Example 45.11.1. Let $k = \mathbf{C}$ and $F = \mathbf{C}$ both be equal to the field of complex numbers. For $X$ smooth projective over $k$ denote $H^{p, q}(X) = H^ q(X, \Omega ^ p_{X/k})$. Let $(H')^*$ be the functor which sends $X$ to $(H')^*(X) = \bigoplus H^{p, q}(X)$ with the usual cup product. This is a classical Weil cohomology theory (insert future reference here). By Proposition 45.7.11 we obtain a $\mathbf{Q}$-linear symmetric monoidal functor $G'$ from $M_ k$ to the category of graded $F$-vector spaces. Of course, in this case for every $M$ in $M_ k$ the value $G'(M)$ is naturally bigraded, i.e., we have

$(G')(M) = \bigoplus (G')^{p, q}(M),\quad (G')^ n = \bigoplus \nolimits _{n = p + q} (G')^{p, q}(M)$

with $(G')^{p, q}$ sitting in total degree $p + q$ as indicated. Now we are going to construct a $\mathbf{Q}$-linear symmetric monoidal functor $G$ to the category of graded $F$-vector spaces by setting

$G^ n(M) = \bigoplus \nolimits _{n = 3p - q} (G')^{p, q}(M)$

We omit the verification that this defines a symmetric monoidal functor (a technical point is that because we chose odd numbers $3$ and $-1$ above the functor $G$ is compatible with the commutativity constraints). Observe that $G(\mathbf{1}(1))$ is still sitting in degree $-2$! Hence by Lemma 45.7.9 we obtain a functor $H^*$, cycle classes $\gamma$, and trace maps satisfying all classical axioms (A), (B), (C), except for possibly the classical axioms (A)(a) and (A)(d). However, if $E$ is an elliptic curve over $k$, then we find $\dim H^{-1}(E) = 1$, i.e., axiom (A)(a) is indeed violated.

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