1. Introduction#
In 1926, Solomon Lefschetz Lef26 gave a formula that relates the number of fixed points of a map to the induced maps on homology.
Definition 1. Suppose \( X \) is a space such that all \( H _ k(X) \) are finitely generated abelian groups, and \( H _ k(X) = 0 \) for all sufficiently large \( X \). For a continuous map \( f \), define its Lefschetz number as \[ \displaystyle \begin{aligned} \Lambda(f) = \sum _ {k \ge 0}^{} (-1)^k \mathrm{Tr}( f _ {\ast k} : H _ k(X) \rightarrow H _ k(X) ). \end{aligned} \] Here, the trace of a endomorphism of a finitely generated abelian group is defined in the following way. First note that we may decompose \( H _ k(X) = \mathbb{Z}^{\oplus r} \oplus \mathrm{Tor} \), and \( f _ {\ast k} \vert _ {\mathrm{Tor}} : \mathrm{Tor} \rightarrow \mathrm{Tor} \). Then \( f _ {\ast k} \) induces a map \( \tilde{f} _ {\ast k} : \mathbb{Z}^{\oplus r} \rightarrow \mathbb{Z}^{\oplus r} \) by taking the quotient by \( \mathrm{Tor} \). The trace of this map is then well-defined.
Theorem 2 (Lefschetz fixed-point formula). Let \( K \) be a finite simplicial complex and let \( X = \lvert K \rvert \) be its realization. For each \( f : X \rightarrow X \) a continuous map with isolated fixed points, \[ \displaystyle \begin{aligned} \sum _ {f(x) = x}^{} I(f,x) = \Lambda(f), \end{aligned} \] where \( I(f,x) \) denotes the index of the fixed point.
For an illustration, we consider the simple case of \( X = [0,1] \) and take an arbitrary function \( f : [0,1] \rightarrow [0,1] \). Because \( X \) has the homotopy type of a point, the right hand side will always be \[ \displaystyle \begin{aligned} \Lambda(f) = \sum _ {k \ge 0}^{} (-1)^k \mathrm{Tr}( f _ {\ast k}) = \mathrm{Tr}( f _ {\ast 0} : \mathbb{Z} \rightarrow \mathbb{Z} ) + \sum _ {k \ge 1}^{} (-1)^k \mathrm{Tr}( f _ {\ast k} : 0 \rightarrow 0) = 1. \end{aligned} \] For a generic differentiable function \( f \), the graph of \( f \) will look more or less like Figure 1. At each fixed point, \( f \) either has derivative less than \( 1 \) or greater than \( 1 \), depending on the local behavior at the point. Note that under a deformation of \( f \), a fixed point with \( f^\prime > 1 \) and a fixed point with \( f^\prime < 1 \) may cancel out. As a consequence, the quantity \[ \displaystyle \begin{aligned} I(f) = \# \{ x : f(x) = x, f^\prime(x) < 1 \} - \# \{ x : f(x) = x, f^\prime(x) > 1 \} \end{aligned} \] is preserved under homotopy. Because \( I(\mathrm{const}) = 1 \) and every \( f \) is homotopic to the constant function, we see that \( I(f) = 1 \) for all \( f \). Hence if we define \( I(f,x) = 1 \) for \( f^\prime(x) < 1 \) and \( I(f,x) = -1 \) for \( f^\prime(x) > 1 \), then we verify Theorem 2 for "nice" functions \( f \).
Figure 1. A generic differentiable from the unit interval to itself
We remark that Theorem 2 is not in its full generality.
Theorem 3 (Dol95, Proposition VII.6.6). Let \( Y \) be an ENR (Euclidean neighborhood retract), i.e., there exists an open set \( V \subseteq \mathbb{R}^n \) and a retract \( X \) of \( V \) such that \( X \cong Y \). Let \( K \subseteq Y \) be a compact subset and \( f : Y \rightarrow K \subseteq Y \) be a continuous function. If \( f \) has isolated fixed points, then \( \sum _ {f(x) = x}^{} I(f, x) = \Lambda(f) \).
Every finite simplicial complex is indeed a ENR, and moreover, every finite CW complex is an ENR. Hence the Lefschetz fixed-point formula applies to a more general category of spaces. In this paper, we will only deal with finite simplicial complexes for simplicity. However, we note that compactness of the space is essential, because the map \( f : \mathbb{R} \rightarrow \mathbb{R}; ; x \mapsto x+1 \) has no fixed point but has \( \Lambda(f) = 1 \neq 0 \).
The goal of this exposition is first to make Theorem 2 precise, i.e., define the index, and then give a full proof of the formula.
2. The weak Lefschetz fixed-point theorem#
We shall first prove an easier version of Theorem 2 that does not require us to consider the index of a fixed point.
Theorem 4 (Weak Lefshetz fixed-point theorem). Let \( K \) be a finite simplicial complex, and let \( X = \lvert K \rvert \) be its realization. Suppose \( f : X \rightarrow X \) does not have any fixed points. Then \( \Lambda(f) = 0 \).
We are going to prove this by using simplicial approximation of \( f \). The idea is that because \( f _ \ast \) is invariant under homotopy, we may deform \( f \) to make it into a simplicial map that is close to \( f \).
Theorem 5 (Simplicial approximation theorem, Hat02, Theorem 2C.1). Let \( K \) and \( L \) be finite simplicial complexes. Let \( d _ L(,,,) \) be a metric on \( \lvert L \rvert \) that gives rise to the correct topology. Then for any continuous map \( f : \lvert K \rvert \rightarrow \lvert L \rvert \) and \( \epsilon > 0 \), there is a simplicial map \( g : S^M K \rightarrow S^N L \) with suitable \( M, N \ge 0 \) such that \( \lvert g \rvert : \lvert K \rvert \rightarrow \lvert L \rvert \) is homotopic to \( f \) and satisfies \( d _ L( f(x), \lvert g \rvert(x) ) < \epsilon \) for all \( x \in \lvert K \rvert \). (Here, \( S^M K \) denotes the \( M \)-th iterated barycentric subdivision of \( K \).)
We introduce the notion of the open star \( \mathrm{st}(\sigma) \) of a simplex \( \sigma \) in \( K \), \[ \displaystyle \begin{aligned} \mathrm{st} _ K(\sigma) = \bigcup _ {\sigma \subseteq \tau \in K} \mathrm{int}(\lvert \tau \rvert). \end{aligned} \]
Proof. } Consider the open cover of \( \lvert L \rvert \) by radius \( \epsilon / 2 \) balls with respect to \( d _ L \). For each simplex \( \sigma \in L \), there is an integer \( N _ \sigma \) such that each simplex of \( S^{N _ \sigma} \sigma \) lies in one of the balls of radius \( \epsilon / 2 \). Then each simplex of \( S^{N _ \sigma} \sigma \) has diameter less than \( \epsilon \). Let \( N = \max _ {\sigma \in L} N _ \sigma \). Then every simplex in \( S^N L \) has diameter less than \( \epsilon \), with respect to the metric \( d _ L \).
Take a metric \( d _ K \) on \( \lvert K \rvert \) that gives the right topology. Because every simplex has at least one vertex, the set \[ \displaystyle \begin{aligned} \{ \mathrm{st} _ {S^N L}(w) : w \text{ is a vertex of } S^N L \} \end{aligned} \] is an open cover of \( \lvert L \rvert \). Then \( \{ f^{-1}(\mathrm{st} _ {S^N L}(w)) : w \text{ vertex of } S^N L \} \) is an open cover of \( \lvert K \rvert \). By the Lebesgue number lemma, there exists a sufficiently small \( \delta > 0 \) such that every set \( S \subseteq \lvert K \rvert \) with diameter less than \( \delta \) with respect to \( d _ K \) is contained in one of the \( f^{-1}(\mathrm{st} _ {S^N L}(w)) \). Take a sufficiently large integer \( M \ge 0 \) such that each simplex of \( S^M K \) has diameter less than \( \delta / 2 \) in \( \lvert K \rvert \) with respect to \( d _ K \). For every \( v \) a vertex of \( S^M K \) and \( x, y \in \mathrm{clos}(\mathrm{st} _ {S^M K}(v)) \), \[ \displaystyle \begin{aligned} d(x,y) \le d(v,x) + d(v,y) < \delta / 2 + \delta / 2 = \delta. \end{aligned} \] Thus \( \mathrm{clos}(\mathrm{st} _ {S^M K}(v)) \) has diameter less than \( \delta \) for each vertex \( v \) of \( S^M K \). It follows that for each vertex \( v \) of \( S^M K \), there exists a vertex \( w \) of \( S^N L \) such that \( f(\mathrm{clos}(\mathrm{st} _ {S^M K}(v))) \subseteq \mathrm{st} _ {S^N L}(w) \).
