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Let $C_*$ be a chain complex such that each $C_i$ is a torsion-free, finite-range abelian group with $C_i=0$ for all $i<0$. Suppose that $C_i=0$ for all $i$ is large enough. The Euler characteristic of $C_*$ chain complex is defined as $$\chi(C_*)=\sum_{i\geq 0}(-1)^iRank(C_i)$$ Prove that $$\chi(C_*)=\sum_{i\geq 0}(-1)^iRank(H_i(C_*))$$

I have to prove that $\sum_{i\geq 0}(-1)^iRank(C_i)=\sum_{i\geq 0}(-1)^iRank(H_i(C_*))$, I think that one way to do this is by showing that $Rank(C_i)=Rank(H_i(C_*))$ but I do not know if this is true in general, in a nutshell, I want to show that the cardinality of the base of any $C_i$ is the same as the cardinality of the basis of the corresponding homology, how can I do this? Thank you

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    $\begingroup$ Don't you mean $\textrm{Rank}(C_i)$ etc.? $\endgroup$ – Lord Shark the Unknown Jun 4 '18 at 4:29
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    $\begingroup$ @user482152 A counter-example to your guess as to how the proof would go: Say $C_* = \cdots {\mathbb Z}\to {\mathbb Z} \cdots$, where the written arrow is multiplication by $2$, and all the unwritten $C_i$ are identically zero. Assuming that rank of a finitely generated abelian group $A$ means the dimension of the (free) quotient $A/A_{\rm tors}$: the rank of$\mathbb Z$ is $1$. On the other hand, the homology of $C_*$ is $\mathbb Z/ 2\mathbb Z$, which has rank $0$. $\endgroup$ – peter a g Jun 4 '18 at 13:21
  • $\begingroup$ @LordSharktheUnknown If that is. $\endgroup$ – user482152 Jun 4 '18 at 16:21
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Your chain complex looks like $$0\to C_n\to C_{n-1}\to\cdots\to C_0\to0$$ where the $C_i$ are nonzero outside this range. Proceed by induction on $n$. Call this complex $\mathbf C$. Let $\mathbf{C}'$ be the subcomplex $$0\to0\to C_{n-1}\to\cdots\to C_0\to0.$$ Then there is a short exact sequence of complexes $$0\to\mathbf{C}'\to\mathbf{C}\to\mathbf{C}''\to0$$ where $C''$ consists just of $C_n$ in dimension $n$. This gives a long exact sequence of homology. This starts $$0\to H_n(\mathbf{C})\to C_n\to H_{n-1}(\mathbf{C'})\to H_{n-1}(\mathbf{C})\to0$$ so that $$\textrm{rank}(H_n(\mathbf{C})) -\textrm{rank}(C_n)+\textrm{rank}(H_{n-1}(\mathbf{C'}))-\textrm{rank}(H_{n-1}(\mathbf{C}))=0.\tag{1}$$ For $k<n-1$ another piece of the long exact sequence is $$0\to H_{k}(\mathbf{C'})\to H_{k}(\mathbf{C})\to0$$ so that $$\textrm{rank}(H_k(\mathbf{C'}))=\textrm{rank}(H_k(\mathbf{C})). \tag{2}$$ From (1) and (2) one gets the relation between Euler characteristics: $$\chi(\mathbf{C})=\chi(\mathbf{C'})+(-1)^n\textrm{rank}(C_n)$$ which gives the inductive step.

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  • $\begingroup$ @Lord Why there is a short exact sequence of complexes? $$0\to\mathbf{C}'\to\mathbf{C}\to\mathbf{C}''\to0$$ what does it mean that $C''$ consists just of $C_n$ in dimension $n$.? $\endgroup$ – user482152 Jun 4 '18 at 16:35
  • $\begingroup$ The complex $\mathbf{C}''$ looks like $\cdots0\to C_n\to0\to\cdots$. $\endgroup$ – Lord Shark the Unknown Jun 4 '18 at 18:08
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$\newcommand{\rank}{\operatorname{rk}}$ Consider the short exact sequences for $i\in \mathbb Z$.

$$0\to Z_i \to C_i \stackrel{d}\to B_{i-1} \to 0$$ $$0\to B_i \to Z_i \to H_i \to 0$$

From them it follows that for each $i\in \mathbb Z$, $$ \rank C_i=\rank Z_i + \rank B_{i-1}$$ $$\rank H_i = \rank Z_i-\rank B_i$$

Then

\begin{align*} \sum_{i\in\mathbb Z}(-1)^i \rank H_i&= \sum_{i\in\mathbb Z}(-1)^i \rank Z_i-\sum_{i\in\mathbb Z}(-1)^i \rank B_i\\ &=\sum_{i\in\mathbb Z}(-1)^i \rank Z_i+\sum_{i\in\mathbb Z}(-1)^{i-1} \rank B_i\\ &=\sum_{i\in\mathbb Z}(-1)^i \rank C_i. \end{align*}

This proof works in general for any map $\chi$ from some class of modules to integers which is additive on short exact sequences, meaning that whenever $$0\to A\to B\to C\to 0$$ is exact, $\chi(A) + \chi(C) = \chi(B)$.

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