Write the residue character exponents as \(a\bmod(q^n-1)\) and \(b\bmod(q^m-1)\). The \(i\)th refined packet has exponent
\[
c_i=a\frac{q^L-1}{q^n-1}+bq^i\frac{q^L-1}{q^m-1}
\pmod{q^L-1}.
\]
Norm lifting multiplies an exponent by the quotient of group orders. Shifting the second character by Frobenius multiplies its exponent by \(q^i\). Multiplication of characters adds the two terms.
For \(q=2,n=4,m=6\), the common modulus is \(2^{12}-1=4095\). Taking \(a=b=1\), the two packets have
\[
c_0=273+65=338,\qquad c_1=273+130=403.
\]
Both are nonzero modulo 4095, so neither packet contributes inertia invariants.
Reading the diagram
The test is \(c_i=0\) in the common modulus. A numerical-looking residue formula is an exact character calculation.