Let \( g : S^M K \rightarrow S^N L \) be such that \( f(\mathrm{clos}(\mathrm{st} _ {S^M K}(v))) \subseteq \mathrm{st} _ {S^N L}(g(v)) \) for all vertices \( v \) of \( S^M K \). To check that this is a simplicial map, it suffices to show that if \( \sigma = [v _ 0, \ldots, v _ m] \) is a simplex in \( S^M K \), then \( g(\{v _ 0, \ldots, v _ m\}) \) is the vertices of a simplex in \( S^N L \). Since \[ \displaystyle \begin{aligned} \bigcap _ {i=0}^{m} \mathrm{st} _ {S^N L}(g(v _ i)) \supseteq \bigcap _ {i=0}^{m} f(\mathrm{clos}(\mathrm{st} _ {S^M K}(v))) \supseteq f(\lvert \sigma \rvert) \neq \emptyset, \end{aligned} \] by definition of \( \mathrm{st} \) there is at least one simplex \( \tau \in S^N L \) such that \( g(v _ 0), \ldots, g(v _ m) \) are vertices of \( \tau \). It follows that \( \{ g(v _ 0), \ldots, g(v _ m)\} \) is the set of vertices of some simplex in \( S^N L \). Hence \( g \) is a simplicial map.
We now claim that \( \lvert g \rvert \) is homotopic to \( f \). We shall explicitly construct the homotopy \( F \) between \( \lvert g \rvert \) and \( f \). For an arbitrary \( x \in \lvert K \rvert \), there exists a simplex \( \sigma \in S^M K \) such that \( x \in \lvert \sigma \rvert \). Write \( \sigma = [v _ 0, \ldots, v _ m] \). Because \( \sigma \subseteq \mathrm{clos}(\mathrm{st} _ {S^M K}(v _ i)) \) for every \( 0 \le i \le m \), \[ \displaystyle \begin{aligned} f(x) \in f(\lvert \sigma \rvert) \subseteq \bigcap _ {i=0}^{m} \mathrm{st} _ {S^N L}(g(v _ i)). \end{aligned} \] It follows that \( f(x) \) and \( \lvert g \rvert(x) \) are both in some simplex that has \( g(v _ 0), \ldots, g(v _ m) \) as vertices. Once \( f(x) \) and \( \lvert g \rvert(x) \) are in some same simplex, their affine linear combination \( \lambda f(x) + (1-\lambda) \lvert g \rvert(x) \) is unambiguously defined. Hence we may define \[ \displaystyle \begin{aligned} F _ \lambda(x) = \lambda f(x) + (1-\lambda) \lvert g \rvert(x). \end{aligned} \] Note that this does not depend on the choice of \( \sigma \). This implies that the maps \( F _ \lambda(x) \) glue well on the boundaries of the simplices, and hence \( F _ \lambda(x) : \lvert K \rvert \rightarrow \lvert L \rvert \) is continuous everywhere. Moreover, \( F _ 0 = \lvert g \rvert \) and \( F _ 1 = f \). This shows that \( f \) and \( \lvert g \rvert \) are homotopic.
Finally, the fact that \( f(x) \) and \( \lvert g \rvert(x) \) are in the same simplex implies that \( d _ L(f(x), \lvert g \rvert(x)) < \epsilon \), because every simplex in \( S^N L \) has diameter less than \( \epsilon \). ▨
Simplicial approximation allows us to use simplicial homology instead of singular homology. Since the chain groups of a finite simplicial complex are free abelian groups of finite rank, we may hope to express the traces of the maps on homology algebraically in terms of the maps on the chain groups.
Lemma 6. Let \( A \), \( B \), \( C \) be finitely generated abelian groups and let the following be a morphism of two short exact sequences.
Proof. Because \( \mathbb{Q} \) is a flat \( \mathbb{Z} \)-module, we may take the tensor with \( \mathbb{Q} \) and obtain a diagram of \( \mathbb{Q} \)-vector spaces.
Now we are dealing with finite dimensional vector spaces. Take a \( \mathbb{Q} \)-basis \( \{a _ 1, \ldots, a _ n\} \) of \( A \otimes \mathbb{Q} \), and let the matrix representation of \( \tilde{\alpha} \) with respect to \( a _ 1, \ldots, a _ n \) be \( M _ \alpha \in \mathcal{M} _ {n \times n}(\mathbb{Q}) \). Then clearly \( \mathrm{Tr} \tilde{\alpha} = \mathrm{Tr} M _ \alpha \). Let \( a _ 1, \ldots, a _ n, b _ 1, \ldots, b _ m \) be a \( \mathbb{Q} \)-basis of \( B \otimes \mathbb{Q} \). By exactness, \( j(b _ 1), \ldots, j(b _ m) \) is a \( \mathbb{Q} \)-basis for \( C \otimes \mathbb{Q} \). Let \( M _ \gamma \in \mathcal{M} _ {m \times m}(\mathbb{Q}) \) be the matrix representing \( \tilde{\gamma} \) with respect to \( j(b _ 1), \ldots, j(b _ m) \). By commutation of the diagram, the matrix representing \( \tilde{\beta} \) with respect to the basis \( a _ 1, \ldots, a _ n, b _ 1, \ldots, b _ m \) will take the form \[ \displaystyle \begin{aligned} M _ \beta = \begin{pmatrix} M _ \alpha & \displaystyle X \\ O & \displaystyle M _ \gamma \end{pmatrix} \in \mathcal{M} _ {(n+m) \times (n+m)}(\mathbb{Q}). \end{aligned} \] Hence \( \mathrm{Tr} \tilde{\beta} = \mathrm{Tr} M _ \beta = \mathrm{Tr} M _ \alpha + \mathrm{Tr} M _ \gamma = \mathrm{Tr} \tilde{\alpha} + \mathrm{Tr} \tilde{\gamma} \). ▨
Proposition 7 (Algebraic Lefschetz formula). Let \( (C _ {\bullet}, \partial) \) be a chain complex of finitely generated abelian groups, and assume that \( C _ n = 0 \) for all sufficiently large \( \lvert n \rvert \). For a chain map \( f : C _ {\bullet} \rightarrow C _ {\bullet} \), \[ \displaystyle \begin{aligned} \sum _ {k}^{} (-1)^k \mathrm{Tr} (f _ k : C _ k \rightarrow C _ k) = \sum _ {k}^{} (-1)^k \mathrm{Tr} (f _ {\ast k} : H _ k(C _ \bullet) \rightarrow H _ k(C _ \bullet) ). \end{aligned} \]
Proof. Note that we have short exact sequences \[ \displaystyle \begin{aligned} 0 \rightarrow \ker \partial _ k \rightarrow C _ k \rightarrow \mathrm{im} \partial _ k \rightarrow 0, \quad 0 \rightarrow \mathrm{im} \partial _ {k+1} \rightarrow \ker \partial _ k \rightarrow H _ k(C _ \bullet) \rightarrow 0, \end{aligned} \] that behaves well respect to the chain maps \( f \). By Lemma 6, we get \[ \displaystyle \begin{aligned} \mathrm{Tr} (f _ k : C _ k \rightarrow C _ k) & \displaystyle= \mathrm{Tr} (f _ k : \ker \partial _ k \rightarrow \ker \partial _ k) + \mathrm{Tr} (f _ {k-1} : \mathrm{im} \partial _ {k} \rightarrow \mathrm{im} \partial _ {k}), \\ \mathrm{Tr} (f _ k : \ker \partial _ k \rightarrow \ker \partial _ k) & \displaystyle= \mathrm{Tr} (f _ k : \mathrm{im} \partial _ {k+1} \rightarrow \mathrm{im} \partial _ {k+1}) + \mathrm{Tr} (f _ {\ast k} : H _ k(C _ \bullet) \rightarrow H _ k(C _ \bullet) ). \end{aligned} \] Adding the two equations times \( (-1)^k \) for all \( k \in \mathbb{Z} \), we get \[ \displaystyle \begin{aligned} \sum _ {k}^{} (-1)^k \mathrm{Tr} (f _ k : C _ k \rightarrow C _ k) & \displaystyle= \sum _ {k}^{} ((-1)^k + (-1)^{k-1}) \mathrm{Tr}(f _ {k-1} : \mathrm{im} \partial _ k \rightarrow \mathrm{im} \partial _ k) \\ & \displaystyle\quad + \sum _ {k} (-1)^k \mathrm{Tr}(f _ {\ast k} : H _ k(C _ \bullet) \rightarrow H _ k(C _ \bullet)) \\ & \displaystyle= \sum _ {k}^{} (-1)^k \mathrm{Tr}(f _ {\ast k} : H _ k(C _ \bullet) \rightarrow H _ k(C _ \bullet)). \end{aligned} \] ▨
We are now ready to prove Theorem 4.
