Past exam of the Lie algebras and representation theory of symmetric groups course of the University of Oxford / 2008 (C2.1, DMAC 2462)

Source: Mathematics release. This five-page Mathematics Part C paper is dated 2 June 2008. Bundle Part C 2007 11.pdf, pages 112–116.

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Define $L_1=L$ and $L_{n+1}=[L,L_n]$, where brackets of subspaces mean their linear span. Each term is an ideal. Inductively, if $L_n$ is an ideal, the Jacobi identity givesso $L_{n+1}$ is an ideal as well. Also $L_{n+1}\subseteq L_n$. Nilpotence means that this descending series eventually reaches zero; class $c$ means $L_{c+1}=0$ with $L_c\ne 0$ for nonzero $L$.

The idealizer here is the normalizer ${\mathrm{Id}}_L(A)=\{x\in L:[x,A]\subseteq A\}$. Since $A$ is a subalgebra it contains its own brackets, so $A\subseteq {\mathrm{Id}}_L(A)$.

Because $L_1⊈A$ and the lower central series terminates, choose $j$ with $L_j⊈A$ but $L_{j+1}\subseteq A$. Take $x\in L_j\setminus A$. Then $[x,A]\subseteq [L_j,L]=L_{j+1}\subseteq A$. Thus $x$ belongs to the idealizer and not to $A$, proving $\boxed{A<{\mathrm{Id}}_L(A)}$.

Start with $I_0=I$ and set $I_{k+1}=[L,I_k]$. Every term lies in $I$, and $I_k\subseteq L_{k+1}$, so the sequence eventually vanishes. Choose its last nonzero term $I_j$. Then $[L,I_j]=0$, hence $0\ne I_j\subseteq I\cap Z(L)$. Therefore $\boxed{I\cap Z(L)\ne \{0\}}$.

Induct on $i$, proving the claim for all $j$ simultaneously. The case $i=1$ is the definition $[L,L_j]=L_{j+1}$. If it holds for $i$, Jacobi givesBoth uses of the induction hypothesis have first index $i$, so the induction is valid. Thus $\boxed{[L_i,L_j]\subseteq L_{i+j}}$ for all positive indices.

Let $M=L_r$ and write $M_k$ for its own lower central series. Part (c) proves inductively that $M_k\subseteq L_{rk}$: it holds at $k=1$, and $[M,M_k]\subseteq [L_r,L_{rk}]\subseteq L_{r(k+1)}$. If $L$ has class less than $rs$, then $L_{rs}=0$, so $M_s=0$. Consequently $\boxed{\mathrm{class}(L_r)<s}$, with the zero algebra understood to have class zero.

The Killing form is ${\kappa }_L(x,y)=\mathrm{tr}({\mathrm{ad}}_{L}x{\mathrm{ad}}_{L}y)$. Symmetry follows from $\mathrm{tr}(AB)=\mathrm{tr}(BA)$. Since $\mathrm{ad}[z,x]=[\mathrm{ad}z,\mathrm{ad}x]$, cyclicity of trace gives the invariance identityIf $x\in I^{\perp }$ and $y\in I$, then $[z,y]\in I$, so the right side is zero. Hence $[z,x]\in I^{\perp }$ for every $z$, proving that $\boxed{I^{\perp }\text{ is an ideal}}$.

For $x\in I$, use a basis of $I$ followed by a complementary basis of $L$. Because $[x,L]\subseteq I$, the matrix of ${\mathrm{ad}}_{L}x$ has blocks $\left(\begin{array}{cc}{\mathrm{ad}}_{I}x& B_x\\ 0& 0\end{array}\right)$. Multiplying two such matrices showsThe restriction statement has both arguments in $I$; it does not identify the two forms for arbitrary elements of $L$ outside $I$.

