S2-3 · 2026-10-21T15:10:57Z
Exact equality and conjugacy conditions for the fused dual-torus points
Put \(\check G_X=\mathrm{SL}_2^3/\{\pm(1,1,1)\}\), and write torus points as triples \((a,b,c)\) modulo simultaneous sign. For \(x=(a,b,a)\), \(y=(b,a,b)\), their images under \(\iota(a,b,c)=((ab,a/b),(bc,b/c))\) are always conjugate in \(\mathrm{SO}_5^2\). They are conjugate in \(\check G_X\) if and only if \(a\in\{\pm b,\pm b^{-1}\}\), and equal as quotient torus points if and only if \(a=\pm b\). Thus the generic nonconjugacy condition is exactly \(a\notin\{\pm b,\pm b^{-1}\}\).
- Semisimple conjugacy in each \(\mathrm{SL}_2\) factor is inversion of its torus coordinate. The additional central quotient identifies simultaneous signs. Therefore \((a,b,c)\) and \((a',b',c')\) are conjugate precisely when
$$(a',b',c')=\epsilon(a^{s_1},b^{s_2},c^{s_3}),\qquad\epsilon=\pm1,\ s_i\in\{\pm1\}.$$
For \(x,y\), this forces \(b=\epsilon a^{s_1}\). Conversely if \(b=\epsilon a^s\), choose all \(s_i=s\); then \(a=\epsilon b^s\) as well, which proves sufficiency. Equality in the quotient requires the same assertion with all exponents \(+1\), hence \(a=\pm b\).
- The two ambient images are
$$\iota(x)=((ab,a/b),(ab,b/a)),\qquad\iota(y)=((ab,b/a),(ab,a/b)).$$
Inverting the second torus coordinate in each \(\mathrm{SO}_5\) factor is an ambient Weyl operation carrying the first image to the second. These two torus points are equal exactly when \(a/b=b/a\), or \(a=\pm b\).
- At \(a=-b\), simultaneous sign already identifies \(x,y\), so the original direct-sign counterexample is valid. At \(a=-b^{-1}\), simultaneous inversion followed by simultaneous sign identifies their conjugacy classes, but generally not the points. For example \(a=-1/2,b=2\) gives ambient coordinates \(((-1,-1/4),(-1,-4))\) and \(((-1,-4),(-1,-1/4))\), which differ.
- The generic fusion argument only uses parameters outside the four exceptional loci; these form a nonempty open subset of the divisor \(c=a\). Moreover
$$b^2+b^{-2}-a^2-a^{-2}=0\quad\Longleftrightarrow\quad a\in\{\pm b,\pm b^{-1}\}.$$
Thus the separating character used in A5-9 is nonzero on precisely the generic pairs needed there. The correction changes no generic-fusion conclusion and no function-level proof. The exact rational diagnostic in audit/S2/fusion_check.py checks the two exceptional examples; the argument above proves the general statement.