Lemma 45.9.7. Assume given (D0), (D1), (D2), and (D3) satisfying (A), (B), and (C). Let $X$ be a nonempty smooth projective scheme over $k$ equidimensional of dimension $d$. Choose a basis $e_{i, j}, j = 1, \ldots , \beta _ i$ of $H^ i(X)$ over $F$. Using Künneth write

\[ \gamma ([\Delta ]) = \sum \nolimits _ i \sum \nolimits _ j e_{i, j} \otimes e'_{2d - i , j} \quad \text{in}\quad \bigoplus \nolimits _ i H^ i(X) \otimes _ F H^{2d - i}(X)(d) \]

with $e'_{2d - i, j} \in H^{2d - i}(X)(d)$. Then $\int _ X e_{i, j} \cup e'_{2d - i, j'} = (-1)^ i\delta _{jj'}$.

**Proof.**
Recall that $\Delta ^* : H^*(X \times X) \to H^*(X)$ is equal to the cup product map $H^*(X) \otimes _ F H^*(X) \to H^*(X)$, see Remark 45.9.3. On the other hand, recall that $\gamma ([\Delta ]) = \Delta _*1$ (Lemma 45.9.6) and hence

\[ \int _{X \times X} \gamma ([\Delta ]) \cup a \otimes b = \int _{X \times X} \Delta _*1 \cup a \otimes b = \int _ X a \cup b \]

by Lemma 45.9.1. On the other hand, we have

\[ \int _{X \times X} (\sum e_{i, j} \otimes e'_{2d -i , j}) \cup a \otimes b = \sum (\int _ X a \cup e_{i, j})(\int _ X e'_{2d - i, j} \cup b) \]

by axiom (B)(b); note that we made two switches of order so that the sign for each term is $1$. Thus if we choose $a$ such that $\int _ X a \cup e_{i, j} = 1$ and all other pairings equal to zero, then we conclude that $\int _ X e'_{2d - i, j} \cup b = \int _ X a \cup b$ for all $b$, i.e., $e'_{2d - i, j} = a$. This proves the lemma.
$\square$

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