The residue algebra is generated by
E = T_3, N = T_11, E^2 = E, N^2 = EN = NE = 0.
So
A ≅ F_3 × F_3[ε]/(ε^2).
There are exactly two simple characters:
chi_v7: E=1, N=0chi_pair: E=0, N=0In the cyclic basis [w, T_3w, T_11w], the residue regular module splits as
A_reg ≅ chi_v7 ⊕ P_pair
with - chi_v7 line spanned by [0, 1, 0] - P_pair plane spanned by [[1, 2, 0], [0, 0, 1]]
So the regular module is not an indecomposable length-3 bridge between the two characters. It is a direct sum of the split branch and the unique non-split self-extension of chi_pair.
| Ext^1(top, bottom) | nonzero Yoneda elements | non-split isomorphism classes | verdict |
|---|---|---|---|
Ext^1(chi_pair, chi_pair) |
2 |
1 |
one-dimensional, unique non-split module |
Ext^1(chi_pair, chi_v7) |
0 |
0 |
split only |
Ext^1(chi_v7, chi_pair) |
0 |
0 |
split only |
Ext^1(chi_v7, chi_v7) |
0 |
0 |
split only |
The only nontrivial residue extension is
Ext^1(chi_pair, chi_pair) ≅ F_3,
while the cross-block groups vanish:
Ext^1(chi_v7, chi_pair) = Ext^1(chi_pair, chi_v7) = 0.
So at residue level there is no genuine indecomposable bridge between the split V7 branch and the glued V1/V5 branch. The node is block-diagonal at the level of simples, and the only non-semisimplicity is the self-extension of the glued pair block.