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For a connected complete Riemannian manifold of dimension $n\ge 2$ and constant sectional curvature $k>0$, its complete simply connected universal cover is the round sphere of radius $k^{-1/2}$. This is the constant-curvature case of Cartan-Ambrose-Hicks theorem. Thus the manifold is a quotient by a finite group of isometries acting freely. In even dimension, the determinant embeds this group in $\{1,-1\}$: every determinant-one orthogonal transformation in odd ambient dimension has eigenvalue one and therefore a fixed point. The only possible nonidentity element is the antipodal involution, giving Real projective space. In odd dimensions, free cyclic scalar actions on complex unit spheres instead produce lens spaces.

Complexification replaces each fiber of a real vector bundle by its tensor product with $\mathbb{C}$ over $\mathbb{R}$. The complexification of a real vector bundle of rank one is a complex line bundle. For the tautological bundle over Real projective space, this complex line bundle squares to the trivial vector bundle.

If $L$ is the complexification of a real vector bundle of the tautological bundle and $\mu =[L]-1$, thenRealizing Real projective space as the circle bundle of the square of the tautological bundle on Complex projective space gives this through the K-theory Gysin sequence of a sphere bundle.

The convex hull of the positive and negative coordinate unit vectors. Its boundary triangulates a sphere, and each face contains at most one member of each antipodal pair of coordinate unit vectors. This makes its barycentric subdivision useful for triangulating Real projective space.

In unit-curvature Real projective space of dimension at least two, Riemannian distance is $d([v],[u])=\arccos |⟨v,u⟩|$ for unit lifts. A unit-speed geodesic has lift $\cos tv+\sin tw$, with $w$ unit and perpendicular to $v$, and therefore ceases to minimize after $\pi /2$. Its normal Jacobi fields with zero initial value are $\sin t$ times parallel vector fields along a curve, giving first conjugate point at time $\pi$. The two times differ in every direction.

For $n\ge 2$, a self-map of $S^n$ factoring through Real projective space ${\mathbb{RP}}^n$ has degree zero when $n$ is even, and can have exactly the even degrees when $n$ is odd. In the odd case the first map lifts through the double covering and the covering has degree two. The dimension-one exception allows every integer, since ${\mathbb{RP}}^1\cong S^1$.

For a closed oriented manifold of dimension $4r+2$, Poincare duality pairs all rational Betti numbers except the middle one, making their total contribution to the Euler characteristic even. The middle cohomology degree is odd, so its nondegenerate Poincare duality pairing is an alternating bilinear form and has even dimension. Hence the Euler characteristic is even. Nonorientable manifolds can have odd Euler characteristic, as Real projective space of even dimension demonstrates.

This is the CW complex obtained as the increasing union of finite Real projective spaces under their standard inclusions. It has one cell in each nonnegative dimension. Its cellular chain complex has boundary multiplication by two in positive even dimensions and zero in odd dimensions. Consequently its integral cohomology is $\mathbb{Z}$ in degree zero, $\mathbb{Z}/2$ in positive even degrees and zero in odd degrees. Its mod-two cohomology is ${\mathbb{Z}}_2$ in every nonnegative degree.

The cellular cochain complex of finite Real projective space consists of alternating maps zero and two. Its groups can be tensored using the Künneth theorem; the integral Künneth torsion polynomial efficiently counts all tensor and Tor functor summands. A product of factors of dimensions two, three and four has free-rank polynomial $1+t^3$ and torsion polynomial $3t^2+3t^3+5t^4+6t^5+5t^6+4t^7+2t^8+t^9$.

Let $L$ be the complexification of a real vector bundle of the tautological bundle over Real projective space ${\mathbb{RP}}^{2n+1}$. For $k\ge 1$, the K-theory Euler class of $L^{\oplus k}$ is $-2^{k-1}([L]-1)$, andThe K-theory Gysin sequence of a sphere bundle gives an extension by $\mathbb{Z}$, which splits as an exact sequence of abelian groups.

For any continuous map between the displayed Real projective spaces, the induced map on cohomology sends the degree-one generator to $\alpha x+\beta y$. Its $2n$th power is $\left(\genfrac{}{}{0ex}{}{2n}{n}\right){\alpha }^n{\beta }^n{x}^n{y}^n$, since all other terms exceed the truncation exponent of one factor. The coefficient is zero modulo two because $\left(\genfrac{}{}{0ex}{}{2n}{n}\right)=2\left(\genfrac{}{}{0ex}{}{2n-1}{n-1}\right)$. Naturality of the evaluation pairing then makes the top induced map on homology zero. This obstruction does not apply to $n=0$, when both spaces are points.

