Past exam of the Lie algebras and representation theory of symmetric groups course of the University of Oxford / 2010 (C2.1, DMAC 2462, DMCO 2462, DMPC 2462)

Source: Mathematics release, bundle Part C 2007 11.pdf, pages 309–312. This combined paper was sat on 8 June 2010, 2.30–5.30pm, for Mathematics, Mathematics and Computer Science, and Mathematics and Philosophy Part C.

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We prove Lie's theorem in the required form by induction on $\dim L$. If $L=0$, any nonzero vector works. Otherwise solvability gives $[L,L]\ne L$, so choose a codimension-one ideal $K\supseteq [L,L]$ and $x\in L\setminus K$. By induction there are $v\ne 0$ and a linear functional $\lambda :K\to \mathbb{C}$ with $av=\lambda (a)v$ for all $a\in K$.

Let $W=\mathrm{span}\{v,xv,x^{2}v,\dots \}$. Choose the cyclic basis $v,xv,\dots ,x^{r-1}v$ up to its first linear dependence. The commutator identity $a{x}^j=xa{x}^{j-1}+[a,x]x^{j-1}$, with $[a,x]\in K$, proves inductively thatHence $W$ is invariant under $K$ and $x$, and every $a\in K$ has trace $r\lambda (a)$ on $W$. In particular ${\mathrm{tr}}_W([x,a])=0$ implies $r\lambda ([x,a])=0$. Characteristic zero gives $\lambda ([x,a])=0$.

The simultaneous weight space $V_{\lambda }=\{w:aw=\lambda (a)w\forall a\in K\}$ is nonzero and stable under $x$, becauseOver $\mathbb{C}$, the restriction of $x$ to this nonzero finite-dimensional space has an eigenvector. That vector is an eigenvector of both $x$ and every $a\in K$, hence of every element of $L$.

Let $\pi :L\to L/I$ be the quotient map. A full flag of ideals in $L/I$ lifts by inverse images to a flag from $I$ to $L$, with dimensions increasing by one. Prepending $0<I$ supplies the required full flag in $L$. Thus supersolubility lifts across a one-dimensional ideal.

Over $\mathbb{C}$, a nonzero soluble $L$ has a one-dimensional ideal: apply part (a) to its adjoint image acting on $L$. A common eigenvector spans an adjoint-invariant line, which is an ideal. The quotient is soluble and has smaller dimension, so induction and the lifting result give a full flag of ideals. Therefore a soluble complex Lie algebra is supersoluble.

Conversely this implication needs no field restriction. If $L$ has a full flag, its first nonzero ideal $I$ is one-dimensional and abelian, and $L/I$ has a full flag. By induction $L/I$ is soluble, so some term of the derived series of $L$ lies in $I$; its next derived term is zero. Hence

Let $A$ be the two-dimensional affine Lie algebra with basis $u,v$ and bracket $[u,v]=v$, and putBoth factors have zero centre, so $Z(L)=0$. The line $\mathbb{C}v$ in the second factor is an ideal, making $L$ PSS. But its quotient onto the nonabelian simple algebra ${\mathfrak{sl}}_2$ is not soluble. A quotient of a soluble algebra would be soluble, so $L$ is not soluble.

If $L$ is abelian and nonzero, $L^{\prime }=0<L$. Otherwise choose $x\notin Z=Z(L)$ such that $Z+Fx$ is an ideal. For each $a\in L$ writeThe coefficient $\lambda (a)$ is uniquely defined and linear. The Jacobi identity givesso $\lambda ([a,b])=0$. If $\lambda \ne 0$, then $L^{\prime }\subseteq \ker \lambda <L$.

If $\lambda =0$, the same equation says $[[a,b],x]=0$ for all $a,b$. Hence $L^{\prime }\subseteq C_L(x)$. Since $x$ is not central, its centralizer is proper. Thus in both casesNo algebraic closure or characteristic restriction was used.

Choose a basis $(u,v)$ with $U=Fu$. The representation matrices areThe one-dimensional modules $U,V/U$ have characters $\lambda ,\mu$. If they coincide, these modules are isomorphic and the first alternative holds.

