Differentials II

 

Recall, given a (locally) closed immersion Z \to X, it comes from a sheaf of ideals, and we can use this sheaf of ideals to construct thickenings of Z.  Doing this in the case of \Delta(X) \subset X\times_Y X for a Y-scheme X (with map f \colon X \to Y), I get a particular sheaf of \mathcal{O}_X-modules, called the sheaf of differentials on X (I’m neglecting to put pushforwards or pullbacks into the notation):

\Omega_{X/Y}^1 := \ker(\mathcal{O}_{P^1} \to \mathcal{O}_X)

In the previous post, it was shown that the projection p_i \colon X\times_Y X \to X give rise to a derivation d\colon \mathcal{O}_X \to .  More generally, for a given sheaf of \mathcal{O}_X modules E, I can ask for derivations D\colon \mathcal{O}_X \to E (i.e. f^{-1}\mathcal{O}_Y linear and satisfies Leibniz rule):

\mbox{Der}_{X/Y}(E)

Now given an \mathcal{O}_X module homomorphism \alpha \colon \Omega_{X/Y} \to E, I get a derivation into E simply by first applying d\colon \mathcal{O}_X \to \Omega_{X/Y}, i.e. \alpha \circ d \in \mbox{Der}_{X/Y}(E).  So there is a natural map 

\hom_{\mathcal{O}_X}(\Omega_{X/Y}, E) \to \mbox{Der}_{X/Y}(E)

This map is an isomorphism in the affine case.  Its also an isomorphism when it comes to deformations…

 Deformations

This whole discussion happens in \mbox{Sch}_Y for some base scheme Y.  Suppose I have g\colon T \to X and a thickening i\colon T \to T'.  Intuitively, a deformation of g is “thickening” or “stretching” of its domain to T', i.e. I want g' \colon T' \to X such that T \xrightarrow{i} T' \xrightarrow{g'} X is a factorization of g:

 \mbox{Dfm}_g(T') = \{g' \colon T' \to X: g = g' \circ i \}

Note:

  • i\colon T \to T' is a closed immersion so defined by an ideal I, then g_* I is a sheaf of \mathcal{O}_X modules.  A
  • s T, T' are topologically the same, a deformation is determined by a map of sheaves, i.e. by a commutative diagram: 

\begin{array}{ccc} \mathcal{O}_X & \xrightarrow{\phi} & g_*\mathcal{O}_{T'} \\ \mbox{} & \searrow & \downarrow \\ \mbox{} & \mbox{} & g_*\mathcal{O}_T \end{array}

  • When T \to T' is a first order thickening there is an action \mbox{Der}_{X/Y}(g_*I) \times \mbox{Dfm}_g(T') \to \mbox{Dfm}_g(T') given by (D, \phi) \mapsto \phi + D=: \phi' where \phi is a sheaf maps that fits into a commutative diagram as above.

For the last bullet point, its clear that \phi' is f^{-1}\mathcal{O}_Y linear, and that \phi = \phi' modulo g_*I i.e. that \phi' fits into a commutative diagram as above.  So it just remains to show \phi' is a ring homomorphism.  Now

\phi'(bb') = \phi(bb') + D(bb') = \phi(b)\phi(b') +\phi(b)D(b') + \phi(b')D(b)

Recall \phi \colon \mathcal{O}_X \to g_*\mathcal{O}_T is what gives g_*\mathcal{O}_T its \mathcal{O}_X module structure.  On the other hand

\phi'(b) \phi'(b') = (\phi(b) + D(b))(\phi(b') + D(b'))

= \phi(b)\phi(b') + \phi(b)D(b') + \phi(b')D(b) + D(b)D(b')

and the last term is zero for a first order thickening.  

I think this is the first time I’ve had a need to mention torsor.  Intuitively, for a group G, a G-torsor is a set S with an action of G such that for any two elements a,b of S there is a unique element of G taking a to b.  In fact any G-torsor is isomorphic to G, but there is no preferred isomorphism.  Ogus’ definiton a G pseudo torsor is a G-set such that G \times s \to S is bijective for every s \in S.  A G torsor is a G pseudo torsor that is nonempty.  I think the point of defining G pseudo torsor is because in some case you might want to define a set of ‘would be’ torsors, but in certain cases you might get the emptyset, for example in our case it turns out the above action makes \mbox{Dfm} into a \mbox{Der} torsor, but in some cases some maps might not have any deformations.  In any case, here is a theorem:

Thm: Given g\colon T \to X and a first order thickening i \colon T \to T' then \mbox{Dfm}_g(T') is a \mbox{Der}_{X/Y}(g_*I) pseudo torsor.

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I’m not going to go through the proof, but here’s an outline.  What needs to be shown is that 

\mbox{Der} \times \mbox{Dfm} \to \mbox{Dfm} \times \mbox{Dfm}

sending (D, g') \mapsto (g' + D, g') is an bijective.  One way to define a free action is to say the above map is injective.  So check the action is free to get injectivity.  The difficult part is surjectivity.  Given g_1',g_2' the task is to find a derivation such that g_1' = g_2' + D.  The g_i' given a map T \to X \times_Y X basically you play with this to get a map of short exact sequences:

\Omega_{X/Y} \to \mathcal{O}_{P^1} \to \mathcal{O}_X

into 

g_*I \to g_*\mathcal{O}_{T'} \to g_*\mathcal{O}_T

More specifically there should be some scheme diagram that gives the latter two vertical arrows, and the universal property of kernel should then give a sheaf map \Omega \to g_*I.  Then precomposing d \colon \mathcal{O}_X \to \Omega should give the desired derivation, this uses something like g_i' \colon T' \rightrightarrows X factorizes as T' \to X\times_Y X \rightrightarrows X.  In effect this constructs an inverse that factorizes through H:= \hom(\Omega, g_*I):

\mbox{Dfm}\times \mbox{Dfm} \to H \times \mbox{Dfm} \to \mbox{Der} \times \mbox{Dfm}

As I’ve outlined \mbox{Der} \times \mbox{Dfm} \to \mbox{Dfm}\times \mbox{Dfm} is bijective, the first map above is injective, and the latter in surjective.  Also it shouldn’t be hard to see that H \to \mbox{Der} is injective, this gives

Cor. With the same assumption as the thm, the natural map 

\hom(\Omega_{X/Y}, g_*I) \to \mbox{Der}_{X/Y}(g_*I) 

h \mapsto h \circ d

is bijective. QED

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Next, maybe some examples.

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