Successive-minima volume deformation and collision avoidance

The vectors $a_1,\dots ,a_n$ form a real basis by [F2], so $x={\sum }_i{x}_i{a}_i$ is a well-defined coordinate representation; $L(t):={\sum }_i{t}_i{a}_i$ is an invertible linear map.

$U=\mathrm{int}(C)$ is open, convex, nonempty, bounded and symmetric ($U=-U$), and $0\in U$; also $\mathrm{int}(tC)=tU$ for every $t>0$.

(Collision avoidance.) Let $x,y\in U$ be distinct with $\mathrm{\Phi }(x)-\mathrm{\Phi }(y)\in 2\mathrm{\Lambda }$, and put $\mu :=(\mathrm{\Phi }(x)-\mathrm{\Phi }(y))\mathrm{/}2\in \mathrm{\Lambda }$. Let $k$ be the largest index with $x_k\mathrm{\ne }{y}_k$.

(No nonzero lattice point in the image.) Let $\nu \in \mathrm{\Phi }(U)\cap \mathrm{\Lambda }$, say $\nu =\mathrm{\Phi }(x)$ with $x\in U$, and suppose $x\mathrm{\ne }0$; let $k$ be the largest index with $x_k\mathrm{\ne }0$.

For $x\in U$ and $j\ge 2$ the slice $F_j(x)$ of the statement is compact and convex (an intersection of the convex set $C$ with the affine subspace $\{z_i=x_i,i\ge j\}$), it contains $x$, and it has positive $(j-1)$-dimensional volume: $B(x,\delta )\subseteq U$ for some $\delta >0$, and the relative ball $B(x,\delta )\cap (x+\mathrm{span}(a_1,\dots ,a_{j-1}))$ lies in $F_j(x)$. For $j=1$, $F_1(x)=\{x\}$.

(The interior has the same volume as the body.) Fix $x_0\in U$ and put $C_k:=(1-1\mathrm{/}k)C+(1\mathrm{/}k)x_0$ for $k\ge 2$; every scale $1-1\mathrm{/}k$ is positive. Each $C_k$ is compact and convex, and $C_k\subseteq U$ by [F8]. To see the sequence is increasing, for $c\in C$ set $c^{\mathrm{\prime }}:=(1-1\mathrm{/}{k}^2)c+(1\mathrm{/}{k}^2)x_0\in C$; then $(1-1\mathrm{/}k)c+(1\mathrm{/}k)x_0=(1-1\mathrm{/}(k+1))c^{\mathrm{\prime }}+(1\mathrm{/}(k+1))x_0$, so $C_k\subseteq C_{k+1}$. For every $z\in U$, the points $c_k:=z+(z-x_0)\mathrm{/}(k-1)$ tend to $z$, so for all sufficiently large $k$ openness of $U$ gives $c_k\in U\subseteq C$ and $z=(1-1\mathrm{/}k)c_k+(1\mathrm{/}k)x_0\in C_k$. Thus $U={\bigcup }_{k\ge 2}{C}_k$, and [F7] and [F6] give $\mathrm{vol}(U)={\lim }_{k\to \mathrm{\infty }}\mathrm{vol}(C_k)={\lim }_{k\to \mathrm{\infty }}(1-1\mathrm{/}k{)}^n\mathrm{vol}(C)=\mathrm{vol}(C)$; the Countable Choice hypothesis of [F6] is supplied by [A1].

By [F5] applied to the compact convex slice $F_j(x)$, its centroid ${\phi }_j(x)$ lies in $F_j(x)\subseteq C$; in particular ${\phi }_1(x)=x\in U$.

For $z\in F_j(x)$ one has $z_i=x_i$ whenever $i\ge j$; hence the $i$-th coordinate of ${\phi }_j(x)$ equals $x_i$ for $i\ge j$, and for $i<j$ the integral defining that coordinate is taken over the fibre of $C$ over $(x_j,\dots ,x_n)$, so it depends only on those coordinates.

${\phi }_j$ is odd: $x\mapsto -x$ maps $U$ onto $U$ and $F_j(x)$ onto $F_j(-x)=-F_j(x)$. On the affine hull of each slice this reflection is an affine isometry whose linear part has determinant of absolute value $1$, so it preserves relative Lebesgue measure by [F6]. Changing variables in the centroid integral therefore gives ${\phi }_j(-x)=-{\phi }_j(x)$. This uses the paired-slice identity $F_j(-x)=-F_j(x)$ and does not require an individual slice $F_j(x)$ to be symmetric.

(Borelness of the centroids.) In the coordinate model of step 1.1 write $C^{\mathrm{\prime }}=L^{-1}(C)$, which is compact because $L$ is a homeomorphism, and write $U^{\mathrm{\prime }}=L^{-1}(U)$. Fix $j\ge 2$ and split $t=(s,\tau )$ with $s\in {\mathbb{R}}^{j-1}$ and $\tau \in {\mathbb{R}}^{n-j+1}$. The functions $V_j(\tau ):={\int }_{{\mathbb{R}}^{j-1}}{\mathbf{1}}_{C^{\mathrm{\prime }}}(s,\tau )ds$ and $M_{j,i}(\tau ):={\int }_{{\mathbb{R}}^{j-1}}{s}_i{\mathbf{1}}_{C^{\mathrm{\prime }}}(s,\tau )ds$ are Borel by Tonelli [F4], with the signed moment split into positive and negative parts; compactness of $C^{\mathrm{\prime }}$ makes their supports bounded. Let $A_j$ be the matrix with columns $a_1,\dots ,a_{j-1}$. The restriction of $L$ to the first $j-1$ coordinates scales intrinsic fibre measure by the constant $J_j=\sqrt{\det (A_j^{\mathsf{T}}{A}_j)}>0$, independent of $\tau$. Thus this factor cancels in the centroid ratios, and for $i<j$ the $i$-th coordinate of ${\phi }_j$ in the $a$-basis is $M_{j,i}(\tau )\mathrm{/}{V}_j(\tau )$. The projection of the open set $U^{\mathrm{\prime }}$ to the $\tau$-coordinates is open, and $V_j(\tau )>0$ there by step 2.1. Hence these ratios are Borel on that open set; extending them by $0$ outside gives globally Borel functions of $\tau$.

$\mathrm{\Phi }$ is odd, because each ${\phi }_j$ is odd by step 4.2; in particular $\mathrm{\Phi }(0)=0$.

For $j>k$ the coordinates $(x_j,\dots ,x_n)$ and $(y_j,\dots ,y_n)$ agree, so ${\phi }_j(x)={\phi }_j(y)$ by step 4.1; hence $\mathrm{\Phi }(x)-\mathrm{\Phi }(y)={\sum }_{j\le k}({\lambda }_j-{\lambda }_{j-1})({\phi }_j(x)-{\phi }_j(y))$ and $\mu ={\sum }_{j\le k}({\lambda }_j-{\lambda }_{j-1})u_j$ with $u_j:=({\phi }_j(x)-{\phi }_j(y))\mathrm{/}2$.

Define $\mathrm{\Phi }(x):={\sum }_{j=1}^n({\lambda }_j-{\lambda }_{j-1}){\phi }_j(x)$ for $x\in U$. In coordinates, for fixed $i$ the coordinates of ${\phi }_j$ with index $i<j$ depend only on $(x_j,\dots ,x_n)$ by step 4.1, while for $j\le i$ the $i$-th coordinate of ${\phi }_j$ equals $x_i$; hence ${\mathrm{\Phi }}_i(x)={\sum }_{j\le i}({\lambda }_j-{\lambda }_{j-1})x_i+{\sum }_{j>i}({\lambda }_j-{\lambda }_{j-1})({\phi }_j(x){)}_i={\lambda }_i{x}_i+{\psi }_i(x_{i+1},\dots ,x_n)$, where ${\psi }_i$ is Borel by step 5.1.