Now use the ordered basis in the question for a central ideal $J$. The images of the $u_j$ in the quotient have adjoint matrix $A$, so the $u$-coordinate block of their images is $A$; arbitrary $J$-components form $B$. Each $v_j$ is central, so its image under every adjoint map is zero. Thus the matrix is exactly $\left(\begin{array}{cc}A& 0\\ B& 0\end{array}\right)$. Products have top-left block $A_x{A}_y$ and zero bottom-right block. Taking traces givesFinally, a nonzero finite-dimensional nilpotent Lie algebra has nonzero centre by (1)(b). Induct on its dimension: the Killing form descends to the quotient by its centre, which is again nilpotent of smaller dimension, and is therefore zero by induction. The zero algebra starts the induction. Apply this to the nilpotent ideal $I$ and then use the restriction identity to obtain $\boxed{{\kappa }_L(x,y)=0\text{ for }x,y\in I}$. This avoids the false general claim that the product of two nilpotent linear maps must be nilpotent.

By irreducible traceless matrix Lie algebra is semisimple, let $R$ be the solvable radical. Lie's theorem triangularizes its action, with finitely many diagonal characters ${\lambda }_i:R\to \mathbb{C}$, and supplies a nonzero common eigenspace for one of them. For $x\in L$, conjugation by $e^{tx}$ preserves $R$ and gives an equivalent representation of it. Thus the multiset of characters ${\lambda }_i\circ e^{t\mathrm{ad}x}$ is constant. Each character varies continuously in $t$ and has values in that finite set of characters, so it is individually constant. Differentiating at zero gives ${\lambda }_i([x,r])=0$ for all $r\in R$.

Choose a common eigenspace $V_{\lambda }\ne 0$. For $v\in V_{\lambda }$,Thus $V_{\lambda }$ is $L$-invariant, so irreducibility makes it all of ${\mathbb{C}}^n$. Every element of $R$ acts as a scalar. Its trace is zero because $L\subseteq {\mathfrak{sl}}_n$, so that scalar is zero. Faithfulness of the matrix inclusion gives $R=0$. Hence $\boxed{L\text{ is semisimple}}$.

One sufficient diagonal-matrix condition for simplicity is that $L$ contain all traceless diagonal matrices. Indeed invariance under their adjoint actions then decomposes $L$ into that diagonal algebra and some matrix-unit spaces $\mathbb{C}{e}_{ij}$. Put a directed edge $j\to i$ when $e_{ij}\in L$. Irreducibility makes the directed graph strongly connected; otherwise a terminal component spans a proper invariant coordinate subspace. Bracketing matrix units along a simple directed path supplies the matrix unit between its endpoints. Strong connectivity therefore supplies every off-diagonal matrix unit, so $L={\mathfrak{sl}}_n$, which is simple. The mere existence of a diagonal matrix with distinct diagonal entries would not be a sufficient condition.

For the specified construction, $H$ has dimension two. Its diagonal adjoint action and the twelve distinct nonzero weights give a direct sum $L=H\oplus {⨁}_{i=1}^{12}\mathbb{C}{g}_i$, of dimension fourteen. The stated bracket condition and the weight-vector relations make it a Lie algebra. A subspace invariant under it is invariant under all the $g_i$ and hence their associative algebra, so it acts irreducibly on ${\mathbb{C}}^7$. The preceding result makes it semisimple.

Use prime-dimensional faithful irreducible module forces simplicity. To see this, decompose the algebra into simple ideals. Complete reducibility for one ideal and commutation with the others make its isotypic component invariant under the whole algebra, hence all of the representation. Writing that component as $W\otimes \mathrm{Hom}(W,V)$ and repeating produces a tensor product of irreducibles for the simple ideals. Faithfulness makes every factor nontrivial, of dimension at least two. Their dimensions multiply to seven, so only one factor is possible. Thus the constructed $L$ is simple.

The diagonal algebra $H$ is toral, and its centralizer inside $L$ is just $H$, since no $g_i$ has zero weight. A maximal toral subalgebra containing $H$ lies in that centralizer, so equals $H$; it is a Cartan subalgebra. Thus the rank is two and the twelve weights are the roots. Pair them as ${\alpha }_{i+6}=-{\alpha }_i$. The opposite-root brackets have weight zero and belong to $H$. For distinct $i,j\le 6$, the sum cannot be zero, and the supplied root-bracket result says their bracket is nonzero exactly when ${\alpha }_i+{\alpha }_j$ is a root. This is equivalent to $-{\alpha }_i-{\alpha }_j$ being a root, proving the requested equivalence for the paired brackets.

