Trace-free Euler characteristic of a stable-pair complex

We first justify the character of the one-dimensional sheaf, including its multiplicities. A power of the radical ideal of the support annihilates $F$, so its finite filtration by those powers has quotients on the reduced support. Remove finitely many intersections of distinct components; each quotient then has a generic rank on each integral support curve. Comparison with its finite normalization has kernel and cokernel supported at finitely many points. On a normalized smooth curve a coherent sheaf of generic rank $r$, modulo its finite-support part, differs in its Grothendieck class from $r\mathcal{O}$ only by finite-support classes: choose a rational generic basis, clear its finitely many poles by an effective divisor, and compare the resulting rank-$r$ inclusion with its finite-length cokernel; the divisor exact sequence compares each twist with $\mathcal{O}$. Adding the filtration shows that in $G_0(X)$ modulo finite-support classes, $[F]$ is the sum of the normalized curve structure-sheaf pushforwards with coefficients equal to its generic lengths. These coefficients are exactly $\beta$.

Apply GRR to a normalization map $j:\tilde{C}\to X$. Its formula $\mathrm{ch}(j_{\ast }{\mathcal{O}}_{\tilde{C}})\mathrm{td}(T_X)=j_{\ast }\mathrm{td}(T_{\tilde{C}})$ has no terms of codimension zero or one on the right, and its codimension-two term is $j_{\ast }[\tilde{C}]=[C]$. The same formula for a closed point has no term below codimension three. Since the Todd class starts with $1$, step 1.1 proves ${\mathrm{ch}}_0(F)={\mathrm{ch}}_1(F)=0$ and ${\mathrm{ch}}_2(F)=\beta$. Write $\mathrm{ch}(F)=\beta +\gamma$ with $\gamma \in A^3(X{)}_{\mathbb{Q}}$.

In $K_0(X)$, $[I^{\bullet }]=[{\mathcal{O}}_X]-[F]$. Hence $\mathrm{ch}(I^{\bullet })=1-\beta -\gamma$ and $\mathrm{ch}((I^{\bullet }{)}^{\vee })=1-\beta +\gamma$. Multiplication and truncation above dimension three give $\mathrm{ch}((I^{\bullet }{)}^{\vee }{\otimes }^{\mathbf{L}}{I}^{\bullet })=1-2\beta$. The identity and trace maps split the endomorphism complex: their composite on ${\mathcal{O}}_X$ is multiplication by the virtual rank of $I^{\bullet }$, which is $1$. Thus the trace-free derived endomorphism class has character $-2\beta$, after subtracting the scalar ${\mathcal{O}}_X$ summand. The trace statement can also be checked on any finite vector-bundle resolution: the evaluation trace of the identity is the alternating sum of its ranks.

GRR for $X\to \mathrm{Spec}\mathbb{C}$ now gives $\chi (R\mathrm{Hom}(I^{\bullet },I^{\bullet }{)}_0)={\int }_X(-2\beta )\mathrm{td}(T_X)=-{\int }_X\beta c_1(T_X)$. Only ${\mathrm{td}}_1$ can contribute because $\beta$ has codimension two on a threefold. Negating gives the claimed formula; its right-hand side is the integer degree of the integral cycle $\beta$ intersected with the Cartier class $c_1(T_X)$. No POT or virtual-cycle comparison is used in this computation.