Proof. } Let \( f : \lvert K \rvert \rightarrow \lvert K \rvert \) be a continuous map without fixed points. Let \( d _ K \) be a metric on \( \lvert K \rvert \) that is gives rise to the right topology. The map \( x \mapsto d _ K(x, f(x)) \) takes values in \( \mathbb{R} _ {>0} \). Because \( \lvert K \rvert \) is compact, the function attains a minimal value at some point. Hence there exists a sufficiently small \( \epsilon > 0 \) such that \( d _ K(x, f(x)) > \epsilon \) for all \( x \in \lvert K \rvert \).
Using the simplicial approximation theorem (Theorem 5), we find barycentric subdivisions \( S^M K \) and \( S^N K \) along with a simplicial map \( g : S^M K \rightarrow S^N K \) such that \( \lvert g \rvert \) is homotopic to \( f \) and \( d _ K(f(x), \lvert g \rvert(x)) < \epsilon / 3 \) for all \( x \in \lvert K \rvert \). Since we may take \( N \) to be as large as we want in the proof, we may assume that every simplex of \( S^N K \) have diameter less than \( \epsilon / 3 \). Again, once \( N \) is fixed, we may take \( M \) to be as large as we want in the proof. Hence without loss of generality, we can assume \( M \ge N \).
Because \( f \) is homotopic to \( \lvert g \rvert \), they induce the same maps on homology and thus their Lefschetz numbers \( \Lambda(f) \) and \( \Lambda(\lvert g \rvert) \) are equal. Moreover, because \( g : S^M K \rightarrow S^N K \) is a simplicial map, the map \( \lvert g \rvert _ {\ast k} : H _ k(\lvert K \rvert) \rightarrow H _ k(\lvert K \rvert) \) is identical to the map \( g _ {\ast k} : H _ k^{\mathrm{sim}}(S^M K) \rightarrow H _ k^{\mathrm{sim}}(S^N K) \) under the identification \( H _ k^{\mathrm{sim}}(S^M K) \cong H _ k^{\mathrm{sim}}(S^N K) \cong H _ k(\lvert K \rvert) \).
This can be formalized in the following way. For a simplicial complex \( K \), let us denote by \( (K) _ \bullet \) the corresponding simplicial chain complex. A simplicial map \( g : S^M K \rightarrow S^N K \) induces a chain map \( g : (S^M K) _ \bullet \rightarrow (S^N K) _ \bullet \), and this further induces a map \( g _ {\ast k} : H _ k^{\mathrm{sim}}(S^M K) \rightarrow H _ k^{\mathrm{sim}}(S^N K) \) on simplicial homology. We now define a new chain map \( s : (S^N K) _ \bullet \rightarrow (S^M K) _ \bullet \) given by \[ \displaystyle \begin{aligned} s : \sigma \mapsto S^{M-N} \sigma. \end{aligned} \] Note that this chain map induces isomorphisms \( s _ {\ast k} : H _ k^{\mathrm{sim}}(S^N K) \cong H _ k^{\mathrm{sim}}(S^M K) \). Hence the composition of the chain maps \( s \circ g : (S^M K) _ \bullet \rightarrow (S^M K) _ \bullet \) give maps on homology \( (s \circ g) _ {\ast k} : H _ k^{\mathrm{sim}}(S^M K) \rightarrow H _ k^{\mathrm{sim}}(S^M K) \) that are equal to \( \lvert g \rvert _ {\ast k} : H _ k(\lvert K \rvert) \rightarrow H _ k(\lvert K \rvert) \) under the natural isomorphism \( H _ k^{\mathrm{sim}}(S^M K) \cong H _ k(\lvert K \rvert) \). Therefore \[ \displaystyle \begin{aligned} \Lambda(\lvert g \rvert) & \displaystyle= \sum _ {k \ge 0}^{} (-1)^k \mathrm{Tr}(\lvert g \rvert _ {\ast k} : H _ k(\lvert K \rvert) \rightarrow H _ k(\lvert K \rvert)) \\ & \displaystyle= \sum _ {k \ge 0}^{} (-1)^k \mathrm{Tr}((s \circ g) _ {\ast k} : H _ k^{\mathrm{sim}}(S^M K) \rightarrow H _ k^{\mathrm{sim}}(S^M K)) \\ & \displaystyle= \sum _ {k \ge 0}^{} (-1)^k \mathrm{Tr}((s \circ g) _ k : (S^M K) _ k \rightarrow (S^M K) _ k) \end{aligned} \] by Lemma 7.
Recall that \( (S^M K) _ k \) is by definition the free abelian group generated by the \( k \)-dimensional simplices of \( S^M K \). Let \( \{\sigma _ 1, \ldots, \sigma _ t\} \) be the set of \( k \)-simplices of \( S^M K \). Let us write \[ \displaystyle \begin{aligned} (s \circ g) _ k (\sigma _ i) = \sum _ {j = 1}^{t} a _ {ij} \sigma _ j \end{aligned} \] for integers \( a _ {ij} \in \mathbb{Z} \). Then the trace of \( (s \circ g) _ k \) can be computed as \( \mathrm{Tr}((s \circ g) _ k : (S^M K) _ k \rightarrow (S^M K) _ k) = \sum _ {i=1}^{t} a _ {ii} \).
We claim that \( a _ {ii} = 0 \) for each \( 1 \le i \le t \). This is equivalent to saying that \( \lvert g(\sigma _ i) \rvert \subseteq \lvert K \rvert \) does not contain \( \lvert \sigma _ i \rvert \subseteq \lvert K \rvert \) as sets. From the way we defined \( g \), \( N \), and \( M \), we obtain for every \( x \in \lvert K \rvert \), \[ \displaystyle \begin{aligned} d _ K(x, \lvert g \rvert(x)) \ge d _ K(x, f(x)) - d _ K(f(x), \lvert g \rvert(x)) > \epsilon - \epsilon/3 = 2\epsilon / 3. \end{aligned} \] Suppose that \( \lvert \sigma _ i \rvert \subseteq \lvert g(\sigma _ i) \rvert \) for some \( i \). Let \( v \) be a vertex of \( \sigma _ i \). Then \( v \in \lvert \sigma _ i \rvert \subseteq \lvert g(\sigma _ i) \rvert \) and also \( g(v) \in \lvert g(\sigma _ i) \rvert \). Because \( g(\sigma _ i) \) is a simplex of \( S^M K \), the diameter of \( \lvert g(\sigma _ i) \rvert \) is less than \( \epsilon / 3 \). This implies that \( d _ K(v, g(v)) < \epsilon / 3 \). This contradicts the inequality \( d _ K(x, \lvert g \rvert(x)) > 2\epsilon / 3 \) for all \( x \in \lvert K \rvert \). Hence \( a _ {ii} = 0 \) for all \( 1 \le i \le t \). As a consequence, \[ \displaystyle \begin{aligned} \mathrm{Tr}((s \circ g) _ k : (S^M K) _ k \rightarrow (S^M K) _ k) = \sum _ {i=1}^{t} a _ {ii} = 0 \end{aligned} \] for all \( k \ge 0 \). Therefore \[ \displaystyle \begin{aligned} \Lambda(f) = \Lambda(\lvert g \rvert) = \sum _ {k \ge 0}^{} (-1)^k \mathrm{Tr}((s \circ g) _ k : (S^M K) _ k \rightarrow (S^M K) _ k) = 0. \end{aligned} \] That is, the Lefschetz number of \( f \) is zero. ▨
3. Index of a fixed point#
Recall the example of the function \( f : [0,1] \rightarrow [0,1] \). When \( x \in X \) is an isolated fixed point of \( f : X \rightarrow X \) for a sufficiently nice topological space \( X \), the index \( i(f, x) \) is completely determined by the local behavior of \( f \). We shall roughly follow the exposition given in section VII.5 of Dol95.
