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The standard CW complex structure on Real projective space has one cell in each dimension from zero to three. Its integral cellular boundary is multiplication by $2$ in even positive degrees and zero in odd degrees. Thus the integral cellular cochain complex for ${\mathbb{RP}}^3$ isin degrees $0,1,2,3$. With coefficients ${\mathbb{F}}_2$, all its differentials vanish, so every one of these four cohomology groups is one-dimensional.

The lift-and-divide construction of the Bockstein homomorphism turns the integral differential $2$ into $1$ modulo $2$. Therefore $\beta :H^1\to H^2$ is an isomorphism, while the maps from degrees $0,2,3$ are zero. The Bockstein cohomology is consequentlyFor comparison, in the mod-two cohomology ring of real projective space ${\mathbb{F}}_2[t]/(t^4)$, $|t|=1$, this says $\beta (t)=t^2$, $\beta (t^2)=0$ and $\beta (t^3)=0$. The last two formulas also follow from the Bockstein derivation rule and the truncation $t^4=0$.

The real tautological line bundle has total space $E=\{(\ell ,v):\ell \in {\mathbb{RP}}^n,v\in \ell \}$ and projection $(\ell ,v)\mapsto \ell$. On the open chart $U_i=\{[x]:x_i\ne 0\}$, the vector $s_i([x])=x/x_i$ is a continuous nowhere-zero section of a vector bundle, independent of the representative of $[x]$. The explicit local trivialization iswith inverse $([x],v)\mapsto ([x],v_i)$. Both maps are continuous and linear on each fiber. The charts cover the base, proving the existence of the required local trivializations.

Let $E\to B$ be a rank-$r$ vector bundle oriented over a commutative ring $R$, with $r\ge 1$. An $R$-orientation is a coherent choice of generator of each fiber's top relative cohomology, or equivalently a Thom class $u_E$. The Thom isomorphism theorem identifies $H^{q-r}(B;R)$ with $H^q(D(E),S(E);R)$ by $a\mapsto {\pi }^{\ast }a⌣u_E$. The disk bundle retracts to $B$, and the relative-to-absolute map becomes multiplication by the Euler class $e(E)$. The pair's long exact sequence therefore becomes the Gysin sequence of a sphere bundle:With real coefficients, the bundle must be oriented in the usual sense. For the computation here use $R={\mathbb{F}}_2$, over which every real vector bundle is oriented, including the nonorientable real tautological line bundle when $n\ge 1$.

The unit sphere bundle of ${\gamma }_n$ is $S^n$: a unit vector determines its line. Put $a=e_2({\gamma }_n)=w_1({\gamma }_n)\in H^1({\mathbb{RP}}^n;{\mathbb{F}}_2)$. For $n\ge 1$, the pullback $H^0({\mathbb{RP}}^n;{\mathbb{F}}_2)\to H^0(S^n;{\mathbb{F}}_2)$ is an isomorphism, so the next connecting homomorphism is zero and multiplication by $a$ is injective on $H^0$. Since the sphere's cohomology vanishes strictly between degrees zero and $n$, exactness makes multiplication by $a$ an isomorphism $H^{q-1}\to H^q$ for $1\le q<n$, and injective for $q=n$. For $q=1$ these assertions also use the just-noted vanishing of the connecting homomorphism from $H^0(S^n)$.

The standard $n$-dimensional CW complex structure of Real projective space gives $H^q=0$ for $q>n$. Consequently the segment $H^n(S^n;{\mathbb{F}}_2)\to H^n({\mathbb{RP}}^n;{\mathbb{F}}_2)\stackrel{⌣a}{\to }{H}^{n+1}({\mathbb{RP}}^n;{\mathbb{F}}_2)=0$ makes $H^n$ at most one-dimensional. The preceding injectivity makes it exactly one-dimensional, generated by $a^n$. This includes $n=1$; $n=0$ is a point separately. We have derived the multiplication, not just the dimensions:

