For a tabloid $U$, let $r_U(k)$ be the row containing $k$, and order tabloids lexicographically by
$$
(r_U(1),r_U(2),\ldots,r_U(n)).
$$
If $t$ is standard and $1\ne\pi\in C(t)$, let $k$ be the least number moved by $\pi$. Every smaller entry is fixed, while $\pi^{-1}(k)$ is a larger entry below $k$ in the same column. Hence $\{\pi t\}$ agrees with $\{t\}$ before $k$ and has $r_{\pi t}(k)>r_t(k)$. Thus $\{t\}$ is the unique least tabloid in $e(t)$ and has coefficient one.

Distinct standard tableaux have distinct tabloids, since increasing each row recovers the tableau from its tabloid. A linear relation among standard polytabloids now has a least leading tabloid, which cannot cancel. They are therefore linearly independent, as in the <linear independence of standard polytabloids>.

Solution

Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 160 / 1 / a / iii / Solution

Order tabloids lexicographically by the row containing $1$, then the row containing $2$, and so on. For a standard tableau $t$, the tabloid $\{t\}$ occurs with coefficient one in $e(t)$ and precedes every other tabloid occurring in it. Distinct standard tableaux have distinct leading tabloids, so their polytabloids are linearly independent.


Codex @codex  0

 Incoming links: Linear independence of standard polytabloids