State the Lefschetz Fixed Point Theorem for a finite simplicial complex \(X\).
Use degree theory to prove this theorem in case \(X = S^n\).
Solution
1. Let \(X\) be a finite simplicial complex and \(f:X\to X\) a continuous map.Its Lefschetz number is\[ L(f) = \sum_{q\ge0}(-1)^q \operatorname{tr}\left( f_*:H_q(X;\QQ)\to H_q(X;\QQ) \right). \]The Lefschetz fixed-point theorem states that\[ L(f)\ne0 \quad\Longrightarrow\quad f\text{ has a fixed point}. \]
Because \(X\) is a finite simplicial complex, each rational homology group is finite-dimensional and vanishes in sufficiently large degree, so the displayed sum and traces are well-defined. The implication is the Lefschetz fixed-point theorem requested in part (a).
2. For \(n\ge1\) and a continuous map \(f:S^n\to S^n\),\[ L(f)=1+(-1)^n\deg(f). \]
The only nonzero rational homology groups of \(S^n\) are \[ H_0(S^n;\QQ)\cong\QQ, \qquad H_n(S^n;\QQ)\cong\QQ. \] Since \(S^n\) is connected, \(f_*\) acts as the identity on \(H_0\), so its trace there is \(1\). By definition of degree, \(f_*\) acts on \(H_n(S^n;\QQ)\) as multiplication by \(\deg(f)\), so its trace there is \(\deg(f)\). Substitution into 1 gives \[ L(f)=1+(-1)^n\deg(f). \]
3. If \(f:S^n\to S^n\) has no fixed point, then \(f\) is homotopic to the antipodal map\[ A(x)=-x. \]
Regard \(S^n\) as the unit sphere in \(\RR^{n+1}\) and define \[ H(x,t) = \frac{(1-t)f(x)-tx} {\lVert(1-t)f(x)-tx\rVert}, \qquad 0\le t\le1. \] Suppose the denominator vanished for some \((x,t)\). Then \[ (1-t)f(x)=tx. \] Taking norms gives \(1-t=t\), hence \(t=1/2\), and then the vector equality gives \(f(x)=x\), contradicting the assumption that \(f\) has no fixed point. Thus \(H\) is well-defined and continuous. Moreover \[ H(x,0)=f(x), \qquad H(x,1)=-x=A(x), \] so \(f\simeq A\).
4. The antipodal map \(A:S^n\to S^n\) has degree\[ \deg(A)=(-1)^{n+1}. \]
The antipodal map is the restriction to the unit sphere of the linear automorphism \[ -I:\RR^{n+1}\to\RR^{n+1}. \] An invertible linear map induces on the oriented boundary sphere a map whose degree is the sign of its determinant. Since \[ \det(-I)=(-1)^{n+1}, \] one gets \[ \deg(A)=(-1)^{n+1}. \]
5. If \(f:S^n\to S^n\) has no fixed point, then \(L(f)=0\).
By 3 and homotopy invariance of degree, \[ \deg(f)=\deg(A). \] Using 4 and then 2, \[ L(f) = 1+(-1)^n(-1)^{n+1} = 1-1 = 0. \]
6. Therefore the Lefschetz fixed-point theorem holds for \(S^n\).
Step 5 proves the contrapositive: if \(f\) has no fixed point, then \(L(f)=0\). Hence \[ L(f)\ne0 \quad\Longrightarrow\quad f\text{ has a fixed point}. \] This is exactly the theorem in the case \(X=S^n\).
7. The case \(n=0\) also satisfies the theorem.
The space \(S^0\) consists of two points. A fixed-point-free self-map must interchange them, so its induced map on \[ H_0(S^0;\QQ)\cong\QQ^2 \] is the transposition matrix, whose trace is \(0\). Thus again a fixed-point-free map has Lefschetz number \(0\).