(\{j,J\}) \right| + \sum\limits_{J \in \binom{[n]}{{k-1}},\; i \in J, \; j \notin J} \left| d_{S}(\{j,J\}) \right| \\&\quad\geq (\sqrt{2} -1)\left(\sum\limits_{J \in \binom{[n]}{{k-1}},\; i,j \notin J} \left| d_{S}(\{j,J\}) \right| + \sum\limits_{J \in \binom{[n]}{{k-1}},\; i \in J, \; j \notin J} \left| d_{S}(\{j,J\}) \right| \right) \\ &\quad =(\sqrt{2} -1) \sum\limits_{J \in \binom{[n]}{{k-1}},\; j \notin J} \left| d_{S}(\{j,J\}) \right|. \end{align*} $$

Finally, the sum in the left-hand side is exactly $F_2(S)\left \langle \sigma _S (i), d_S(i)\right \rangle ,$ and by Lemma 5.2, it is equal to $F_2(S) |v_i|^2.$ Similarly, we have $(\sqrt {2} - 1) F_2(S) |v_j|^2$ in the right-hand side of the last inequality. Dividing by $F_2(S),$ we obtain $|v_{i}|^2 \geq (\sqrt {2} - 1) |v_{j}|^2.$  ▪

As mentioned in the Introduction, Corollary 1.5 is a consequence of Theorem 1.4 and McMullen’s symmetric formula (1.3).

Proof of Corollary 1.5

Let $H_q$ be the orthogonal complement of $H_{n-q}$ in $\mathbb {R}^n.$ Let $v_i$ and $v^{\prime }_i$ be the projections of the vector $e_i$ onto $H_{n-q}$ and $H_q,$ respectively. Clearly, $|v_i|^2 + |v^{\prime }_i|^2 = 1.$ By Theorem 1.4, we conclude that $|v^{\prime }_i|^2$ is at most $1/(\sqrt {2} - 1)$ and at least $(\sqrt {2} -1)$ of the average squared length of the projections of the standard basis, which is $ q/n.$ Therefore,

$$ \begin{align*} 1 \geq \frac{m_n}{M_n} \geq \sqrt{\frac{1 - \frac{1}{\sqrt{2} -1 } \frac{q}{n}}{1 - (\sqrt{2} -1)\frac{q}{n}}}, \end{align*} $$

which tends to 1 as n tends to infinity. ▪