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Let $D^b_m(X)$ (resp $D^b(X)$) denote the derived category of mixed Hodge modules (resp. constructible sheaves) on a complex variety $X$. Let

$rat\colon D^b_m(X)\to D^b(X)$

be the `forgetful' functor. This is t-exact for the perverse t-structure on the right. Write $MHM(X)$ for the abelian category of mixed Hodge modules on $X$. Then $MHM(pt)$ is the category of graded polarizable mixed Hodge structures, and $rat\colon MHM(pt) \to VectorSpaces$ is the evident forgetful functor.

Now let $M,N\in D^b_m(X)$. Set

$\mathcal{H}om(M,N) = \Delta^!(\mathbb{D}M \boxtimes N)$,

where $\Delta\colon X\to X\times X$ is the diagonal map, and $\mathbb{D}$ is Verdier duality.

Let $a\colon X \to pt$ be the evident map. Then

$rat ( H^0(a_*\mathcal{H}om(M,N))) = H^0(rat(a_*\mathcal{H}om(M,N))) = Hom(rat(M), rat(N))$

and in this way we get a Hodge structure on $Hom(rat(M),rat(N))$. All functors are derived.

My question: If $M,N$ are pure pointwise pure (see Geordie Williamson's comment below), then is the induced structure on $Hom(rat(M), rat(N))$ pure?

My gut answer is no (even if $X$ is complete, the $\Delta^!$ should be messing weights up), but it would make me happier if the answer is yes!

If the answer is no, under what additional conditions (other than requiring $X$ to be smooth and complete plus $M,N$ being the `constant' sheaf) can the answer be converted to yes?

I guess one could also ask the same sort of question for mixed $\ell$-adic sheaves. But I am even less familiar with that setting.

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    $\begingroup$ It's not clear to me either right now. As for a sufficient condition, if $X$ is smooth and $M,N$ are locally constant, then it should be fine, because the internal Hom can be constructed more naively for variations of pure Hodge structures. $\endgroup$
    – Donu Arapura
    Commented Oct 6, 2012 at 12:44
  • $\begingroup$ @Donu Arapura: Do you know whether the corresponding statement for $\ell$-adic sheaves is true? $\endgroup$
    – Reladenine Vakalwe
    Commented Oct 6, 2012 at 17:14
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    $\begingroup$ In the cases I know best (IC's on flag varieties) the statement is true by pointwise purity. (The local global spectral sequence degenerates for weight reasons, eg. in the BGG argument mentioned in your other question "About an argument in Koszul duality..."). Hence to have a counterexample one needs to consider morphisms between non pointwise pure sheaves. However this is easy: take a non pointwise pure IC and consider hom to or from a skyscraper sheaf. $\endgroup$
    – Geordie Williamson
    Commented Oct 8, 2012 at 12:30
  • $\begingroup$ @Geordie Williamson: I am slightly worried that the local to global degeneration for weight reasons that you mention uses that Hom between the ICs is pure. No? Let $X=X_0 \supset ⋯\supset X_1$ be the filtration by closed subspaces corresponding to the stratification. Let $v_k:X_k \to X$ be inclusion. The degeneration is obtained by looking $Hom(v^∗_k M,−)$ applied to $i_∗i^!v^!_k N \to v^!_k N \to j_∗j^!v^!_k N$ (I hope what my $i$ and $j$ are is clear). Now for degeneration we want the connecting map in the long exact to be zero. Without weights (or parity vanishing) I dont see how to get it $\endgroup$
    – Reladenine Vakalwe
    Commented Oct 8, 2012 at 15:39
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    $\begingroup$ Search for "What's an example of whose stalks are pure but not pointwise pure?" for an example of non pointwise purity. I learnt a nice example from Luca Migliorini: take a family of elliptic curves with smooth total space and some singular fibres. Then the decomposition theorem says that the direct image of the constant sheaf on the total space breaks into its cohomology sheaves (on a curve IC's are shifts of sheaves). Now it is not difficult to see that the "middle" summand (coming from the $H^1$ of the elliptic curves in the family) cannot be pointwise pure. $\endgroup$
    – Geordie Williamson
    Commented Oct 8, 2012 at 16:49

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