6
$\begingroup$

As the question title asks for, how do others visualize local complete intersection morphisms? My experiment in asking people in real life didn't pan out, so I'm consulting the MO algebraic geometry community. Bonus points for pictures!

$\endgroup$
  • 1
    $\begingroup$ Hmm, of all the classes of morphisms out there, I find this one of the most intuitive... The definition is entirely geometric! $\endgroup$ – R. van Dobben de Bruyn Oct 10 '18 at 17:49
  • 1
    $\begingroup$ A morphism is lci iff it factors as a regular embedding and a smooth morphism. You can visualize a regular embedding as a subscheme which admits a tubular neighborhood, and a smooth morphism as one which is locally a projection with smooth fibers. Then you can think of lci as precisely the condition that the morphism has a well behaved normal bundle. $\endgroup$ – Dan Petersen Oct 10 '18 at 19:40
3
$\begingroup$

As Dan Petersen said the comments, an lci morphism $f:X\to Y$ is precisely one that factors Zariski-locally as $X\to Y\times \mathbf{A}^n \to Y$, where $X\to Y\times \mathbf{A}^n$ is a regular embedding (on affines, this means that you are looking at a closed subscheme defining by the vanishing of an ideal generated by a regular sequence) and $Y\times \mathbf{A}^n \to Y$ is the projection map. A regular embedding should be thought of as a closed subscheme which has a tubular neighborhood (every closed immersion of smooth schemes is a regular embedding), so an lci morphism is precisely a morphism for which it is reasonable to talk about the normal bundle.

This can be made more precise. Namely, an lci morphism $f:X\to Y$ can be characterized as a morphism such that the relative cotangent complex $L_{Y/X}$ is perfect, and has Tor-amplitude in $[-1,0]$. The latter statement means that for every quasicoherent sheaf $\mathscr{F}$ on $Y$, the derived tensor product $L_{Y/X}\otimes^\mathbf{L} \mathscr{F}$ has cohomology concentrated in degrees $-1$ and $0$. (If your definition of an lci morphism is as a morphism which is locally given by quotienting by a regular sequence, then one direction of the above statement is proved as Tag 08SL.) In any case, the conormal bundle $N_{Y/X}$ can now be defined as $L_{Y/X}[-1]$, so that the normal bundle of $f$ is $N_{Y/X}^\vee = L_{Y/X}^\vee [-1]$. Since $f$ is an lci morphism, this is actually a vector bundle over $Y$, so it makes sense to call this the normal bundle. This should make sense: if $f:Y = \mathrm{Spec}(B) \to X = \mathrm{Spec}(A)$ is a surjective morphism and $I = \ker(A\to B)$, then $H^1(L_{Y/X}) = I/I^2$.

$\endgroup$
  • $\begingroup$ An lci morphism is precisely a morphism for which it is reasonable to talk about the virtual tangent bundle. It will have a normal bundle if it is an lci closed immersion (AKA regular closed immersion). If it is not a closed immersion it may still have smoething in degree 0 and will just be some 2 term complex. $\endgroup$ – user123627 Oct 12 '18 at 20:11
  • $\begingroup$ Also Im not sure about your Zariski local factorization, youll also have some etale map in there in general (regular embedding followed by etale followed by affine space projection) $\endgroup$ – user123627 Oct 12 '18 at 20:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.