# Can non-split extension be isomorphic to the split one as objects

Is it possible to have a non-split short exact sequences of vector bundles (on some smooth variety) $$0\rightarrow V_1 \rightarrow V_2 \rightarrow V_3 \rightarrow 0$$. Such that $$V_2\cong V_1\oplus V_3$$ as vector bundles?

• What do you mean by 'as objects'? – Jef Nov 17 '20 at 18:49
• As vector bundles. Not as extensions (which obviously is impossible according to the assumptions). – user127776 Nov 17 '20 at 18:50
• See also mathoverflow.net/questions/163041/… . – LSpice Nov 17 '20 at 19:24
• You can take a non-split short exact sequence on $\mathbb{P}^1$ of the form $$0 \to \mathcal{O} \to \mathcal{O} \oplus \mathcal{O}(n) \to \mathcal{O}(n) \to 0.$$ It exists as soon as $n \geq 2$, as one can check by computing the Ext$^1$ group. – Francesco Polizzi Nov 17 '20 at 19:26
• @FrancescoPolizzi : but for example, there is a non-split sequence $0 \to \mathcal O \to \mathcal O(1)^{\oplus 2} \to \mathcal O(2) \to 0$ corresponding to a non-zero element in $Ext^1(\mathcal O(2), \mathcal O)$. So given a non-zero class in $Ext^1(\mathcal O, \mathcal O(n))$ do you know that the middle term will be given by $\mathcal O \oplus \mathcal O(n)$ ? – Nicolas Hemelsoet Nov 17 '20 at 20:06

$$\newcommand{\cO}{\mathcal{O}}$$Consider exact sequence of trivial vector bundles $$0\to\cO\xrightarrow{\left(\begin{matrix}x \\ y\end{matrix}\right)}\cO\oplus\cO\xrightarrow{\left(\begin{matrix}y & -x\end{matrix}\right)}\cO\to 0$$ on $$X=\mathbb{A}^2_{x,y}\setminus\{0\}$$. One checks easily that it is exact on stalks (but the same sequence on $$\mathbb{A}^2$$ is not exact at $$(x,y)=(0,0)$$). It is the pullback of $$0\to\cO(-1)\to \cO\oplus\cO\to \cO(1)\to 0$$ from $$\mathbb{P}^1$$. A splitting of this sequence would be a pair of functions $$f_1,f_2\in H^0(\mathbb{A}^2\setminus\{0\},\cO)=k[x,y]$$ such that $$yf_1-xf_2=1$$ but there is no such pair.
On the positive side, any such sequence has to be split if $$X$$ is proper. In this case the spaces $$Hom(V_3, V_1)$$ and $$Hom(V_3, V_3)$$ are finite-dimensional over $$k$$. Applying $$Hom(V_3,-)$$ to this exact sequence we get a left exact sequence of finite-dimensional vector spaces $$0\to Hom(V_3, V_1)\to Hom(V_3,V_2)\to Hom(V_3,V_3)$$ The dimension of the vector space in the middle $$Hom(V_3,V_2)\simeq Hom(V_3, V_1\oplus V_3)$$ is equal to the sum of the dimensions of first and third terms. Therefore, the sequence has to be exact on the right and, in particular, the identity $$Id_{V_3}\in Hom(V_3, V_3)$$ lifts to a morphism $$Hom(V_3, V_2)$$ that gives a splitting.
• Yes my $X$ is proper. Your argument is super-smart. So there are no examples of such thing on projective line. This refutes one of the comments above. Thanks. – user127776 Nov 17 '20 at 20:11