Let $X$ be a smooth projective irreducible variety of pure dimension $d$ over an algebraically closed field of positive characteristic $p$. Let $\ell \not = p$ be a prime number. There is an isomorphism $\mathrm{Tr}: H^{2d}(X,\mathbb Q_{\ell}) \xrightarrow{\sim} \mathbb Q_{\ell}(-d)$, and cup-product define a pairing $$\mathrm{Tr}(x\cup y): H^i(X,\mathbb Q_{\ell}) \times H^{2d-i}(X,\mathbb Q_{\ell}) \to \mathbb Q_{\ell}(-d).$$ Poincaré duality states that this pairing is perfect.
In particular, for $i = d$ we obtain a bilinear form on $H^d(X,\mathbb Q_{\ell})$. In La conjecture de Weil: I (1974), Deligne writes that if $d$ is odd, this pairing is alternating (cf. paragraph 2.6).
I don't understand where this property comes from. I thought that cup product on $H^d(X,\mathbb Q_{\ell})$ was commutative, ie. $x \cup y = y \cup x$. If the form were alternating, it would be in particular skew-symmetric, ie. $\mathrm{Tr}(x \cup y) = - \mathrm{Tr}(y \cup x)$. This is impossible if cup-product really is commutative. What would be the explanation?
As with singular cohomology, the cup product is graded commutative, not commutative. This means that for $x\in H^p(X,\mathbb Q_\ell)$ and $y\in H^q(X,\mathbb Q_\ell)$, $$x\cup y=(-1)^{pq}y\cup x\in H^{p+q}(X,\mathbb Q_\ell).$$