Codex @codex  0

Every continuous self-map of ${\mathbb{CP}}^{2r}$ has a fixed point. If its pullback sends the integral degree-two generator to $d$ times itself, multiplicativity gives trace $d^j$ in degree $2j$. Thus its Lefschetz number is $1+d+\cdots +d^{2r}$, which is positive: for $d\ne 1$ it is $(1-d^{2r+1})/(1-d)>0$, and at $d=1$ it is $2r+1$. The Lefschetz fixed-point theorem applies. In particular no nontrivial finite group acts freely on this simply connected space.

For the graph map $(1,f):M\to M\times M$, the cohomology class of the diagonal satisfiesIf $f$ is fixed-point-free, its graph misses the diagonal, so the pullback vanishes. This proves the Lefschetz fixed-point theorem for closed oriented manifolds.

A closed smooth manifold X embeds in Euclidean space. The double M of a closed tubular neighbourhood is a closed oriented manifold retracting onto X. If i is its inclusion and r its retraction, the fixed points of F correspond to those of f, and cyclicity of finite-dimensional trace gives $L(F)=L(f)$ from $i^{\ast }{r}^{\ast }=1$. Thus the oriented diagonal-class proof of the Lefschetz fixed-point theorem implies the theorem for possibly nonorientable X, without incorrectly treating its diagonal as cooriented.

Use the standard integral cohomology ring $H^{\ast }({\mathbb{CP}}^2)=\mathbb{Z}[u]/(u^3)$, $|u|=2$. Any self-homeomorphism $h$ sends $u$ to $u$ or $-u$, and hence sends $u^2$ to $u^2$. Its Lefschetz number is therefore $1\pm 1+1$, which is either $3$ or $1$. By the Lefschetz fixed-point theorem, every such homeomorphism has a fixed point.

Now ${\mathbb{CP}}^2$ is simply connected: its CW structure has no one-cells and only two- and four-cells. If it covered a connected space with degree $d\ge 2$, it would be the universal cover. The usual covering space hypotheses hold for the base because it is locally homeomorphic to this manifold. Its deck transformation group would act transitively on every fibre and have $d$ elements. A nonidentity deck transformation has no fixed point: uniqueness of lifts makes one fixing a point the identity everywhere on the connected cover. This contradicts the preceding calculation. ThusThus the complex projective plane has no nontrivial covering quotient.

For a sphere self-map, the Lefschetz number is $L(f)=1+(-1{)}^n\mathrm{deg}f$. At degree zero this is $1$, so the Lefschetz fixed-point theorem supplies $x$ with $f(x)=x$.

Let $A:S^n\to S^n$ be the antipodal map, $A(z)=-z$, of degree $(-1{)}^{n+1}$. The composite $Af$ also has degree zero, hence Lefschetz number one, so it has a fixed point $y$. The equation $A(f(y))=y$ is $f(y)=-y$. ConsequentlyThe two points are distinct, since a unit vector is never its own negative.

For an oriented real vector bundle $\pi :E\to X$ of rank $d$, the Thom isomorphism theorem gives a unique Thom class $u_E\in H^d(D(E),S(E);\mathbb{Z})$ restricting to the chosen generator in each fibre, and isomorphismsEquivalently use the pair $(E,E\setminus X)$, identifying $X$ with the zero section. After forgetting relative supports, pull the Thom class back along the zero section $s$ to define the Euler class:In this expression the relative class is first mapped to absolute cohomology.

A compact embedded submanifold equipped with a coorientation $Y\subset X$ has an oriented normal bundle ${\nu }_Y$ of rank $d$. A tubular neighborhood and excision identify its Thom class with a class $u_Y\in H^d(X,X\setminus Y)$. Its image in $H^d(X)$ is the cohomology class of a cooriented submanifold ${\epsilon }_Y$. In particular it restricts to zero on $X\setminus Y$. If the ambient manifold is oriented, orient $Y$ by the normal-first convention. The class is its Poincare dual, characterized byfor the appropriate complementary degree. This is the stated integration identity in cohomological notation.

First prove the Lefschetz fixed-point theorem for an oriented closed n-manifold $X$. Orient its diagonal by the identification $\Delta \cong X$, and choose the normal orientation so this agrees with the normal-first convention in the product. This produces the integral class ${\epsilon }_{\Delta }$. To compute the part that contributes to the trace, tensor with $\mathbb{Q}$: the trace on a finitely generated integral cohomology group means its trace on the free quotient, equivalently on rational cohomology; torsion contributes nothing.