A nematic director is a unit vector specifying the local unoriented axis of a nematic liquid crystal. Head-tail symmetry identifies $\mathbf{n}$ with $-\mathbf{n}$, so the order-parameter space is a Real projective space rather than a sphere.

For $n\ge 2$, a nonsurjective self-map of Real projective space ${\mathbb{RP}}^n$ has zero mod-two degree of a map between closed manifolds. In the mod-two cohomology ring of real projective space, the nonzero degree-one class $a$ has $a^n\ne 0$. If pullback fixed $a$, it would fix $a^n$, contradicting zero degree. Thus the map is trivial on the fundamental group and lifts to $S^n$. A point missed by the original map has both its inverse images missed by the lift. The lift therefore takes values in a punctured sphere, which is contractible, and is null-homotopic. Composing its contraction with the covering projection proves the conclusion.

An odd map between spheres is a continuous map $f:S^n\to S^m$ satisfying $f(-x)=-f(x)$. Thus it is equivariant for the two antipodal maps. It descends to a continuous map of Real projective spaces.

The outward-normal-first volume form on the unit sphere pulls back under the antipodal map by $(-1{)}^{n+1}$. For $n\ge 1$, the sphere is connected, and its orientation descends to the twofold quotient exactly when this deck transformation preserves it. Thus Real projective space is orientable precisely in odd positive dimensions; the zero-dimensional case is one point and is also orientable.

Work over $\mathbb{Q}$ and put $b_i=\dim H^i(M;\mathbb{Q})$. For a closed orientable six-dimensional manifold, Poincare duality gives $b_i=b_{6-i}$, soThe remaining Betti number is even. Indeed, the cup product pairingis nondegenerate by duality and skew-symmetric by graded commutativity, since $(-1{)}^{3\cdot 3}=-1$. If its dimension were odd, a matrix $A$ for it would satisfy $\det A=\det (-A^T)=-\det A$, forcing $\det A=0$ in characteristic zero, a contradiction. Thus $b_3$ is even andThis is the Euler characteristic parity in dimensions congruent to two modulo four. If the manifold has several components, apply the argument to each component and add.

The orientability assumption is essential. The Real projective space ${\mathbb{RP}}^6$ has one cell in each dimension from zero to six, givingIt is nonorientable: its double cover $S^6$ has the orientation-reversing antipodal deck transformation, of degree $(-1{)}^7=-1$.

For the construction, ${\mathbb{CP}}^3$ is a closed orientable real six-dimensional manifold with cells in dimensions $0,2,4,6$, hence Euler characteristic four. The product $S^3\times S^3$ is also closed and orientable, with Euler characteristic $\chi (S^3{)}^2=0$. The six-sphere has Euler characteristic two.

In dimension six, removing an open ball subtracts one from the Euler characteristic. To see this, write $M$ as the punctured manifold joined to a closed ball along $S^5$, whose Euler characteristic is zero, and apply the cellular inclusion-exclusion formula. Gluing two punctured manifolds along $S^5$ therefore givesA connected sum of $r$ copies of ${\mathbb{CP}}^3$ and $s$ copies of $S^3\times S^3$ haswith the empty sum interpreted as $S^6$. Given any even integer $n$, takeBoth are nonnegative integers and the displayed Euler characteristic equals $n$. Hence every even integer is the Euler characteristic of a closed oriented six-manifold, including zero and negative integers; the construction is connected.

A finite CW complex can be defined without assuming separation properties in advance. Start with a finite discrete set $X^0$. Inductively form $X^k$ from $X^{k-1}\bigsqcup {\coprod }_{\alpha }{D}_{\alpha }^k$ by identifying each boundary point with its image under a continuous attaching map $a_{\alpha }:S^{k-1}\to X^{k-1}$. Only finitely many disks are attached in total, and $X=X^N$ for some finite $N$. The interiors of the disks are its cells. Its weak topology of a CW complex is the final topology of the characteristic maps: a subset is closed exactly when its inverse image in every characteristic disk is closed. In this finite construction it is the topology obtained by the successive quotients. Closure-finiteness is automatic because there are only finitely many cells.

