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We call a sequence of $L_\infty$-algebras (weak) maps $$0\to L\xrightarrow{f} M\xrightarrow{g} N\to 0$$ is exact if it is exact on the the underlying chain complexes level.

Thought I don't know whether this is a good notion. A trivial case is exact sequence of Lie algebras. It is clear that given a surjective map of Lie algebras, the kernel is a Lie algebra as well.

My naive question: if we are only given $L_\infty$-algebras $$M\xrightarrow{g} N\to 0$$ being exact, can we find a $L_\infty$-algebra $L$ fitting the longer sequence above?

I expect the answer to this naive question is no, but I do not have an example. If the answer is no, then how do one sensibly fix it? I heard someone talked about homotopy fibre of $L_\infty$-algebras, does it work for this case?

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In these settings it is usually more pertinent to speak about "exact triangles" than "exact sequences". –  user36931 Aug 1 '13 at 3:22
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You probably want to take the homotopy fibre rather than the literal fibre. The literal fibre (as chain complexes) I imagine wouldn't carry an $L_\infty$-algebra structure. –  David Roberts Aug 1 '13 at 10:39
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I'm being a little lazy, it's probably on the nLab. Urs Schreiber is on holidays, or doubtless he would turn up and give you a detailed answer. –  David Roberts Aug 1 '13 at 10:40
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Indeed, I am on vacation and on a super-slow connection, therefore withouth hyperlinks: the nLab entry titled "model structure for L-infinity algebras" has a little section on recognition of homotopy fiber sequences of L-infinity algebras for the case that the rightmost algebra is abelian. –  Urs Schreiber Aug 1 '13 at 13:17
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