Definition 8. Let \( V \subseteq \mathbb{R}^n \) be an open set containing the origin. Consider a continuous map \( f : V \rightarrow \mathbb{R}^n \) with \( f(0) = 0 \) and assume that \( 0 \) is the only fixed point of \( V \). Then \( \mathrm{id} _ V - f : (V, V \setminus 0) \rightarrow (\mathbb{R}^n, \mathbb{R}^n \setminus 0) \) is a map of pairs, and \( H _ n(V, V \setminus 0) \cong H _ n(\mathbb{R}^n, \mathbb{R}^n \setminus 0) \cong \mathbb{Z} \) by excision. Define the index \( I(f, 0) \) of \( f \) at \( 0 \) by \[ \displaystyle \begin{aligned} (\mathrm{id} _ V - f) _ {\ast n} : H _ n(V, V \setminus 0) \cong H _ n(\mathbb{R}^n, \mathbb{R}^n \setminus 0) \xrightarrow{\times I(f,0)} H _ n(\mathbb{R}^n, \mathbb{R}^n \setminus 0). \end{aligned} \]
Proposition 9 (Dol95, Proposition VII.5.9). Let \( V _ 1 \subseteq \mathbb{R}^{n _ 1} \) and \( V _ 2 \subseteq \mathbb{R}^{n _ 2} \) be open sets, each containing the origins. Let \( f _ 1 : V _ 1 \rightarrow \mathbb{R}^{n _ 2} \) and \( f _ 2 : V _ 2 \rightarrow \mathbb{R}^{n _ 1} \) be continuous maps with \( f _ 1(0) = 0 \), \( f _ 2(0) = 0 \). Assume that \( f _ 2 \circ f _ 1 : f _ 1^{-1}(V _ 2) \rightarrow \mathbb{R}^{n _ 1} \) has a unique fixed point \( 0 \). Then \( f _ 1 \circ f _ 2 : f _ 2^{-1}(V _ 1) \rightarrow \mathbb{R}^{n _ 2} \) also has a unique fixed point \( 0 \). Moreover, \( I(f _ 1 \circ f _ 2, 0) = I(f _ 2 \circ f _ 1, 0) \).
Proof. Because \( f _ 1(0) = 0 \) and \( f _ 2(0) = 0 \), it is clear that \( 0 \in \mathbb{R}^{n _ 1} \) is a fixed point of \( f _ 2 \circ f _ 1 \) and \( 0 \in \mathbb{R}^{n _ 2} \) is a fixed point of \( f _ 1 \circ f _ 2 \). Let \( x \) be a fixed point of \( f _ 1 \circ f _ 2 \), i.e., \( f _ 1 \circ f _ 2(x) = x \). Then \( f _ 2 \circ f _ 1 \circ f _ 2(x) = f _ 2(x) \), and by the uniqueness of the fixed point of \( f _ 2 \circ f _ 1 \), we obtain \( f _ 2(x) = 0 \). Then \( x = f _ 1 \circ f _ 2(x) = f _ 1(0) = 0 \).
This shows that \( f _ 1 \circ f _ 2 \) also has a unique fixed point, \( 0 \). This shows that \( I(f _ 1 \circ f _ 2, 0) \) and \( I(f _ 2 \circ f _ 1, 0) \) are well-defined. Let us now show that \( I(f _ 1 \circ f _ 2, 0) = I(f _ 2 \circ f _ 1, 0) \). Define a new map \[ \displaystyle \begin{aligned} g : V _ 1 \times V _ 2 \rightarrow \mathbb{R}^{n _ 1 + n _ 2}; \quad (x _ 1, x _ 2) \mapsto (f _ 2(x _ 2), f _ 1(x _ 1)). \end{aligned} \] Because \( (x _ 1, x _ 2) = (f _ 2(x _ 2), f _ 1(x _ 1)) \) implies that \( x _ 1 \) is a fixed point of \( f _ 2 \circ f _ 1 \) and \( x _ 2 \) is a fixed point of \( f _ 1 \circ f _ 2 \), we see that \( g \) also has a unique fixed point \( 0 \in \mathbb{R}^{n _ 1 + n _ 2} \).
Let us compute \( I(g, 0) \). Consider a homotopy \( A _ t \) given by \[ \displaystyle \begin{aligned} A _ t : (x _ 1, x _ 2) \mapsto ( (1-t) f _ 2(x _ 2) + t f _ 2 \circ f _ 1(x _ 1), f _ 1(x _ 1)). \end{aligned} \] We note that for each \( t \), \( A _ t \) has unique fixed point \( 0 \). Thus \( \mathrm{id} - A _ t : (V, V \setminus 0) \rightarrow (\mathbb{R}^n, \mathbb{R}^n \setminus 0) \) for all \( t \). This shows that \( I(g, 0) = I(A _ 0, 0) = I(A _ 1, 0) \), because homotopic maps induce the same maps on homology. Consider now the homotopy \( B _ t \) given by \[ \displaystyle \begin{aligned} B _ t : (x _ 1, x _ 2) \mapsto (f _ 2 \circ f _ 1(x _ 1), (1-t) f _ 1(x _ 1)). \end{aligned} \] Again, every \( B _ t \) has unique fixed point \( 0 \) and hence \( I(A _ 1, 0) = I(B _ 0, 0) = I(B _ 1, 0) \). So computing \( I(g, 0) \) reduces to computing \( I(B _ 1, 0) \).
The map \( \mathrm{id} - B _ 1 \) is given explicitly by \[ \displaystyle \begin{aligned} \mathrm{id} - B _ 1 : (x _ 1, x _ 2) \mapsto (x _ 1 - f _ 2 \circ f _ 1(x _ 1), x _ 2). \end{aligned} \] That is, it can be considered as the product of the two maps \( x _ 1 \mapsto x _ 1 - f _ 2 \circ f _ 1(x _ 1) \) and \( x _ 2 \mapsto x _ 2 \). By the K"unneth formula, we have the diagram
Proposition 10. Let \( X \) be a topological space and let \( x _ 0 \in X \) be a fixed point of a continuous map \( f : X \rightarrow X \). Suppose that \( x _ 0 \) is an isolated fixed point. Let \( U \) be a neighborhood of \( x _ 0 \) such that \( x _ 0 \) is the only fixed point of \( f \) in \( U \). Suppose there exists an embedding \( i : U \hookrightarrow V \subseteq \mathbb{R}^n \) where \( V \) is an open set of \( \mathbb{R}^n \) and \( i(x _ 0) = 0 \), along with a retract \( r : V \rightarrow U \). Then the map \( i \circ f \circ r : r^{-1}(f^{-1}(U) \cap U) \rightarrow \mathbb{R}^n \) has a unique fixed point \( 0 \). Moreover, the index \( I(i \circ f \circ r, 0) \) is independent of the choice of \( U \), \( V \), \( i \), and \( r \), i.e., depends only on \( f \) and \( x _ 0 \).
Proof. Suppose that \( i \circ f \circ r(x) = x \) for some \( x \in r^{-1}(f^{-1}(U) \cap U) \). Then \( r \circ i = \mathrm{id} \) implies that \( r(x) = r \circ i \circ f \circ r(x) = f \circ r(x) \). Thus \( r(x) \in U \) is a fixed point of \( f \). It follows that \( r(x) = x _ 0 \) and thus \( x = i \circ f(x _ 0) = i(x _ 0) = 0 \). Hence \( 0 \) is the unique fixed point of \( i \circ f \circ r \).
Now consider two such choices of \( U, V, i, r \). Let \( i _ j : U _ j \rightarrow V _ j \subseteq \mathbb{R}^{n _ j} \) be an embedding and \( r _ j : V _ j \rightarrow U _ j \) a retract. Since all the maps \( i _ j, r _ j, f \) are continuous, the compositions of those maps are defined on some open neighborhood of either \( x _ 0 \in X \), \( 0 \in \mathbb{R}^{n _ 1} \), or \( 0 \in \mathbb{R}^{n _ 2} \). Since the index \( I \) only depends on the local behavior of a function near the fixed point, we may compose the maps as we want, as long as they make sense.
From Proposition 9 we obtain \[ \displaystyle \begin{aligned} I(i _ 1 \circ f \circ r _ 1, 0) & \displaystyle= I(i _ 1 \circ f \circ (r _ 2 \circ i _ 2) \circ r _ 1, 0) = I((i _ 1 \circ f \circ r _ 2) \circ (i _ 2 \circ r _ 1), 0) \\ & \displaystyle= I((i _ 2 \circ r _ 1) \circ (i _ 1 \circ f \circ r _ 2), 0) = I(i _ 2 \circ f \circ r _ 2, 0). \end{aligned} \] Thus the index \( I(i \circ f \circ r, 0) \) only depends on \( f \) and \( x \). ▨
This independence allows us to define the index of a fixed point generally.
