----- > [!proposition] Proposition. ([[computing the Euler characteristic of a vector space complex using homology]]) > Let $k$ be a [[field]] and consider a [[chain complex of modules|complex]] of $k$-[[vector space|vector spaces]] and [[linear map|linear maps]]: $V_{\bullet}: 0 \to V_{N} \xrightarrow{\ \ \alpha_{N}\ \ } V_{N-1} \xrightarrow{\alpha_{N-1}} \cdots \xrightarrow{\ \ \ \alpha_{2}\ \ \ } V_{1} \xrightarrow{ \ \ \ \alpha_{1 \ \ \ }} V_{0} \to 0.$ > The [[Euler characteristic of a complex of vector spaces|Euler characteristic]] of $V_{\bullet}$ may be computed as $\chi(V_{\bullet})=\sum_{i=0}^{N}(-1)^{i} \ \text{dim }H_{i}(V_{\bullet}),$ > where $H_{i}(V_{\bullet})$ denotes the $i^{th}$ [[(co)homology of a complex|homology]] $\frac{\text{ker }\alpha_{i}}{\text{im }\alpha_{i+1}}$ of $V_{\bullet}$. In particular, if $V_{\bullet}$ is [[exact sequence|exact]], then $\chi(V_{\bullet})=0$. > ^proposition > [!NOTE] Remark. > Though it might not be obvious from the theorem statement, the only notable result needed to prove it is the [[Rank-Nullity theorem]] (!). In particular, the Euler characteristic is independent of the choice of boundary map. > [!proof]- Proof. ([[computing the Euler characteristic of a vector space complex using homology]]) > ~ > > There is nothing to show if $N=0$ [^1]. > \ > **Base case.** Suppose $N=1$, so that $V_{\bullet}$ looks like $0 \to V_{1} \xrightarrow{\alpha_{1}} V_{0} \to 0.$The homologies are $H_{1}(V_{\bullet})=\text{ker } \alpha_{1}$ and $H_{0}(V_{\bullet})=\frac{V_{0}}{\text{im }\alpha_{1}}$. The following [[chain complex of modules|sequences]] are [[short exact sequence|short]] [[exact sequence|exact]]: $\begin{align} > 0 \to& \text{ker }\alpha_{1} \to V_{1} \to \text{im }\alpha_{1} \to 0 \\ > 0 \to& \text{ im }\alpha_{1} \to V_{0} \to \frac{V_{0}}{\text{im }\alpha_{1}} \to 0; > \end{align}$ > > thus [[Rank-Nullity theorem|rank-nullity]] implies $\begin{align} > \text{dim }V_{1}= & \text{dim ker }\alpha_{1} + \text{dim im }\alpha_{1} = \text{dim }H_{1}(V_{\bullet}) + \text{dim im }\alpha_{1} \\ > \text{dim }V_{0} = & \text{dim im } \alpha_{1} + \text{dim } \frac{V_{0}}{\text{im }\alpha_{1}} = \text{dim im }\alpha_{1} + H_{0}(V_{\bullet}). > \end{align}$ > Now, $\begin{align} > \chi(V_{\bullet})= & -\text{dim }V_{1} + \text{dim }V_{0} \\ > = & -\big(\text{dim }H_{1}(V_{\bullet}) + \text{dim im }\alpha_{1}\big) + \text{dim im }\alpha_{1} + H_{0}(V_{\bullet}) \\ > = & - \text{dim }H_{1}(V_{\bullet}) + \text{dim }H_{0}(V_{\bullet}) > \end{align}$ > as required. > > **Inductive step.** Suppose the result holds for 'shorter' sequences. Given a [[chain complex of modules|complex]] $V_{\bullet}: 0 \to V_{N} \xrightarrow{\ \ \alpha_{N}\ \ } V_{N-1} \xrightarrow{\alpha_{N-1}} \cdots \xrightarrow{\ \ \ \alpha_{2}\ \ \ } V_{1} \xrightarrow{ \ \ \ \alpha_{1 \ \ \ }} V_{0} \to 0.