Let's assume we have two short exact sequence of vector bundles on a smooth variety $X$ over a field. $l_1: 0\rightarrow V_1 \rightarrow V \rightarrow V_2 \rightarrow 0$ and $l_2: 0\rightarrow V_3 \rightarrow V \rightarrow V_4 \rightarrow 0$. Let's pullback everything to $X\times \mathbb{A}^2$. Now we have a map from $V_1\oplus V_3$ to $V$ which at each component is the inclusion. Consider the pushout of this map along the map from $V_1\oplus V_3$ to itself that is multiplication by $x$ on first component and $y$ on the second one. Note that $x$ and $y$ correspond to the variables of the $\mathbb{A}^2$. Let's call this pushout $M$. $M$ is a coherent sheaf on $X\times \mathbb{A}^2$ that is a vector bundle on $X\times (\mathbb{A}^2\setminus \{(0,0)\})$. Around $X\times\{(0,0)\}$ things get complicated. Still it is easy to write down $M|_{x=0}$ or $M|_{y=0}$. Each of these restriction is the direct sum of two coherent sheaves. One of them is a vector bundle that is either $V_1$ or $V_3$ pulled back to $X\times \mathbb{A}^1$. The other summand is something that is a little bit similar to the deformation of a submodule of $V_2$ or $V_4$ to a direct sum. But you can write down them exactly. My question is what is the restriction of dual of $M$, i.e. $M^{\vee}$, to either of the lines $x=0$ or $y=0$. You can assume that $M^{\vee}$ is a vector bundle on $X\times \mathbb{A}^2$ if that helps.

You can show that any vector bundle on $X\times (\mathbb{A}^2\setminus \{(0,0)\})$ extends uniquely to a vector bundle on $X\times \mathbb{A}^2$. The unique extension is given by the double dual of the pushforward or in our case it is $M^{\vee \vee}$. My ultimate goal is to understand $M^{\vee \vee}|_{x=0}$ and $M^{\vee \vee}|_{y=0}$. I have a guess for these restrictions but I am not able to prove them precisely.

My guess: $M^{\vee \vee}|_{x=0}\cong W_1\oplus W_2$ and $M^{\vee \vee}|_{y=0}=W_3\oplus W_4$. Where all $W_i$ s are deformations of a subbundle to the split exact sequence. The kernels appearing in $W_1$ and $W_3$ are the same (something related to $V_1\cap V_2$) and cokernels appearing in $W_2$ and $W_4$ are the same (something related to $V/(V_1+V_2)$). But I don't know how to approach the problem and actually calculate them or verify my guess.


By your definition there is an exact sequence $$ 0 \to M^\vee \to V^\vee \to V_1^\vee \otimes \mathcal{O}/x \oplus V_3^\vee \otimes \mathcal{O}/y \to 0\tag{*} $$ on $X \times \mathbb{A}^2$ (where $V$ and $V_i$ denote the pullbacks of the same named bundles). Restricting to $X \times \{x = 0\}$ one obtains an exact sequence $$ 0 \to V_1^\vee \otimes \mathcal{O}/x \to M^\vee \otimes \mathcal{O}/x \to V^\vee \otimes \mathcal{O}/x \to V_1^\vee \otimes \mathcal{O}/x \oplus V_3^\vee \otimes \mathcal{O}/(x,y) \to 0. $$ The morphism $V^\vee \otimes \mathcal{O}/x \to V_1^\vee \otimes \mathcal{O}/x$ is the restriction of the morphism dual to $V_1 \to V$, therefore it is surjective and its kernel is $V_2^\vee \otimes \mathcal{O}/x$. This means that $M^\vee\vert_{x = 0}$ is an extension of the kernel of the morphism $$ V_2^\vee\vert_{x = 0} \to V_3^\vee\vert_{x = y = 0} $$ (the morphism is induced by the composition $V_3 \to V \to V_2$) by the bundle $V_1^\vee\vert_{x = 0}$.

EDIT. Let me explain how the sequence $(*)$ is obtained.

First, note that there is an exact sequence $$ 0 \to V_1 \oplus V_3 \stackrel{(x,y)}\to V_1 \oplus V_3 \to V_1 \otimes \mathcal{O}/x \oplus V_3 \otimes \mathcal{O}/y \to 0. $$ Therefore, by definition of pushout there is an exact sequence $$ 0 \to V \to M \to V_1 \otimes \mathcal{O}/x \oplus V_3 \otimes \mathcal{O}/y \to 0. $$ Dualizing it one obtains $(*)$.

  • $\begingroup$ Thanks for you answer. How did you get the first exact sequence? I know dualizing is left exact and turns the pushout into a pullback. I'm not sure how the surjectivity of your short exact sequence works. By $V_3^{\vee}|_{x=y=0}$ do you mean pukkback from $x=y=0$ to $x=0$? because one is defined on $x=0$ and the other one on $x=y=0$. $\endgroup$ – user127776 Nov 18 '20 at 19:09
  • $\begingroup$ I added an explanation about the first sequence. By $V_3^\vee\vert_{x = y = 0}$ I mean the pullback with respect to the composition $X \times \{x = y = 0\} \to X \times \mathbb{A}^2 \to X$. Note that this map is an isomorphism, so this is essentially you original bundle $V_3$. $\endgroup$ – Sasha Nov 18 '20 at 19:57
  • $\begingroup$ Alright. But then, the two vector bundles $V_2^{\vee}|_{x=0}$ and $V_3^{\vee}|_{x=y=0}$ are not defined on the same space. How is there a map between them? $\endgroup$ – user127776 Nov 18 '20 at 20:05
  • $\begingroup$ This is the morphism from the first to the pushforward of the second. $\endgroup$ – Sasha Nov 18 '20 at 20:35
  • $\begingroup$ Oh Ok thanks. You mentioned that you added an explanation for the first sequence. (I cannot find it though, not even in the history of the post!) I understand that the morphism to each direct summand in the first ses is surjective. I don't understand why that map to the direct sum of both is also surjective, $\endgroup$ – user127776 Nov 18 '20 at 22:28

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