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.


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