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

Question

$\DeclareMathOperator\Hom{Hom}$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 $H$.

Answer 1

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.

Answer 2

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