$ > Consider then the truncation $V_{\bullet}': 0 \to V_{N-1} \xrightarrow{\alpha_{N-1}} \cdots \xrightarrow{\alpha_{2}} V_{1} \xrightarrow{\alpha_{1}}V_{0} \to 0.$ > Then $\chi(V_{\bullet})=(-1)^{N} \text{ dim }V_{N} + \chi(V_{\bullet}')=(-1)^{N} \text{ dim }V_{N} + \sum_{i=0}^{N-1}(-1)^{i} \text{dim }H_{i}(V_{\bullet}').$ > and $H_{i}(V_{\bullet}) = H_{i}(V_{\bullet}') \text{ for }0 \leq i \leq N-2$ > while $H_{N-1}(V_{\bullet}')=\text{ker }\alpha_{N-1}, H_{N-1}(V_{\bullet})=\frac{\text{ker }\alpha_{N-1}}{\text{im }\alpha_{N}}, H_{N}(V_{\bullet})=\text{ker }\alpha _{N}.$ > Now we proceed not too differently than the base case. [[Rank-Nullity theorem]] says $\text{dim }V_{N}=\text{dim }H_{N}(V_{\bullet})+ \text{dim im }\alpha_{N},$ > and also (recall $\text{dim ker }\alpha_{N-1}=H_{N-1}(V_{\bullet}')$ ) $\text{dim }H_{N-1}(V_{\bullet}')=\text{dim im }\alpha_{N} + \text{dim }\frac{\text{ker }\alpha_{N-1}}{\text{im }\alpha_{N}}=\text{dim im }\alpha_{N} + \text{dim }H_{N-1}(V_{\bullet}).$ > (this is like the base case but we include into kernel rather than vector space itself. See iPad notes, or Aluffi, if confused). > > Now $\begin{align}\chi(V_{\bullet}) = & (-1)^{N} \text{dim }V_{N} + \chi(V_{\bullet}') \\= & (-1)^{N} \text{dim }V_{N} + (-1)^{N-1} H_{N-1}(V_{\bullet}') + \sum_{i=0}^{N-2} (-1)^{N} H_{i}(V_{\bullet}') \\= & (-1)^{N} \text{dim }H_{N}(V_{\bullet}) + (-1)^{N} \text{dim im }\alpha_{N} + (-1)^{N-1} H_{N-1}(V_{\bullet}') + \sum_{i=0}^{N-2} H_{i}(V_{\bullet}) \\ = & (-1)^{N} \text{dim }H_{N}(V_{\bullet}) + \cancel{ (-1)^{N} \text{dim }H_{N-1}(V_{\bullet}') } \overbrace{- (-1)^{N}\text{dim }H_{N-1}(V_{\bullet})}^{+ (-1)^{N-1} \text{dim }H_{N-1}(V_{\bullet})} \\+ & \cancel{ (-1)^{N-1} H_{N-1}(V_{\bullet}') } + \sum_{i=0}^{N-2} (-1)^{N} H_{i}(V_{\bullet}) \end{align}$ > which is the precisely the result. (For some reason Aluffi's work looks a lot cleaner here than mine, but oh well.) > ----- #### [^1]: Needlessly thorough explanation: if $N=0$ then the chain looks like $\cdots \to 0 \to V_0 \to 0 \to \cdots$ which means the task is just to compute $H_{0}(V_{\bullet})=\frac{\text{ker }(V_{0} \to 0)}{\text{im }(0 \to V_{i})}=\frac{V_{0}}{\{ 0 \}}=V_{0}$ and we are done since $\chi(V_{\bullet})=\text{dim }V_{0}=\text{dim }H_{0}(V_{\bullet})$ because $V_{0}$ literally equals $H_{0}(V_{\bullet})$. ---- #### References > [!backlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM [[]] > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ``` > [!frontlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM outgoing([[]]) > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ```