# Dimensions of vector spaces in an exact sequence

I've read the following formula in wikipedia:

Given finite dimensional vector spaces $V_i$ and an exact sequence $\cdots\rightarrow V_i\rightarrow V_{i+1}\rightarrow\cdots$, we have

$$\sum_{n\in 2\mathbb{Z}}\dim V_n = \sum_{n\in 2\mathbb{Z}+1}\dim V_n$$

Is there a name for this theorem? Could anyone please tell me where to find a proof of this in the literature?

Thank you!

• the Euler characteristic of the complex is $0$ Dec 10 '12 at 11:41
• @user8268: maybe I'm wrong but isn't the Euler characteristic of a complex the alternated sum of lengths of the homologies of the complex? here the complex is just acyclic so the homologies are all trivial. Dec 10 '12 at 12:28

Let us prove the version of the statement in the Wikipedia entry:

If $$0 \to V_1 \to V_2 \cdots \to V_r\to 0$$ is an exact sequence of finite dimensional vector spaces then $$\sum_{i=1}^{r} (-1)^i \dim V_i=0.$$

Let $$f_0$$ be the first map in the sequence, $$f_i$$ be the map from $$V_i$$ to $$V_{i+1}$$, etc. By the Rank Nullity theorem, we have $$\dim V_i = \dim\ker f_i + \dim \operatorname{im} f_i.$$ Thus the left hand side is

$$\sum_{i=1}^{r} (-1)^i \dim\ker f_i+\sum_{i=1}^{r} (-1)^i \dim\operatorname{im} f_i.$$

Now by the defining property of an exact sequence, $$\operatorname{im} f_i = \ker f_{i+1}.$$ Place that information into one of the sums, and the two sums then cancel out.

Note that in order for the series in question to converge, the sequence must be of the form in this answer, perhaps with extra $$0$$'s.

• the same argument works for exact sequences that are not bounded, as the ones in the question. In particular, this statement does not work only for the dimesion of vector spaces but for any function $L:\mathfrak A\to \mathbb R_{\geq 0}\cup\{\infty\}$, where $\mathfrak A$ is an abelian category, such that $L(A)=L(B)+L(C)$ whenever $0\to B\to A\to C\to 0$ is a s.e.s. in $\mathfrak A$. (Examples: composition length, torsion-free rank for modules over left Ore domains, log of the cardinality in Abelian groups, ...) Dec 10 '12 at 12:38
• Thank you all - that was more easy than I thought... Dec 10 '12 at 13:21
• I think each $V_i$ should be $f_i$ in the final displayed equation. I tried to edit but the modification wasn’t enough characters for math.se. Mar 11 '19 at 3:22

For a finite exact sequence of finite dimensional vector spaces, the claimed equality is also provable by induction. For the induction anchor ($$n = 1$$), observe that if $$0 \rightarrow V_1 \rightarrow 0$$ exact, then the second arrow is injective, and thus $$\dim(V_1) = 0$$, such that the claimed equality holds. For the induction step, the exact sequence in question is (up to adjustments for $$n = 2$$ or $$n = 3$$): $$0 \rightarrow V_1 \rightarrow V_2 \rightarrow V_3 \rightarrow \ldots \rightarrow V_n \rightarrow 0$$ from which the following, shorter sequence can be naturally constructed (by considering the domain of the the third arrow, from $$V_2$$ to $$V_3$$, modulo its kernel, which is, by exactness of the sequence in question, isomorphic to $$V_1$$, and thus identified with $$V_1$$): $$0 \rightarrow V_2 \mathbin{/} V_1 \rightarrow V_3 \rightarrow \ldots \rightarrow V_n \rightarrow 0$$ which is still exact, because the second arrow is now an injective map, while it has, by construction, the same image in $$V_3$$ as the corresponding arrow of the original exact sequence. Combining $$\dim(V_2 \mathbin{/} V_1) = \dim(V_2) - \dim(V_1)$$ and the induction hypothesis yields the claimed equality.