Here $u_1=(x-y)\mathrm{/}2\in U$ by step 1.2, and $u_j=({\phi }_j(x)+(-{\phi }_j(y)))\mathrm{/}2\in C$ for $j\ge 2$ by step 3.1 and [F8]. The weights ${\lambda }_j-{\lambda }_{j-1}$ are nonnegative and sum to ${\lambda }_k$, with positive first weight ${\lambda }_1$ on the interior point $u_1$. Repeated application of [F8] (or induction on the finite number of terms) puts the convex combination $\mu \mathrm{/}{\lambda }_k$ in $U$, that is $\mu \in \mathrm{int}({\lambda }_{k}C)$ by step 1.2.

By steps 6.1 and 5.2 and [F3] applied with $a_i={\lambda }_i>0$ to the Borel set $E=U^{\mathrm{\prime }}$, the image $\mathrm{\Phi }(U)$ is Borel and the conjugate ${\mathrm{\Phi }}^{\mathrm{\prime }}=L^{-1}\circ \mathrm{\Phi }\circ L$ satisfies $\mathrm{vol}({\mathrm{\Phi }}^{\mathrm{\prime }}(U^{\mathrm{\prime }}))=({\prod }_i{\lambda }_i)\mathrm{vol}(U^{\mathrm{\prime }})$; conjugating by the invertible linear map $L$ and using [F6] gives $\mathrm{vol}(\mathrm{\Phi }(U))=({\prod }_i{\lambda }_i)\mathrm{vol}(U)$.

The $a_k$-coordinate of $\mu$ is ${\lambda }_k(x_k-y_k)\mathrm{/}2\mathrm{\ne }0$: for $j\le k$ the $a_k$-coordinate of ${\phi }_j(x)-{\phi }_j(y)$ is $x_k-y_k$ by step 4.1. Every vector in either $\mathrm{span}\{a_1,\dots ,a_{k-1}\}$ or $\mathrm{span}\{a_j:{\lambda }_j<{\lambda }_k\}$ has zero $a_k$-coordinate, so $\mu$ lies in neither span.

Since $\mathrm{\Phi }(0)=0$ by step 5.2 and $\mathrm{\Phi }(0)={\sum }_j({\lambda }_j-{\lambda }_{j-1}){\phi }_j(0)$ with ${\phi }_j(0)=0$ by step 4.2, the same computation as steps 5.3 and 6.2 with $y=0$ gives $\nu ={\sum }_{j\le k}({\lambda }_j-{\lambda }_{j-1}){\phi }_j(x)\in \mathrm{int}({\lambda }_{k}C)$, and the $a_k$-coordinate of $\nu$ is ${\lambda }_k{x}_k\mathrm{\ne }0$.

Combining steps 7.1 and 2.2 gives $\mathrm{vol}(\mathrm{\Phi }(U))=({\prod }_i{\lambda }_i)\mathrm{vol}(C)$, which is clause 3.

But $\mu \in \mathrm{\Lambda }\cap \mathrm{int}({\lambda }_{k}C)$, so clause 3 of [F2] forces $\mu \in \mathrm{span}\{a_j:{\lambda }_j<{\lambda }_k\}$, contradicting step 7.2. Hence no two distinct points of $\mathrm{\Phi }(U)$ differ by an element of $2\mathrm{\Lambda }$.

Again clause 3 of [F2] would put $\nu$ in $\mathrm{span}\{a_j:{\lambda }_j<{\lambda }_k\}$, contradicting the nonzero $a_k$-coordinate. Hence $x=0$, and since $\mathrm{\Phi }(0)=0$ the only lattice point of $\mathrm{\Phi }(U)$ is $0$, which is clause 4 together with step 8.2.

Clause 1 is step 4.1 with step 5.1, clause 2 is steps 6.1 and 5.2, clause 3 is step 8.1, and clause 4 is steps 8.2 and 9.1.