Otherwise choose $h$ with $\delta (h)\ne 0$, where $\delta =\lambda -\mu$. Replacing $v$ by $v-a(h)u/\delta (h)$ makes the upper-right entry of $\rho (h)$ zero. The representation identity on commutators now gives, for every $x\in H$,Iteration yields $a((\mathrm{ad}h{)}^{r}x)=\delta (h{)}^{r}a(x)$. Nilpotence of $H$ implies $(\mathrm{ad}h{)}^{r}x=0$ for sufficiently large $r$, by its lower central series. Since $\delta (h)\ne 0$, it follows that $a(x)=0$ for all $x$. The new line $W=Fv$ is therefore a submodule, andThe argument works over the stated arbitrary field, including positive characteristic.

The flag makes every composition factor one-dimensional, described by a character $\lambda :H\to F$. The Jordan–Hölder theorem for modules for finite-length modules says that the multiset of simple composition factors is independent of the chosen composition series.

Here is a construction valid over every $F$. For $h\in H$, every generalized eigenspace of $\rho (h)$ is an $H$-submodule. Indeed $\mathrm{ad}(h)$ is nilpotent on $H$, say of exponent $c$, andIf $v$ is killed by $(\rho (h)-aI{)}^{N_0}$, take $N\ge N_0+c$. Terms with $j\ge c$ vanish by adjoint nilpotence, and all other terms kill $v$. This proves invariance of the generalized eigenspace.

The flag also makes the characteristic polynomial of every $\rho (h)$ split over $F$. Whenever two composition characters differ, choose an $h$ on which they differ and split the module into the distinct generalized eigenspaces of $\rho (h)$. Repeat within each summand until all its composition factors have the same character. Every nontrivial split lowers dimensions, so the process terminates. Finally combine summands carrying the same character. Jordan–Hölder ensures that the resulting distinct blocks have distinct factor types, givingThis recursive method does not assume that a single element separates every character, which could fail over a finite field.

Writing matrices in $2\times 2$ blocks, the defining relation givesThe trace is automatically zero. The relation is linear, and if $as=-s{a}^T$, $bs=-s{b}^T$, thenThus $L$ is closed under the commutator and is a Lie subalgebra of ${\mathfrak{sl}}_4(F)$.

Its diagonal subalgebra isPut ${\epsilon }_1(D(a,b))=a$, ${\epsilon }_2(D(a,b))=b$. Since $[D,E_{ij}]=(D_{ii}-D_{jj})E_{ij}$, the required block decomposition isThe four lines have respective characters ${\epsilon }_1-{\epsilon }_2$, $-{\epsilon }_1+{\epsilon }_2$, ${\epsilon }_1+{\epsilon }_2$, $-{\epsilon }_1-{\epsilon }_2$. They are distinct and nonzero because $\mathrm{char}F\ne 2$. The two-dimensional summand $H$ has two trivial composition factors. These five submodules therefore have exactly the factor-grouping properties of part (b). Their dimensions sum to $6$, the dimension obtained from four free entries in $A$ and one each in the skew-symmetric blocks $B,C$.

A reduced crystallographic root system $\Phi$ is a finite spanning subset of a real Euclidean vector space, excluding zero, such that $\Phi \cap \mathbb{R}\alpha =\{\pm \alpha \}$, each reflection $s_{\alpha }(v)=v-2(v,\alpha )\alpha /(\alpha ,\alpha )$ preserves $\Phi$, and all Cartan integers $⟨\beta ,\alpha ⟩=2(\beta ,\alpha )/(\alpha ,\alpha )$ are integers. A base is a vector-space basis $\Delta \subset \Phi$ in which each root has integer coordinates either all nonnegative or all nonpositive. The Weyl group is generated by the root reflections.

For nonparallel roots put $p=⟨\beta ,\alpha ⟩$ and $q=⟨\alpha ,\beta ⟩$. Then $pq=4{\cos }^2\theta$ is an integer strictly below $4$ and at least zero. ThereforeIf $(\alpha ,\beta )>0$, the positive integers $p,q$ have product $1,2$ or $3$, so at least one is one. If $p=1$, reflection gives $\beta -\alpha \in \Phi$ and hence its negative. If $q=1$, reflection directly gives $\alpha -\beta \in \Phi$. This proves the positive-inner-product assertion.

