Fixed Point Index and the Lefschetz Theorem: Examples

The Lefschetz number is $2$. Every rotation is homotopic to the identity through rotations. Since $H_0(S^2;\mathbb{Q})=H_2(S^2;\mathbb{Q})=\mathbb{Q}$ and the other groups vanish by [F1], the trace formula gives $L(\mathrm{i}\mathrm{d})=1+1=\chi (S^2)=2$, and homotopy invariance gives $L(R)=2$. Equivalently, $R_{\ast }$ is the identity on $H_0$ and multiplication by $\mathrm{deg}R=1$ on $H_2$, so $L(R)=1+1$.

The fixed points of a generic rotation. In coordinates $z$ on $S^2$ centred at the north pole the rotation is $z\mapsto e^{i\theta }z$ for $\theta \mathrm{\notin }2\pi \mathbb{Z}$; fixed points solve $(e^{i\theta }-1)z=0$, so $z=0$, and near the south pole the same computation in the chart $w=1\mathrm{/}z$ shows that only the south pole is fixed. At each pole the displacement has invertible differential $1-e^{\pm i\theta }\mathrm{\ne }0$, so both fixed points are nondegenerate and by [F2] each has index $\mathrm{sign}\det (1-e^{\pm i\theta }\cdot {\mathrm{i}\mathrm{d}}_{{\mathbb{R}}^2})=+1$ (the linear map is a positive multiple of a rotation). Hence $I(R)=1+1=2=L(R)$, in agreement with the index formula. For the half-turn $\theta =\pi$, the same two poles are the entire fixed set and $I-DR=2I$ on each tangent plane, so each is nondegenerate of index $+1$. For $\theta \in 2\pi \mathbb{Z}$ the map is the identity and fixes all of $S^2$; this fixed set is not isolated, so only the Lefschetz number $2$ is asserted directly.

The trace computation. By [F1] the only nonzero rational homology groups are $H_0$ and $H_n$, with $f_{\ast }=\mathrm{i}\mathrm{d}$ on $H_0$ and $f_{\ast }=d$ on $H_n$; the defining alternating sum therefore has exactly the two terms $(-1{)}^0\mathrm{tr}(\mathrm{i}\mathrm{d})=1$ and $(-1{)}^{n}d$, so $L(f)=1+(-1{)}^{n}d$.

Consequences. For $n=1$ this is $1-d$, vanishing exactly for the degree-one circle maps such as rotations; for $n=2$ it is $1+d$, and this is nonzero exactly when $d\mathrm{\ne }-1$, so the fixed-point theorem forces a fixed point in that case; the degree-$-1$ antipodal map has no fixed points and Lefschetz number $1-1=0$, the standard sharpness example. When $f$ is smooth with nondegenerate fixed points the same number is the index sum by [L1], for instance for an integer $d\ge 2$ the map $z\mapsto z^d$ extends smoothly to $S^2$, with the coordinate expression $w\mapsto w^d$ at infinity. Its fixed points are $0$, $\mathrm{\infty }$, and the $d-1$ solutions of $z^{d-1}=1$. The derivative is zero at $0$ and infinity, and is complex multiplication by $d$ at those roots; hence $I-Df$ has positive real determinant at all $d+1$ points and every local index is $+1$ by The index of a nondegenerate fixed point is the sign of det(I-Df). A nonzero finite target value has $d$ regular preimages of positive orientation, proving the asserted degree $d$ by Regular-value formula for degree.

No fixed points for $a\mathrm{\ne }0$. A point $x\in T^2$ is fixed by $T_a$ exactly when $a=x-x=0$ in ${\mathbb{R}}^2\mathrm{/}{\mathbb{Z}}^2$; hence for nonzero $a$ the translation is fixed-point-free.

The Lefschetz number vanishes. The family $t\mapsto T_{ta}$, $t\in [0,1]$, is a homotopy from the identity to $T_a$, so $L(T_a)=L(\mathrm{i}\mathrm{d})=\chi (T^2)$ by [L1]. The standard CW structure of [F1] has cell count $1-2+1=0$, and by [F2] this is $\chi (T^2)$; hence $L(T_a)=0$. Thus a vanishing Lefschetz number coincides here with the complete absence of fixed points, which is exactly why the example does not contradict the theorem but exhibits its sharpness.

The indices. The fixed point equation on $\mathbb{C}$ is $z^2=0$, so $0$ is the only finite fixed point and it is isolated; in the chart $w=1\mathrm{/}z$ at infinity, the fixed point equation is $w^2\mathrm{/}(w+1)=w$, i.e. $w=0$, so $\mathrm{\infty }$ is the other fixed point. By [F1] the two indices are $2$ and $+1$, so the geometric Lefschetz number is $I(f)=2+1=3$; the point $0$ is degenerate, so the determinant formula The index of a nondegenerate fixed point is the sign of det(I-Df) does not apply to it, and the value $2$ comes from the explicit degree computation of the displacement $-z^2$ on a small circle.

The Lefschetz number agrees. The expression $w\mapsto w^2\mathrm{/}(w+1)$ is smooth near infinity, so the polynomial extends smoothly. The finite value $1$ has exactly two distinct preimages solving $z^2+z=1$, namely $(-1\pm \sqrt{5})\mathrm{/}2$; at each the derivative is nonzero complex multiplication by $1+2z$, of positive real determinant. Infinity maps to infinity and is not a preimage of $1$. Hence $1$ is a regular value and Regular-value formula for degree gives degree $2$, so by [F2] $L(f)=1+2=3$. Hence $I(f)=3=L(f)$ even though the fixed point at $0$ is degenerate: the index formula holds for isolated fixed points and does not require nondegeneracy, which is exactly the content of Lefschetz-Hopf index formula.

The fixed points. A point $\theta$ is fixed by $g$ exactly when $\epsilon \sin \theta =0$ in $\mathbb{R}\mathrm{/}2\pi \mathbb{Z}$, i.e. exactly when $\theta \in \{0,\pi \}$. Both solutions are nondegenerate because $g^{\mathrm{\prime }}(\theta )=1+\epsilon \cos \theta$ is $1+\epsilon$ at $0$ and $1-\epsilon$ at $\pi$, neither equal to $1$ for $0<\epsilon <1$; their indices are $\mathrm{sign}(1-(1+\epsilon ))=-1$ at $0$ and $\mathrm{sign}(1-(1-\epsilon ))=+1$ at $\pi$. So $f$ has fixed points and its index sum is $(-1)+(+1)=0$.

The periodic function $\epsilon \sin \theta$ makes $g_t(\theta )=\theta +t\epsilon \sin \theta$ a well-defined homotopy of circle maps from the identity to $f$. By [L1], $L(f)=L(\mathrm{i}\mathrm{d})$, and [F1] gives the two identity traces one in degrees zero and one, so $L(f)=1-1=0$. Thus $L(f)=0$ while $f$ has the two fixed points found in step 1.1.

The refutation. The displayed statement asserts that a vanishing Lefschetz number forces the map to be fixed-point-free; $f$ has $L(f)=0$ and the fixed points $0,\pi$, so the implication fails. The example is the curved version of the standard caution that $L$ is only a signed count: the nonvanishing of $L$ guarantees a fixed point, but its vanishing merely allows fixed points to cancel, as here with indices $-1$ and $+1$.