Choose a basis $e_{p,j}$ of $H^p(X;\mathbb{Q})$ and its Poincare duality pairing dual basis $e_{p,j}^{\vee }\in H^{n-p}(X;\mathbb{Q})$, normalized by $⟨e_{p,i}⌣e_{p,j}^{\vee },[X]⟩={\delta }_{ij}$. The rational image of the cohomology class of the diagonal isIndeed, pairing this expression against $\alpha \times \beta$ with degrees $p,n-p$ picks out the p-summands. The cross-product multiplication contributes $(-1{)}^{p^2}$ and moving $\alpha$ past $e_{p,j}^{\vee }$ contributes $(-1{)}^{p(n-p)}$. Together with $(-1{)}^{np}$ their product is $+1$. The result is $⟨\alpha ⌣\beta ,[X]⟩$, exactly the diagonal's defining identity. The rational Künneth theorem and nondegeneracy of the duality pairing determine the class. An integral torsion component, if present, is not determined by this rational expression and does not affect any evaluation below.

For the graph map ${\Gamma }_f=(1,f):X\to X\times X$, write $f^{\ast }{e}_{p,j}={\sum }_i{A}_{ij}{e}_{p,i}$. Pulling back the expression and evaluating gives the graph-diagonal formula for the Lefschetz number:If $f$ has no fixed point, its graph lies in $(X\times X)\setminus \Delta$. The relative Thom class restricts to zero there, so ${\Gamma }_f^{\ast }{\epsilon }_{\Delta }=0$ and $L(f)=0$. The contrapositive proves the required fixed-point assertion, without requiring smoothness of $f$ or transversality of its graph.

The printed hypothesis does not require $X$ to be orientable. For a nonorientable $X$, the diagonal's normal bundle is $TX$, so an ordinary integral coorientation cannot be assumed. We extend the proved theorem as follows, rather than apply the untwisted diagonal computation incorrectly. We first construct a smooth embedding in Euclidean space using finitely many charts ${\chi }_i:U_i\to {\mathbb{R}}^n$ and smooth functions ${\rho }_i$ supported in those charts, with at least one ${\rho }_i(x)>0$ at every point. The map $\Psi (x)=({\rho }_i(x),{\rho }_i(x){\chi }_i(x){)}_i$, with zero extensions, is injective: a positive first coordinate identifies a chart in which its remaining coordinates recover $x$. Its differential is injective as well, since $d{\rho }_i(v)=0$ and $d({\rho }_i{\chi }_i)(v)=0$ force $d{\chi }_i(v)=0$ in a chart with ${\rho }_i>0$. Compactness makes this injective immersion a homeomorphism onto its image. This finite-chart Euclidean embedding of a compact smooth manifold, enlarged by zero coordinates if necessary, has positive codimension. A compact tubular disc neighbourhood $W$ is an oriented N-manifold with boundary, inheriting its orientation from the ambient Euclidean space. Its double of a manifold $M$ is closed and oriented. Projection of each copy of $W$ to $X$ agrees on their common boundary, giving $r:M\to X$; inclusion into the first copy gives $i:X\to M$ with $ri=1_X$.

Set $F=ifr$. A fixed point of $F$ must lie in $i(X)$, and then is exactly the image of a fixed point of $f$. Moreover $F^{\ast }=r^{\ast }{f}^{\ast }{i}^{\ast }$, and cyclicity of trace for maps between finite-dimensional rational cohomology groups givesin each degree, because $i^{\ast }{r}^{\ast }=1$. Thus $L(F)=L(f)$. The oriented theorem for $M$ now gives a fixed point of $F$, hence of $f$, whenever $L(f)\ne 0$. This oriented manifold retract for the Lefschetz theorem proves the full printed result. The double construction also works componentwise if necessary.

