Manuscript IX-B-ST@1 · IX-B-ST.tex · section · f6af1523

Numerical involution

Regularized association and the missing sharp operation › Numerical involution · lines 1446–1479

Let $τ (x, y, x^{'}) = (x^{'}, y, x)$ , and define

⟨ U^{†}, Φ ⟩ = ⟨ U, Φ \circ τ ⟩ .

The explicit formulae give

(T_{ε}^{ρ})^{†} = T_{ε}^{ρ}, r^{†} = r .

Indeed $D_{ε} (- z) = D_{ε} (z)$ , and exchange swaps the two mixed factors. The strong finite-part limit then gives

(u_{μ}^{ρ})^{†} = u_{μ}^{ρ} .

For $O = a D$ , put

O^{†} U = (O (U^{†}))^{†} .

Then

(O u_{μ}^{ρ})^{†} = O^{†} u_{μ}^{ρ} .

An $x^{0}$ -derivative is exchanged with an $x^{'0}$ -derivative; a descendant need not itself be exchange invariant. These are distributional symmetry identities, not a quantum algebra involution theorem.

Displayed formulas in this passage

unlabeled · line 1449
$⟨ U^{†}, Φ ⟩ = ⟨ U, Φ \circ τ ⟩ .$
unlabeled · line 1454
$(T_{ε}^{ρ})^{†} = T_{ε}^{ρ}, r^{†} = r .$
unlabeled · line 1461
$(u_{μ}^{ρ})^{†} = u_{μ}^{ρ} .$
unlabeled · line 1465
$O^{†} U = (O (U^{†}))^{†} .$
unlabeled · line 1470
$(O u_{μ}^{ρ})^{†} = O^{†} u_{μ}^{ρ} .$