Lecture 19

Solution: Let $\lambda,\mu,\nu\vdash n$. Clearly

for all $1\leq m\leq|\lambda|$, so $\lambda\unlhd\lambda$. If $\lambda\unlhd\mu$ then

for all $1\leq m\leq\min\{|\lambda|,|\mu|\}$ which implies

for all $1\leq m\leq\min\{|\lambda|,|\mu|\}$ and $\lambda\unrhd\mu$. Lastly suppose also that $\mu\unlhd\nu$, then

for all $1\leq m\leq\min\{|\lambda|,|\mu|,|\nu|\}$ so

for all $1\leq m\leq\min\{|\lambda|,|\nu|\}$ and $\lambda\unlhd\nu$. Thus $\unlhd$ is a partial order.

Solution: Let $[t],[s],[u]\in\mathbf{YTD}^{\lambda}$. Note that as $[t]$ has all elements in the same row as $[t]$ we have $[t]\not\prec[t]$.

Let $[t]\prec[s]$, then for some $1\leq i\leq n$ we have that for every $j<i$ $[t]$ and $[s]$ agree, in terms of rows, and $i$ appears on a lower row in $[t]$ than $[s]$. As it is not on a higher row in $[t]$ and the rows agree for all $j<i$ we have $[s]\not\prec[t]$.

Suppose also that $[s]\prec[u]$, then for some $1\leq k\leq n$ we have that for every $l<k$ $[s]$ and $[u]$ agree, in terms of rows, and $k$ appears on a lower row in $[s]$ than $[u]$. For all $j\leq\min\{i,k\}$, $j$ is on the same row in both $[t]$ and $[u]$. If $\min\{i,k\}=i$ then $i$ is on the same row in $[u]$ as it is $[s]$ which is higher than $[t]$ hence $[t]\prec[u]$. Otherwise $k$ is on a higher row in $[u]$ than it is in $[s]$ and $k$ is on the same row in $[t]$ as it is in $[s]$ so $[t]\prec[u]$.

Lastly let $[t]\neq[s]$. Then clearly there exists a smallest $1\leq i\leq n$ such that the row $i$ is on in each differs. Thus it agrees for $j<i$ and we have $[t]\prec[s]$ or $[s]\prec[t]$.

Solution: Suppose $[s]\neq[t]$. Then there exists at least one $i$ that is in a different row in $[s]$ than in $[t]$. As $s$ and $[s]$ share the same set of entries in each row and so do $t$ and $[t]$ we have $s\neq t$. Now suppose $[s]=[t]$ so the rows in each are the same. As $s$ and $t$ are standard tableau there is only one possible way to arrange each row of $[s]$, that is in ascending order left to right. Thus each row of $s$ and $t$ is the same and $s=t$.

Solution: