[proofplan] We determine the diagonal class $\delta = D_{M \times M}^{-1}[\Delta(M)] \in H^d(M \times M; \mathbb{Q})$ by expanding it in the Künneth basis $\{a_i \otimes b_j\}$ and solving for the coefficients $C_{ij}$. We evaluate $(b_k \otimes a_\ell) \smile \delta$ on $[M \times M]$ in two ways: algebraically using the Künneth formula, which isolates $C_{k\ell}$ up to a sign; and geometrically using the intersection interpretation of the diagonal, which forces $C_{k\ell} = (-1)^{|a_k|} \delta_{k\ell}$. [/proofplan] [step:Expand $\delta$ in the Künneth basis] Let $M$ be a compact $\mathbb{Z}$-oriented $d$-manifold. By the Künneth theorem over $\mathbb{Q}$, \begin{align*} H^*(M \times M; \mathbb{Q}) \cong H^*(M; \mathbb{Q}) \otimes_\mathbb{Q} H^*(M; \mathbb{Q}). \end{align*} Let $\{a_i\}$ be a homogeneous basis for $H^*(M; \mathbb{Q})$ and $\{b_i\}$ the dual basis under the cup product pairing: $\langle a_i, b_j \rangle := (a_i \smile b_j)[M] = \delta_{ij}$. Such a dual basis exists because the cup product pairing is non-singular by the [Non-Singularity of the Cup Product Pairing](/theorems/2293) (with $R = \mathbb{Q}$, where $H_*(M; \mathbb{Q})$ is automatically free). Note that $|b_i| = d - |a_i|$. Since $\{a_i \otimes b_j\}$ is a basis for $H^*(M \times M; \mathbb{Q})$, we can write \begin{align*} \delta = \sum_{i,j} C_{ij}\, a_i \otimes b_j \end{align*} for unique coefficients $C_{ij} \in \mathbb{Q}$. We must show $C_{ij} = (-1)^{|a_i|} \delta_{ij}$. [/step] [step:Evaluate $(b_k \otimes a_\ell) \smile \delta$ on $[M \times M]$ algebraically] Fix indices $k$ and $\ell$. The cup product in $H^*(M \times M; \mathbb{Q})$ satisfies the Künneth sign rule: \begin{align*} (\alpha_1 \otimes \alpha_2) \smile (\beta_1 \otimes \beta_2) = (-1)^{|\alpha_2||\beta_1|} (\alpha_1 \smile \beta_1) \otimes (\alpha_2 \smile \beta_2). \end{align*} Applying this with $\alpha_1 = b_k$, $\alpha_2 = a_\ell$, $\beta_1 = a_i$, $\beta_2 = b_j$: \begin{align*} (b_k \otimes a_\ell) \smile (a_i \otimes b_j) = (-1)^{|a_\ell||a_i|} (b_k \smile a_i) \otimes (a_\ell \smile b_j). \end{align*} Evaluating on $[M \times M] = [M] \otimes [M]$: \begin{align*} \bigl((b_k \otimes a_\ell) \smile (a_i \otimes b_j)\bigr)[M \times M] = (-1)^{|a_\ell||a_i|} (b_k \smile a_i)[M] \cdot (a_\ell \smile b_j)[M]. \end{align*} By graded commutativity, $(b_k \smile a_i)[M] = (-1)^{|b_k||a_i|}(a_i \smile b_k)[M] = (-1)^{|b_k||a_i|} \delta_{ik}$. Similarly, $(a_\ell \smile b_j)[M] = \delta_{\ell j}$. Substituting: \begin{align*} \bigl((b_k \otimes a_\ell) \smile \delta\bigr)[M \times M] &= \sum_{i,j} C_{ij} (-1)^{|a_\ell||a_i| + |b_k||a_i|} \delta_{ik} \delta_{\ell j} \\ &= C_{k\ell} \cdot (-1)^{|a_\ell||a_k| + |b_k||a_k|}. \end{align*} [guided] The Künneth sign rule for the cup product on a product space inserts a sign $(-1)^{|\alpha_2||\beta_1|}$ when commuting the "cross" factors. This is because the cup product on $M \times M$ is the composition of the external cross product with the diagonal, and the cross product picks up a Koszul sign. The evaluation on $[M \times M]$ then reduces to the individual evaluations on $[M]$ via the Künneth formula