# Does there exist a faithful exact embedding of $D^b(\dim(N)) \to D^b(\dim(N-1))$

$$\DeclareMathOperator\Hom{Hom}$$I am trying to show that if $$X,Y$$ are nice schemes with $$\dim(X) > \dim(Y)$$ there is no faithful FM transform $$\Phi_{K}: D^b(X) \to D^b(Y)$$. Does someone have a proof of this result?

Below are some attempts and comments I've made trying to prove this.

Morally correct proof idea 0: Define dimension of a derived category and show faithful functors do not decrease it. Idk anything about those.

Morally correct proof idea 1: we expect the induced map $$\Hom^{dim(X)}(p,p) \to \Hom_Y^{\dim(X)}(\Phi(p),\Phi(p))$$ to be noninjective for a generic point $$p$$ (i.e I mean skyscrapers above).

Warnings:

• For idea 1, we need to use the indice $$dim(X)$$ and not less; for example for a surface to a curve we could sum three random maps so that the derivative doesn't vanish in all at the same time.
• If $$X$$ is a curve and $$Y$$ a point, $$K$$ is $$\mathcal{O}_{p \times Y}$$, then the map will be an isomorphism on $$\Hom^{1}(p,p)$$ (but not on $$\Hom^{1}(p',p')$$)
• As an amusing fact; maps coming from geometric closed embeddings don't always give faithful functors, though if the domain is a curve it is true.

For the case of curve, we can indeed easily prove this strategy works via writing $$K$$ as a sum of bundles and torsions, and choosing $$p'$$ not in the torsions.

Already for the case of $$X$$ a surface and $$Y$$ a curve I'm having trouble. Before I write some junk, the real problem is higher ext's have more and more annoying descriptions.

Denote by $$\pi_l,\pi_r$$ the projections $$X \times Y \to X,Y$$, thus we factor our map as $$\Hom_X^{\dim(X)}(p,p) \to^{\alpha^*} \Hom^{\dim(X)}_{X \times Y}(K \otimes \pi_l^*(p), K \otimes \pi_l^*(p)) \to^{\alpha_*} \Hom^{\dim(X)}_{Y}(\pi_* K \otimes \pi_l^*(p), \pi_* K \otimes \pi_l^*(p))$$

• The genericity of $$p$$ can appear in the objects above being the same, but the morphism $${\alpha^*}$$ being different. This is what happens on a curve to a point.

• An immoral proof that only works for projective (while the claim should be true affinely as well) would be to somehow count the "size" of Homs which should be larger on the surface side. But pushforward can blowup the size; just the pushforward to a point records the cohomologies which grow as a polynomial degree in the dimension.

Now for my junk attempt, we write the kernel $$K$$ wlog as a complex of line bundles so that each map between $$L_i,L_j$$ vanishes along a hypersurface $$H \subset X \times Y$$, then we probably want $$p$$ to be s.t $$p \times Y$$ is generic; it doesn't intersect any $$H \cap H'$$ nor tangent to any $$H$$.

• What is your question? Nov 16, 2023 at 19:51
• @Sasha Sorry clarified, I am asking for a proof that there is no such faithful exact functor Nov 16, 2023 at 20:24
• What does FM mean? Nov 16, 2023 at 20:55
• "Fourier-Mukai", maybe? But even that doesn't make much grammatical sense without the word "transform". Nov 16, 2023 at 21:02
• MathJax note: since MathJax has no notion of a preamble, it cannot swallow blank space in what we think of as the preamble, as LaTeX would. Therefore, the \$ ending the math block in which @YCor \DeclareMathOperator{\Hom}{Hom}'d (edit) must immediately abut the following text; your edit putting a following blank space in the source creates blank space in the rendered post. I edited again to put it back as @‍YCor had it. Nov 17, 2023 at 0:59

If you're not assuming full you can use Prop. 2.3 in

Canonaco, Alberto; Stellari, Paolo, Non-uniqueness of Fourier-Mukai kernels, Math. Z. 272, No. 1-2, 577-588 (2012). ZBL1282.14033.

to prove the case $$dim(Y)=1$$ in the case of smooth projective varieties already.

• Thanks AT0, sorry for not responding earlier. I'm still unsatisfied since I think the proof idea should go through in all dimensions Nov 20, 2023 at 10:32

This is answered by Noah Olander in Fully Faithful Functors and Dimension.

• I'm not assuming full, only faithful! (and I don't understand why people in this derived sheaves business always assume both full and faithful, do you know why? ) Nov 16, 2023 at 20:30
• @user135743 Full implies faithful for smooth projective varieties. A fully faithful triangulated functor can be shown to be a FM transform with a unique kernel (for projective schemes and derived cats of perfect complexes)
– AT0
Nov 16, 2023 at 22:58