Among rank-two simple root systems, $A_2,B_2,G_2$ have respectively six, eight and twelve roots. ThereforeChoose simple roots $\alpha$ short and $\beta$ long. In $G_2$, $\beta +\alpha$ and $\beta +2\alpha$ are roots. Renumber with $g_1\in L_{\alpha }$, $g_3\in L_{\beta }$, $g_4\in L_{\beta +\alpha }$ and $g_5\in L_{\beta +2\alpha }$, retaining opposite partners in the other six positions. Root spaces are one dimensional, and the allowed bracket equality gives

Use the convention $A_{ij}=⟨{\alpha }_j,{\alpha }_i^{\vee }⟩=2({\alpha }_i,{\alpha }_j)/({\alpha }_i,{\alpha }_i)$. For a finite reduced crystallographic root system, diagonal entries are two and off-diagonal entries belong to $\{0,-1,-2,-3\}$, with $A_{ij}{A}_{ji}\in \{0,1,2,3\}$. Transposing this convention merely transposes the Cartan matrix.

The $\alpha$-string through $\beta$ has the consecutive roots $\beta -p\alpha ,\dots ,\beta +q\alpha$, where $p,q\ge 0$ and $p-q=⟨\beta ,{\alpha }^{\vee }⟩$. For distinct simple roots ${\alpha }_i,{\alpha }_j$, the negative step is absent, so $p=0$ and $q=-A_{ij}$.

The matrix determines each simple reflection on the simple-root basis:It therefore determines their generated Weyl group as linear transformations. Every root is in the Weyl orbit of a simple root. For a positive nonsimple root $\beta =\sum c_i{\alpha }_i$, some $(\beta ,{\alpha }_i)$ is positive because $\sum c_i(\beta ,{\alpha }_i)=(\beta ,\beta )>0$. Its reflection subtracts a positive integer multiple of ${\alpha }_i$ and lowers height. It stays positive unless the root is simple, since the other simple-root coefficients are unchanged and at least one is positive. Induction on height reaches a simple root; negative roots follow by sign. Hence $\Phi =W\Delta$, which is completely reconstructed by those reflection matrices. This determines the system up to root-system isomorphism, rather than fixing an absolute length scale; irreducible components may be rescaled.

For positive roots for a chain Cartan matrix, symmetry and adjacency imply all simple roots have equal length; normalize their squared lengths to two. Adjacent inner products are minus one and all others zero. Since every root is Weyl-conjugate to a simple root, every root also has squared length two. Write a positive root as $\beta ={\sum }_{i=1}^l{c}_i{\omega }_i$ with nonnegative integers $c_i$, and set $c_0=c_{l+1}=0$. ThenThere must be exactly two nonzero integer differences, one $+1$ and one $-1$. Nonnegativity and the two zero endpoint values imply that the coefficients are one on one consecutive interval and zero elsewhere.

Conversely a single simple root is a root. If ${\omega }_i+\cdots +{\omega }_j$ is a root and $j<l$, its pairing with ${\omega }_{j+1}^{\vee }$ is minus one. Reflecting in ${\omega }_{j+1}$ therefore adds that simple root, making the next longer interval a root. Induction constructs every interval. Thus

Use the convention that insertion returns the insertion tableau $P$ first and the recording tableau $Q$ second. Successive row insertions of $3,4,1,6,7,5,2$ giveFor example inserting five bumps six to the second row, and the final insertion of two bumps four from the first row and six from the second, creating the third row. The newly created boxes at these steps receive recording labels six and seven respectively.

The deletion input is this pair with the two tableaux interchanged. The symmetry property of RSK therefore gives the inverse permutation. One can also reverse-insert the boxes in descending recording-label order. The result isIndeed the displayed second row lists the positions of one through seven in the original permutation.