Definition 11. Let \( X \) be a topological space and let \( x _ 0 \in X \) be an isolated fixed point of a continuous map \( f : X \rightarrow X \). Assume that there exists a neighborhood of \( x _ 0 \) that is homeomorphic to a retract of an open subset of some Euclidean space. Then we define the index \( I(f, x _ 0) \) of \( f \) at \( x _ 0 \) to be \( I(i \circ f \circ r, 0) \) as in Proposition 10.
Note that if \( X = V \subseteq \mathbb{R}^n \) is an open set and \( 0 \) is the unique fixed point of \( f : V \rightarrow \mathbb{R} \), then the two definitions of \( I(f,0) \) agree. This is because we may choose \( i = r = \mathrm{id} \) to compute \( I(f, 0) \) for the second definition.
In the case when \( X \) is a finite simplicial complex, every point has a neighborhood that is a retract of an open subset of a Euclidean space. Hence if \( X \) is a finite simplicial complex, then the index of an isolated fixed point is always well-defined. More generally, this is true if \( X \) is an ENR (Euclidean neighborhood retract).
Example 1. Consider \( X = S^1 \cong [0,1] / \{0,1\} \). Take \( f : X \rightarrow X; ; x \mapsto x^2 \). Then \( f \) has one fixed point \( x = 0 = 1 \), since \( 0 < x^2 < x < 1 \) for \( 0 < x < 1 \). Let us compute the index at \( x _ 0 = 0 \). Using the identification \( S^1 \cong \mathbb{R} / \mathbb{Z} \), we may write \[ \displaystyle \begin{aligned} f(x) = \begin{cases} (x+1)^2 - 1 = 2x - x^2 & \displaystyle \text{if } -0.1 < x \le 0 \\ x^2 & \displaystyle \text{if } 0 \le x < 0.1. \end{cases} \end{aligned} \] Then we can use \( V = (-0.1, 0.1) \subseteq \mathbb{R} \) and \( i = r = \mathrm{id} \).
The map \( \mathrm{id} - f : V \rightarrow \mathbb{R} \) is given by \[ \displaystyle \begin{aligned} (\mathrm{id} - f)(x) = \begin{cases} - x + x^2 \ge 0 & \displaystyle \text{if } -0.1 < x \le 0 \\ x - x^2 \ge 0 & \displaystyle \text{if } 0 \le x < 0.1. \end{cases} \end{aligned} \] Thus \( (\mathrm{id} - f)(x) \ge 0 \) for \( x \in (-0.1, 0.1) \) with equality if and only if \( x = 0 \). It follows that \( f : (V, V \setminus 0) \rightarrow (\mathbb{R}, \mathbb{R} \setminus 0) \) factors through \( (\mathbb{R} _ {\ge 0}, \mathbb{R} _ {> 0}) \) which has \( H _ 1(\mathbb{R} _ {\ge 0}, \mathbb{R} _ {> 0}) = 0 \). It follows that \( I(f, 0) = 0 \).
Example 2. Now let us look at the space \( X = S^2 \cong \mathbb{P} _ {\mathbb{C}}^1 \). Consider the map \( f : x \mapsto x / (x+1) \). If \( x = \infty \), then \( f(\infty) = 1 \), and if \( x \in \mathbb{C} \), then \( f(x) = x / (x+1) = x \) if and only if \( x = 0 \). Thus \( x _ 0 = 0 \in \mathbb{C} \) is the only fixed point of \( f \). Again, let us compute \( I(f, 0) \). We have a natural identification \( \mathbb{C} \cong \mathbb{R}^2 \). Let us use \( V = \Delta(0, 0.5) \), the disc of radius \( 0.5 \) centered at \( 0 \). The map \( \mathrm{id} - f \) is described as \[ \displaystyle \begin{aligned} (\mathrm{id} - f)(z) = z - \frac{z}{z+1} = \frac{z^2}{z+1}. \end{aligned} \] As \( z \) rotates around the origin with the parametrization \( z = 0.3 e^{2 \pi i \theta} \) for \( 0 \le \theta \le 1 \), the image \( z^2 / (z+1) \) goes around the origin twice, i.e., the winding number of the curve \( 0.09 e^{4 \pi i \theta} / (1 + 0.3 e^{2 \pi i \theta}) \) for \( 0 \le \theta \le 1 \) is \( 2 \). This shows that \( I(f, 0) = 2 \).
Theorem 12. For \( j = 1, 2 \), let \( X _ j \) be a space with \( x _ j \in X _ j \) an isolated fixed point of \( f _ j : X _ j \rightarrow X _ j \), and suppose there exists a neighborhood of \( x _ j \in X _ j \) a retract of an open set in Euclidean space. Let \( X _ 3 = (X _ 1 \amalg X _ 2) / \{x _ 1, x _ 2\} \), and define \( f _ 3 : X _ 3 \rightarrow X _ 3 \) as \( f(x) = f _ 1(x) \) if \( x \in X _ 1 \) and \( f(x) = f _ 2(x) \) if \( x \in X _ 2 \). Write \( x _ 3 = [\{x _ 1, x _ 2\}] \in X _ 3 \). Then \( I(f _ 3, x _ 3) = I(f _ 1, x _ 1) + I(f _ 2, x _ 2) - 1 \).
Proof. Let \( x _ j \in U _ j \) be a neighborhood with an embedding \( i _ j : U _ j \rightarrow V _ j \subseteq \mathbb{R}^{n _ j} \) and a retract \( r _ j : V _ j \rightarrow U _ j \). Then \( I(f _ j, x _ j) \) is defined as \( I(f _ j, x _ j) = I(i _ j \circ f _ j \circ r _ j, 0) \). Let us write \( n = n _ 1 + n _ 2 \). Note that we may assume \( n _ 1, n _ 2 \ge 2 \), since we may embed in a space large as we want.
Using these, we can construct an embedding and a retract for a neighborhood of \( x _ 3 \in X _ 3 \). Consider the retract \[ \displaystyle \begin{aligned} R : \mathbb{R}^{n} \rightarrow (\mathbb{R}^{n _ 1} \times 0) \cup (0 \times \mathbb{R}^{n _ 2}); \quad (\vec{v} _ 1, \vec{v} _ 2) \mapsto \begin{cases} (\vec{v} _ 1, 0) \cdot \frac{\lvert \vec{v} _ 1 \rvert - \lvert \vec{v} _ 2 \rvert}{\lvert \vec{v} _ 1 \rvert} & \displaystyle \lvert \vec{v} _ 1 \rvert \ge \lvert \vec{v} _ 2 \rvert, \\ (0, \vec{v} _ 2) \cdot \frac{\lvert \vec{v} _ 2 \rvert - \lvert \vec{v} _ 1 \rvert}{\lvert \vec{v} _ 2 \rvert} & \displaystyle \lvert \vec{v} _ 1 \rvert \le \lvert \vec{v} _ 2 \rvert. \end{cases} \end{aligned} \] Let \( U _ 3 = U _ 1 \cup U _ 2 \subseteq X _ 3 \) and define \[ \displaystyle \begin{aligned} i _ 3 : U _ 3 \rightarrow (V _ 1 \times 0) \cup (0 \times V _ 2); \quad x \in U _ 1 \mapsto (i _ 1(x), 0), ; x \in U _ 2 \mapsto (0, i _ 2(x)), \end{aligned} \] \( V _ 3 = R^{-1}((V _ 1 \times 0) \cup (0 \times V _ 2)) \), and \[ \displaystyle \begin{aligned} r _ 3 : V _ 3 \rightarrow U _ 3; \quad (\vec{v} _ 1, \vec{v} _ 2) \mapsto \begin{cases} r _ 1(\vec{v} _ 1 \cdot \frac{\lvert \vec{v} _ 1 \rvert - \lvert \vec{v} _ 2 \rvert}{\lvert \vec{v} _ 1 \rvert}) \in U _ 1 & \displaystyle \lvert \vec{v} _ 1 \rvert \ge \lvert \vec{v} _ 2 \rvert, \\ r _ 2(\vec{v} _ 2 \cdot \frac{\lvert \vec{v} _ 2 \rvert - \lvert \vec{v} _ 1 \rvert}{\lvert \vec{v} _ 2 \rvert}) \in U _ 2 & \displaystyle \lvert \vec{v} _ 1 \rvert \le \lvert \vec{v} _ 2 \rvert. \end{cases} \end{aligned} \] It is clear that \( i _ 3 \) is an embedding and \( r _ 3 \) is a retract so that \( r _ 3 \circ i _ 3 = \mathrm{id} \). Also we have \[ \displaystyle \begin{aligned} i _ 3 \circ f _ 3 \circ r _ 3 : (\vec{v} _ 1, \vec{v} _ 2) \mapsto \begin{cases} (i _ 1 \circ f _ 1 \circ r _ 1(\vec{v} _ 1 \cdot \frac{\lvert \vec{v} _ 1 \rvert - \lvert \vec{v} _ 2 \rvert}{\lvert \vec{v} _ 1 \rvert}), 0) & \displaystyle \lvert \vec{v} _ 1 \rvert \ge \lvert \vec{v} _ 2 \rvert, \\ (0, i _ 2 \circ f _ 2 \circ r _ 2(\vec{v} _ 2 \cdot \frac{\lvert \vec{v} _ 2 \rvert - \lvert \vec{v} _ 1 \rvert}{\lvert \vec{v} _ 2 \rvert}) ) & \displaystyle \lvert \vec{v} _ 1 \rvert \le \lvert \vec{v} _ 2 \rvert. \end{cases} \end{aligned} \] From this explicit formula, it is clear that \( (0,0) \) is the only fixed point of \( i _ 3 \circ f _ 3 \circ r _ 3 \). For convenience, let us write \( g _ j = i _ j \circ f _ j \circ r _ j \). Our goal is to show \( I(g _ 3, 0) = I(g _ 1, 0) + I(g _ 2, 0) - 1 \).
Note that \( \mathrm{id} - g _ 3 \) can also be written as \[ \displaystyle \begin{aligned} \mathrm{id} - g _ 3 : (\vec{v} _ 1, \vec{v} _ 2) \mapsto \begin{cases} ((\mathrm{id} - g _ 1) (\vec{v} _ 1 \cdot \frac{\lvert \vec{v} _ 1 \rvert - \lvert \vec{v} _ 2 \rvert}{\lvert \vec{v} _ 1 \rvert}) + \frac{\lvert \vec{v} _ 2 \rvert}{\lvert \vec{v} _ 1 \rvert} \cdot \vec{v} _ 1, \vec{v} _ 2) & \displaystyle \lvert \vec{v} _ 1 \rvert \ge \lvert \vec{v} _ 2 \rvert, \\ (\vec{v} _ 1, (\mathrm{id} - g _ 2) (\vec{v} _ 2 \cdot \frac{\lvert \vec{v} _ 2 \rvert - \lvert \vec{v} _ 1 \rvert}{\lvert \vec{v} _ 2 \rvert}) + \frac{\lvert \vec{v} _ 1 \rvert}{\lvert \vec{v} _ 2 \rvert} \cdot \vec{v} _ 2) & \displaystyle \lvert \vec{v} _ 1 \rvert \le \lvert \vec{v} _ 2 \rvert. \end{cases} \end{aligned} \] Because we are only interested in the local behavior of \( \mathrm{id} - g _ 3 \) near the origin, we may scale the whole map and assume that \( B^{n _ 1} \times B^{n _ 2} \) is in the domain, where \( B^n \subseteq \mathbb{R}^n \) is the closed unit ball. Note that there is a deformation retract \( r : \mathbb{R}^{n} \setminus 0 \rightarrow \partial(B^{n _ 1} \times B^{n _ 2}) \) given by multiplying an appropriate positive scalar. Then \( I(g _ 3, 0) \) is the map \( (r \circ (\mathrm{id} - g _ 3)) _ {\ast (n - 1)} : H _ {n - 1}(\partial(B^{n _ 1} \times B^{n _ 2})) \rightarrow H _ {n - 1}(\partial(B^{n _ 1} \times B^{n _ 2})) \).
We note that \( (r \circ (\mathrm{id} - g _ 3)) \vert _ {S^{n _ 1-1} \times S^{n _ 2-1}} = \mathrm{id} \). Hence we get a sequence of pairs arising from the map \( (r \circ (\mathrm{id} - g _ 3)) : (\partial(B^{n _ 1} \times B^{n _ 2}), S^{n _ 1 - 1} \times S^{n _ 2 - 1}) \rightarrow (\partial(B^{n _ 1} \times B^{n _ 2}), S^{n _ 1-1} \times S^{n _ 2-1}) \).
From the exact sequence, we see that \( H _ {n-1}(\partial(B^{n _ 1} \times B^{n _ 2}), S^{n _ 1 - 1} \times S^{n _ 2-1}) \cong \mathbb{Z}^2 \), and the basis elements must be the fundamental classes of \( B^{n _ 1} \times S^{n _ 2-1} \) and \( S^{n _ 1-1} \times B^{n _ 2} \). Because \( (r \circ (\mathrm{id} - g _ 3)) \vert _ {B^{n _ 1} \times S^{n _ 2-1}} \) only depends on \( g _ 2 \) and not on \( g _ 1 \), the diagonal entry on the \( [B^{n _ 1} \times S^{n _ 2-1}] \) depends only on \( g _ 2 \). Likewise, the diagonal entry on the \( [S^{n _ 1-1} \times B^{n _ 2}] \) only depends on \( g _ 1 \). Thus \( I(g _ 3, 0) \) is of the form \( (\text{sth depending only on } g _ 1) + (\text{sth depending only on } g _ 2) \). To actually compute it, we go back to the original setting.
Because \( g _ 1 \) is determined (up to some choice) by \( f _ 1 \) and \( g _ 2 \) is determined by \( f _ 2 \), we see that \( I(f _ 3, x _ 3) \) is the sum of a function of \( f _ 1 \) and a function of \( f _ 2 \). Let us write \( f _ 3 = f _ 1 \vee f _ 2 \) and \( I(f _ 1 \vee f _ 2, x _ 3) = A(f _ 1) + B(f _ 2) \). Let \( f _ 1^\prime : X _ 1 \rightarrow x _ 1 \) be the constant map, and let \( f _ 2^\prime : X _ 2 \rightarrow x _ 2 \) be the constant map. We have \[ \displaystyle \begin{aligned} I(f _ 1 \vee f _ 2, x _ 3) & \displaystyle= (A(f _ 1) + B(f _ 2^\prime)) + (A(f _ 1^\prime) + B(f _ 2)) - (A(f _ 1^\prime) + B(f _ 2^\prime)) \\ & \displaystyle= I(f _ 1 \vee f _ 2^\prime, x _ 3) + I(f _ 1^\prime \vee f _ 2, x _ 3) - I(f _ 1^\prime \vee f _ 2^\prime, x _ 3). \end{aligned} \] Because \( f _ 1 \vee f _ 2^\prime : X _ 3 = X _ 1 \vee X _ 2 \rightarrow X _ 1 \) is a composition of \( f _ 1 \) and the retract \( \mathrm{id} _ {X _ 1} \vee X _ 2 \), the indices of \( x _ 1 = x _ 2 = x _ 3 \) are the same for \( f _ 1 \vee f _ 2^\prime \) and \( f _ 1 \). Hence \( I(f _ 1 \vee f _ 2^\prime, x _ 3) = I(f _ 1, x _ 1) \). Likewise \( I(f _ 1^\prime \vee f _ 2, x _ 3) = I(f _ 2, x _ 2) \) and \( I(f _ 1^\prime \vee f _ 2^\prime, x _ 3) = I(f _ 1^\prime, x _ 1) = 1 \). Therefore \( I(f _ 3, x _ 3) = I(f _ 1, x _ 1) + I(f _ 2, x _ 2) - 1 \). ▨
4. The Lefschetz fixed-point formula#
As an intermediate step, we are first going to show that if all fixed points of \( X \) has index \( 0 \), then \( \Lambda(f) = 0 \).
Theorem 13. Let \( K \) be a finite simplicial complex and let \( X = \lvert K \rvert \) be its realization. For a continuous map \( f : X \rightarrow X \) with isolated fixed points, if \( I(f,x) = 0 \) for every fixed point \( x \in X \), then \( \Lambda(f) = 0 \).
Lemma 14. Let \( B^n = \{ x \in \mathbb{R}^n : \lvert x \rvert \le 1 \} \) be the closed ball, and let \( f : B^n \rightarrow r \cdot B^n \) be a continuous map such that \( f(0) = 0 \) is the only fixed point of \( f \). If \( I(f,0) = 0 \), then there exists a homotopy \( F _ t : B^n \rightarrow (2+r) B^n \) such that \( f = F _ 0 \) and \( f(x) = F _ t(x) \) if \( \lvert x \rvert = 1 \) and \( F _ 1 \) has no fixed points.
Proof. Let us explicitly construct such a homotopy \( F _ t \). First we define \[ \displaystyle \begin{aligned} A _ t : B^n \rightarrow r B^n; \quad x \mapsto \begin{cases} \lvert x \rvert f(x / \lvert x \rvert) & \displaystyle \lvert x \rvert \ge 1-t \\ (1-t) f(x / (1-t)) & \displaystyle \lvert x \rvert \le 1-t. \end{cases} \end{aligned} \] This gives a homotopy between \( A _ 0 = f \) and \( A _ 1 \).
Since \( I(f,0) = 0 \), we have \( (\mathrm{id} - f) : (B^n, B^n \setminus 0) \rightarrow (\mathbb{R}^{n}, \mathbb{R}^n \setminus 0) \) inducing \( (\mathrm{id} - f) _ {\ast n} = 0 \). Since we have isomorphisms \( H _ n(B^n, B^n \setminus 0) \cong H _ {n-1}(B^n \setminus 0) \cong H _ {n-1}(S^{n-1}) \) for \( 0 < t \le 1 \), we see that the map \[ \displaystyle \begin{aligned} g : S^{n-1} \rightarrow (1 + r) B^n \setminus 0; \quad x \mapsto (\mathrm{id} - f)(x) \end{aligned} \] has degree \( 0 \). Furthermore, because \( (1 + r) B^n \setminus 0 \) has homotopy type \( S^{n-1} \), we have \( \pi _ {n-1}((1+r) B^n \setminus 0) \cong \mathbb{Z} \). That is, the degree of the map determines the homotopy class. This shows that there is a homotopy \( G _ t : S^{n-1} \rightarrow (1+r) B^n \setminus 0 \) such that \( G _ 0 = g \) and \( G _ {1/2}(x) = e _ 1 \) is the constant map, where \( e _ 1 = (1, 0, \ldots 0) \in \mathbb{R}^n \). Let \( G _ t(x) = e _ 1 \) for all \( 1/2 \le t \le 1 \).
Using this homotopy, we now construct \[ \displaystyle \begin{aligned} B _ t : B^n \rightarrow (2+r) B^n; \quad x \mapsto \begin{cases} \lvert x \rvert f(x / \lvert x \rvert) & \displaystyle \lvert x \rvert \ge t \\ \lvert x \rvert (x / \lvert x \rvert - G _ {t - \lvert x \rvert}(x / \lvert x \rvert)) & \displaystyle \lvert x \rvert \le t. \end{cases} \end{aligned} \] This gives a homotopy between \( B _ 0 = A _ 1 \) and \( B _ 1 \).
Again construct a homotopy for \( 0 \le t \le 1/2 \), \[ \displaystyle \begin{aligned} C _ t : B^n \rightarrow (2+r) B^n; \quad x \mapsto \begin{cases} x - \lvert x \rvert G _ {1 - \lvert x \rvert}(x / \lvert x \rvert) & \displaystyle \lvert x \rvert \ge t, \\ x - t e _ 1 & \displaystyle \lvert x \rvert \le t. \end{cases} \end{aligned} \] This is indeed a homotopy between \( C _ 0 = B _ 1 \) and \( C _ {1/2} \) because \( G _ t(x) = e _ 1 \) for \( 1/2 \le t \le 1 \). Moreover, \( C _ {1/2} \) does not have fixed points because \( G _ {1 - \lvert x \rvert}(x / \lvert x \rvert) \neq 0 \) and \( x - e _ 1 / 2 \neq x \). Thus concatenating \( A, B, C \) give the homotopy \( F \). ▨
Proof of Theorem 13. Let \( x _ 1, \ldots, x _ m \) be the fixed points of \( f : X \rightarrow X \). Because \( X \) is the realization of a finite simplicial complex, for each \( j \) we may choose sufficiently small neighborhoods \( U _ j \) of \( x _ j \), along with an embedding \( i _ j : U _ j \hookrightarrow \mathrm{int}(B^{n _ j}) \) and a deformation retract \( r _ j : \mathrm{int}(B^{n _ j}) \rightarrow U _ j \). We may also assume that \( U _ 1, \ldots, U _ m \) are pairwise disjoint.
Let us form a new space \( X^\prime \) by gluing all the balls \( \mathrm{int}(B^{n _ j}) \) to \( U _ j \) along the embeddings \( i _ j \). More precisely, \[ \displaystyle \begin{aligned} X^\prime = X \amalg \coprod _ {j=1}^{m} \mathrm{int}(B^{n _ j}) \Big/ (x \sim i _ j(x)). \end{aligned} \] Then there is a natural embedding \( i : X \hookrightarrow X^\prime \), and also a deformation retract \[ \displaystyle \begin{aligned} r : X^\prime \rightarrow X; \quad x \mapsto \begin{cases} x & \displaystyle \text{if } x \in X, \\ r _ j(x) & \displaystyle \text{if } x \in B^{n _ j}. \end{cases} \end{aligned} \] This is indeed a deformation retract because the homotopies for the deformation retracts \( r _ j \) glue to give a homotopy for the retract \( r \).
Denote \( \tilde{f} = i \circ f \circ r : X^\prime \rightarrow X^\prime \). Because \( X \) is a deformation retract of \( X^\prime \), the maps \( i \) and \( r \) induce isomorphisms \( i _ \ast : H _ \bullet(X) \rightarrow H _ \bullet(X^\prime) \) and \( r _ \ast : H _ \bullet(X^\prime) \rightarrow H _ \bullet(X) \) on homology. Moreover, they are inverse maps. Hence \( \Lambda(f) = \Lambda(i \circ f \circ r) = \Lambda(\tilde{f}) \).
Now note that there is again a finite simplicial complex structure on \( X^\prime \). Also the fixed points of \( \tilde{f} \) are still \( x _ 1, \ldots, x _ m \), and \( I(\tilde{f}, x _ j) = I(f, x _ j) = 0 \). For each \( j \), there exists a sufficiently small \( 0 < \epsilon < 0.1 \) such that \( \tilde{f}(\epsilon B^{n _ j}) \subseteq 0.5 B^{n _ j} \) because \( \tilde{f}(0 \in B^{n _ j}) = 0 \in B^{n _ j} \) and \( \tilde{f} \) is continuous. Using Lemma 14, we can continuously deform \( \tilde{f} \) and change it into a new function \( \tilde{f} _ 1 \) that is identical to \( \tilde{f} \) when restricted to \( X^\prime \setminus B^{n _ j} \) but has no fixed point in \( B _ {n _ j} \). Applying this process for each \( j \), we obtain a map \( \tilde{f}^\prime : X^\prime \rightarrow X^\prime \), homotopic to \( \tilde{f} \), but that has no fixed points at all. Theorem 4 then implies \( \Lambda(\tilde{f}^\prime) = 0 \). Therefore \( \Lambda(f) = \Lambda(\tilde{f}) = \Lambda(\tilde{f}^\prime) = 0 \). ▨
Proof of Theorem 12. Let \( x _ 1, \ldots, x _ m \) be the fixed points of \( f : X \rightarrow X \). As in the proof of Theorem 13, we are going to resolve the fixed points by applying appropriate operations on the space.
Recall that in Example 1 and 2, we have given explicit examples of fixed points with indices \( 0 \) and \( 2 \). Let us name the maps as \( \rho _ 1 : S^1 \rightarrow S^1 \) with fixed point \( p _ 1 \in S _ 1 \), and \( \rho _ 2 : S^2 \rightarrow S^2 \) with fixed point \( p _ 2 \in S^2 \). Note that \[ \displaystyle \begin{aligned} F _ t : S^1 \rightarrow S^1; \quad x \mapsto x - t x^2, \qquad G _ t : S^2 \rightarrow S^2; \quad z \mapsto \frac{z}{tz + 1} \end{aligned} \] shows that \( \rho _ 1 \) and \( \rho _ 2 \) are homotopic to the identity map.