To prove that an $\alpha$-string has no gaps, suppose consecutive occurring integer positions $a<b$ in $\beta +t\alpha$ had $b-a\ge 2$. If $(\beta +a\alpha ,\alpha )<0$, apply the previous result to $\beta +a\alpha$ and $-\alpha$ to obtain the missing root at $a+1$. Thus this inner product must be nonnegative. Similarly the absence of a root at $b-1$ forces $(\beta +b\alpha ,\alpha )\le 0$. But the latter inner product exceeds the former by $(b-a)(\alpha ,\alpha )>0$, impossible. Thus every intermediate integer position occurs.

Write the complete finite string as $\beta -r\alpha ,\dots ,\beta +s\alpha$. Reflection maps position $t$ to $-p-t$, so it interchanges the two endpoints. HenceHere $r,s$ are the maximal endpoint distances, not arbitrary shorter distances mentioned in the existence assertion.

Use the Cartan-matrix convention $C_{ij}=⟨{\alpha }_i,{\alpha }_j⟩$. The displayed matrix is realized bywith the standard orthonormal coordinates. Closure under the two simple reflections givesThe positive roots are ${\alpha }_1,{\alpha }_2,{\alpha }_1+{\alpha }_2,2{\alpha }_1+{\alpha }_2$. Conversely the standard fact that every root is Weyl-conjugate to a simple root, or successive height reduction by simple reflections, shows these two reflection orbits exhaust the roots. This is the rank-two system $B_2$, also called $C_2$ after exchanging the length convention. With a transposed Cartan-matrix convention, interchange which simple-root label is short; the geometric system and Weyl group are unchanged.

The reflection in $e_2$ flips the second coordinate, while the reflection in $e_1-e_2$ swaps the two coordinates. These generate all eight signed permutation matrices in two coordinates. Thuswhere $D_8$ means the symmetry group of a square.

For a Young diagram cell $(i,j)$, its hook length is one plus the number of cells to its right and below. The hook-length formula isApplying it to the two shapes givesThe shapes here require $n\ge 4$.

A hook tabloid remembers its first-row set and the ordered singleton entries below it. The proposed map keeps that first-row set and merges the remaining entries into one unordered row. Thus it is well-defined on tabloids. Permuting entries before merging gives the same tabloid as merging before permuting, so it is a module homomorphism. Every target tabloid is reached by putting its second-row entries, in any order, into the singleton rows. Therefore it is surjective.

For $r=2$, the strictly dominating partitions of $\lambda =(n-2,1,1)$ are exactly $(n),(n-1,1),(n-2,2)$. Indeed dominance forces the first part to be at least $n-2$, and the possibilities for the remaining sum give precisely those shapes or $\lambda$ itself.

In characteristic zero, Maschke's theorem makes the permutation module semisimple, and Young's rule gives one copy of $S^{\lambda }$ and only strictly dominating additional shapes. The trivial multiplicity is one, since a transitive permutation module has a one-dimensional invariant space. For the standard representation $S^{(n-1,1)}$, Frobenius reciprocity identifies its multiplicity with the dimension of fixed vectors under the row stabilizer $S_{n-2}$. In its realization $\{(z_1,\dots ,z_n):\sum z_i=0\}$, those fixed vectors have the form $(c,\dots ,c,a,b)$ with $(n-2)c+a+b=0$, a two-dimensional space. Thus its multiplicity is two.

Finally $\dim M^{\lambda }=n(n-1)$. Subtracting the dimensions of one $S^{\lambda }$, one trivial module and two standard modules leaves $n(n-3)/2$, exactly one $S^{(n-2,2)}$. ThereforeThe additional multiplicities are $1,2,1$ respectively. If the printed weak summation index is read literally alongside its separate $S^{\lambda }$, then $m_{\lambda ,\lambda }=0$; the total multiplicity of $S^{\lambda }$ is one.

A tabloid is the row-equivalence class $\{t\}$ of a tableau filled bijectively with $1,\dots ,n$: order inside each row is forgotten, but the rows remain distinguished. The permutation module $M^{\lambda }$ has these tabloids as an $F$-basis, with ${\Sigma }_n$ permuting their entries. Let $C_t$ be the column stabilizer of $t$ and defineThe Specht module $S^{\lambda }$ is the submodule spanned by all polytabloids $e_t$.

