Lower bound for the dimension of the space of deformations $\mathrm{Defor}(f : X \to Y)$ in relative setting

Question

Let $f: X \to Y$ a morphism between smooth varieties over alg. closed field of characteristic zero. It is known that the deformation theory in relative setting of $f$ is encoded in the cohomology of the trunicated cotangent complex

$$ \mathbb{L}_f = \tau_{\ge -1}\left[f^*\Omega_Y \to \Omega_X\right] $$

concentrated in degrees $[-1,0]$. The Ext sheaves $\mathcal{Ext}^i(\mathbb{L}_f, \mathcal{O}_X) $ are then what is in literature often called the $\mathcal{T}^i$-functors.

The local-to-global spectral sequence assures in favourable situations that $H^i\mathcal{Ext}^j(\mathbb{L}_f, \mathcal{O}_X)) $ converge to Ext groups $\text{Ext}^{i+j}(\mathbb{L}_f, \mathcal{O}_X) $.

My question is how to see that if we assume that $\mathrm{Ext}^0(\mathbb{L}_f, \mathcal{O}_X) = 0$, then there exist a lower bound

$$ \dim \mathrm{Defor}(f : X \to Y) \geq \mathrm{ext}^1(\mathbb{L}_f, \mathcal{O}_X) - \mathrm{ext}^2(\mathbb{L}_f, \mathcal{O}_X) $$

This extimation suggests that there exist certain exact sequence between the Ext groups such that $\mathrm{Defor}(f : X \to Y)$ can be naturaly embedded in one term there. Can it be explicitly written down how the defomations 'sit' there? Motivation: This estimation was used here and I would like understand the reason why this estimation works here.

Note that I'm only interested in relative setting. The case $Y= \text{Spec}(k)$ is well understood: the deformations (precisely first order deformations) of a nonsingular variety $X $ over $k$ can be identified with group $H^1(X,\mathcal T_X)$. The question is how the story changes in relative setting and especially how to get the bound estimation from above?

Answer 1

This is too long for a comment.

You need some sort of hypothesis to get the existence of a versal deformation space for morphisms $f$. The most common hypothesis is that $X$ is proper over your field $k$.

In that case, Schlessinger gives a versal formal deformation over Spf of a power series $k$-algebra, $k[[x_1,\dots,x_n]]/I$, where $I$ is an ideal in $\langle x_1,\dots,x_n \rangle^2$. The integer $n$ equals the dimension that you denote $\text{ext}^1(\mathbb{L}_f,\mathcal{O}_X)$, i.e., $\text{dim}_k \mathbb{E}\text{xt}^1_{\mathcal{O}_X}(\mathbb{L}_f,\mathcal{O}_X)$.

If $f_1,\dots,f_r$ is a minimal set of generators of $I$, then $I/\langle x_1,\dots,x_n\rangle I$ is a free $k$-vector space with basis $\overline{f}_1,\dots,\overline{f}_r$. Use Krull's Intersection Theorem to see that for sufficiently large integers $e$, the quotient $k$-vector space $$\frac{\langle x_1,\dots,x_n \rangle^{e+1}+ I}{\langle x_1,\dots,x_n \rangle^{e+1}+ \left( \langle x_1,\dots,x_n \rangle I\right) }$$ also has basis $\overline{f}_1,\dots,\overline{f}_r$. Now consider the formal deformation over Spec of the quotient $k$-algebra $k[[x_1,\dots,x_n]]/\left(\langle x_1,\dots,x_n \rangle^{e+1}+ I\right)$ of $k[[x_1,\dots,x_n]]/\left(\langle x_1,\dots,x_n \rangle^{e+1} + \left( \langle x_1,\dots,x_n \rangle I \right)\right)$ that does not extend to a deformation over any intermediate quotient.

The deformation theory result you note then gives an injection of $k$-vector spaces from the $r$-dimensional quotient vector space above to the $k$-vector space dual of $\mathbb{E}\text{xt}^2_{\mathcal{O}_X}(\mathbb{L}_f,\mathcal{O}_X)$. Thus, $r$ is less than or equal to the $k$-vector space dimension $\text{ext}^2(\mathbb{L}_f,\mathcal{O}_X)$ of this $k$-vector space (it can be strictly less, e.g., for deformations of an Abelian variety of dimension $>1$ over $Y=\text{Spec}\ k$). By the Krull Hauptidealsatz, the dimension of $k[[x_1,\dots,x_n]]/I$ is at least $n-r$.