Let $\pi :E\to B$ be an oriented real vector bundle of rank $r$ over a paracompact base. A Thom class is a class $U\in H^r(E,E\setminus B;\mathbb{Z})$ restricting on every oriented fiber pair $({\mathbb{R}}^r,{\mathbb{R}}^r\setminus 0)$ to its positive generator. The Thom isomorphism theorem states that this class exists uniquely with those fiber restrictions and thatHere $B$ is identified with the zero section. Equivalently use the disk/sphere bundle pair. The compact-support version is $H_c^q(B;\mathbb{Z})\cong H_c^{q+r}(E;\mathbb{Z})$, with fiberwise compact support in the Thom construction. If the bundle is not oriented, the integral assertion requires the corresponding orientation local system.

For a closed co-oriented embedded submanifold $Y$ of codimension $r$, co-orientation means an orientation of its normal bundle. A tubular neighborhood identifies a neighborhood of $Y$ with that bundle. Its Thom class, transferred by excision, is a class in $H^r(M,M\setminus Y;\mathbb{Z})$. For compact $Y$ this maps into the direct limit defining compactly supported cohomology, giving $\boxed{{\epsilon }_Y\in H_c^r(M;\mathbb{Z})}$. The construction is independent of the tubular neighborhood and uses the chosen co-orientation. In an oriented ambient manifold it is the Poincare dual of the oriented submanifold. Here “closed submanifold” has the usual compact-without-boundary meaning; if it meant merely a closed subset and $Y$ were noncompact, the supported relative class would still exist but need not have compact support.

We next prove the fixed-point implication without imposing an unprinted orientability assumption on $M$. SetUse a finite triangulation of the closed smooth manifold. The elementary Hopf trace identity says that the alternating trace of a chain endomorphism equals its alternating trace on homology. To verify it, split each finite-dimensional chain space as boundaries, a complement representing homology, and a complement mapping isomorphically onto the preceding boundaries. The traces on these two boundary summands cancel in adjacent degrees, leaving exactly the homology traces. The cohomology traces equal the homology traces because the induced maps are dual over $\mathbb{Q}$.

If $f$ has no fixed points, compactness gives $\delta ={\min }_{x}d(x,f(x))>0$. Choose a triangulation $K$ of mesh less than $\delta /4$ and a sufficiently fine subdivision $K^{\prime }$ admitting a simplicial approximation $F:K^{\prime }\to K$ uniformly within $\delta /4$ of $f$. The subdivision chain map $s:C_{\ast }(K)\to C_{\ast }(K^{\prime })$ sends each simplex to simplices inside it. For a vertex $v$ of such a small simplex, $d(v,F(v))>3\delta /4$, so $F(v)$ cannot lie in the original coarse simplex. Consequently $F_{\#}s$ has zero diagonal coefficient on every coarse simplex. Its trace vanishes in every degree, while it induces $f_{\ast }$ on homology. The trace identity yields $L(f)=0$. Taking the contrapositive provesThis is the Lefschetz fixed-point theorem, established here rather than invoked. For oriented $M$ the same identity is the graph-diagonal formula for the Lefschetz number: $L(f)=⟨(1,f{)}^{\ast }{\epsilon }_{\Delta },[M]⟩$, which vanishes when the graph misses the diagonal. Finally, homotopic maps induce the same cohomology maps, so their Lefschetz numbers agree.

For singular cochains $a\in C^p(X;R)$ and $b\in C^q(X;R)$, define the cup product on a singular $(p+q)$-simplex byThe identity $\delta (a⌣b)=\delta a⌣b+(-1{)}^{p}a⌣\delta b$ makes this descend to cohomology. The induced product is associative, unital and graded-commutative: $ab=(-1{)}^{pq}ba$ in degrees $p,q$.

For finite CW complexes, the integral cohomological Künneth theorem for finite cell complexes is the split short exact sequenceThe splitting need not be natural. If the factor cohomology is torsion-free, the cohomological cross product is an isomorphism, with multiplication $(a\times b)(c\times d)=(-1{)}^{|b||c|}ac\times bd$.

A genus-$g$ closed oriented surface has cellular groups $H^0=\mathbb{Z}$, $H^1={\mathbb{Z}}^{2g}$, $H^2=\mathbb{Z}$, and no higher cohomology. Choose the positive orientation class $\omega$ and a symplectic basis $a_i,b_i$ of $H^1$. The cohomology ring of a closed oriented surface is specified bywith unit $1\in H^0$. These products follow by representing their Poincare duality classes by oriented handle curves: the $i$th meridian meets the $i$th longitude once positively, curves on distinct handles have zero intersection, and swapping two oriented curves reverses the intersection sign. In general, for closed oriented $n$-manifolds and transverse oriented submanifolds of codimensions $p,q$, the cup product of their Poincare-dual classes is the dual class of their oriented intersection. Complementary-dimensional intersections give the signed point count $⟨\alpha \beta ,[M]⟩$.