Let $f^{\lambda }$ be the number of standard tableaux of shape $\lambda$. First the associated polytabloids are linearly independent. Order tabloids lexicographically by reading each row after sorting its entries. For a column-increasing tableau $t$, the term $\{t\}$ in $e_t$ has coefficient one; every other tabloid term has larger row word. To see this, take the first row changed by a nontrivial column permutation. Earlier rows are fixed entry by entry, since each entry belongs to one original column. The changed row can only receive entries from lower positions of those increasing columns, so after sorting it is lexicographically larger. In a linear relation between standard polytabloids, choose the smallest row word with a nonzero coefficient. Its own tabloid coefficient cannot be cancelled by a larger standard polytabloid, a contradiction. Thus $\dim S^{\lambda }\ge f^{\lambda }$.

Work first over $\mathbb{C}$. The RSK bijection between permutations and pairs of standard tableaux of the same shape gives ${\sum }_{\lambda ⊢n}(f^{\lambda }{)}^2=n!$. The assumed pairwise nonisomorphic irreducible Specht modules, and the decomposition of the regular representation with irreducible multiplicities equal to their dimensions, give ${\sum }_{\lambda }(\dim S^{\lambda }{)}^2\le n!$. Combining the two inequalities forces equality term by term: $\dim S^{\lambda }=f^{\lambda }$. Hence the independent standard polytabloids are a basis.

The same conclusion holds over every characteristic-zero field. All polytabloid coordinates in the tabloid basis are integers, so the ranks of their coordinate matrices are the same over any characteristic-zero field as over $\mathbb{C}$. In particular the dimension equality and independence transfer without requiring an embedding of the whole field into $\mathbb{C}$.

The equality ${\sum }_{\lambda }(\dim S^{\lambda }{)}^2=n!$ exhausts the dimension of the regular representation. There is no room for another irreducible constituent, since it would contribute its positive squared dimension. ThusEach has the standard-polytabloid basis just proved. The usual split characteristic-zero statement follows over any characteristic-zero field because these integral Specht models remain irreducible after extension to an algebraic closure and already exhaust all dimensions there. In the literal wording of the source, polytabloids themselves are vectors; it is the Specht modules with those bases that form the family of representations.

If the cycle partition contains a part $r$, the Murnaghan–Nakayama rule isA border strip is a connected skew Young diagram with no $2\times 2$ block; its height is its number of occupied rows minus one. The remaining boxes must be a partition diagram. The recursion starts with character one for the empty diagram at the empty cycle type, and an empty sum is zero. Repeated removal for all cycle lengths computes the character value.

Write the hook cycle partition as $\mu =(k,1^{n-k})$, $1\le k\le n$. At the identity, the representation dimension is $n-1$: a standard tableau of shape $(2,1^{n-2})$ is determined by the entry in its top-right box, any one of $2,\dots ,n$.

For $2\le k\le n-2$, the only removable border strip of length $k$ is the vertical bottom strip. Its height is $k-1$, and the remaining hook has size $n-k$ and dimension $n-k-1$. A strip of length $n-1$ would leave the top-left box but disconnect the removed top-right box from the removed column, so none exists and the value is zero. For $k=n$, the whole hook is a border strip of height $n-2$ and contributes $(-1{)}^{n-2}$. These cases combine asIn particular the formula includes zero at $k=n-1$ and the full-cycle sign at $k=n$.

The two partition diagrams are transposes, so their characteristic-zero Specht modules differ by tensoring with the sign representation. A transposition has sign minus one. ConsequentlyThe middle character is $n-3$, since the standard representation is the point-permutation representation minus its trivial line, and a transposition fixes $n-2$ points.

The permutation module $M^{\lambda }$ has a basis consisting of row tabloids of shape $\lambda$: tableaux are identified when each row has the same set of entries. A permutation acts by applying it to every entry, and the action is extended linearly over $F$. Its dimension is $n!/{\prod }_i{\lambda }_i!$.

