Past exam of the mathematics course of the University of Cambridge / 2009 / iii / Paper 4 / 3

For integer partitions $\lambda ,\mu$ of $n$, padded by zero parts, the dominance order on partitions isA $\lambda$-tableau fills the Young diagram of $\lambda$ with $1,\dots ,n$, each once. A tabloid $[t]$ remembers the sets of entries in each row, ignoring their order within that row. The Young permutation module $M^{\lambda }=\mathbb{C}{X}_{\lambda }$ has these tabloids as basis, with $S_n$ acting by permuting the entries.

The row and column stabilizers $R_t,C_t$ independently permute entries within the rows and columns of $t$. The column antisymmetrizer is ${\kappa }_t={\sum }_{c\in C_t}\mathrm{sgn}(c)c$, and $e_t={\kappa }_t[t]$ is the associated polytabloid. Since $C_{gt}=g{C}_t{g}^{-1}$ and conjugation preserves permutation sign,Thus the Specht module is the submoduleEvery tableau is $gt$ for a fixed $t$, so any single $e_t$ generates it under $S_n$. Also $C_t\cap R_t=\{1\}$, since a permutation preserving both rows and columns must fix every cell. The tabloids $[ct]$ for $c\in C_t$ are consequently distinct, and $e_t$ has coefficient $1$ at $[t]$; in particular $e_t\ne 0$.

We use the following precise column antisymmetrizer facts. If some row of a $\mu$-tableau $s$ contains two entries from one column of $t$, the transposition interchanging them pairs equal tabloids with opposite signs in ${\kappa }_t[s]$, so ${\kappa }_t[s]=0$. Otherwise each row of $s$ meets each column of $t$ at most once. The first $r$ rows of $s$ therefore contain at most $\min (r,{\lambda }_j^{\prime })$ entries from column $j$ of $t$, whenceThis states and explains dominance from a nonzero column antisymmetrizer: ${\kappa }_t[s]\ne 0$ implies $\lambda ⊵\mu$.

When $\mu =\lambda$, the nonzero column antisymmetrizer criterion says more: either ${\kappa }_t[s]=0$, or a column permutation takes $[s]$ to $[t]$, and ${\kappa }_t[s]=\pm e_t$. Thus ${\kappa }_t{M}^{\lambda }=\mathbb{C}{e}_t$. This follows by matching the row positions within each column when all intersections have size at most one; the equal row and column sizes force the Ferrers incidence pattern.

For the inner product, use the positive Hermitian extension of the orthonormal tabloid basis, linear in its first argument. It has the supplied self-adjointness of ${\kappa }_t$. Its orthogonal complement of $S^{\lambda }$ agrees with that obtained from the complex tabloid bilinear form, since the spanning polytabloids have real coefficients. Because the coefficient of $[t]$ in $e_t$ is $1$, self-adjointness and the one-dimensional image give the useful identityIndeed, the scalar multiplying $e_t$ is the $[t]$-coefficient of ${\kappa }_{t}u$, which is $⟨{\kappa }_{t}u,[t]⟩=⟨u,{\kappa }_t[t]⟩=⟨u,e_t⟩$.

Suppose first that $U$ is not contained in $(S^{\lambda }{)}^{\perp }$. Since polytabloids span $S^{\lambda }$, there exist $u\in U$ and a tableau $t$ with $⟨u,e_t⟩\ne 0$. The column antisymmetrizer identity from the preliminaries givesTherefore $e_t\in U$. Because $U$ is an $S_n$-submodule, it contains all translates $g{e}_t=e_{gt}$, and these span the Specht module. Hence $S^{\lambda }\subseteq U$. If the initial supposition fails, $U\subseteq (S^{\lambda }{)}^{\perp }$. This proves the James submodule theoremNow let $W$ be a submodule of $S^{\lambda }$. Apply the dichotomy to $W\subseteq M^{\lambda }$. Either $W=S^{\lambda }$, or $W\subseteq S^{\lambda }\cap (S^{\lambda }{)}^{\perp }$. Positivity of the Hermitian form gives $S^{\lambda }\cap (S^{\lambda }{)}^{\perp }=\{0\}$: a vector in this intersection has squared norm zero. The same intersection is zero for the bilinear form by the agreement of complements explained above. Since $S^{\lambda }\ne 0$, the only submodules are $0$ and itself. ThereforeThis use of characteristic zero and nondegeneracy is why the conclusion does not automatically extend to every modular Specht module.

