For a bounded complex $M$ of finite-dimensional $k$-vector spaces we define its Euler characteristic as $$ \chi=\sum_{n\in \mathbb{Z}} (-1)^n\dim(M_n) $$ In particular, if complex is exact then its Euler characteristic is zero. Is the converse true?
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1How long have you thought about this? The euler characteristic knows nothing about the boundary maps! – Martin Brandenburg Jan 21 '14 at 09:29
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1Evidently, not for long enough. – Jimmy R Jan 21 '14 at 15:56
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I'd like to mention a positive result. As Martin Brandenburg pointed out, the Euler characteristic knows nothing about the boundary maps, so we might as well define the Euler characteristic of a bounded graded vector space (i.e. graded by $\mathbb Z$ with all spaces null but finitely many).
We say such a bounded graded vector space $V_.$ is:
- potentially exact if you can endow it with differentials that turn it into an exact complex.
- stably potentially exact if there is a potentially exact bounded graded vector space $V'_.$ such that $V_. \oplus V_.'$ is potentially exact.
Theorem: A bounded graded vector space is stably potentially exact if and only if its Euler characteristic is zero.
For a proof, see Grayson, Algebraic K-theory via binary complexes, lemma 5.4. It is stated and proved in greater generality there.
Bruno Stonek
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