An odd map between spheres descends to $\bar{f}:{\mathbb{RP}}^n\to {\mathbb{RP}}^m$. The odd maps pull back the real tautological line bundle argument gives ${\gamma }_n\cong {\bar{f}}^{\ast }{\gamma }_m$: with $x$ a unit vector the fiber map is $tx\mapsto tf(x)$, unchanged when $(x,t)$ is replaced by $(-x,-t)$. Naturality of the first Stiefel–Whitney class gives ${\bar{f}}^{\ast }b=a$. If $n>m$, the relation $b^{m+1}=0$ would imply $a^{m+1}=0$ in a ring where this power is nonzero. Hence $\boxed{n\le m}$. The case $n=0$ is immediate, and the same argument rules out $m=0<n$.

For the map with separate oddness, pass to $\bar{g}:{\mathbb{RP}}^n\times {\mathbb{RP}}^n\to {\mathbb{RP}}^n$. Let $a,b$ be the degree-one classes from its two domain factors and $c$ the target class. There is an isomorphismgiven on fibers by $(sx)\otimes (ty)\mapsto stg(x,y)$; flipping either unit representative leaves this map well defined. The First Stiefel–Whitney class of a tensor product of real line bundles gives ${\bar{g}}^{\ast }c=a+b$. By the Künneth theorem,Pulling back $c^{n+1}=0$ shows that $(a+b{)}^{n+1}=0$. The two end monomials already vanish, while every interior monomial $a^j{b}^{n+1-j}$, $1\le j\le n$, is a distinct nonzero basis element. Therefore all interior binomial coefficients in row $N=n+1$ are even. To identify such rows, write $N={\sum }_{i\in I}{2}^i$. In ${\mathbb{F}}_2[t]$ repeated squaring gives $(1+t{)}^N={\prod }_{i\in I}(1+t^{2^i})$. If $I$ has at least two elements, the coefficient of $t^{2^{\min I}}$ is one and its exponent is strictly between zero and $N$, a contradiction. Thus $N$ is a power of two, as in binomial coefficients with even interior terms, andThis is the necessary cohomological obstruction to separately odd sphere multiplication; it does not assert existence in every dimension of this form.

Use ${\mathbb{F}}_2$ coefficients throughout. Write $h=c_1(\mathcal{O}(1))\mathrm{mod}2$ on Complex projective space and $a=w_1({\gamma }_d)$ on Real projective space. We use the standard cohomology ring of complex projective space ${\mathbb{F}}_2[h]/(h^{d+1})$, with $|h|=2$, and the real projective calculation from the preceding solution. The complex projective calculation follows, for example, from its one cell in each even dimension together with the fact that transverse projective hyperplanes represent the powers of $h$; $d$ such hyperplanes have one intersection point.

For the standard inclusion $i:{\mathbb{RP}}^d\hookrightarrow {\mathbb{CP}}^d$, the restricted complex tautological line is ${\gamma }_d{\otimes }_{\mathbb{R}}\mathbb{C}$. As a real vector bundle it is ${\gamma }_d\oplus {\gamma }_d$, whose total Stiefel–Whitney class is $(1+a{)}^2=1+a^2$. The Stiefel–Whitney class of the underlying real bundle of a complex line identifies its degree-two class with its First Chern class reduced modulo two. Passing to the dual line defining $h$ changes only a sign, which disappears modulo two. Thus mod-two restriction from complex to real projective space gives $i^{\ast }h=a^2$. In particular the answer for dimension two is the ring mapIt sends $1$ to $1$, is an isomorphism in degree two, and sends $h^2$ to zero.

