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Lefschetz number
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The Lefschetz number of a self-map $f:K \rightarrow K$ of a finite cell complex $K$ is the alternating sum of the traces (the trace is the sum of the diagonal elements of the matrix) of the homology maps of $f$: $$\lambda_{f}=\sum_{n}(-1)^{n}Trace(f_{\ast n}),$$ where $f_{\ast n}:H_{n}(M)\rightarrow H_{n}(M)$ is induced by $f$.
The formula looks very similar to that of the Euler characteristic. In fact, the latter is equal to the Lefschetx number of the identity function of $K$: $$\chi (K) = \lambda _{Id}.$$
Lefschetz fixed point theorem. If $$\lambda _f \ne 0$$ then $f$ has a fixed point: $$f(a)=a.$$
One commonly known application... The atmosphere form one "thick" layer around the Earth and the air moves within this layer. Assuming that this layer is closed (does not gradually comes to nothing), what we have is a "thick" sphere. Its Euler characteristic is still $2$, thick or not. Now you can think of the wind as a continuous transformation of this solid. This map is homotopic to the identity map and, therefore, has Lefschetz number equal to the Euler characteristic, i.e., $2$. Hence there is a fixed point. With some extra work you can show that this implies that the vector field has a stationary point. At that point the wind speed is $0$, and not just the horizontal component. That point might, but does not have to, be a cyclone.
Generally, we study
- Fixed Point Problem: If $M$ is a manifold and $f:M\rightarrow M$ is a map, what can be said about the set of points $x\in M$ such that $f(x)=x$?
The Lefschetz number only tells us that this set is non-empty. The converse isn't true!
Another issue is whether we can remove the fixed points by a homotopy.
The proof is similar to that of the Euler-Poincare formula -- via manipulating the dimensions of all subspaces (cycles and boundaries) involved. Let's just copy it here for now.
A couple of facts from linear algebra:
- Fact 1. If $M, L$ are vector spaces and $A \colon M \rightarrow L$ is a linear operator, then
- Fact 2. If $Y$ is a subspace of vector space $X$, then
$$\dim X / Y = \dim X - \dim Y.$$
Now there are four vector spaces involved in the computation of the Betti numbers of $K$: the chain group, the cycle group, the boundary group, and the homology group. These are their dimensions: $$\begin{array}{l} c_k = \dim C_k(K),\\ z_k = \dim Z_k(K), \\ b_k = \dim B_k(K), \\ \beta_k = \dim H_k(K). \end{array}$$
Recall that for the boundary operator $$\partial_k \colon C_k(K) \rightarrow C_{k-1}(K), $$ we define $$Z_k(K) = \ker \partial_k,$$ and $$B_{k-1}(K) = \hspace{3pt} im \hspace{3pt} \partial_k.$$
Facts $1$ and $2$ imply as a corollary: $$\dim M - \dim \ker A = \dim im \hspace{3pt} A.$$
Applying this to the boundary operator above, we obtain: $$\dim C_k(K) - \dim ker \hspace{3pt} \partial_k = \dim im \hspace{3pt} \partial_k,$$ or:
Lemma 1. $$c_k - z_k = b_{k-1}.$$
Recall that $$H_k(K) = Z_k(K) / B_k(K).$$ Then, from Fact $2$, it follows
Lemma 2. $$\beta_k = z_k - b_k.$$
Finally, suppose $n$ is the highest dimension of a cell in $K$. Then $$\begin{array}{l} \beta_0 - \beta_1 + \beta_2 - \ldots + (-1)^k \beta_n, {\rm \hspace{3pt} substitute \hspace{3pt} from \hspace{3pt} Lemma \hspace{3pt}}2 \\ &= (z_0 - b_0) - (z_1 - b_1) + (z_2 - b_2) - \ldots + (-1)^n(z_n - b_n) \\ &= z_0 - b_0 - z_1 + b_1 + z_2 - b_2 - \ldots + (-1)^n z_n - (-1)^n b_n, {\rm \hspace{3pt} substitute \hspace{3pt} from \hspace{3pt} Lemma \hspace{3pt}} 1 \\ &= z_0 - (c_1 - z_1) - z_1 + (c_2 - z_2) + z_2 - (c_3 - z_3) - \ldots + (-1)^n z_n - (-1)^n(c_{n+1} - z_{n+1}), {\rm \hspace{3pt} cancel \hspace{3pt}} z's \\ &= z_0 - c_1 + c_2 - c_3 - ... + (-1)^{n+1}c_{n+1} - (-1)^n z_{n+1} \\ &= c_0 - c_1 + c_2 - c_3 - \ldots + 0 - 0 \\ &= \chi(K). {\rm \hspace{3pt} \blacksquare} \end{array}$$
