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Suppose $f:X\to Y$ is a map of smooth complex algebraic varieties. There is a pushforward functor

$f_\ast : D(X) \to D(Y)$

on the derived category of $D$-modules. This certainly does not preserve coherence of $D$-modules in general. However, if $f$ is proper then the pushforward preserves coherence, $t$-structure, and purity. There is also natural bound on the singular support of the pushforward.

The situation for the pullback functor seems to be entirely analogous, but where the word "proper" is replaced by "smooth". However the pullback, the results for smooth pullbacks can be generalized to the case of $f^{+} M$, when $M$ is non-characteristic for $f$. This is a certain condition on the singular support of $M$ (see for the definition).

Question: Is there a notion for pushforwards analogous to non-characteristic for pullbacks?

It looks like the words "elliptic pair" might be relevant here, but I don't see exactly how this fits the analogy (this may be because I don't fully understand the definitions...).

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  • $\begingroup$ Slightly confused about a certain point, why is pushforward along a proper map preserving t-structure? Regarding the question, there is a good estimate on singular support of $f_*M$ if $f_{\pi}\colon f_d^{-1} Ch(M)\to T^*Y$ is finite (I hope the notation is self-explanatory. See Kashiwara's D-Modules and microlocal calculus section 4.7. This condition is analogous to the non-characteristic condition. $\endgroup$
    – Reladenine Vakalwe
    Commented Dec 20, 2012 at 0:52
  • $\begingroup$ Thanks for pointing that out - I have edited the question. I'll think about your answer. $\endgroup$
    – Sam Gunningham
    Commented Dec 20, 2012 at 1:38
  • $\begingroup$ @SamGunningham Maybe this is naive but isn't being "proper on the support of $M$" the correct analogy? $\endgroup$
    – Saal Hardali
    Commented Jul 29, 2017 at 22:42
  • $\begingroup$ @SaalHardali the pushforward does not preserve the t-structure under those conditions, and also the estimate on singular support is only a containment rather than an equality. I believe that the correct condition is the one mentioned in Reladenine Vakalwe's comment above. This includes the case when $f$ is a closed embedding for example (which is most directly analogous to being smooth). $\endgroup$
    – Sam Gunningham
    Commented Jul 30, 2017 at 2:48

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