# The Unapologetic Mathematician

## The Euler Characteristic of an Exact Sequence Vanishes

Naturally, one kind of linear map we’re really interested in is an isomorphism. Such a map has no kernel and no cokernel, and so its index is definitely zero. If it weren’t clear enough already, this shows that isomorphic vector spaces have the same dimension!

But remember that in abelian categories we’ve got diagrams to chase and exact sequences to play with. And these have something to say about the matters at hand.

First, remember that a linear map whose kernel vanishes looks like this in terms of exact sequences:

$\mathbf{0}\rightarrow V\xrightarrow{f}W$

And one whose cokernel vanishes looks like this:

$V\xrightarrow{f}W\rightarrow\mathbf{0}$

So an isomorphism is just an exact sequence like this:

$\mathbf{0}\rightarrow V\xrightarrow{f}W\rightarrow\mathbf{0}$

And then we have the equation

$-\dim(V)+\dim(W)=0$

Yes, I’m writing the negative of the index here, but there’s a good reason for it.

Now what if we have a segment of an exact sequence:

$...\rightarrow U\xrightarrow{f}V\xrightarrow{g}W\rightarrow...$

Considering the map $f$ allows us to break up $V$ as $\mathrm{Im}(f)\oplus\mathrm{Cok}(f)$ (since short exact sequences split). On the other hand, considering the map $g$ allows us to break up $V$ as $\mathrm{Ker}(g)\oplus\mathrm{Coim}(g)$. Exactness tells us that $\mathrm{Im}(f)=\mathrm{Ker}(g)$, which gives us the isomorphism $V\cong\mathrm{Im}(f)\oplus\mathrm{Coim}(g)$.

Now the rank-nullity theorem says that $\mathrm{Im}(f)\cong\mathrm{Coim}(f)$, and similarly for all other linear maps. So we get $V\cong\mathrm{Coim}(f)\oplus\mathrm{Im}(g)$ — which expresses $V$ as the direct sum of one subspace of $U$ and one of $W$. And each of those vector spaces has another part to hand off to the vector space on its other side, and so on!

What does this mean? It says that if we look at every other term of an exact sequence and take their direct sum, the result is the same whether we look at the odd or the even terms. More explicitly, let’s say we have a long exact sequence

$\mathbf{0}\rightarrow V_n\xrightarrow{f_n}V_{n-1}\xrightarrow{f_{n-1}}...\xrightarrow{f_2}V_1\xrightarrow{f_1}V_0\rightarrow\mathbf{0}$

Then we can decompose each term as either $V_k\cong\mathrm{Im}(f_{k+1})\oplus\mathrm{Coim}(f_k)$ or $V_k\cong\mathrm{Coim}(f_{k+1})\oplus\mathrm{Im}(f_k)$ — one for the even terms and the other for the odd terms. Then direct-summing everything up we have an isomorphism

$\displaystyle\bigoplus\limits_{\substack{0\leq k\leq n\\k\mathrm{~even}}}V_k\cong\bigoplus\limits_{\substack{0\leq k\leq n\\k\mathrm{~odd}}}V_k$

which tells us that

$\displaystyle\dim\left(\bigoplus\limits_{\substack{0\leq k\leq n\\k\mathrm{~even}}}V_k\right)-\dim\left(\bigoplus\limits_{\substack{0\leq k\leq n\\k\mathrm{~odd}}}V_k\right)=0$

But since direct sums add dimensions this means

$\displaystyle\sum\limits_{\substack{0\leq k\leq n\\k\mathrm{~even}}}\dim\left(V_k\right)-\sum\limits_{\substack{0\leq k\leq n\\k\mathrm{~odd}}}\dim\left(V_k\right)=0$

And now we can just combine these sums:

$\displaystyle\sum\limits_{k=0}^n(-1)^k\dim\left(V_k\right)=0$

Which generalizes the formula we wrote above in the case of an isomorphism. This alternating sum we call the “Euler characteristic” of a sequence, and we’ll be seeing a lot more of that sort of thing in the future. But here the major result is that for exact sequences we always get the value zero.

In fact, this amounts to a far-reaching generalization of the rank-nullity theorem. And that theorem, of course, is essential to the proof. Yet again we see this pattern of “bootstrapping” our way from a special case to a larger theorem. Despite some mathematicians being enamored of reductio ad absurdum, this induction from special to general has to be one of the most useful tools we keep running across.

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Posted by John Armstrong | Algebra, Linear Algebra

## 2 Comments	»

1. [...] the Euler characteristic we find [...]

Pingback by | July 24, 2008 | Reply

2. [...] sum satisfies inclusion-exclusion because of the Mayer-Vietoris sequence (and the fact that the Euler characteristic of an exact sequence vanishes), and it obviously satisfies the cardinality axiom, so by homotopy invariance it must be the Euler [...]

Pingback by | June 10, 2011 | Reply

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