Past exam of the mathematics course of the University of Cambridge / 2017 / iii / Paper 103 / 5

Let $\lambda ,\mu ⊢n$ each be a partition of an integer. Pad the lists of parts with zeros and use the dominance order on partitionsThe total sums agree, so only finitely many inequalities matter. This is a partial order: reflexivity and transitivity hold term by term, and equality of all partial sums forces equality of every part, proving antisymmetry. For example $(2,2)⊴(3,1)$. It compares concentration in the top rows, whereas diagram inclusion compares individual cells and is a different relation.

A single-box up-move from row $j$ to row $i<j$ changes the partial sum through row $h$ byEvery change is nonnegative, so $\lambda ⊴\mu$ in the dominance order on partitions. Transitivity gives the same conclusion along any chain of up-moves. Thus the existence of the indicated chain implies dominance. A chain of length zero handles equality; each positive-length move must still leave a partition of an integer.

Suppose $\lambda ⊴\mu$ in the dominance order on partitions and $\lambda \ne \mu$. Let $i$ be the first unequal row. Then ${\lambda }_i<{\mu }_i$. PutWe have $D_i>0$, and eventually $D_h=0$. Let $j>i$ be its first return to zero. Then $D_h\ge 1$ for $i\le h<j$ and ${\lambda }_j>{\mu }_j$.

Move one box from row $j$ to row $i$. Increasing row $i$ is legal: if $i>1$, then ${\lambda }_{i-1}={\mu }_{i-1}\ge {\mu }_i>{\lambda }_i$. Decreasing row $j$ is also legal. If ${\lambda }_j={\lambda }_{j+1}$, then ${\mu }_{j+1}\le {\mu }_j<{\lambda }_j={\lambda }_{j+1}$, forcing $D_{j+1}<0$, a contradiction. Also ${\lambda }_j>0$. Thus the new list $\nu$ is a partition of an integer.

The new deficits equal $D_h-1$ on $i\le h<j$ and $D_h$ elsewhere, so $\lambda ⊴\nu ⊴\mu$. The quantity ${\sum }_r|{\lambda }_r-{\mu }_r|$ decreases by two, since the changed rows were a deficit and a surplus. Repeating must terminate at $\mu$. HenceThe construction proves both termination and preservation of the intermediate dominance inequalities.

It suffices to handle one single-box up-move, and then use the chain already constructed. Suppose the affected row sizes are $a={\lambda }_i$ and $b={\lambda }_j$, $i<j$. Since $\lambda$ is a partition of an integer, $a\ge b\ge 1$. Put $N=a+b$. Over the complex numbers, the two-row Young permutation module decomposition givesA zero second row is omitted. Therefore

Let $K$ be the product of the symmetric groups of the unaffected rows and the symmetric group $S_N$ on the union of the affected rows. The two relevant Young subgroups lie in $K$. Transitivity of induced representations expresses $M^{\lambda }$ and $M^{\nu }$ by inducing the preceding two-row modules, tensored with the trivial representations of the unaffected factors, from $K$ to $S_n$. Induction preserves this direct sum, sofor an actual group representation $L_{\lambda ,\nu }$, not merely a difference of characters. Iterate along the chain and take the direct sum of the induced complements. The resulting complement can be realized as an invariant subspace of $M^{\lambda }$, either through these isomorphisms or by Maschke's theorem. ThusIf $\lambda =\mu$, use the zero complement. The characteristic-zero hypothesis is essential to this use of the irreducible two-row decomposition.