Given a short exact sequence of vector bundles on a projective variety, after tensoring with an $\mathcal{O}(n)$ with high $n$ that makes all terms globally generated (so that taking global sections becomes an exact functor), we can lift short exact sequence to a short split exact sequence of trivial bundles. This means that there is a short split exact sequence of vector bundles that surjects to our initial SES. Now the by taking kernels we get a new SES. My question is which short exact sequences admit a finite resolution by split short exact sequences? Does the original SES needs to be split or most SES admit such a resolution?


1 Answer 1


Every short exact sequence $0\to X\stackrel{\alpha}{\to}Y\stackrel{\beta}{\to}Z\to0$ is the quotient of a split short exact sequence by a split short exact subsequence. This answer is copied from my answer on math.stackexchange.

$$\require{AMScd}\begin{CD} @.0@.0@.0\\ @.@VVV@VVV@VVV\\ 0@>>>0@>>>X@>1>>X@>>>0\\ @.@VVV@VV\begin{pmatrix}1\\\alpha\end{pmatrix}V@VV\alpha V\\ 0@>>>X@>\begin{pmatrix}1\\0\end{pmatrix}>>X\oplus Y@>\begin{pmatrix}0&1\end{pmatrix}>>Y@>>>0\\ @.@VV1V@VV\begin{pmatrix}\alpha&-1\end{pmatrix}V@VV-\beta V\\ 0@>>>X@>\alpha>>Y@>\beta>>Z@>>>0\\ @.@VVV@VVV@VVV\\ @.0@.0@.0 \end{CD}$$

  • $\begingroup$ This technically answer my question so I accept it, although doesn't solve my problem! $\endgroup$
    – user127776
    Jan 3, 2021 at 1:40

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