Here is a direct proof of both requested topological properties, without presupposing that arbitrary attaching maps give a triangulable space. Suppose the compact previous stage $Y$ embeds by $e:Y\hookrightarrow {\mathbb{R}}^m$, and attach one $k$-cell, $k\ge 1$. Extend the boundary coordinate map radially byIt is continuous at zero because the image of the sphere is bounded. Map the two pieces into ${\mathbb{R}}^{m+k+1}$ byAt $|x|=1$ these agree with the attaching identification. They therefore induce a continuous map on the quotient. Its restriction to the old stage is injective. An interior point has positive last coordinate $t=1-|x|$, and the preceding coordinate block recovers $x$ by division by $t$. Hence it cannot coincide with an old point or another interior point. The induced map is injective.

The quotient is compact, being the image of a compact disjoint union. A continuous injection into Euclidean space is Hausdorff, by inverse images of disjoint separating neighborhoods; it is also a homeomorphism onto its image because closed subsets of the compact source have compact, hence closed, images in the Hausdorff target. Finite discrete $X^0$ embeds in $\mathbb{R}$, and induction over the finite list of cells proves $X$ is Hausdorff and its weak topology is exactly an induced Euclidean subspace topology. This is Euclidean embedding by finite cell attachment; no optimal ambient dimension is needed.

For Real projective space, filter by ${\mathbb{RP}}^0\subset \cdots \subset {\mathbb{RP}}^n$. The complement of ${\mathbb{RP}}^{k-1}$ in ${\mathbb{RP}}^k$ is an affine $k$-cell. A characteristic disk is a hemisphere of $S^k$, whose boundary is attached by the antipodal quotient $S^{k-1}\to {\mathbb{RP}}^{k-1}$. Thus there is one cell in every dimension $0,\dots ,n$.

The incidence number is the sum of the two sheet contributions. Fixing the orientation of one sheet gives degree one; the other differs by the antipodal map on $S^{k-1}$, whose degree is $(-1{)}^k$. Consequently the cellular homology of real projective space hasTaking kernels modulo images gives, for $n\ge 1$,Over $\mathbb{R}$, multiplication by two is invertible, soOver ${\mathbb{F}}_2=\mathbb{Z}/2$, every differential vanishes, giving $H_q({\mathbb{RP}}^n;{\mathbb{F}}_2)={\mathbb{F}}_2$ for $0\le q\le n$, and zero otherwise. For $n=0$ the space is a point and only degree-zero homology is present, with the chosen coefficient group.

By part (b), $M=S^n/\Gamma$, where the group of deck transformations acts freely by round isometries. Every round isometry is the restriction of an orthogonal transformation of ${\mathbb{R}}^{n+1}$: it preserves the inner product $⟨x,y⟩=\cos d(x,y)$, so its images of an orthonormal basis determine its linear extension.

Suppose $n$ is even. Every $A\in SO(n+1)$ has eigenvalue one. Indeed, the nonreal eigenvalues come in conjugate pairs with product one, while the remaining real eigenvalues are $\pm 1$. Their number is odd and their product is one, forcing at least one $+1$. Such a transformation fixes a unit vector. Freeness of the deck transformation action consequently implies that the determinant map $\Gamma \to \{\pm 1\}$ has trivial kernel. Thus $|\Gamma |\le 2$.

If $\Gamma$ is nontrivial with generator $A$, then $A^2=I$. Its eigenvalues are $\pm 1$, and freeness excludes the eigenvalue $+1$. Hence $A=-I$, the antipodal involution. We obtainThis proves the even-dimensional part of the classification of complete positive constant-curvature manifolds.

The conclusion fails in odd dimensions. For $n=2m-1\ge 3$, identify $S^{2m-1}$ with the unit sphere in ${\mathbb{C}}^m$. Scalar multiplication by $\zeta =e^{2\pi i/r}$, with $r\ge 3$, generates a freely acting cyclic group of isometries: ${\zeta }^{j}z=z$ for nonzero $z$ forces ${\zeta }^j=1$. The quotient is complete with constant sectional curvature one and has fundamental group $C_r$, so it is neither the simply connected sphere nor Real projective space with fundamental group $C_2$. In particular $S^3/C_3$ is a lens space counterexample.