Let $0\ne f:S^{\lambda }\to M^{\mu }$ be a module homomorphism. Since $S^{\lambda }$ is simple, $f$ is injective. Fix a $\lambda$-tableau $t$; its nonzero polytabloid $e_t$ then has $f(e_t)\ne 0$. The sign sum in the column group satisfiesbecause every coefficient in the square receives $|C_t|$ equal contributions. Thus ${\kappa }_t{e}_t=|C_t|e_t$, and equivariance givesThe column antisymmetrizer therefore acts nontrivially on $M^{\mu }$, so some basis tabloid $[s]$ has ${\kappa }_t[s]\ne 0$. The stated dominance from a nonzero column antisymmetrizer yieldsFor $\mu =\lambda$, the same calculation forces $f(e_t)=|C_t{|}^{-1}{\kappa }_{t}f(e_t)$ into the one-dimensional space $\mathbb{C}{e}_t$. Write $f(e_t)=c{e}_t$. Since $e_t$ generates the Specht module, equivariance gives $f(g{e}_t)=cg{e}_t$ for every $g$, so $f$ is $c$ times the inclusion $S^{\lambda }\hookrightarrow M^{\lambda }$. That inclusion is nonzero, henceAs a useful consequence, if $S^{\lambda }\cong S^{\mu }$, their inclusions into $M^{\lambda },M^{\mu }$ give dominance both ways. Antisymmetry of the dominance order on partitions then gives $\lambda =\mu$, so different partitions produce nonisomorphic simple modules.

The preceding parts give a simple Specht module for each partition and show that different partitions give nonisomorphic simples. The conjugacy classes of $S_n$ are indexed by partitions of $n$, namely cycle types. We use the standard finite-group character theorem: the number of irreducible complex characters equals the number of conjugacy classes, and they form an orthonormal basis of complex class functions. Consequently the ${\chi }^{\lambda }$ are the complete list of irreducible characters of $S_n$.

By Maschke's theorem, the Young permutation module $M^{\mu }$ decomposes as a direct sum of these simple modules. For a semisimple complex module, Schur lemma makes the multiplicity of a simple module equal to the dimension of the homomorphism space from that simple; character orthogonality identifies the same number with the character inner product. HencePart (ii) makes $a_{\lambda \mu }=0$ unless $\lambda ⊵\mu$, and gives $a_{\mu \mu }=1$. Taking characters provesThus the permutation-character decomposition is triangular in the dominance order on partitions, with diagonal multiplicity one. These are the dominance and diagonal-multiplicity consequences of Young's rule, obtained here without assuming its full multiplicity formula.

The permutation character ${\xi }^{\mu }(g)$ is the number of $\mu$-tabloids fixed by $g$, since the trace of a permutation matrix counts fixed basis elements. It is therefore an integer. Arrange the partitions in decreasing lexicographic order. If $\lambda$ strictly dominates $\mu$, the first part at which they differ is larger for $\lambda$, so this ordering places $\lambda$ before $\mu$.

For the first partition $(n)$, there is just one tabloid, and ${\chi }^{(n)}={\xi }^{(n)}=1_{S_n}$ is integer-valued. At a general partition $\mu$, the decomposition in (iii) can be rearranged asBy induction, all characters in the sum are already integer-valued; the permutation character and all coefficients are also integral. Thus ${\chi }^{\mu }$ is integer-valued. Finite induction proves that symmetric-group characters are integer-valued:The diagonal coefficient one is essential: it lets us solve for ${\chi }^{\mu }$ without division, preserving integrality.

Take $g=(12\cdots n)$. Since $n\equiv 3(\mathrm{mod}4)$ is odd, its sign is $(-1{)}^{n-1}=1$, so $g\in A_n$. Define $s$ to fix $1$ and reverse the other cyclic labels: it interchanges $2$ with $n$, $3$ with $n-1$, and so on. Then $sg{s}^{-1}=g^{-1}$, whileThe centralizer of an $n$-cycle in $S_n$ is $⟨g⟩$: a commuting permutation is determined by the image of one point, and that image fixes the corresponding power of the cycle. All its elements are even, because $g$ is even. Any other conjugator taking $g$ to $g^{-1}$ differs from $s$ by a centralizer element and therefore remains odd. Thus $g$ and $g^{-1}$ are not conjugate in $A_n$. This is the parity obstruction described by inversion of an odd cycle in an alternating group.

For any complex character of a finite group, a unitary realization givesIf every $\theta \in \mathrm{Irr}(A_n)$ had real value at $g$, all these characters would take equal values on $g$ and $g^{-1}$. But irreducible characters separate conjugacy classes: they form a basis of class functions, so equality of every irreducible value would also give equality of the class indicator functions. The two distinct conjugacy classes cannot have that property. ConsequentlyIn other words, these alternating groups are not ambivalent groups, even though all symmetric-group irreducible characters are integer-valued.