# 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.