What is a pure Hodge structure of integer weight $n$?
Wikipedia defines a pure Hodge structure of integer weight $n$ to be an abelian group $H_\mathbb{Z}$ equipped with a direct sum decomposition (as complex vector spaces) of its complexification $H_\mathbb{C} := H_\mathbb{Z} \otimes_\mathbb{Z} \mathbb{C}$ as $H_\mathbb{C} = \oplus_{p,q \in \mathbb Z,\, p+q = n}H^{p,q}$ "with the property that $H^{p,q}$ is the complex conjugate vector space of $H^{q,p}$ for all $p,q \in \mathbb Z$."
If taken literally, the part in quotes means in particular that on the diagonal where $p = q$, we have $H^{p,p} = \overline{H^{p,p}}$, so that $z\cdot h = \bar z \cdot h$ for all $z \in \mathbb C$ and $h \in H^{p,p}$. This implies that $H^{p,p} = 0$. But I know there are supposed to be important examples of Hodge structures where $H^{p,p} \neq 0$. So the part in quotes is apparently not to be taken literally.
My best guess as to how to interpret the part in quotes is that, in order to specify a Hodge structure on $H_\mathbb{Z}$, in addition to the choice of a direct sum decomposition $H_\mathbb{C} = \oplus H^{p,q}$, we are also required to choose an isomorphism of complex vector spaces $\phi : H^{p,q} \cong \overline{H^{q,p}}$. This seems like a natural interpretation of the definition, but one thing still bothers me, again concerning the diagonal. Namely, if this is the correct interpretation, then we are choosing an isomorphism $\phi: H^{p,p} \cong \overline{H^{p,p}}$ for each $p \in \mathbb Z$. But surely we don't want to allow an arbitrary choice of isomorphism in this case? For instance, I'd think we want to require that $\phi(x) = x$ for $x \in H_\mathbb Z \cap H^{p,p}$, maybe? At any rate, under this interpretation there seems to be more to say about the definition than what is said on wikipedia, so I mistrust this interpretation because surely in this case wikipedia would say these things?
Moreover, in the case where $H$ comes from the cohomology of a complex projective variety, say, I think the isomorphism $\phi : H^{p,q} \cong \overline{H^{q,p}}$ is supposed to be given locally by $f dz_1 \wedge \cdots \wedge dz_p \wedge d\bar w_1 \cdots \wedge \bar w_q \mapsto \overline{f dz_1 \wedge \cdots \wedge dz_p \wedge d\bar w_1 \cdots \wedge \bar w_q}$, but I'm not sure — for instance, it might make sense to take $(-1)^{pq}$ times this map. This could be particularly relevant when $p = q$.
Best Answer
I disagree with both of the ways you read the part in quotes, and I disagree that your first interpretation is "taking it literally."
The action of complex conjugation on $H$ is via the action on the $\mathbb{C}$ in $H_\mathbb{Z} \otimes \mathbb{C}$. It is possible that there is some $v \neq 0 \in H^{p, p}$ such that $\overline{v} \in H^{p, p}$. This should not be interpreted as saying that $zv = \overline{z}v$ for all $z \in \mathbb{C}$, which you are right is not possible for $v \neq 0$.