# Tuesday April 7th > (Missing first few minutes.) Take $I_{q+1}$ to be the minimal $I$ such that $\mfm_q I_q \subset I \subset I_1$ and $\xi_q$ lifts to $S/I$. :::{.claim} Such a minimal $I$ exists, i.e. if $I, I'$ satisfy the two conditions then $I \intersect I'$ does as well. So $I, I'$ are determined by their images $v, v'$ in the vector space $I_q \tensor k$. ::: So enlarge either $v$ or $v'$ such that $v + v' = I_q \tensor k$ but $v \intersect v'$ is the same. We can thus assume that $I + I' = I_q$, and so \[ S / I \intersect I' = S/I \cross_{S/I_q} S/I' \] which by H1 yields a map \[ F(S/I\intersect I') &\to F(S/I) \cross_{F(S/I_q)} F(S/I') \] So $I\intersect I'$ satisfies both conditions and thus a minimal $I_{q+1}$ exists. Let $\xi_{q+1}$ be a lift of $\xi_q$ over $S/I_{q+1}$ (noting that there may be many lifts). ## Showing Miniversality :::{.claim} Define $R = \directlim R_q$ and $\xi = \directlim \xi_q$, the claim is that $(R, \xi)$ is miniversal. ::: We already have $h_R \mapsvia{\xi} F$ and thus $t_R \mapsvia{\cong}t_F$ is fulfilled. We need to show formal smoothness, i.e. for $A' \to A$ a small thickening, suppose we have a lift \begin{tikzcd} & & h_R \ar[d, "\xi"] \\ h_a \ar[rru, "n"] \ar[rr, bend right] \ar[r] & h_{A'} \ar[r] & F \end{tikzcd} If we have a $u'$ such that commutativity in square 1 holds (?) then we can form a lift $u'$ satisfying commutativity in both squares 1 and 2. We can restrict sections to get a map $F(A') \to F(A)$ and using representability obtain $h_R(A') \to h_R(A)$. Combining H1 and H2, we know $t_F$ acts transitively on fibers, yielding \begin{tikzcd} t_R \selfmap \ar[d, "\cong"] & u'\in h_R(A') \ar[r] \ar[d] \ar[r] & u\in h_R(A) \ar[d]\\ t_F \selfmap & \eta' \in F(A') \ar[r] \ar[r] & \eta \in F(A) \\ \end{tikzcd} Then $u' \mapsto u$ is equivalent to (1), and $u' \mapsto \eta'$ is equivalent to (2). Let $\eta_0$ be the image of $u'$ and define $\eta' = \eta_0 + \theta, \theta \in t_F$ then $u' = u' + \theta, \theta \in t_R$. So we can modify the lift to make these agree. Thus it suffices to show \begin{tikzcd} A' \ar[r] & A & R_q \ar[l] \\ S \ar[u, "v"] \ar[r] & \ar[lu, dotted, "\exists_? u'"] \ar[u, "u"] \ar[ur] & \end{tikzcd} We get a diagram of the form \begin{tikzcd} S \ar[d] \ar[r, "w"] & A' \cross_A R_1 \ar[r] \ar[d, "{ \pi_2, \text{small} }"] & A' \ar[d, "{ \text{small} }"] \\ R \ar[r] & R_q \ar[r] & A \end{tikzcd} :::{.observation} \envlist - $S \to R_q$ is surjective. - $\im(w) \subset A' \cross_A R_1$ is a subring, so either - $\im(w) \mapsvia{\cong} R_q$ if it doesn't meet the kernel, or - $\im(w) = A' \cross_A R_q$ In case (a), this yields a section of the middle map and we'd get a map $R_q \to A'$ and thus the original map we were after $R \to A$. ::: So assume $w$ is surjective and consider \begin{tikzcd} 0 \ar[r] & I \ar[r] & S \ar[r] & A' \cross_A R_q \ar[r]\ar[d, "\text{small}"] & 0 \\ & & & R_q & \end{tikzcd} and we have $\mfm_S I_1 \subset I \subset I_q$ where the second containment is because $I$ a quotient of $R_q$ factors through $S/I$ and the first is because $S/I$ is a small thickening of $R_q$. But $\xi_q$ lifts of $S/I$, and we have \[ \xi \in F(S/I) \surjects \xi = \xi' \cross \xi_q ? .\] Therefore $I_{q+1} \subset I$ and we have a factorization \begin{tikzcd} S \ar[rr] \ar[dr, dotted] & & S/I \\ & R_{q+1}\ar[ur, dotted] & \end{tikzcd} Recall that we had ![](figures/image_2020-04-07-13-17-11.png) \todo[inline]{Image to diagram} where the diagonal map $u'$ gives us the desired lift, and thus \begin{tikzcd} R \ar[r] \ar[rr, bend left] & R_{q+1} \ar[r] & A' \end{tikzcd} exists. This concludes showing miniversality. ## Part of Proof To finish, we want to show that H4 implies that the map on sections $h_R \mapsvia{\xi} F$ is bijective. \begin{tikzcd} & & & h_R \ar[d, "\xi"] \\ & h_A \ar[rr, bend right, "\eta"] \ar[rru, bend left, "u"] & h_{A'} \ar[r, "\eta'"] \ar[ru, "\exists ! u'"] & F \end{tikzcd} where the map $\xi$ is "formal etale", which will necessarily imply that it's a bijection over all artinian rings. So we just need to show formal étaleness. We have a diagram \begin{tikzcd} t_R \selfmap u'\in h_R(A') \ar[r] \ar[d] & u\in h_R(A) \ar[d] \\ t_F \selfmap \eta' \in h_R(A') \ar[r] & \eta \in h_R(A) \end{tikzcd} where $u'$ exists by smoothness. Assume that are two $u', u''$, then $u' = u'' + \theta$ and $\im(u') = \im(u'') + \theta \implies \theta = 0$ and thus $u' = u''$. ## Revisiting Goals We originally had two goals: 1. Given a representable moduli functor (such as the Hilbert functor), we wanted to understand the local structure by analyzing the deformation functor at a given point. 