Now assume $k\ge 1$ and put $M={\mathbb{CP}}^{2k}$, $N={\mathbb{RP}}^{2k}$, $U=M\setminus N$, with inclusions $i:N\hookrightarrow M$ and $j:U\hookrightarrow M$. The real dimensions are $4k$ and $2k$, respectively. A tubular neighborhood and excision identify the cohomology of $(M,U)$ with that of the normal disk bundle relative to its sphere bundle. Every real normal bundle is oriented over ${\mathbb{F}}_2$, so the Thom isomorphism theorem givesUnder this identification the relative-to-absolute map is the cohomological Gysin map of an embedding $i_{!}$. The exact sequence of the pair is thereforeHere $\delta$ denotes the ordinary connecting homomorphism followed by the inverse of the isomorphism supplied by the Thom isomorphism theorem.

To compute $i_{!}$, use mod-two Poincare duality and its evaluation formulaFor $0\le r\le k$, take $b=a^{2r}$ and $z=h^{k-r}$. The right side is $⟨a^{2k},[N{]}_2⟩=1$. Since the target group is generated by $h^{k+r}$ and $⟨h^{2k},[M{]}_2⟩=1$, this provesOdd-degree classes have zero image because $M$ has no odd-degree cohomology. Exactness now givesFor precision, below degree $2k-1$ the restriction map is an isomorphism. In degree $2k-1$ the next map is $i_{!}:H^0(N)\to H^{2k}(M)$, an isomorphism, so the complement group is zero. In even degrees $2k+2r$, $0\le r<k$, the preceding $i_{!}$ is an isomorphism and the following odd-degree $i_{!}$ is zero; consequently $\delta :H^{2k+2r}(U)\to H^{2r+1}(N)$ is an isomorphism. The intervening odd groups and the top group $H^{4k}(U)$ vanish.

It remains to establish multiplication. Put $x=j^{\ast }h$. Since $h^k=i_{!}1$, exactness gives $x^k=0$. Choose $y\in H^{2k}(U)$ with $\delta y=a$, which is possible by the just-computed isomorphism. Naturality of the relative cup product, and the module property of the Thom isomorphism theorem, giveThus $\delta (x^{r}y)=a^{2r+1}$ for $0\le r<k$. Each class $x^{r}y$ is nonzero and generates its upper-half group, while $1,x,\dots ,x^{k-1}$ generate the lower-half groups. Also $y^2=0$ because $H^{4k}(U)=0$. These classes account for every group, so there are no further relations:By the field-coefficient Künneth theorem this is precisely the cohomology ring of ${\mathbb{CP}}^{k-1}\times S^{2k}$. For $k=1$ the relation $x=0$ leaves the sphere's ring. The calculation identifies rings and does not claim a homotopy equivalence of the spaces; $k=0$ is outside the expression involving ${\mathbb{CP}}^{k-1}$.

The two Lie groups have the same local infinitesimal structure, but different global topology. This difference determines which Lie algebra representations integrate to representations of each group.

An element of the SU(2) group is a unitary matrix of determinant one. Orthogonality of its columns and its determinant give the unique formThus its group manifold is the unit three-sphere in ${\mathbb{C}}^2\simeq {\mathbb{R}}^4$: this is SU(2) as the three-sphere. In particular it is compact, connected and simply connected. In terms of the Pauli matrices one may also write $U=a_{0}I-i\mathbf{a}\cdot \mathbf{\sigma }$ with real coefficients satisfying $a_0^2+{\mathbf{a}}^2=1$.

Differentiate $U(t{)}^{†}U(t)=I$ and $\det U(t)=1$ at $U(0)=I$. The tangent space consists of traceless skew-Hermitian matrices:The Lie bracket of a Matrix Lie group is the matrix commutator. With $T_a=-i{\sigma }_a/2$, the Pauli matrix multiplication law givesThis derives the SU(2) Lie algebra as a three-dimensional real Lie algebra. The Hermitian physics generators $J_a={\sigma }_a/2$ instead obey $[J_a,J_b]=i{ϵ}_{abc}{J}_c$; they are $i$ times the skew-Hermitian tangent generators, so these are consistent conventions.

The SO(3) group consists of real orthogonal matrices with determinant one. Differentiating $R(t{)}^{T}R(t)=I$ at the identity givesThe determinant condition gives no additional infinitesimal constraint because a skew-symmetric matrix already has trace zero. Define $L_a\mathbf{v}={\mathbf{e}}_a\times \mathbf{v}$. The cross product identity impliesTherefore the SO(3) Lie algebra has the same structure constants and $T_a\mapsto L_a$ is a Lie algebra isomorphism.

The global relation is the Adjoint double cover from SU(2) to SO(3). For $V=\mathbf{v}\cdot \mathbf{\sigma }$, define $R(U)$ byConjugation preserves the real space of traceless Hermitian matrices and its inner product $\frac{1}{2}\mathrm{tr}(VW)=\mathbf{v}\cdot \mathbf{w}$. Hence $R(U)$ is orthogonal. Continuity and connectedness, together with $R(I)=I$, put it in $SO(3)$. Composition of conjugations makes $R$ a group homomorphism. If $R(U)=I$, then $U$ commutes with every Pauli matrix, hence is scalar; unitarity and determinant one leave precisely $U=\pm I$. The differential sends $T_a$ to $L_a$, so it is an isomorphism. More concretely,induces rotation through angle $\theta$ about $\mathbf{n}$, by the Rodrigues rotation formula. Every three-dimensional rotation has such an axis and angle, proving surjectivity. ConsequentlyThe matrices $U$ and $-U$ are antipodal points on the three-sphere, so the SO(3) group manifold is Real projective space ${\mathbb{RP}}^3$. Equivalently, the closed axis-angle ball $\Vert \theta \mathbf{n}\Vert \le \pi$ has opposite boundary points identified. The fundamental group is ${\pi }_1(SO(3))\simeq {\mathbb{Z}}_2$, whereas ${\pi }_1(SU(2))=0$. A $2\pi$ rotation lifts from $I$ to $-I$; a $4\pi$ rotation returns to $I$. Thus the covering is the universal cover and the groups are not globally isomorphic.

For representation theory, specify finite-dimensional complex continuous representations. Compactness permits an invariant Hermitian inner product, obtained by averaging against Haar measure, and therefore complete reducibility. Complexifying either real Lie algebra gives the sl2 Lie algebra. Its finite-dimensional irreducibles are indexed by $n\in {\mathbb{Z}}_{\ge 0}$, have highest weight $n$, and have dimension $n+1$. By integration of a Lie-algebra representation, since $SU(2)$ is simply connected, every such Lie algebra representation integrates uniquely. The resulting homogeneous polynomial representation of SU2 isIn the spin angular momentum notation $j=n/2$, its Hermitian $J_3$ eigenvalues are $j,j-1,\dots ,-j$. The central matrix $-I$ acts on the symmetric power by $(-1{)}^n$, so descent of an SU(2) representation to SO(3) occurs exactly when $n$ is even. HenceFor a reducible representation, every summand must satisfy the descent condition. The spin-one-half doublet is a genuine representation of the covering group but does not define a single-valued representation of $SO(3)$; the spin-one triplet does and is its vector representation. The distinction is topological, rather than a difference in their isomorphic Lie algebras.

Triangulate the sphere by the boundary complex of the cross-polytope with simplicial vertices $\pm e_0,\dots ,\pm e_n$: a face contains at most one simplicial vertex from each antipodal pair. Take its barycentric subdivision. Its simplicial vertices are the nonempty faces $F$ of the original complex, and its simplices are strictly nested chains $F_0\subset \cdots \subset F_k$.

Form a quotient complex whose simplicial vertices are pairs $\{F,-F\}$, and whose simplices are the images of these chains. No chain contains both $F$ and $-F$. Moreover a chain of face-orbits lifts uniquely up to simultaneous negation: after choosing a largest face, every smaller face has at most one representative contained in it, because an original face contains no antipodal simplicial vertices. Thus the quotient is an honest simplicial complex, without two different simplices with the same simplicial vertex set. Its realization is homeomorphic to the sphere modulo its antipodal action. This provides a triangulation of Real projective space ${\mathbb{RP}}^n$. The subdivision avoids the duplicate-simplex problem of naively identifying simplicial vertices in the unsubdivided cross-polytope boundary.

