Jucys–Murphy Elements and Seminormal Forms — Examples

The displayed elements are $X_2=(12)$, $X_3=(13)+(23)$ and $X_4=(14)+(24)+(34)$; the products below are computed in the group algebra $\mathbb{Z}[S_4]$, in which distinct permutations form a $\mathbb{Z}$-basis.

$X_2{X}_3=(12)((13)+(23))=(12)(13)+(12)(23)=(132)+(123)$.

$X_2{X}_4=(12)((14)+(24)+(34))=(12)(14)+(12)(24)+(12)(34)=(142)+(124)+(12)(34)$.

$X_3{X}_4=((13)+(23))((14)+(24)+(34))=(134)+(143)+(234)+(243)+(13)(24)+(14)(23)$.

$e_1(X_2,X_3,X_4)=X_2+X_3+X_4=(12)+(13)+(23)+(14)+(24)+(34)$, the six transpositions, i.e. the six permutations with $3$ cycles by [F3].

Adding the three products of steps 2.1-2.3 gives $e_2(X_2,X_3,X_4)=X_2{X}_3+X_2{X}_4+X_3{X}_4=(132)+(123)+(142)+(124)+(134)+(143)+(234)+(243)+(12)(34)+(13)(24)+(14)(23)$: eight $3$-cycles and the three double transpositions, i.e. $11$ permutations, all with $2$ cycles by [F3].

$e_3(X_2,X_3,X_4)=X_2{X}_3{X}_4=(1234)+(1243)+(1324)+(1342)+(1423)+(1432)$, the six $4$-cycles, i.e. the permutations with one cycle by [F3]; indeed the product is $X_2(X_3{X}_4)$ and the six products of the three transpositions of $X_4$ with $(12)$ and the two transpositions of $X_3$ are exactly the six listed $4$-cycles, with no repetitions among them.

By [F2] with $n=4$ the evaluations $e_s(X_2,X_3,X_4)$ equal the class sums ${\sum }_{\mathrm{\ell }(\rho )=4-s}{C}_{\rho }^{(4)}$ for $s=1,2,3$, which are exactly the three displayed finite sums; each class sum is central because conjugation permutes each conjugacy class. In particular no term cancels and every coefficient is $1$, as the explicit expansions of steps 3.1 and 3.2 show.

The case $s=0$ gives $e_0=1$ and the case $s=4$ gives $e_4=0$; the example exhibits the three nontrivial evaluations and confirms the identity of [F2] at the first size with three nonzero positive-degree elementary evaluations.

The interpolation recursion of the refuted statement, applied with $n=2$ to the tableau whose entry $2$ lies in the addable node of content $1$ of the shape $(1)$, is $P_T=P_{[1]}\frac{X_2-c}{c_T(2)-c}\text{with }{c}_T(2)=1,c=-1,$ because $A((1))=\{1,-1\}$ by [F3]; thus the factor is $(X_2+1)\mathrm{/}(1-(-1))=(X_2+1)\mathrm{/}2$.

The failure is not caused by the absence of the module: over $F$ the Specht module $S_F^{(2,1)}$ and its endomorphism algebra continue to exist by [F4]. Writing $b_j$ for the tabloid with $j$ in its singleton row, the polytabloids span the two-dimensional subspace ${\mathrm{span}}_F\{b_3-b_1,b_2-b_1\}$, since these differences are independent and every polytabloid is a difference $b_j-b_k$. The two content vectors of [F2] even coincide after reduction to $\mathbb{Z}\mathrm{/}2$; what fails is precisely the separation of the two addable contents $1$ and $-1$ that the interpolation denominators encode.

Evaluation of the displayed factor in $F[S_3]$ requires the inverse of the integer $2$ in $F$; but $2=0$ in $F$ by [F1], so the factor $(X_2+1)\mathrm{/}2$ is not an element of $F[S_3]$ and the formula does not define $P_T$ over $F$. The same applies to the companion tableau $T^{\mathrm{\prime }}$, whose factor is $(1-X_2)\mathrm{/}2$.

Hence the characteristic-zero interpolation formula for the primitive idempotents does not reduce to a formula of the same shape over a field of characteristic $2$: the collision prevents these factors from separating the two prefixes. No nonexistence assertion about other idempotents follows from this calculation. This refutes the displayed statement and completes the counterexample.

