Remark. Support parity, group degree and sign sectors [fdip-003P]
AGENTDRAFTED
For the four-cycle over a nonzero commutative ring with invertible
two, three decompositions record different structures. The
support grading separates fields supported on
\(\{0,2\}\) and \(\{1,3\}\). The one-step shift
\(Tf(i)=f(i+1)\) exchanges these supports, so it is odd, whereas
\(P=T^2\) and \(E_\pm =h(I\pm P)\) preserve them and are even.
A single half-turn sign sector contains both
\(b_{\varepsilon ,0}=\delta _0+\varepsilon \delta _2\) and
\(b_{\varepsilon ,1}=\delta _1+\varepsilon \delta _3\), which have
different support parities. A minus eigenvalue does not mean odd support.
The canonical group grading of \(k[C_2]\) has
components \(k\,1\) of degree \(1_G\) and \(k\,g\) of degree \(r\).
Their multiplication follows the group law, in particular \(g^2=1\).
Over a ring these are rank-one free summands. The representation sends
\(g\) to support-even \(P\); it therefore does not preserve these
particular gradings under \(r\mapsto 1\in \mathbb Z/2\mathbb Z\).
The projector ideals are different again: \(e_-^2=e_-\ne 0\), while
an odd homogeneous component would have its square in the even component.
The direct product, the group grading and the support grading cannot be
interchanged. These are the distinct splitting levels
applied to the concrete action.
The ordinary adjoint uses the counting pairing
\(b(f,u)=\sum _i f_i u_i\), with identity coefficient involution.
Delta vectors recover all coefficients, so the pairing is perfect.
Reindexing a finite sum shows that every permutation action is an
isometry, hence
\[b(\rho (\sigma )f,u)=b(f,\rho (\sigma ^{-1})u).\]
Uniqueness of the ordinary adjoint gives
\(\rho (\sigma )^\dagger =\rho (\sigma ^{-1})\); in particular
\(T^\dagger =T^{-1}\) and \(P^\dagger =P\). There is no complex
conjugation or Koszul sign in this calculation.
The corresponding ordinary group-algebra operation is
\[\left (\sum _g a_g[g]\right )^*=\sum _g a_g[g^{-1}].\]
It is \(k\)-linear, fixes scalars, squares to the identity and reverses
products, since \((gh)^{-1}=h^{-1}g^{-1}\) and all sums are finite.
For an isometric representation with a perfect pairing, taking the
adjoint of its finite linear extension therefore gives
\(\widehat \rho (z^*)=\widehat \rho (z)^\dagger \).
The isometry premise matters: the infinite cyclic group acting on
\(\mathbb R\) by multiplication by \(2\) has adjoint \(2\) and inverse
\(1/2\), so this identity fails for that representation. The counting
action satisfies the premise.
The group basis and inversion are as in
[sengupta2010representations, Sections 3.1-3.2, pp. 39-41];
the grading and adjoint comparisons above are direct calculations.
The shift calculation supplies the support-degree and
adjoint identities. Clifford parity of a quadratic quotient is a further
structure; these ordinary formulas introduce neither a graded star nor
a Hermitian extension.