I don't think this proof guarantees that the transition functions are holomorphic, no?

@Meow I wanted to take the pushout along the maps $\eta\rightarrow X,X_{mdR}$. Sorry, I got distracted; I'll try to make some time to think about this tomorrow.

Ah OK, I misunderstood. I assumed by meromorphic connection you simply meant a $\mathcal{D}$-module which is intermediate extended, but you actually want that plus the data of a lattice. Let me think and see if I can say anything about that case. I am tempted to just take the coproduct of $X_{mdR}$ and $X$ over $\eta$ - let me think if that works...

That's fair; the point I was trying to make was that every field has a list of "standard adjunctions", and that in practice these sorts of questions are often answered by comparing to those "standard adjunctions." But I see that my comment could easily be interpreted as "this way (via pushforwards & pullbacks) is the best way to think about it."

@DavidRoberts I'm afraid I don't see how this is related to my comments. I agree there is a fairly concrete description - it is provided by my first comment, no? I wrote it down in the form of a sequence of exercises because the question asked for insight into the thought processes involved, and I thought that doing those exercises was the easiest way to gain that insight.

It may be helpful to explain what your background is. For me, coming from an algebraic geometry/sheaf-theory background, I look at this and I see that you're asking for the right adjoint of a pullback, so the answer should look like a pushforward. But different fields talk about similar notions in different language, and so the most helpful answer may be from someone who speaks your language. (BTW, most but not all category-theorists do have in mind explicit models of their constructions.)

A sequence of exercises that may give insight into the process of "reification": 1. For an object $X$ in $\operatorname{Set}/A$ and an element $a\in A$, show that the preimage of $a$ may be identified with $\operatorname{Hom}_{\operatorname{Set}/A}{a,X}.$ 2. Use this description and the definition of right adjoint to compute the fibers of the dependent product.

I am tempted to argue as follows: $X$ has a toric resolution $Y\rightarrow X.$ By the decomposition theorem, the intersection cohomology of $X$ is a direct summand of the cohomology of $Y$, and as you note, the cohomology of $Y$ is Hodge-Tate.

Indeed, I misread the question and thought you were lifting the category of all bundles. But now I think there is a different issue. You write "the same is true for the morphisms..." but that amounts to choosing a section of $\Gamma(\mathcal{O}(m-n))\rightarrow\Gamma_C(\mathcal{O}(m-n)).$ This map is not surjective in general so I am skeptical of the claim that there is always a lifting of the morphisms. Am I missing something?

Literally an answer, but IMO less interesting: This is not possible because the classes in K-theory of such bundles (once you fix a slope) do not generate the K-group.

This is not literally an answer, but I think you should look into the notion of a Bridgeland stability condition. (Essentially, I am saying that the right derived-categorical notion to capture this situation is that of a stabillity condition, not of a t-structure.)

Unfortunately this is not possible in general: The only bundles you can get from restrictions from the ambient projective space have K-theory classes a linear combination of the classes of $\mathcal{O}$ and $L$.

Ah, I see. In that case, I think the answer is yes, they do satisfy the Hodge-Riemann bilinear relations. I learned about this from de Cataldo-Migliorini's "The decomposition theorem, perverse sheaves and the topology of algebraic maps", but they cite some papers of Saito as the original source.

My understanding (but it's been a long time since I've thought about any of this): It does admit a pure Hodge structure, but there is no good cup product on intersection cohomology, so I'm not sure how to make sense of "satisfies the Hodge-Riemann bilinear relations".

Here is a counterexample to uniqueness: Take $G=\mathbb{Z}/2\mathbb{Z}$, with simple irreps denoted by $1$ and $-1$. Note that $(1\oplus -1)\otimes(1\oplus -1)\cong (1\oplus -1)\otimes(1\oplus 1).$

I'm a little worried about the last sentence: Isn't it possible that $\tilde{\Delta}$ intersects $Z$ non-transversely and therefore may have zero intersection number?

I think just thinking about rationality of the cubic surface (or in @JoeSilverman's formulation the curve) isn't particularly useful: they are rational, for instance in your case via the parametrization $n=\frac{p^2}{q^2}$. The more relevant fact is that there is no nontrivial section of the map to $\mathbb{P}^1$, given by forgetting $p$ and $q$.

Do you know of such a polynomial for say, $n=6$?