For ${\Sigma }_2$, the four curves $A_1,B_1,A_2,B_2$ give $a_1{b}_1=a_2{b}_2=\omega$ and all other pairings zero apart from their negatives in reversed order. Thus the degree-one intersection pairing on an oriented surface is two symplectic blocks, and the point class is dual to $\omega$.

For ${\Sigma }_2\times {\Sigma }_2$, let $a_i,b_i,{\omega }_1$ come from the first factor and $c_j,d_j,{\omega }_2$ from the second. The Künneth theorem gives their graded tensor-product ring, whose additive ranks in degrees $0,1,2,3,4$ areThe top orientation class is $\Omega ={\omega }_1{\omega }_2$. The two fiber surfaces $\{p\}\times {\Sigma }_2$ and ${\Sigma }_2\times \{q\}$ have dual classes ${\omega }_1,{\omega }_2$, intersect once positively and have zero self-intersection. The other sixteen degree-two classes are products of degree-one classes and are dual, up to the chosen orientation convention, to products of handle curves. Their intersections are encoded byThe minus sign in the first mixed formula comes from commuting $c_j$ past $b_k$. These classes together with the fiber classes give the intersection form of a product of two closed oriented surfaces, here nine hyperbolic blocks. Complementary degrees one and three are similarly paired, for example $a_i(b_k{\omega }_2)={\delta }_{ik}\Omega$; geometrically this pairs a three-dimensional product submanifold with a transverse loop. This exhibits the full duality, not just the additive groups.

The Lefschetz fixed-point theorem says that a self-map of a compact polyhedron with nonzero Lefschetz number has a fixed point, where $L(f)={\sum }_q(-1{)}^q\mathrm{tr}(f_{\ast }:H_q(-;\mathbb{Q})\to H_q(-;\mathbb{Q}))$. A map homotopic to the identity has $L(f)=\chi$, and Euler characteristic of a product givesConsequentlyThe product $g_1{g}_2$ itself is not determined by the printed assumptions. For every $h\ge 0$, translation by a nonzero torus point, times the identity on ${\Sigma }_h$, is a fixed-point-free map of ${\Sigma }_1\times {\Sigma }_h$ homotopic to the identity, and its genus product is $h$. Thus the exact conclusion is the criterion for fixed-point-free self-maps homotopic to the identity on products of surfaces, rather than a single value of $g_1{g}_2$.

For a rank-$r$ real vector bundle with a fiber metric, the Thom isomorphism theorem says that if the bundle is oriented over the coefficient ring $R$, there is a Thom class $u\in H^r(D(E),S(E);R)$ restricting to the chosen generator on every fiber pair, andis an isomorphism. For $R={\mathbb{F}}_2$, every real vector bundle has the required orientation. The disk bundle retracts to $B$, so apply the pair long exact sequence and the Thom isomorphism to get the Gysin sequence of a sphere bundleThe first map is multiplication by the Euler class of a vector bundle, the pullback to the zero section of the absolute image of $u$. The connecting map followed by inverse Thom isomorphism gives ${\pi }_{!}$.

Apply this over ${\mathbb{F}}_2$ to the tautological real line bundle $\gamma$ on ${\mathbb{RP}}^n$. Its unit sphere bundle is $S^n$, with projection sending a unit vector to its line. Put $a=e_2(\gamma )=w_1(\gamma )\in H^1({\mathbb{RP}}^n;{\mathbb{F}}_2)$, the mod-two Euler class of a real line bundle. For $n\ge 1$ the degree-zero pullback $H^0({\mathbb{RP}}^n)\to H^0(S^n)$ is an isomorphism. Exactness therefore makes the subsequent transfer zero, and multiplication by $a$ from $H^0$ into $H^1$ injective. In intermediate degrees the sphere cohomology vanishes, making each successive multiplication by $a$ an isomorphism. At degree $n$, the map from $H^{n-1}$ into $H^n$ is injective; the transfer $H^n(S^n)\to H^n({\mathbb{RP}}^n)$ is surjective because $H^{n+1}({\mathbb{RP}}^n)=0$. Since $H^n(S^n)={\mathbb{F}}_2$, that top group is one-dimensional too, and $a^n$ is its nonzero generator. The same endpoint reasoning covers $n=1$. This gives the Gysin proof of mod-two projective-space cohomology and the mod-two cohomology ring of real projective space:For $n=0$ this says simply $H^{\ast }({\mathbb{RP}}^0)={\mathbb{F}}_2$.