Here the two singleton rows form an ordered pair of distinct entries, soA standard tableau of the same shape must put one in its top-left corner. Choose two entries from $2,\dots ,n$ for the two remaining boxes of the first column, in increasing order; all other entries then fill the top row in increasing order. Thus the standard-polytabloid basis gives

Write $v_{ij}$ for the tabloid with second-row entry $i$ and third-row entry $j$. Equivariance gives $\varphi (v_{ij})=w_i+w_j$ for every distinct pair. For a tableau $t$, the only nontrivial column antisymmetrizer is on the three entries in its first column. In its alternating sum, pair each column permutation with the permutation obtained by interchanging the two singleton-row positions. Their images under $\varphi$ are the same unordered sum of two $w$'s, while their signs are opposite. Hence $\varphi (e_t)=0$ for every polytabloid, proving $\boxed{S^{(n-2,1,1)}\subseteq \ker \varphi }$.

For distinct $i,j,k$,Each summand is an image and two is invertible in characteristic zero. Thus every basis vector of $M^{(n-1,1)}$ is in the image, proving surjectivity. rank-nullity theorem then gives

The stabilizer of an ordered pair is the subgroup ${\Sigma }_{n-2}$ fixing both entries, so the permutation module is ${\mathrm{Ind}}_{{\Sigma }_{n-2}}^{{\Sigma }_n}1$. Induce in two stages through ${\Sigma }_{n-1}$. The first stage is $S^{(n-1)}\oplus S^{(n-2,1)}$. The characteristic-zero induction branching rule for a symmetric group adds one box in every possible partition position, with multiplicity one. Thus the two second-stage decompositions areBy transitivity of induction this yieldsThe dimensions are $1$, $2(n-1)$, $n(n-3)/2$ and $(n-1)(n-2)/2$, whose sum is $n(n-1)$, checking the decomposition also at $n=4$.

Give $M^{\lambda }$ the invariant symmetric bilinear form for which its tabloid basis is orthonormal. DefineIn characteristic $p>0$, this quotient is nonzero and simple when $\lambda$ is $p$-regular, meaning that no part occurs $p$ or more times. For other partitions it is zero, so the label “simple module” includes the nonzero/$p$-regular qualification.

The James submodule theorem states that for every $F{\Sigma }_n$-submodule $U\subseteq M^{\lambda }$, either $S^{\lambda }\subseteq U$ or $U\subseteq (S^{\lambda }{)}^{\perp }$. Applied to submodules of $S^{\lambda }$, it says that every proper submodule lies in the radical of the restricted form, giving the simplicity of the displayed nonzero quotient.

Choose the two standard tableauxand use their polytabloids $e_1,e_2$, with column-vector coordinates. Writing $[ab\mid cd]$ for the tabloid with those two row sets givesThe coefficients at $[12\mid 34]$ and $[13\mid 24]$ prove independence over every field. Applying the generators directly givesand $(34)$ acts in the same way as $(12)$. Their span is invariant under the generators, contains $e_1$, and contains its whole orbit, hence is the Specht module. Thus it is a basis without needing a characteristic-zero argument. The matrices areThese integral matrices may be reduced in the chosen characteristic.

For the characteristic-three Specht module for the two-by-two partition, reduce the matrices modulo three. The vector $e_1+e_2$ is fixed by every displayed transposition: $(12)$ and $(34)$ send it to $e_1-2e_2=e_1+e_2$, and $(23)$ interchanges its two terms. Thus its span is a proper nonzero trivial submodule.

On the quotient impose $e_2=-e_1$. Each generating transposition then acts as minus one, so the quotient is the sign representation. Consequentlyand the two composition factors are trivial and sign, each once. In the $D$ notation they are $D^{(4)}$ and $D^{(2,2)}$ respectively. Indeed the restricted tabloid form has Gram matrix $\left(\begin{array}{cc}4& 2\\ 2& 4\end{array}\right)$, whose characteristic-three radical is precisely the trivial line, so the simple quotient is $D^{(2,2)}$.

The extension does not split: the minus-one eigenspace of $\rho (12)$ is the line through $e_2$, but $\rho (23)$ sends that line to the line through $e_1$. Thus there is no sign submodule complementing the trivial line, consistent with the failure of semisimplicity.