If a module homomorphism $\varphi :M^{\lambda }\to M^{\mu }$ is nonzero on $S^{\lambda }$, it is nonzero on some spanning $e_t$. Equivariance gives $0\ne \varphi (e_t)={\kappa }_t\varphi (\{t\})$, so ${\kappa }_t{M}^{\mu }\ne 0$. The allowed criterion yields $\lambda \ge \mu$.

In characteristic zero, Maschke's theorem gives an equivariant projection $M^{\lambda }\to S^{\lambda }$. If $S^{\lambda }\cong S^{\mu }$, compose that projection with the isomorphism and the inclusion into $M^{\mu }$. The preceding argument gives $\lambda \ge \mu$. Reversing the roles gives $\mu \ge \lambda$, so antisymmetry of dominance givesFor a map defined only on a Specht module in arbitrary characteristic, dominance need not hold. Take $F={\mathbb{F}}_2$, $n=2$, $\lambda =(1,1)$ and $\mu =(2)$. The first Specht module is the one-dimensional span ofThe transposition fixes $e$, so this is the trivial module. The one-row permutation module $M^{(2)}$ is also trivial and one-dimensional. Sending $e$ to its single tabloid is a nonzero equivariant map, yet $(1,1)\ngeqq (2)$. In this characteristic such a map need not extend to the surrounding permutation module; that is exactly where the earlier proof ceases to apply.

Let ${\pi }^{\lambda }$ be the permutation character of $M^{\lambda }$. Young's rule gives ${\pi }^{\lambda }={\chi }^{\lambda }+{\sum }_{\mu >\lambda }{K}_{\mu \lambda }{\chi }^{\mu }$, with integer Kostka multiplicities. This dominance-triangular system has diagonal entries one, so recursive subtraction expresses every irreducible character as an integer linear combination of permutation characters. The latter count fixed tabloids, proving that all ${\chi }^{\lambda }(g)$ are integers.

For a $k$-cycle, the Murnaghan–Nakayama rule reduces towhere height is the number of occupied rows minus one, and $f^{\nu }$ is the degree of the remaining Specht character at the identity. In particular a full $n$-cycle has character zero unless $\lambda =(n-r,1^r)$ is a hook, in which case the value is $(-1{)}^r$.

There is more than one solution: the trivial representation $\lambda =(6)$ has character $1$ everywhere and the sign representation $\lambda =(1^6)$ has character $\pm 1$ everywhere. ThusIn fact they are the only ones. The full-cycle formula in the preceding discussion forces any candidate to be a hook $(6-r,1^r)$. For $r=1$, the standard character is the number of fixed points minus one, so it vanishes on cycle type $(5,1)$; its sign twist $r=4$ does too. For $r=2$, the hook representation is the exterior square of the standard representation and has character $\frac{1}{2}(\chi (g{)}^2-\chi (g^2))$. At cycle type $(4,1,1)$ both standard-character values are one, so this character is zero. Its sign twist $r=3$ also vanishes there. Only $r=0,5$ remain.

Irreducible character orthogonality gives $⟨\chi ,\chi ⟩=1$. Under the proposed vanishing condition, only the identity contributes, soBut $\chi (1)$ is an integer dimension and $720$ is not a square. Thus no such partition exists.

The full-cycle formula says that precisely the nonhook shapes have zero value on a 6-cycle. These are $(4,2),(3,3),(3,2,1),(2,2,1,1),(2,2,2)$.

Use the standard two-row permutation-character rule ${\pi }^{(6-r,r)}={\chi }^{(6)}+{\chi }^{(5,1)}+\cdots +{\chi }^{(6-r,r)}$. A transposition has four fixed points, so ${\chi }^{(5,1)}=3$. It fixes $\left(\genfrac{}{}{0ex}{}{4}{2}\right)+1=7$ two-subsets, giving ${\chi }^{(4,2)}=7-1-3=3$. It fixes $\left(\genfrac{}{}{0ex}{}{4}{3}\right)+\left(\genfrac{}{}{0ex}{}{4}{1}\right)=8$ three-subsets, giving ${\chi }^{(3,3)}=8-1-3-3=1$.

Conjugating a partition tensors its character with the sign character. Hence the values on a transposition for the conjugate shapes are $-3$ and $-1$. The shape $(3,2,1)$ is self-conjugate, so its character equals its own sign twist and is zero on every odd permutation. The complete candidate table is therefore:

Thus the unique answer is