Lens-space correction-term formula for L(p,1)

The alpha-gamma diagram, and the diagram with those two attaching curves reversed, each have one intersection in each Spin${}^c$ structure. Their infinity differentials are zero: the Maslov index of a self-disk in a torsion Spin${}^c$ structure is $2n_z$, so cannot be one. The minus lattice in each summand is therefore exactly $U{\mathbb{F}}_2[U][x_i,0]$, and its plus image has bottom $[x_i,0]$. Its correction term is the absolute grading of that generator, with no unknown exponent shift.

For the negative-trace map from $S^3$ to $-L(p,1)$, the fixed triangle has index and basepoint multiplicity zero and count one. It sends the sphere generator at exponent zero to $[x_i,0]$. The four-dimensional data are $c_1^2=-k^2\mathrm{/}p$, $\chi =1$ and $\sigma =-1$. Hence its degree is $D_i=(1-k^2\mathrm{/}p)\mathrm{/}4$, and $d(-L(p,1),i)=D_i$. This computes the negative orientation directly.

The cyclically reordered triangle gives the actual fixed-structure map from $L(p,1)$ to $S^3$ and sends $[x_i,0]$ to the sphere generator at exponent zero. It is the same oriented triple four-manifold with the other sphere input filled, so it has the same degree $D_i$. Consequently $0=\mathrm{gr}[x_i,0]+D_i$, and $d(L(p,1),i)=-D_i=((2i-p{)}^2-p)\mathrm{/}(4p)$. This argument uses an actual turned-around triangle map rather than a general grading-duality assertion.

The characteristic labels under orientation reversal or conjugation are paired by $k\mapsto -k$, equivalently $i\mapsto p-i$ modulo $p$. Their squares agree, so the two computations give exactly the asserted negation for the opposite orientation with the conjugate label. For $p=1$ the same computation has $k=-1$ and both correction terms zero.