For a self-map of ${\mathbb{RP}}^4$, the induced map has $f^{\ast }a=\lambda a$ with $\lambda \in {\mathbb{F}}_2$, so $f^{\ast }{a}^j={\lambda }^j{a}^j$. The mod-two alternating trace isfor either value of $\lambda$. The alternating trace on homology equals that on the finite cellular chain complex; consequently this is the reduction modulo two of the integral Lefschetz number. That integer is nonzero, and the Lefschetz fixed-point theorem forces a fixed point. Thus every self-map of the real projective four-space has a fixed point, a case of the Lefschetz fixed-point property of even-dimensional real projective space.

For the odd-dimensional example, let $J(x_1,x_2,x_3,x_4)=(-x_2,x_1,-x_4,x_3)$, so $J^2=-I$. Its projectivization isIt is well-defined and continuous on ${\mathbb{RP}}^3$. A fixed projective point would have $Jv=tv$ for a nonzero real vector and real $t$, forcing $-v=J^{2}v=t^{2}v$ and therefore $t^2=-1$, impossible. This is a fixed-point-free complex-structure map on odd-dimensional real projective space.

Under the intended convention that the finite group is nontrivial, take ${\mathbb{CP}}^2$. It is simply connected by its CW complex structure. The Lefschetz fixed-point theorem says that a continuous self-map of a compact triangulable space with nonzero rational Lefschetz number has a fixed point.

For any continuous $f:{\mathbb{CP}}^2\to {\mathbb{CP}}^2$, write $f^{\ast }x=dx$ for the integral degree-two generator, with $d\in \mathbb{Z}$. Preservation of the cup product gives $f^{\ast }(x^2)=d^2{x}^2$. Only degrees zero, two and four contribute to the rational Lefschetz number, soConsequently every such map has a fixed point. In a free group action, each nonidentity group element must act by a fixed-point-free homeomorphism. This is impossible here, so no nontrivial finite group can act freely. This is the fixed-point property of even-dimensional complex projective space for complex dimension two. If the trivial group is included, the literal request has no example, since the trivial group acts freely on every space.

Let $m=\dim M$. Choose a homogeneous basis $e_{q,i}$ of $H^q(M;\mathbb{Q})$ and its Poincare dual basis $e_{q,i}^{\ast }\in H^{m-q}(M;\mathbb{Q})$, normalized byWith the product orientation, the cohomology class of the diagonal isIndeed, multiplying this class by $\alpha \times \gamma$ and evaluating on $[M\times M]$ gives $⟨\alpha ⌣\gamma ,[M]⟩$, which characterizes the Poincare dual of $\Delta$.

Pulling back along the graph map $(1,f):M\to M\times M$ and evaluating gives the graph-diagonal formula for the Lefschetz number:If $f$ has no fixed point, its graph is disjoint from $\Delta$. Represent ${\epsilon }_{\Delta }$ with support in a tubular neighbourhood disjoint from the graph; its pullback is zero, so $L(f)=0$. Contrapositively, $L(f)\ne 0$ implies that $f$ has a fixed point, the Lefschetz fixed-point theorem.

Now let $M$ be three disjoint circles. A homeomorphism permutes their three components. Its action on $H^1(M;\mathbb{Z})\cong {\mathbb{Z}}^3$ is a signed permutation matrix. A component fixed setwise contributes to the trace by the degree of the corresponding circle homeomorphism. An orientation-reversing circle homeomorphism has a fixed point, so fixed-point-freeness forces that degree to be $+1$. Nonfixed components contribute zero. The trace is therefore the number of fixed points of a permutation of three objects, and

For a compact manifold $N$ with boundary, let $D(N)=N_{+}{\cup }_{\partial N}{N}_{-}$ be its double and define $F$ by applying $f$ on both copies. The Mayer–Vietoris sequence for this decomposition is natural under $F$. Alternating traces in a finite-dimensional exact sequence sum to zero, so the two copies of $N$ contribute twice and their intersection contributes with the opposite sign:This is the Lefschetz number of a doubled map.