2. We want to use representability of the deformation functors to get global representability of the original functor. :::{.question} What can we now deduce about the local structure of functors using their deformation theory? ::: :::{.fact title="1"} Any two hulls $h_R \to F$ are isomorphic but not canonically. We can lift maps at every finite level and induct up, which is an isomorphism on tangent spaces and thus an isomorphism. The sketch: use smoothness to get the map, and the tangent space condition will imply the full isomorphism. ::: :::{.fact title="3"} Suppose that $F$ has an obstruction theory (not necessarily strong). This implies there exists a hull $h_R \mapsvia \xi F$. The obstruction theory of $F$ *gives* an obstruction theory of $h_R$: given $A' \to A$ a small thickening, we need a functorial assignment \[ t_R = \mathrm{def} \selfmap h_R(A') \to h_R(A) \mapsvia{\mathrm{obs}} \mathrm{obs} \\ \mathrm{def} \selfmap F(A') \to F(A) \mapsvia{\mathrm{obs}} \mathrm{obs} \] where there are vertical maps with equality on the edges. ![Vertical maps](figures/image_2020-04-07-13-35-00.png) By formal smoothness, $\eta'$ lifts to some $\xi'$, but using the transitivity of the action of the tangent space can fix this. We already had an obstruction theory of $R$, since we can always find a quotient \[ I \to S = k[[t_R\dual]] \surjects R \] and $h_K$ has an obstruction theory - $\mathrm{def} = t_R = \qty{\mfm_R/\mfm_R^2}\dual$ - $\mathrm{obs} = \qty{I/\mfm_S I}\dual$ ::: :::{.fact title="proof can be found in FGA"} Any other obstruction theory $(\mathrm{def}', \mathrm{obs}')$ of $h_R$ admits an injection $\qty{I/\mfm_S I}\dual \injects \mathrm{obs}'$. ::: Combining these three facts, we conclude the following: If $F$ has an obstruction theory $\mathrm{def}(F), \mathrm{obs}(F)$, then $F$ has a miniversal family $h_R \mapsvia \xi F$ with $R = S/ I$ a quotient of the formal power series ring over some ideal, where $S = k[[t_F\dual]]$. It follows that $\dim(I/\mfm_S I) \leq \dim \mathrm{obs}(F)$, and thus the minimal number of generators of $I$ (equal to the LHS by Nakayama) is bounded by the RHS. Thus \[ \dim_k \mathrm{def}(F) \geq \dim R \geq \dim \mathrm{def}(F) - \dim \mathrm{obs}(F) .\] In particular, if $\dim(R) = \dim \mathrm{def}(F) - \dim \mathrm{obs}(F)$, then $R$ is a complete intersection. If $\dim(R) = \dim \mathrm{def}(R)$, the ideal doesn't have any generators, and $R \cong S$. In particular, if $\mathrm{obs}(F) = 0$, then $R \cong S$ is isomorphic to this power series ring. Finally, if $F$ is the deformation functor for a global representable functor, then $R = \hat{\OO}_{\mfm, p}$ is the completion of this local ring and the same things hold for this completion. Thus regularity can be checked on the completion. So if you have a representable functor with an obstruction theory (e.g. the Hilbert Scheme) with zero obstruction, then we have smoothness at that point. If we know something about the dimension at a point relative to the obstruction, we can deduce information about being a local intersection. So the deformation tells you the dimension of a minimal smooth embedding, and the obstruction is the maximal number of equations needed to cut it out locally. :::{.remark} The content here: see Hartshorne's *Deformation Theory*. The section in FGA is in less generality but has many good examples. See "Fundamental Algebraic Geometry". See also representability of the Picard scheme. :::