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Given $f:X\to Y$ a morphism of schemes (or stacks if it's not harder), I am interested in a geometric reformulation of the condition that the functor $f^*:D^b(Coh(Y))\to D^b(Coh(X))$ is full. I can only find full and faithful appearing together in the literature, and I need to extricate the two conditions. Does anyone know a simple formulation, or a good reference?

Intuitively, asking for $f^*$ to be full seems alot like asking that anytime you have a sheaf $F$ on $Y$, and a section of it defined only on $X$ (i.e. a section of $f^*F$), it can be extended to a section on all of $Y$. And so that would seem to indicate that the image of $f$ should have codimension-two complement. However, that intuition only really applies to underived $f^*$, and maybe deriving $f^*$ eliminates the codimension 2 requirement? Also, this intution is assuming that f is mono, so that $f^*$ is just restriction, which I don't think is true a priori.

As a side-note: I'd be interested in the same question (geometric characterization of fullness) for $f_*$ and $f^!$.`

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If they are affine, fullness might imply that f is a closed immersion by considering $hom(\mathcal{O}_Y, \mathcal{O}_Y) \to hom(\mathcal{O}_X, \mathcal{O}_X)$ ? – name Jul 13 '12 at 18:32
The link you talk about in 1. is the case $f_*$ is full and faithful. – name Jul 13 '12 at 18:33
Thanks for the correction! I got thrown by the word "restriction" in the statement of the theorem. – David Jordan Jul 14 '12 at 1:40
If they are smooth and projective the work of Bondal and Orlov on Fourier-Mukai transformations might give something...but I'm not sure, and hesitate to add this comment in case its a red herring. – name Jul 14 '12 at 8:41
In the projective case (if the variety admits a very ample line bundle $L$) then I think you can recover the (a) homogeneous coordinate ring using global sections of the $L^n$. You might be able to deduce something from this? – name Jul 14 '12 at 9:01

1 Answer 1

up vote 7 down vote accepted

In case $X, Y$ are smooth, $ f^* $ is full if and only if it is full and faithful. This is explained in the introduction of arXiv:1101.5931 (by Canonaco-Orlov-Stellari), which also studies when this implication holds more generally. Thus the pull-back is full and faithful if and only if $Rf_* \mathcal{O}_X = \mathcal{O}_Y$; for example, when $f$ is birational.

Here is a short explanation of their argument for the case of pull-back functors: $f^*$ is full and faithful if and only if $\mathrm{Ext}^i_{Y}(f^* O_{x_1}, f^* O_{x_2})$ is

  1. $\mathbb C$ for $x_1 = x_2$, $i = 0$,
  2. 0 for $i \notin [0, \mathrm{dim} Y]$, and
  3. 0 for $x_1 \neq x_2$.

(This is due to Bondal-Orlov and Bridgeland.)

Since $f^* $ is full, the 2nd and 3rd condition are automatically satisfied. Thus it will be full and faithful if and only if $f^* O_x \neq 0$ for all $x \in X$.

Pick $x \in X$ such that $f^* O_x$ is non-zero. Note that $\mathrm{Hom}^{\bullet}_Y(f^* O_x, f^* O_x) = \mathrm{Hom}^{\bullet}_X(O_x, f_* f^* O_x)$ is a quotient of $\mathrm{Hom}^{\bullet}_X(O_x, O_x)$, and $f_* f^* O_x$ is supported at $x$. From this one can show that $O_x$ is a sheaf, and in fact isomorphic to $O_x$; all we need is that it's Chern character (in cohomology) is non-zero. But the Chern character of $f_* f^* O_x$ is independent of $x$, so $f^* O_x \neq 0$ for all $x \in X$.

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This is incredibly helpful, thanks! – David Jordan Jul 19 '12 at 21:09
Is it true that if $f^*$ is fully faithfull, then the subcategory $f^*(D^b(Coh(Y)))$ is admissible in $D^b(Coh(X))$, i.e. $f^*$ has a left adjoint? – Nullstellensatz Aug 31 at 16:22

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