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

Example

Take $n = 2$ and $m = 6$ , and let $a : = 1 / \sqrt{ι (2)}$ , so that $u : = (a, a) \in R^{2}$ has $∥ u ∥_{2} = \sqrt{a^{2} + a^{2}} = 1$ . Define $v : 6 \to R^{2}$ 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 $j \neq i$ is an ordered basis of $F^{n}$ ; hence $\dim_{F} F^{n} = n$ , and $F^{0}$ is the zero space with basis $\emptyset$ and dimension $0$ ). Every $∥ v_{i} ∥_{2}$ 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 $ι (2)$ .

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

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 $ι (3) + ι (2) \sqrt{ι (2)} > ι (4)$ . So that ordering has a partial sum of norm strictly greater than $ι (2)$ , and the theorem is saying something.

One step of the descending construction. With $b^{6}$ the identity of $6$ and $μ_{j}^{6} : = ι (4) / ι (6)$ for $j < 6$ , the pair $(b^{6}, μ^{6})$ 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

$Λ = {μ \in [0, 1]^{6} : \sum_{j < 6} μ_{j} v_{j} = 0, \sum_{j < 6} μ_{j} = ι (3)},$

and $μ : = (1, \frac{1}{ι (2)}, 0, 1, \frac{1}{ι (2)}, 0)$ 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 $4 \leq k - 1 = 5$ , 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 $μ^{5} = (1, \frac{1}{ι (2)}, 0, 1, \frac{1}{ι (2)})$ .

Facts & Assumptions

Given: The list $v : 6 \to R^{2}$ above, with $a = 1 / \sqrt{ι (2)}$ and $u = (a, a)$ ; the partial sums $s_{k} = \sum_{j < k} v_{j}$ of the identity ordering; the vector $μ = (1, 1 / ι (2), 0, 1, 1 / ι (2), 0)$ .

[L1]

$∥ y ∥_{2} = \sqrt{y_{0}^{2} + y_{1}^{2}}$ on $R^{2}$ , and it is a norm (The $p$ -norms $∥ x ∥_{p}$ for rational $p \geq 1$ , and $∥ x ∥_{\infty}$ , The Euclidean inner product $⟨ x, y ⟩ = \sum_{k < n} x_{k} y_{k}$ on $R^{n}$ , A norm on a real vector space, the induced metric, and the dictionary with the metric axioms, Each $∥ \cdot ∥_{p}$ is a norm on $R^{n}$ , and the induced metrics are exactly $d_{1}$ , $d_{2}$ and $d_{\infty}$ of the published metric-spaces page).

[L2]

Square roots: $\sqrt{c}$ is the unique nonnegative $s$ with $s^{2} = c$ ; $(\sqrt{ι (2)})^{2} = ι (2)$ ; and for $c, d \geq 0$ , $c \leq d$ exactly when $\sqrt{c} \leq \sqrt{d}$ (Square roots exist: a unique $\sqrt{a} \geq 0$ with $(\sqrt{a})^{2} = a$ ; the positives are ${x^{2} : x \neq 0}$ , Squaring is monotone on the nonnegatives, Integer powers $a^{m}$ ).

[L3]

Canonical naturals are positive and strictly increasing and carry sums to sums and products to products (The canonical natural $ι (n) = n \cdot 1_{F}$ of a field, Canonical naturals are positive and strictly increasing); inverses of positives are positive (Inverses of positives are positive, and reciprocation reverses order).

[L4]

Finite sums in $R^{2}$ are computed pointwise, with the recursion $\sum_{j < k + 1} y_{j} = \sum_{j < k} y_{j} + y_{k}$ (The standard list $e : n \to F^{n}$ with $e_{i} (i) = 1_{F}$ and $e_{i} (j) = 0_{F}$ for $j \neq i$ is an ordered basis of $F^{n}$ ; hence $\dim_{F} F^{n} = n$ , and $F^{0}$ is the zero space with basis $\emptyset$ and dimension $0$ clause 1, Linear combination of a finite list, and the span $span (S)$ as the smallest linear subspace containing $S$ , Laws of finite sums and finite products, Finite sums and finite products, by recursion).

[L5]

Steinitz's polygonal confinement theorem, its notion of an admissible pair, and its support bound (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 $n$ , Injection, surjection, bijection).

Verification

technique · direct

1.1

$a^{2} = 1 / ι (2)$ by [L2], so $∥ u ∥_{2}^{2} = a^{2} + a^{2} = 1$ and $∥ u ∥_{2} = 1$ ; also $∥ e_{0} ∥_{2} = ∥ e_{1} ∥_{2} = 1$ and $∥ - y ∥_{2} = ∥ y ∥_{2}$ . Hence $∥ v_{i} ∥_{2} = 1$ for every $i < 6$ .

L1L2L3

1.2

$\sum_{i < 6} v_{i} = (e_{0} - e_{0}) + (- e_{1} + e_{1}) + (u - u) = 0$ , computing coordinatewise.

1.3

For the ordering $e_{0}, e_{1}, u, - e_{0}, - e_{1}, - u$ the third partial sum is $(1 + a, 1 + a)$ , whose norm squared is $ι (2) (1 + a)^{2} = ι (2) + ι (2) ι (2) a + ι (2) a^{2} = ι (3) + ι (2) \sqrt{ι (2)}$ , using $ι (2) a = ι (2) / \sqrt{ι (2)} = \sqrt{ι (2)}$ .