Take unit-curvature Real projective space ${\mathbb{RP}}^n=S^n/\{\pm 1\}$ with $n\ge 2$, and any point $p=[v]$. It is complete with positive sectional curvature. A unit-speed geodesic through $p$ lifts toSince the two representatives of an endpoint are antipodal, Riemannian distance in the quotient is the smaller of the two spherical distances:This equals $t$ for $0\le t\le \pi /2$ and is strictly smaller just afterwards. Thus the Riemannian cut point occurs at time $\pi /2$ in every direction.

Along a curvature-one geodesic, every normal Jacobi field with initial value zero is $J(t)=\sin t{P}_{t}a$, where $P_t$ is parallel transport and $a\perp \dot{\gamma }(0)$. Its first further zero is $t=\pi$. The quotient map is a local isometry, so the Jacobi field equation and conjugacy times are unchanged. HenceThis is cut and conjugate times in round real projective space.

Choose real constants $a_0<a_1<\cdots <a_m$ and define the smooth homogeneous quotientIt is unchanged by nonzero rescaling, so it is well defined on Real projective space. On the unit sphere its differential vanishes on the tangent space precisely when $Ax$ is proportional to $x$, where $A=\mathrm{diag}(a_0,\dots ,a_m)$. The entries are distinct, so the only critical projective lines are $p_i=[e_i]$. In the affine chart $x_i=1$, write $u_j=x_j/x_i$ for $j\ne i$. ThenIts Hessian at zero is diagonal with entries $2(a_j-a_i)$. Exactly $i$ are negative, and none vanish. Thus this is a Morse function with one critical point $p_i$ of index $i$ for every $0\le i\le m$.

Choose the quotient of the round metric on the unit sphere. The downward gradient flow has sphere equation ${\dot{x}}_j=-2(a_j-f(x))x_j$ and projective solutionNormalization to the unit sphere supplies the common positive factor. As $t\to +\infty$, the smallest index with nonzero coordinate dominates; as $t\to -\infty$, the largest dominates. ConsequentlyThese are its unstable manifolds and stable manifolds. If $j\ge k$, the tangent directions in $W^u(p_j)$ allow coordinates $0,\dots ,j$, and those in $W^s(p_k)$ allow $k,\dots ,m$, modulo the common projective line. Together they span the full tangent space at every intersection. If $j<k$ the intersection is empty. Thus the flow is a Morse-Smale gradient flow, with no generic perturbation needed.

Orient each unstable manifold using its affine coordinates. The Morse-Smale complex has $C_i=\mathbb{Z}{p}_i$; its differential counts oriented unparametrized connecting trajectories of index difference one. A trajectory from $p_i$ to $p_{i-1}$ has only coordinates $x_{i-1},x_i$ nonzero, and there are exactly two, distinguished by the sign of $x_{i-1}/x_i$. Each sign branch is a single orbit modulo time translation.

To calculate their signs rather than merely count two, use the characteristic closed disk of the unstable cell: the closed upper hemisphere of $S^i$ with $x_i\ge 0$, followed by projection onto ${\mathbb{RP}}^i$. Its interior is $W^u(p_i)$, and its boundary map is the antipodal double cover $S^{i-1}\to {\mathbb{RP}}^{i-1}$. After collapsing ${\mathbb{RP}}^{i-2}$, its degree on the $(i-1)$-cell is the sum of the degrees of the two sheets. Comparing the sheets by the antipodal map on $S^{i-1}$ gives relative orientation $(-1{)}^i$: it is induced by $-I$ on ${\mathbb{R}}^i$, whose determinant is $(-1{)}^i$. Thus the two trajectory signs add to $1+(-1{)}^i$, up to the choice of generator orientations. The projective quadratic Morse complex is thereforeIn particular ${\partial }_1=0$, as required for the connected one-dimensional skeleton. The signed trajectory count agrees with this attaching-map incidence calculation, so it computes integral rather than only mod-two homology.

For $m\ge 1$, taking kernels modulo images givesFor example, an odd interior-degree generator has zero outgoing differential but twice itself is the image from the next degree; an even positive-degree differential is injective. At the top there is no incoming differential, leaving a copy of $\mathbb{Z}$ exactly when $m$ is odd. For $m=0$ the space is one point and only $H_0=\mathbb{Z}$ remains. This recovers the cellular homology of real projective space by explicitly constructing its Morse-Smale complex and signs.