Lemma 10.115.1. Let $n \in \mathbf{N}$. Let $N$ be a finite nonempty set of multi-indices $\nu = (\nu _1, \ldots , \nu _ n)$. Given $e = (e_1, \ldots , e_ n)$ we set $e \cdot \nu = \sum e_ i\nu _ i$. Then for $e_1 \gg e_2 \gg \ldots \gg e_{n-1} \gg e_ n$ we have: If $\nu , \nu ' \in N$ then

$(e \cdot \nu = e \cdot \nu ') \Leftrightarrow (\nu = \nu ')$

Proof. Say $N = \{ \nu _ j\}$ with $\nu _ j = (\nu _{j1}, \ldots , \nu _{jn})$. Let $A_ i = \max _ j \nu _{ji} - \min _ j \nu _{ji}$. If for each $i$ we have $e_{i - 1} > A_ ie_ i + A_{i + 1}e_{i + 1} + \ldots + A_ ne_ n$ then the lemma holds. For suppose that $e \cdot (\nu - \nu ') = 0$. Then for $n \ge 2$,

$e_1(\nu _1 - \nu '_1) = \sum \nolimits _{i = 2}^ n e_ i(\nu '_ i - \nu _ i).$

We may assume that $(\nu _1 - \nu '_1) \ge 0$. If $(\nu _1 - \nu '_1) > 0$, then

$e_1(\nu _1 - \nu '_1) \ge e_1 > A_2e_2 + \ldots + A_ ne_ n \ge \sum \nolimits _{i = 2}^ n e_ i|\nu '_ i - \nu _ i| \ge \sum \nolimits _{i = 2}^ n e_ i(\nu '_ i - \nu _ i).$

This contradiction implies that $\nu '_1 = \nu _1$. By induction, $\nu '_ i = \nu _ i$ for $2 \le i \le n$. $\square$

Comment #2830 by Dario Weißmann on

Typo in the proof: "For suppose that ... " should read "Suppose that ... "

Comment #2929 by on

Google:"For I will consider my Cat Jeoffry."

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