For connected oriented closed manifolds of dimension $n$, the degree of a map between oriented manifolds is the integer determined bywhere the brackets are the chosen fundamental classes. Functoriality gives $\mathrm{deg}(g\circ f)=\mathrm{deg}g\mathrm{deg}f$.

Give $S^n$ its boundary orientation as the boundary of the unit ball in ${\mathbb{R}}^{n+1}$. The antipodal map is the restriction of $-I$. Its ambient determinant is $(-1{)}^{n+1}$, and it takes the outward normal at $x$ to the outward normal at $-x$. Thus its effect on the boundary orientation has this same sign, givingThe original PDF has $A:S^n\to S^n$, correcting the extra prime on the target in the TeX transcription.

For even dimension $n=2k\ge 2$, cellular homology of real projective space gives $H_n({\mathbb{RP}}^n;\mathbb{Z})=0$. Thus the composite map on top homology factors through zero, andZero is attained by constant maps.

For odd dimension $n=2k+1\ge 3$, Real projective space is orientable: the antipodal deck transformation has degree $+1$. Orient it so that the double covering map $q:S^n\to {\mathbb{RP}}^n$ has degree $2$. Since $S^n$ is simply connected, the lifting criterion for a covering space gives $f=q\circ \tilde{f}$ for a map $\tilde{f}:S^n\to S^n$. Hencewhich is even. Every even integer occurs: collapse the complement of an oriented embedded disk in ${\mathbb{RP}}^n$ to obtain a degree-one map $g:{\mathbb{RP}}^n\to S^n$, and choose $\tilde{f}$ of any prescribed integer degree $m$. Maps of all integer degrees on spheres are obtained, for example, by suspending the circle maps $z\mapsto z^m$. Setting $f=q\circ \tilde{f}$ gives degree $2m$. ThusThese are the degrees of maps factoring through real projective space. If $k=0$ is permitted in the odd-dimensional clause, there is an exception: ${\mathbb{RP}}^1\cong S^1$, so every integer degree occurs. The lifting argument requires dimension at least two. The initial connected-manifold degree definition excludes $S^0$; with the usual reduced-homology definition for self-maps of $S^0$, a factorization through the one-point ${\mathbb{RP}}^0$ has degree zero.

Finally suppose $\mathrm{deg}f=p$ is prime. Fix any prime $q\ne p$ and work over $F={\mathbb{F}}_q$. We claim that $f^{\ast }:H^r(N;F)\to H^r(S^n;F)$ is injective. If $a\ne 0$, Poincare duality supplies $b\in H^{n-r}(N;F)$ with $⟨a⌣b,[N{]}_F⟩\ne 0$. Naturality of the cup product and evaluation givesbecause $p$ is invertible in $F$. Hence $f^{\ast }a\ne 0$, proving cohomological injectivity of a map of invertible degree.

The intermediate cohomology of the sphere is zero, so $H^r(N;F)=0$ for $0<r<n$. Over a field, the universal coefficient theorem for cohomology identifies this with the dual of $H_r(N;F)$, so these homology groups vanish as well. The universal coefficient theorem for homology then givesEach integral group is a finitely generated abelian group, by the permitted finite CW complex model. Its decomposition can have no free summand, since that would survive modulo $q$, and no torsion summand divisible by any prime $q\ne p$. Thus it is a finite $p$-primary group. There are only finitely many intermediate degrees, so the exponents of their finite cyclic summands have a common bound $p^a$. Taking $a\ge 1$ also covers the case in which all the groups vanish. ThereforeThis proves that a prime-degree sphere map forces primary torsion, including one uniform exponent for all the degrees.

