On pages 17 and 18 of the following document: https://www.math.tifr.res.in/~sujatha/ihes.pdf, we find the following paragraph:
Let $ \mathbb{Q} \mathrm{HS}$ be the category of pure Hodge structures over $\mathbb{Q}$. There is a functor: $$ \mathcal{R}_{ \mathrm{Hodge} } \colon \mathop{\mathrm{Mot}}^{ \bullet }_{ \mathrm{num} } ( k, \mathbb{Q} ) \to \mathbb{Q} \mathrm{HS} $$ and the Hodge conjecture is equivalent to the assertion that $\mathcal{R}_{ \mathrm{Hodge} }$ is fully faithful.
My questions are:
How is the Hodge realization functor $ \mathcal{R}_{ \mathrm{Hodge} }$ explicitly defined? And how to prove explicitly that the Hodge conjecture is equivalent to the assertion that $\mathcal{R}_{\mathrm{Hodge}}$ is fully faithful?
Thanks in advance for your help.
So here is a recap of how the category of motives is constructed :
The last two steps are universal constructions. This means that to define a monoidal functor $Mot_\sim(k)\rightarrow \mathcal{A}$, we just need a monoidal functor $R$ from the category of correspondance to $\mathcal{A}$ as long as $\mathcal{A}$ has kernels and images of projectors and such that $R(\mathbb{L})$ is invertible in $\mathcal{A}$.
And this is the case for the Hodge realization :
It is important to know the following fact about the category of motives : $$\operatorname{Hom}((h(X),p,m),(h(Y),q,n))=pCorr^{n-m}_\sim(X,Y)q$$ where $Corr^k_\sim(X,Y)=Z^{\operatorname{dim}X+k}(X\times Y)/\sim$.
In particular, $\operatorname{Hom}(\mathbb{Q},(h(X),\operatorname{id},*))=Z^*(X)/\sim$.
On the other hand, in the category of Hodge structure $\operatorname{Hom}(\mathbb{Q},H^*(X))$ are the Hodge classes.
If the Hodge realization functor is full, we can deduce in particular that every Hodge classes comes from an algebraic cycle which is the statement of the Hodge conjecture.
The converse is not more difficult, if the Hodge conjecture hold, every Hodge classes in $H^*(X\times Y)$ is algebraic, but Hodge classes are in $H^*(X\times Y)$ are in bijection with morphisms of Hodge structures $H^*(X)\rightarrow H^*(Y)$. Thus the realization functor is full.
(The faithfulness just comes from the choice of the equivalence $\sim$ and has nothing to do with the Hodge conjecture)