Proposition. The positive CAR product has half-staircase weight [fcap-0036]
AGENTDRAFTED
Let \(K\) be a field in which \(2\) is invertible. Begin with \(N=2\)
and put \(v_+=d_{01}\). The relations are
\[d_{01}^2=d_{10}^2=0,\qquad d_{00}^2=d_{11}^2=1,
\qquad d_{01}d_{10}+d_{10}d_{01}=2.\]
The only raising operator is
\[D(E_{01})=\frac 12(d_{00}d_{01}+d_{01}d_{11}),\]
and each summand kills \(v_+\). For the diagonal operators, the same CAR
relations give
\[D(E_{00})v_+=\frac 32v_+,\qquad
D(E_{11})v_+=\frac 12v_+.\]
Thus \(v_+\) has weight \((3/2,1/2)\), and
\(D(I)v_+=2v_+\). Subtracting the central correction
\((N/2)\operatorname {Tr}\) leaves \((1/2,-1/2)\). In characteristic
zero this is the usual \(\mathfrak {sl}_2\) value \(\rho \).
For general \(N\), order the indices as
\(0<1<\cdots <N-1\) and form the lexicographically ordered product
\[v_+=\prod _{0\leq i<j<N}^{\mathrm {lex}}d_{ij}.\]
Every positive matrix unit is \(Q_{\mathrm {tr}}\)-null, and distinct
positive matrix units are polar-orthogonal. Their Clifford generators
therefore anticommute. PBW linear independence shows that the ordered
product is nonzero over a field.
Let \(i<j\). In
\[D(E_{ij})v_+=\frac 12\sum _kd_{ik}d_{kj}v_+,\]
a term with \(k<j\) vanishes because the positive generator \(d_{kj}\)
already occurs in \(v_+\). If \(k\geq j\), then \(i<k\); first
anticommute \(d_{ik}\) past \(d_{kj}\), using \(i\ne j\), and the repeated
positive factor \(d_{ik}\) makes the term zero. Hence every raising
operator annihilates \(v_+\).
For a diagonal generator,
\[D(E_{ii})v_+=\frac 12\sum _kd_{ik}d_{ki}v_+.\]
The contributions for \(k<i\), \(k=i\), and \(k>i\) are respectively
\(0\), \(v_+\), and \(2v_+\). Therefore
\[D(E_{ii})v_+
=\frac 12\bigl (1+2|\{k:k>i\}|\bigr )v_+.\]
For \(i\in \operatorname {Fin}(N)\), the weight and its quadratic-lift
part are
\[\lambda _i=N-i-\frac 12,\qquad
\lambda _i-\frac N2=\frac {N-1}{2}-i,\qquad
\sum _i\lambda _i=\frac {N^2}{2}.\]
The last equality agrees with the direct central calculation
\(D(I)=N^2/2\).
TauCeti defines this ordered product and proves its nonvanishing,
raising equations, and diagonal weight. See the exact pinned
product and PBW nonvanishing
and
half-staircase theorems.
Nonvanishing needs a field; the highest-vector statement adds invertible
\(2\). Identifying the subtracted weight with the rational staircase
requires characteristic zero.
[panyushev2001exterior, Section 2, Example 2.5(1), preprint p. 8]
identifies, over an algebraically closed field of characteristic zero,
the reduced adjoint spin module with highest weight \(\rho \). The
subtracted weight above is exactly that semisimple staircase. The
added \((N/2)\operatorname {Tr}\) character records the reductive center of
this normal-ordered left action. This comparison does not attribute the
selected product or the full left-regular CAR module to Panyushev, and
it makes no irreducibility or whole-module decomposition claim.