Steinitz's confinement bound realised on an explicit list of six unit vectors in ${\mathbb{R}}^2$ summing to zero

Take $n=2$ and $m=6$, and let $a:=1\mathrm{/}\sqrt{\iota (2)}$, so that $u:=(a,a)\in {\mathbb{R}}^2$ has $\parallel u{\parallel }_2=\sqrt{a^2+a^2}=1$. Define $v:6\to {\mathbb{R}}^2$ by

$v_0:=e_0,v_1:=-e_1,v_2:=u,v_3:=-e_0,v_4:=e_1,v_5:=-u,$

with $e_0=(1,0)$ and $e_1=(0,1)$ (The standard list $e:n\to F^n$ with $e_i(i)=1_F$ and $e_i(j)=0_F$ for $j\mathrm{\ne }i$ is an ordered basis of $F^n$; hence ${\dim }_F{F}^n=n$, and $F^0$ is the zero space with basis $\mathrm{\varnothing }$ and dimension $0$). Every $\parallel v_i{\parallel }_2$ is $1$ and ${\sum }_{i<6}{v}_i=0$, so Steinitz's polygonal confinement theorem: finitely many vectors of norm at most $1$ summing to $0$ can be ordered so that every partial sum has norm at most $n$ applies and asserts an ordering all of whose partial sums have norm at most $\iota (2)$.

A good ordering. The identity ordering works: the partial sums $s_k={\sum }_{j<k}{v}_j$ are

$s_0=(0,0),s_1=(1,0),s_2=(1,-1),s_3=(1+a,a-1),s_4=(a,a-1),s_5=(a,a),s_6=(0,0),$

with norms $0$, $1$, $\sqrt{\iota (2)}$, $\sqrt{\iota (3)}$, $\sqrt{\iota (2)-\sqrt{\iota (2)}}$, $1$, $0$, each at most $\iota (2)$.

A bad ordering, which exceeds the bound. Reordering as $e_0,e_1,u,-e_0,-e_1,-u$ gives the third partial sum $(1+a,1+a)$, whose norm squared is $\iota (3)+\iota (2)\sqrt{\iota (2)}>\iota (4)$. So that ordering has a partial sum of norm strictly greater than $\iota (2)$, and the theorem is saying something.

One step of the descending construction. With $b^6$ the identity of $6$ and ${\mu }_j^6:=\iota (4)\mathrm{/}\iota (6)$ for $j<6$, the pair $(b^6,{\mu }^6)$ is admissible at $k=6$ in the sense of the proof of Steinitz's polygonal confinement theorem: finitely many vectors of norm at most $1$ summing to $0$ can be ordered so that every partial sum has norm at most $n$. At the next stage the feasible set is

$\mathrm{\Lambda }=\{\mu \in [0,1{]}^6:{\sum }_{j<6}{\mu }_j{v}_j=0,{\sum }_{j<6}{\mu }_j=\iota (3)\},$

and $\mu :=(1,\frac{1}{\iota (2)},0,1,\frac{1}{\iota (2)},0)$ lies in it with exactly two coordinates strictly between $0$ and $1$, which is the least possible. Its support is $\{0,1,3,4\}$, of size $4\le k-1=5$, so the support bound holds with room; dropping the coordinate $j_0=5$ gives the admissible pair at $k=5$ with $b^5=(0,1,2,3,4)$ and ${\mu }^5=(1,\frac{1}{\iota (2)},0,1,\frac{1}{\iota (2)})$.

• The bound $\iota (n)$ is not attained here. The largest partial-sum norm of the good ordering is $\sqrt{\iota (3)}$, comfortably below $\iota (2)$. The theorem asserts existence of an ordering below $\iota (n)$ and claims no sharpness, and this example makes no claim about the optimal constant either.

• What the bad ordering shows. Without the theorem there is no reason to expect any ordering to stay bounded independently of $m$: the third partial sum of the bad ordering already exceeds $\iota (2)$, and lists of many unit vectors summing to $0$ can be ordered so that a partial sum has norm of order $m$.

• Why one step of the construction is traced. An example that only asserted the bound would say nothing about how it is obtained. The step above exhibits the object the proof actually manipulates — a feasible vector of coefficients with as few fractional coordinates as possible — and checks the support bound $\mathrm{\#}\mathrm{supp}\le k-1$ that the descending construction turns on.

The two polygonal paths. In each panel the dashed circle is the Euclidean circle of radius $\iota (2)$ centred at $0$, and the arrows join successive partial sums. The left panel uses the $s_k$ above; on the right, $t_k$ denotes the partial sums of the bad ordering $e_0,e_1,u,-e_0,-e_1,-u$. In particular, $t_3=(1+a,1+a)$ lies outside the circle. The plotted coordinates use $a=1\mathrm{/}\sqrt{\iota (2)}$; the verification above establishes the norm comparisons independently of this illustration.