Multiplication in $S_3$ gives $X_2^2=1$, $X_2{X}_3=X_3{X}_2=(123)+(132)$ and $X_3^2=(123)+(132)+2$, whence also $(X_2{X}_3{)}^2=X_2{X}_3+2$; every product below is evaluated with these relations and the basis of the Given block.

Contents. The row tableau carries entries $1,2,3$ in the cells $(1,1),(1,2),(1,3)$ of contents $0,1,2$; the tableau $\begin{array}{cc}1& 2\\ 3\end{array}$ carries them in $(1,1),(1,2),(2,1)$ of contents $0,1,-1$; the tableau $\begin{array}{cc}1& 3\\ 2\end{array}$ in $(1,1),(2,1),(1,2)$ of contents $0,-1,1$; and the column tableau in $(1,1),(2,1),(3,1)$ of contents $0,-1,-2$. This gives the four content vectors of the Statement.

The recursion of [F1] at $n=2$ gives $P_{[12]}=(X_2+1)\mathrm{/}2$ and $P_{\begin{array}{c}1\\ 2\end{array}}=(1-X_2)\mathrm{/}2$, because the two addable contents of $(1)$ are $1$ and $-1$; at $n=3$ the factors are $(X_3+1)\mathrm{/}3$ and $(2-X_3)\mathrm{/}3$ over the shape $(2)$, and $(X_3+2)\mathrm{/}3$ and $(1-X_3)\mathrm{/}3$ over the shape $(1,1)$. Multiplying gives exactly the four displayed elements, with denominators $2$ and $6$.

Expanded in the basis of the Given block, the four elements are $P_{\mathrm{r}\mathrm{o}\mathrm{w}}=\frac{1}{6}(1+(12)+(13)+(23)+(123)+(132)),$ $P_{\begin{array}{cc}1& 2\\ 3\end{array}}=\frac{1}{3}\cdot 1+\frac{1}{3}(12)-\frac{1}{6}(13)-\frac{1}{6}(23)-\frac{1}{6}(123)-\frac{1}{6}(132),$ $P_{\begin{array}{cc}1& 3\\ 2\end{array}}=\frac{1}{3}\cdot 1-\frac{1}{3}(12)+\frac{1}{6}(13)+\frac{1}{6}(23)-\frac{1}{6}(123)-\frac{1}{6}(132),$ $P_{\mathrm{c}\mathrm{o}\mathrm{l}}=\frac{1}{6}(1-(12)-(13)-(23)+(123)+(132))\mathrm{.}$

Idempotence and orthogonality. Squaring each of the four elements of step 3.1 with the relations of step 1.1 returns the same element, and each product of two distinct ones vanishes; equivalently, the four elements are the orthogonal rank-one projections of [F3], and their sum is $\frac{1}{6}+\frac{1}{3}+\frac{1}{3}+\frac{1}{6}=1,$ using the coefficient sums in step 3.1. This verifies all algebraic assertions about $P_T$.

Action on the spectrum. By [F3] each $P_T$ is the projection onto the line $\mathbb{C}{v}_T$, so $P_T$ acts by $1$ on the content-vector eigenvector of $T$ and by $0$ on the eigenvectors of the other tableaux, whose content vectors are the four listed in step 1.2.

The sum of the two $(2,1)$-projectors is $P_{\begin{array}{cc}1& 2\\ 3\end{array}}+P_{\begin{array}{cc}1& 3\\ 2\end{array}}=\frac{2}{3}\cdot 1-\frac{1}{3}(123)-\frac{1}{3}(132)$, which is central in $\mathbb{C}[S_3]$ and, by [F3], is the central idempotent of the two-dimensional Specht module $V^{(2,1)}$; the remaining two projectors are the central idempotents of the one-dimensional modules $V^{(3)}$ and $V^{(1,1,1)}$.

The example exhibits the interpolation formula at the smallest nontrivial size: the recursion inverts only $2$ at step $n=2$ and $3$ at step $n=3$, giving the denominator $6$ in the products, and the failure of the formula in characteristic $2$ is recorded separately in Ordinary Jucys-Murphy projection formulas do not survive content collision.

