On an exact sequence of complex vector spaces

Question

Let us consider the following exact sequence of complex vector spaces:

$0 \to A \to B \to C \to D \to E \to.....\to 0$, where $E$ and the later vector spaces are not necessarily zero.

By rank nullity theorem one can say $\text{dim}(B) \leq \text{dim}(A) +\text{dim}(C)$, can we extend this inequality involving $A,B,C,D$ as something like $\text{dim}(B)+ \text{dim}(D)\leq \text{dim}(A) +\text{dim}(C)$?

I know that one can equate the complete sequence with alternating signs, but I'm looking for an inequality only involving first $4$ spaces.

Any remark from anyone is welcome.

Answer 1

We are given the following information:

1. There is an injective map $f:A \to B$ (so that $\dim f(A) = \dim A$),
2. There is a map $g:B \to C$ whose kernel is $f(A)$,
3. There is a map $h:C \to D$ whose kernel is $g(B)$.

Applying the rank-nullity theorem to $g,h$ yields $$ \dim(B) = \dim \ker g + \dim g(B) = \dim(A) + \dim g(B)\\ \dim(C) = \dim \ker h + \dim h(C) = \dim g(B) + \dim h(C). $$ Subtracting these equations yields $$ \dim(B) - \dim(C) = \dim(A) - \dim h(C) \geq \dim(A) - \dim(D) \implies\\ \dim(B) + \dim(D) \geq \dim(A) + \dim(C). $$

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In general, we can extend this result as follows. As you have correctly noted, the long exact sequence $0\to A_1 \to A_2 \to \cdots \to A_n \to 0$ yields the equality $$ \dim(A_1) - \dim(A_2) + \cdots + (-1)^{n-1} \dim(A_n) = 0 $$ Let $f_i$ denote the map from $A_i \to A_{i+1}$. Truncating this sequence yields the "left exact" sequence $0\to A_1 \to A_2 \to \cdots \to A_k$. With this, we have the equality $$ \dim(A_1) - \dim(A_2) + \cdots + (-1)^{k-1}\dim f_{k-1}(A_{k-1}) = 0. $$ Using the fact that $\dim f_{k-1}(A_{k-1}) \leq \dim(A_k)$ yields the kind of inequality that you're after. For $k = 2p+1$, we get $$ \sum_{i=1}^p \dim A_{2i} \leq \sum_{i=0}^p \dim A_{2i+1}. $$ For $k = 2p$, we get $$ \sum_{i=1}^p \dim A_{2i} \geq \sum_{i=0}^{p-1} \dim A_{2i+1}. $$