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Let $X$ be a smooth, compact, oriented manifold. There is the diagonal embedding$$X \overset{\Delta}\hookrightarrow X \times X.$$Since the normal bundle to $X$ in $X \times X$ is oriented, $\Delta$ determines a cohomology class $\text{cl}(\Delta) \in H^n(X \times X, \mathbb{Q})$. Let us choose biorthogonal bases $\{e_a^i\}_a$ of $H^i(X, \mathbb{Q})$ and $\{g_{n - 1}^a\}_a$ of $H^{n - i}(X, \mathbb{Q})$ (biorthogonal with respect to the cup product pairing).

My question is, how do I see that$$\text{cl}(\Delta) = \sum_{i, a} (-1)^i\left(e_a^i \otimes g_{n - i}^a\right)?$$

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  • $\begingroup$ You probably want $\cup$ instead of $\otimes$. It might help to use Poincaré duality and think of intersecting the Poincaré dual cycles. Remember that $(V\times W)\cap\Delta \cong V\cap W$. $\endgroup$ Oct 1, 2016 at 2:18

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Element $\text{cl}(\Delta)$ has some decomposition in the tensor product $$ \text{cl}(\Delta) = \sum_{i, a} c_{i, a} \left(e_a^i \otimes g_{n - i}^a\right), $$ now let's compute product $(g_{n-j}^b \otimes e_b^j) \text{cl}(\Delta)$. From one hand, using $(V\otimes W)\cap\text{cl}(\Delta) \cong V\cap W$, this is $(g_{n-j}^b \cap e_b^j) = 1$, from the other hand, using our decomposition, only terms with $i=j$ and $a=b$ contribute, so it is $c_{i, a} (-1)^{i^2}$. Thus, $c_{i, a} = (-1)^i$.

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