for the fundamental class: $[M \times M] = [M] \otimes [M]$. The dual basis property $\langle a_i, b_j \rangle = \delta_{ij}$ collapses the double sum to a single term, but we need to account for the graded commutativity sign when writing $(b_k \smile a_i)[M]$. Since $a_i \smile b_k = (-1)^{|a_i||b_k|} b_k \smile a_i$ and $\langle a_i, b_k \rangle = (a_i \smile b_k)[M] = \delta_{ik}$, we get $(b_k \smile a_i)[M] = (-1)^{|a_i||b_k|} \delta_{ik}$. [/guided] [/step] [step:Evaluate $(b_k \otimes a_\ell) \smile \delta$ on $[M \times M]$ geometrically] The diagonal class $\delta = D_{M \times M}^{-1}[\Delta(M)]$ is the Poincaré dual of the diagonal $\Delta(M) \subset M \times M$. By the [Poincaré Dual of a Submanifold](/theorems/2295), for any cohomology class $\omega \in H^d(M \times M; \mathbb{Q})$, the evaluation $\omega[\Delta(M)]$ equals $(\omega \smile \delta)[M \times M]$ (up to the identification $D_{M \times M}(\omega) \frown \delta$). More precisely, the defining property of the Poincaré dual gives \begin{align*} \bigl((b_k \otimes a_\ell) \smile \delta\bigr)[M \times M] = (b_k \otimes a_\ell)[\Delta(M)]. \end{align*} The diagonal embedding $\Delta: M \to M \times M$ sends $x \mapsto (x, x)$, so \begin{align*} (b_k \otimes a_\ell)[\Delta(M)] = \Delta^*(b_k \otimes a_\ell)[M] = (b_k \smile a_\ell)[M] \end{align*} since $\Delta^*(\pi_1^*b_k \smile \pi_2^*a_\ell) = b_k \smile a_\ell$ (as $\pi_i \circ \Delta = \operatorname{id}_M$). By graded commutativity and the dual basis property: \begin{align*} (b_k \smile a_\ell)[M] = (-1)^{|b_k||a_\ell|}(a_\ell \smile b_k)[M] = (-1)^{|b_k||a_\ell|} \delta_{\ell k}. \end{align*} [guided] The geometric evaluation uses the fundamental property of Poincaré duality: the Poincaré dual of a submanifold $N$ is the unique cohomology class $\eta$ such that $(\omega \smile \eta)[M] = \omega[N]$ for all $\omega$ of complementary degree. Here $N = \Delta(M) \subset M \times M$, and $\eta = \delta$. Pulling back via the diagonal is the key step: any class $\alpha \otimes \beta$ on $M \times M$ pulls back to $\alpha \smile \beta$ on $M$, because $\Delta^*(\pi_1^*\alpha \smile \pi_2^*\beta) = (\pi_1 \circ \Delta)^*\alpha \smile (\pi_2 \circ \Delta)^*\beta = \alpha \smile \beta$. [/guided] [/step] [step:Solve for $C_{k\ell}$ by equating the two evaluations] Equating the results of the two evaluations: \begin{align*} C_{k\ell} \cdot (-1)^{|a_\ell||a_k| + |b_k||a_k|} = (-1)^{|b_k||a_\ell|} \delta_{\ell k}. \end{align*} For $\ell \neq k$, the right side is zero, so $C_{k\ell} = 0$. For $\ell = k$: \begin{align*} C_{kk} \cdot (-1)^{|a_k|^2 + |b_k||a_k|} = (-1)^{|b_k||a_k|}. \end{align*} Dividing both sides by $(-1)^{|b_k||a_k|}$ (which is $\pm 1$): \begin{align*} C_{kk} \cdot (-1)^{|a_k|^2} = 1. \end{align*} Since $|a_k|$ is a non-negative integer, $|a_k|^2 \equiv |a_k| \pmod{2}$ (because $n^2 - n = n(n-1)$ is always even). Therefore $(-1)^{|a_k|^2} = (-1)^{|a_k|}$, giving \begin{align*} C_{kk} = (-1)^{|a_k|}. \end{align*} Substituting $C_{ij} = (-1)^{|a_i|} \delta_{ij}$ into the expansion: \begin{align*} \delta = \sum_i (-1)^{|a_i|} a_i \otimes b_i. \end{align*} [/step]