Infinitesimal neighborhood and Ext group

Question

$\DeclareMathOperator\Ext{Ext}$Let $X$ be a smooth projective complex variety and $\iota\colon Y\subset X$ be a smooth closed subvariety. It is well-known that there is a spectral sequence $$E_2^{p,q}=H^p(\bigwedge\nolimits^q N_{Y/X})\Rightarrow \Ext^{p+q}(\iota_*\mathcal{O}_Y, \iota_*\mathcal{O}_Y).$$

Question: If this spectral sequence degenerates at $E_2$-page and we take $i=2$, then how to describe explicitly the embedding $$H^1(N_{Y/X})\to \Ext^2(\iota_*\mathcal{O}_Y, \iota_*\mathcal{O}_Y)?$$

My guess is that for any $p\in H^1(N_{Y/X})$, its image in $\Ext^2$ is given by the composition $$\iota_*\mathcal{O}_Y\to \iota_*N^*_{Y/X}[1]\to \iota_*\mathcal{O}_Y[2]$$ where the first map $\iota_*\mathcal{O}_Y\to \iota_*N^*_{Y/X}[1]$ corresponds to the second infinitesimal neighborhood of $Y$ in $X$ cut out by $I^2_Y$ and the second map is given by the dual of $p$ shifted by $1$.

The above guess looks natural, however, I do not know how to prove this rigorously. Is there any reference for this?

Answer 1

There is no such map. In fact, the spectral sequence gives you a filtration on $\mathrm{Ext}^2(i_*\mathcal{O}_Y, i_*\mathcal{O}_Y)$ with three factors: $$ \mathrm{Ext}^2(\mathcal{O}_Y, \mathcal{O}_Y),\quad \mathrm{Ext}^1(\mathcal{O}_Y, N_{Y/X}),\quad \mathrm{Hom}(\mathcal{O}_Y, \wedge^2N_{Y/X}), $$ so the space $H^1(N_{Y/X})$ is just the factor in the middle.

Even in the simpler case, where $Y$ is a divisor in $X$, you only have an exact sequence $$ \dots \to \mathrm{Ext}^2(\mathcal{O}_Y, \mathcal{O}_Y) \to \mathrm{Ext}^2(i_*\mathcal{O}_Y, i_*\mathcal{O}_Y) \to H^1(N_{Y/X}) \to \dots, $$ so if the first map is injective and the second map is surjective (which happens when the spectral sequence degenerates), so that this becomes a short exact sequence, it doesn't have a canonical splitting.