Let us look at \( x _ j \in X \). Attach a copy of \( (S^1, p _ 1) \) to \( X \) at \( x _ j \) and call the new space \( X^\prime \). Also extend the map \( f \) to \( f^\prime : X^\prime \rightarrow X^\prime \) by \[ \displaystyle \begin{aligned} f^\prime : x \mapsto \begin{cases} f(x) & \displaystyle \text{if } x \in X, \\ \rho _ 1(x) & \displaystyle \text{if } x \in S^1. \end{cases} \end{aligned} \] The new map \( f^\prime \) does not change the set of fixed points, since \( p _ 1 \) is the only fixed point of \( \rho _ 1 \). However, by Theorem 12, the index of \( x _ j \) changes to \( I(f^\prime, x _ j) = I(f, x _ j) + 0 - 1 = I(f, x _ j) - 1 \). That is, this operation that changes \( (X, f) \) to \( (X^\prime, f^\prime) \) decreases one of the fixed point indices by \( 1 \).
We also see through Mayer–Vietoris that \( H _ k(X^\prime) \cong H _ k(X) \) for \( k \neq 1 \) and \( H _ 1(X^\prime) \cong H _ 1(X) \oplus \mathbb{Z} \). Because the Mayer–Vietoris sequence is natural, we moreover obtain \( f^\prime _ {\ast 1} = f _ {\ast 1} \oplus (\rho _ 1) _ {\ast 1} \). This shows that \( \mathrm{Tr}(f^\prime _ {\ast 1}) = \mathrm{Tr}(f _ {\ast 1}) + \mathrm{Tr}((\rho _ 1) _ {\ast 1}) = \mathrm{Tr}(f _ {\ast 1}) + 1 \) because \( \rho _ 1 \) is homotopic to the identity. Thus \( \Lambda(f^\prime) = \Lambda(f) - 1 \), because \( (-1)^1 = -1 \). This shows that attaching the \( S^1 \) operation decreases \( \Lambda \) by \( 1 \), and also decreases one of the fixed point indices by \( 1 \).
Let us now look at the operation of attaching a \( (S^2, p _ 2) \) on \( X \) at \( x _ j \). Because \( I(\rho _ 2, p _ 2) = 2 \), it is clear by Theorem 12 that \( I(f^\prime, x _ j) = I(f, x _ j) + 1 \). On the other hand, Mayer–Vietoris tells us that \( H _ k(X^\prime) \cong H _ k(X) \) for \( k \neq 2 \) and \( H _ 2(X^\prime) \cong H _ 2(X) \oplus \mathbb{Z} \). Because \( (\rho _ 2) _ {\ast 2} : H _ 2(S^2) \rightarrow H _ 2(S^2) \) is the identity map since \( \rho _ 2 \) is again homotopic to the identity map, the traces are related by \( \mathrm{Tr}(f^\prime _ {\ast 2}) = \mathrm{Tr}(f _ {\ast 2}) + 1 \). Thus \( \Lambda(f^\prime) = \Lambda(f) + (-1)^2 = \Lambda(f) + 1 \). Hence the operation of attaching a \( S^2 \) increases \( \Lambda \) by \( 1 \) and also increases one of the fixed point indices by \( 1 \).
Attaching a circle and a sphere to the given space
Given an arbitrary map \( f : X \rightarrow X \) with fixed points \( x _ 1, \ldots, x _ m \), we may apply the operations of attaching \( S^1 \) and \( S^2 \) finitely many times to obtain a map \( \tilde{f} : \tilde{X} \rightarrow \tilde{X} \) with fixed point indices \( I(\tilde{f}, x _ 1) = \cdots = I(\tilde{f}, x _ m) = 0 \). Because \( \Lambda \) changes the same way as the index, the Lefschetz number of \( \tilde{f} \) will be given by \[ \displaystyle \begin{aligned} \Lambda(\tilde{f}) = \Lambda(f) - \sum _ {j=1}^{m} I(f, x _ j). \end{aligned} \] Note that attaching a \( S^1 \) or a \( S^2 \) to a simplicial complex gives a simplicial complex. Then \( \tilde{X} \) will be homeomorphic to a realization of a finite simplicial complex. By Theorem 13, \( \Lambda(\tilde{f}) = 0 \). This implies \[ \displaystyle \begin{aligned} \Lambda(f) & \displaystyle= \sum _ {j=1}^{m} I(f, x _ j). \end{aligned} \] ▨
5. Applications#
The Lefschetz fixed-point formula has a lot of consequences. One of the most obvious corollaries is the Brouwer fixed-point theorem.
Corollary 15 (Brouwer fixed-point theorem). Let \( B^n \subseteq \mathbb{R}^n \) be the closed ball. Every continuous map \( f : B^n \rightarrow B^n \) has a fixed point.
Proof. We compute \( \Lambda(f) \). Clearly \( B^n \) has a simplicial complex structure, and because it is contractible, \( H _ k(B^n) = 0 \) for \( k \neq 0 \) and \( H _ 0(B^n) = \mathbb{Z} \). Also it is obvious that \( f _ {\ast 0} : H _ 0(B^n) \rightarrow H _ 0(B^n) \) is the identity map. Hence \( \Lambda(f) = 1 \). Therefore \( f \) has a fixed point. ▨
In the case \( f = \mathrm{id} _ X \), the Lefschetz number can be computed as \[ \displaystyle \begin{aligned} \Lambda(f) = \sum _ {k \ge 0}^{} (-1)^k \mathrm{Tr}(f _ {\ast k} : H _ k(X) \rightarrow H _ k(X)) = \sum _ {k \ge 0}^{} (-1)^k \mathrm{rank}(H _ k(X)) = \chi(X). \end{aligned} \] Thus if \( f \) is homotopic to the identity and has isolated fixed points, the number of fixed point, counted with multiplicity, is \( \chi(X) \).
Corollary 16 (Poincaré–Hopf theorem). Let \( M \) be a compact smooth manifold and let \( V \) be a smooth vector field on \( M \). Suppose that \( V \) vanishes on finitely many points. Then \[ \displaystyle \begin{aligned} \sum _ {V(x) = 0}^{} \mathrm{index} _ x(V) = \chi(M), \end{aligned} \] where \( \mathrm{index} _ x(V) \) is defined by choosing a neighborhood \( B^n \cong U _ x \ni x \) and letting \( \mathrm{index} _ x(V) = \deg (S^{n-1} \rightarrow S^{n-1}; , x \mapsto V(x) / \lvert V(x) \rvert) \).
Sketch of proof. Let \( \varphi(x, t) \) be the vector flow generated by \( V \), so that \( (\partial / \partial t) \varphi(x, t) = V(\varphi(x, t)) \). If \( V(x) = 0 \), then \( \varphi(x, t) = x \). Because \( M \) is compact, for a sufficiently small \( t _ 0 \) the map \( \varphi _ {t _ 0} : M \rightarrow M \) has fixed point only at the points where \( V \) vanish. Using the fact that every manifold is an ENR (Euclidean neighborhood retract), we may apply the Lefschetz fixed-point formula. \( \varphi(x, t) \) gives a homotopy between the identity map and \( \varphi _ {t _ 0} \). This shows that \( \Lambda(\varphi _ {t _ 0}) = \chi(M) \). Moreover, \( V(x) \cdot t _ 0 \) is approximately \( \varphi _ {t _ 0} - \mathrm{id} \) and hence the index \( \mathrm{index} _ x(V) \) is \( I(\varphi _ {t _ 0}, x) \). It follows that \[ \displaystyle \begin{aligned} \sum _ {V(x)=0}^{} \mathrm{index} _ x(V) & \displaystyle= \sum _ {\varphi _ {t _ 0}(x) = x}^{} I(\varphi _ {t _ 0}, x) = \Lambda(\varphi _ {t _ 0}, x) = \chi(M). \end{aligned} \] ▨
Corollary 17 (Hairy ball theorem). There is no non-vanishing vector field on \( S^2 \).
Proof. The Euler characteristic of \( S^2 \) is \( \chi(S^2) = 2 \neq 0 \). Thus there is at least one fixed point. ▨
References#
[Dol95] Albrecht Dold, Lectures on algebraic topology, Classics in Mathematics, Springer-Verlag, Berlin, 1995, Reprint of the 1972 edition. MR 1335915
[Hat02] Allen Hatcher, Algebraic topology, Cambridge University Press, Cambridge, 2002. MR 1867354
[Lef26] Solomon Lefschetz, Intersections and transformations of complexes and manifolds, Trans. Amer. Math. Soc. 28 (1926), no. 1, 1–49. MR 1501331