Finally let $N$ be a pair of pants. If $f$ is fixed-point-free, so are its double $F$ and its boundary restriction. The Lefschetz fixed-point theorem and the displayed identity give $L(F)=L(f{|}_{\partial N})=0$, hence $L(f)=0$. Since $N$ is connected and $H^i(N;\mathbb{Q})$ is nonzero only for $i=0,1$,Suppose the boundary permutation $\sigma$ had a fixed component. The restriction there is a fixed-point-free circle homeomorphism and thus has degree $+1$, forcing $f$ to preserve the surface orientation. The homology action of a pair-of-pants homeomorphism would then have trace $\#\mathrm{Fix}(\sigma )-1$, which is $2$ for the identity permutation and $0$ for a transposition. Neither is $1$. Therefore $\sigma$ has no fixed component; a permutation of three objects with no fixed point is a three-cycle. Thus

For a continuous self-map $f$ of a finite CW complex, its Lefschetz number over $\mathbb{Q}$ isThe Lefschetz fixed-point theorem states that $L(f)\ne 0$ implies that $f$ has a fixed point. Consequently every fixed-point-free self-map has Lefschetz number zero.

The space in the question is a cyclic point-gluing of two-spheres. Before gluing, the $n$ spheres have total Euler characteristic $2n$. Each of the $n$ point identifications lowers the Euler characteristic by one, soEquivalently, $X\simeq S^1\vee {\bigvee }_{j=1}^n{S}^2$, with rational homology dimensions $1,1,n$ in degrees zero, one, and two.

The Lefschetz fixed-point theorem says that a continuous self-map $f$ of a compact triangulable space has a fixed point whenever its Lefschetz numberis nonzero. In the smooth nondegenerate case, the Lefschetz-Hopf fixed-point theorem expresses this number as the sum of the local fixed-point indices.

Identify $H_1({\mathbb{R}}^2/{\mathbb{Z}}^2;\mathbb{Z})$ with ${\mathbb{Z}}^2$ by sending $v\in {\mathbb{Z}}^2$ to the class of the loop $t\mapsto tv\mathrm{mod}{\mathbb{Z}}^2$. The image of this loop under $f$ lifts from $0$ to the path $t\mapsto \tilde{f}(tv)$, whose endpoint is $\tilde{f}(v)=A(v)$. Its homology class is therefore $A(v)$. Thus $A=f_{\ast }$ under this identification, and in particular $A$ is a homomorphism represented by an integer matrix.

The expansion inequality implies, after taking a derivative, thatfor every tangent vector. The supplied linear-algebra fact shows that every eigenvalue of $D{\tilde{f}}_x$ has modulus at least $\mu >1$, so $1$ is not an eigenvalue. Every fixed point of $f$ is consequently nondegenerate and contributes local index $+1$ or $-1$.

On the torus, the induced maps on $H_0,H_1,H_2$ have traces $1,\mathrm{tr}A,\det A$. HenceThe original expansion inequality applied to lattice vectors gives $|Av|\ge \mu |v|$ for $v\in {\mathbb{Z}}^2$, and homogeneity and density extend it to all $v\in {\mathbb{R}}^2$. If ${\lambda }_1,{\lambda }_2$ are the possibly complex eigenvalues of $A$, then $|{\lambda }_j|\ge \mu$, soSince every fixed point contributes an index of absolute value one, the Lefschetz-Hopf formula gives

Its homology and chain formulas areFor Lefschetz fixed-point theorem, argue contrapositively: after barycentric subdivision, a fixed-point-free map has a simplicial approximation such that no simplex meets its image. Every diagonal coefficient on chains is then zero, so $L(f)=0$.

The antipodal map on $S^n$ has degree $(-1{)}^{n+1}$. For even $n$ this is $-1$, whereas the identity has degree $1$, so they are not homotopic.

If the tangent field $v$ were nowhere zero, thenwould stay on the sphere because $x\cdot v(x)=0$, and would homotope the identity to the antipodal map. This is impossible for even $n$, so $v$ has a zero.