# Differentials III

This is just a few examples of the stuff in the previous two posts.

### Affine Stuff

Set up:

1. ring $A$ and $B = A[t_1, \dotsc, t_n]$
2. $Y = \mbox{Spec} A$, $X = \mbox{Spec} B$ with affine morphism $X \to Y$

Claim: $\Omega_{X/Y} = \widetilde{\Omega_{B/A}}$.  Recall, for $I = \ker (B\otimes_A B \to B)$,

$\Omega_{B/A} = I/I^2$

Its not hard to see that $I$ is generated  by $1\otimes t_i - t_i\otimes 1 = \delta t_i$ so $\Omega_{B/A}$ is generated by $dt_i$ in particular

$\Omega_{B/A} = \oplus_{i=1}^nBdt_i \cong B^n$

Hence for $\widetilde{M}$ an $\mathcal{O}_X$ module,

$\mbox{Der}_{X/Y}(\widetilde{M}) = \mbox{Der}_{B/A}(M) = \hom(\Omega_{B/A}, M)$

### Deformation Stuff

Keeping the same notation recall, from the first post on differentials, that $D(M) = \mbox{Spec}_X(\widetilde{B\oplus M})$ is a first order thickening of X.  In particular, there is a diagram

$\begin{array}{ccc} X & \xleftarrow{a} & D(M) \\ \mbox{} & \nwarrow & \uparrow \\ \mbox{} & \mbox{} & X \end{array}$

So the map $a$ is a deformation of the identity.  The above diagram has a corresponding sheaf diagram:

$\begin{array}{ccc} \mathcal{O}_X & \xrightarrow{a^\#} & \mathcal{O}_{D(M)} \\ \mbox{} & \searrow & \downarrow \\ \mbox{} & \mbox{} & \mathcal{O}_X \end{array}$

By the previous discussion about deformations, if $D \colon \mathcal{O}_X \to M$ is any derivation, then there is another commutative sheaf diagram

$\begin{array}{ccc} \mathcal{O}_X & \xrightarrow{a^\# + D} & \mathcal{O}_{D(M)} \\ \mbox{} & \searrow & \downarrow \\ \mbox{} & \mbox{} & \mathcal{O}_X \end{array}$

So this gives another deformation of the identity.  As $\mbox{Dfm}_{id}(D(M))$ is nonempty (because it contains $a$) its actually a $\mbox{Der}_{X/Y}(M)$ torsor so its isomorphic to $\mbox{Der}_{X/Y}(M)$:

$\mbox{Dfm}_{id}(D(M)) \cong \mbox{Der}_{X/Y}(M)$ (1)

In the case $M = \widetilde{M}$ for some B-module $M$, then the previous affine calculation shows

$\mbox{Dfm}_{id}(D(M)) \cong M^n$

### Example with VE

Let $Y$ a scheme and $E$ qcoh on $Y$, set $X = \mathbb{V}E = \mbox{Spec}_Y S^\bullet E$.  Let $\pi \colon X\to Y$ be the map to the base.  Recall there is a universal arrow $E \to S^\bullet E$ which is the same as $E \to \pi_*\mathcal{O}_{\mathbb{V}E}$ which is the same as

$\pi^*E \to \mathcal{O}_{\mathbb{V}E}$ (2)

The point here is to show

$\pi^*E \cong \Omega_{\mathbb{V}E/Y}$

Notice by (2) and the universal map $d \colon \mathcal{O}_{\mathbb{V}E} \to \Omega_{\mathbb{V}E/Y}$ there is natural map $\pi^*E \to \Omega_{\mathbb{V}E/Y}$, to show its an isomorphism it suffices (by Yoneda) to show

$\hom_{\mathcal{O}_{\mathbb{V}E}}(\Omega_{\mathbb{V}E/Y}, M) \cong \hom_{\mathcal{O}_{\mathbb{V}E}}(\pi^*E, M)$

for every $\mathcal{O}_{\mathbb{V}E}$ module M.  So, I compute

$\begin{array}{rl} \hom_{\mathcal{O}_X}(\Omega_{X/Y},M) & \cong \mbox{Der}_{X/Y}(M) \\ \mbox{} & \cong \mbox{Dfm}_{id}(D(M)) \\ \mbox{} & \cong \hom_{\mathcal{O}_{\mathbb{V}E}}(\mathcal{O}_{\mathbb{V}E}, M) \\ \mbox{} & \cong \hom_{\mathcal{O}_X}(\pi^*E,M) \end{array}$

The second isomorphism uses (1), the next isomorphism (I’m speculating) is justified as follows: a deformation of the identity $\mathbb{V}E \to mathbb{V}E$ corresponds to a sheaf map $a^\#$ and a commutative sheaf diagram

$\begin{array}{ccc} \mathcal{O}_X & \xrightarrow{a^\#} & \mathcal{O}_X \oplus M \\ \mbox{} & \searrow & \downarrow \\ \mbox{} & \mbox{} & \mathcal{O}_X \end{array}$

Commutativity of the diagram means that $a^\# = id + b$ where $id \colon \mathcal{O}_X \to \mathcal{O}_X$, and $b \colon \mathcal{O}_X \to M$, the map $a^\# \mapsto b$ is the correspondence in the second to last isomorphism.  The last isomorphism comes from the universality of (2).

There is also a statement about $\mathbb{P}E$, namely

Thm: Let $E$ qcoh on $Y$, $X = \mathbb{P}E$, $I$ the ideal of the diagonal $\Delta \colon X \to X\times_Y X$, then there is a natural isomorphism

$\Omega_{\mathbb{P}E/Y} = I/I^2 \cong H(-1) = \hom(\mathcal{O}_{\mathbb{P}E}(1), H)$

————————————–

H is some hyperplane that makes the thm make sense.  But I didn’t take good notes, so no proof.