L1L2L3

1.4

The vector $μ = (1, 1 / ι (2), 0, 1, 1 / ι (2), 0)$ lies in $Λ$ : its values lie in $[0, 1]$ ; $\sum_{j < 6} μ_{j} = 1 + 1 / ι (2) + 0 + 1 + 1 / ι (2) + 0 = ι (3)$ ; and $\sum_{j < 6} μ_{j} v_{j} = (v_{0} + v_{3}) + (1 / ι (2)) (v_{1} + v_{4}) = 0 + 0 = 0$ .

L3L4

2.1

The partial sums of the identity ordering are as displayed, by the recursion of [L4] applied coordinatewise.

step 1.2L4

2.2

Since $\sqrt{ι (2)} > 1$ and $ι (2) > 0$ , the quantity of step 1.3 exceeds $ι (3) + ι (2) = ι (5) > ι (4)$ , so that partial sum has norm strictly greater than $ι (2)$ : a bad ordering really does break the bound.

step 1.3L2L3

2.3

The pair $(b^{6}, μ^{6})$ with $b^{6}$ the identity of $6$ and $μ_{j}^{6} = ι (4) / ι (6)$ is admissible at $6$ : the values lie in $[0, 1]$ , $\sum_{j < 6} μ_{j}^{6} v_{j} = (ι (4) / ι (6)) \sum_{j < 6} v_{j} = 0$ , and $\sum_{j < 6} μ_{j}^{6} = ι (6) \cdot ι (4) / ι (6) = ι (4)$ .

step 1.2L3L4L5

2.4

No element of $Λ$ has fewer than two strictly fractional coordinates. If none were fractional, $μ$ would be a ${0, 1}$ -vector with $\sum_{j} μ_{j} = ι (3)$ , hence with support of size $3$ , and $\sum_{j \in supp} v_{j} = 0$ ; the support cannot contain a pair ${0, 3}$ , ${1, 4}$ or ${2, 5}$ , since the remaining single vector would then have to be $0$ while all six are nonzero, so it contains exactly one index from each pair and the sum is $\pm e_{0} \pm e_{1} \pm u$ , whose second coordinate is $\pm 1 \pm a$ and whose first is $\pm 1 \pm a$ , and $∣ \pm 1 \pm a ∣ \neq 0$ because $a \neq 1$ (indeed $a^{2} = 1 / ι (2) \neq 1$ ). If exactly one coordinate were fractional, say with value $t$ , then $\sum_{j} μ_{j}$ would be $t$ plus a canonical natural and could not equal $ι (3)$ .

step 1.4L1L2L3

3.1

Their norms squared are $0$ , $1$ , $ι (2)$ , $(1 + a)^{2} + (a - 1)^{2} = ι (2) + ι (2) a^{2} = ι (3)$ , $a^{2} + (a - 1)^{2} = ι (2) - ι (2) a$ , $ι (2) a^{2} = 1$ and $0$ .

step 2.1L1L2L3

3.2

So $μ$ of step 1.4 is a minimiser, its support is ${0, 1, 3, 4}$ of size $4$ , and $4 \leq 5 = k - 1$ at $k = 6$ : the support bound of [L5] holds, and a coordinate with value $0$ exists, for instance $j_{0} = 5$ .

step 1.4step 2.4L5

4.1

Each of these is at most $ι (4)$ : the largest is $ι (3)$ , and $ι (2) - ι (2) a \leq ι (2)$ because $a > 0$ . So every partial sum of the identity ordering has norm at most $\sqrt{ι (4)} = ι (2)$ , and the identity ordering realises the bound of [L5].

step 3.1L2L3L5

4.2

Deleting position $5$ gives $b^{5} = (0, 1, 2, 3, 4)$ and $μ^{5} = (1, 1 / ι (2), 0, 1, 1 / ι (2))$ , with $\sum_{j < 5} μ_{j}^{5} = ι (3)$ and $\sum_{j < 5} μ_{j}^{5} v_{b^{5} (j)} = 0$ : an admissible pair at $k = 5$ .

step 1.4step 3.2L4L5

5.1

Steps 4.1, 2.2 and 4.2 give, in turn, an ordering realising the bound, an ordering violating it, and one traced step of the descending construction with its support bound checked.

step 4.1step 2.2step 4.2∎

Remarks

The bound $ι (n)$ is not attained here. The largest partial-sum norm of the good ordering is $\sqrt{ι (3)}$ , comfortably below $ι (2)$ . The theorem asserts existence of an ordering below $ι (n)$ 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 $ι (2)$ , 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 $# supp \leq k - 1$ that the descending construction turns on.

Depends on

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Dependency tree · two levels

84 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Levy-Steinitz theorem (Wikipedia) (standard reference, not scraped)
Ernst Steinitz (Wikipedia) (standard reference, not scraped)
T. Oertel, J. Paat and R. Weismantel, A Colorful Steinitz Lemma with Applications to Block Integer Programs (standard reference, not scraped)
T. Banakh, A Simple Inductive Proof of the Levy-Steinitz Theorem (standard reference, not scraped)

Steinitz's confinement bound realised on an explicit list of six unit vectors in R2R2 summing to zero