Additivity with respect to a long exact sequence of

Question

Suppose we have a long exact sequence of finite dimensional vector spaces $$ \cdots\longrightarrow A^i\longrightarrow X^i \rightarrow B^i\rightarrow A^{i+1}\longrightarrow X^{i+1}\longrightarrow B^{i+1}\longrightarrow \cdots$$ how can deduce that $$\sum_{i\ge 0}(-1)^{i}\dim X^i = \sum_{i\ge 0}(-1)^{i}\dim A^i +\sum_{i\ge 0}(-1)^{i}\dim B^i $$ (Remark: We suppose that the sums are well defined). It seems like we will use the fact that $$\sum_{i\ge 0}(-1)^{i}\dim X^i +\sum_{i\ge 0}(-1)^{i}\dim A^i +\sum_{i\ge 0}(-1)^{i}\dim B^i =0$$ but this will give that $$\sum_{i\ge 0}(-1)^{i}\dim X^i =-(\sum_{i\ge 0}(-1)^{i}\dim A^i+\sum_{i\ge 0}(-1)^{i}\dim B^i )$$ Instead of $$\sum_{i\ge 0}(-1)^{i}\dim X^i =\sum_{i\ge 0}(-1)^{i}\dim A^i+\sum_{i\ge 0}(-1)^{i}\dim B^i $$ where does the minus sign go?

Answer 1

Your formula $\sum_{i \geq 0} (-1)^i \dim X^i + \sum_{i \geq 0} (-1)^i \dim A^i + \sum_{i \geq 0} (-1)^i \dim B^i = 0$ is incorrect due to the incorrect indexing. In this sum, the sign of $\dim A^i$ should be opposite to the sign of $\dim X^i$ as they are adjacent in the sequence, but as you wrote it they are the same. For example, let's take the case that all of these vector spaces vanish for $i \geq 1$. Then we have $0 \longrightarrow A^0 \longrightarrow X^0 \longrightarrow B^0 \longrightarrow 0$ exact (I'm assuming you mean to extend the sequence on the left by $0$). Your formula would say that $\dim A^0 + \dim X^0 + \dim B^0 = 0$, which only holds if all three are trivial. The correct formula, of course, is $\dim A^0 - \dim X^0 + \dim B^0 = 0$.

With that in mind, the correct alternating sum is $\dim A^0 - \dim X^0 + \dim B^0 - \dim A^1 + \dim X^1 - \dim B^1 + \dots = 0$. From this, we see that the correct formula is $\sum_{i \geq 0} (-1)^{3i} \dim A^i + \sum_{i \geq 0} (-1)^{3i + 1} \dim X^i + \sum_{i \geq 0} (-1)^{3i+2} \dim B^i = 0$. Simplifying this, we get $\sum_{i \geq 0} (-1)^i \dim A^i + \sum_{i \geq 0} (-1)^{i + 1} \dim X^i + \sum_{i \geq 0} (-1)^i \dim B^i = 0$. From this, we do indeed get $\sum_{i \geq 0} \dim (-1)^i A^i + \sum_{i \geq 0} \dim (-1)^i B^i = \sum_{i \geq 0} (-1)^i \dim X^i$.