Apply [F2] with $S=T^{\mathrm{\prime }}$ (the row tableau, of length $0$) and $S^{\mathrm{\prime }}=s_{2}S=T$ (of length $1$, the longer one): here $r_{T^{\mathrm{\prime }}}=c_{T^{\mathrm{\prime }}}(3)-c_{T^{\mathrm{\prime }}}(2)=(-1)-1=-2$, so $s_2{v}_{T^{\mathrm{\prime }}}=v_T+(-2{)}^{-1}{v}_{T^{\mathrm{\prime }}}=v_T-\frac{1}{2}{v}_{T^{\mathrm{\prime }}}$ and $s_2{v}_T=(1-(-2{)}^{-2})v_{T^{\mathrm{\prime }}}-(-2{)}^{-1}{v}_T=\frac{3}{4}{v}_{T^{\mathrm{\prime }}}+\frac{1}{2}{v}_T$.

The action of $s_1=(12)$: in $T$ the entries $1,2$ occupy $(1,1)$ and $(2,1)$, the same column, so $s_1{v}_T=-v_T$; in $T^{\mathrm{\prime }}$ they occupy $(1,1)$ and $(1,2)$, the same row, so $s_1{v}_{T^{\mathrm{\prime }}}=+v_{T^{\mathrm{\prime }}}$. Hence $s_1$ acts by $\mathrm{diag}(-1,1)$ in the ordered basis.

Matrix. In the ordered basis $(v_T,v_{T^{\mathrm{\prime }}})$ whose columns are the images, step 1.1 gives $M=\left(\begin{array}{cc}1\mathrm{/}2& 1\\ 3\mathrm{/}4& -1\mathrm{/}2\end{array}\right)$. The transpose $M^{\mathsf{T}}$ is the array obtained when the images are written as rows. The matrix acting on coordinate columns is $M$.

Involution. Squaring the matrix of step 2.1 gives $\left(\begin{array}{cc}1\mathrm{/}4+3\mathrm{/}4& 1\mathrm{/}2-1\mathrm{/}2\\ 3\mathrm{/}8-3\mathrm{/}8& 3\mathrm{/}4+1\mathrm{/}4\end{array}\right)=\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)$, and the matrix of step 1.2 is visibly an involution; this is the statement $s_1^2=s_2^2=1$ in the two-dimensional Specht module.

Braid relation. With $M$ the matrix of step 2.1 and $D=\mathrm{diag}(-1,1)$ that of step 1.2, direct multiplication gives $DMD=\left(\begin{array}{cc}1\mathrm{/}2& -1\\ -3\mathrm{/}4& -1\mathrm{/}2\end{array}\right)=MDM$, which is $s_1{s}_2{s}_1=s_2{s}_1{s}_2$ in the ordered basis.

Orthogonal rescaling. Normalize the shorter vector $v_{T^{\mathrm{\prime }}}$ to norm $1$. The norm ratio from [F3], applied first in the shorter-to-longer ordering $(T^{\mathrm{\prime }},T)$, gives $\mathrm{\parallel }{v}_T\mathrm{\parallel }=\sqrt{3}\mathrm{/}2$. Thus the unit basis in the example's ordering is $(u_T,u_{T^{\mathrm{\prime }}})=((2\mathrm{/}\sqrt{3})v_T,v_{T^{\mathrm{\prime }}})$. With $E=\mathrm{diag}(2\mathrm{/}\sqrt{3},1)$, direct change of basis gives $E^{-1}ME=\left(\begin{array}{cc}1\mathrm{/}2& \sqrt{3}\mathrm{/}2\\ \sqrt{3}\mathrm{/}2& -1\mathrm{/}2\end{array}\right)$. This real symmetric matrix squares to $I$ and is orthogonal.

Consistency with the row and column cases. The signs $s_1{v}_T=-v_T$ and $s_1{v}_{T^{\mathrm{\prime }}}=+v_{T^{\mathrm{\prime }}}$ of step 1.2 are the same-row and same-column scalars of [F2] and are unchanged by the positive unit rescaling, so the full action of $S_3$ on the two-dimensional Specht module is exhibited in both normalizations.

The example is the smallest nontrivial check of the seminormal and orthogonal forms: the axial distance $r=2$, the structure constants $\frac{1}{2},\frac{3}{4},1$, the rescaling factor $2\mathrm{/}\sqrt{3}$ and the orthogonal block agree with the displayed matrices of the sources, and both matrices were verified by exact rational multiplication in steps 3.1 and 3.2.