Center of the Auslander Algebra

Let

M = B0 ⊕ R ⊕ B3

for the pair node

R = Z_3[eta]/(eta^2 - 3 eta).

Exact center

Any endomorphism of M commuting with the vertex idempotents must be diagonal:

(lambda0, phi, lambda3)

with lambda0 ∈ Z_3, lambda3 ∈ Z_3, and phi ∈ End_R(R)=R.

Write phi as multiplication by (a,b) ∈ R ⊂ Z_3 × Z_3. Commuting with the branch maps i0,q0,i3,q3 forces

lambda0 = a,
lambda3 = b.

So the center is exactly

Z(End_R(B0 ⊕ R ⊕ B3)) ≅ R.

For the full singular Hecke order

O_sing ≅ Z_3(V7) × R,

the same argument gives

Z(End_{O_sing}(Z_3(V7) ⊕ B0 ⊕ R ⊕ B3)) ≅ O_sing.

So the Hecke node is not just represented by the CM generator algebra: it is recovered exactly as its center.

Residue center

Brute-force enumeration in the 8-dimensional residue Auslander algebra shows that the center is 2-dimensional over F_3, with basis

1 = e0 + eR + e3,
t.

Since t^2 = 0, this residue center is

F_3[t]/(t^2),

which is exactly the dual-number shadow of the pair node.

This is the closest finite shadow so far: the residue quiver algebra remembers the full node through its center, and the center reduces to the same dual numbers that appeared in the syndrome algebra.