As usual for Complex projective space and Real projective space, homogeneous coordinates are represented by nonzero vectors. Componentwise complex conjugation descends to the diffeomorphismIt is well defined because $\overline{\lambda z}=\overline{\lambda }\overline{z}$, and ${\tau }^2=I$. Every standard manifold chart is invariant under $\tau$, with ${\phi }_i\circ \tau =\sigma \circ {\phi }_i$. Part (d) and the gluing of the Fubini-Study form therefore imply ${\tau }^{\ast }{\omega }_{FS}=-{\omega }_{FS}$.

A point $[z]$ in the fixed-point set satisfies $\overline{z}=\lambda z$. Applying complex conjugation again gives $|\lambda |=1$. Choose $a\ne 0$ with $a/\overline{a}=\lambda$; then $\overline{az}=az$, so $[z]$ has a real representative. Conversely every real representative determines a point in the fixed-point set. Hence $\mathrm{Fix}(\tau )={\mathbb{RP}}^n$, and part (c) gives

Let $E$ have rank $r\ge 1$, let $p:\mathbb{P}(E)\to X$ be its projectivization of a real vector bundle, and let $\lambda \subset p^{\ast }E$ be its tautological bundle. Put $u=w_1(\lambda )\in H^1(\mathbb{P}(E);{\mathbb{F}}_2)$. The Projective bundle definition of Stiefel–Whitney classes starts from the isomorphismIn particular, $p^{\ast }$ is injective. Expand $u^r$ uniquely in this basis. The coefficients define the Stiefel–Whitney classes through the relationSet $w_0(E)=1$ and $w_i(E)=0$ for $i>r$; for a rank-zero vector bundle set $w(E)=1$. Uniqueness also proves naturality of the Stiefel–Whitney classes under a pullback vector bundle.

To prove the Whitney product formula for Stiefel–Whitney classes, apply the splitting principle for real vector bundles simultaneously to $E$ and $E^{\prime }$. Its injective pullback on mod-two cohomology follows by iterating the Projective bundle definition of Stiefel–Whitney classes. It suffices to compute for a split vector bundle $E={⨁}_{i=1}^r{L}_i$. Put $x_i=w_1(L_i)$.

On $\mathbb{P}(E)$, the coordinate projections of the tautological inclusion $\lambda \hookrightarrow p^{\ast }E$ give sections of the real line bundles ${\lambda }^{\ast }\otimes p^{\ast }{L}_i$ with no common zero. The mod-two Euler class of a real line bundle is its first Stiefel–Whitney class, and the First Stiefel–Whitney class of a tensor product of real line bundles gives $u+p^{\ast }{x}_i$. Multiplication of Euler classes under a direct sum, and vanishing of the Euler class of a vector bundle with a nowhere-zero section, therefore giveComparing with the unique monic relation defining the Stiefel–Whitney classes shows that $w_j(E)$ is the $j$th elementary symmetric polynomial in the $x_i$. Thus $w(E)={\prod }_i(1+x_i)$. A direct sum joins the two lists, so injectivity of the splitting pullback gives

For the immersion obstruction, use the notation $g=(f,f)$: the printed $f\times f$ here means the map from one copy of Real projective space into the product. Write $x=w_1({\gamma }_{\mathbb{R}})$ and let $z$ be the mod-two reduction of $c_1({\gamma }_{\mathbb{C}}^{\ast })$. The mod-two cohomology ring of real projective space and the cohomology ring of complex projective space giveNontriviality in degree two forces $f^{\ast }z=x^2$. Necessarily $k\ge 1$, since the degree-two cohomology of ${\mathbb{CP}}^0$ vanishes.

The standard tangent bundle identities for Real projective space and Complex projective space areFor the underlying real vector bundle of a complex line bundle, its total Stiefel–Whitney class is $1+c_1\mathrm{mod}2$. HenceIf $g$ is homotopic to an immersion $j$, its normal bundle $\nu$ has rank $4k-(2k+1)=2k-1$, andThe Whitney product formula for Stiefel–Whitney classes and homotopy invariance of cohomology yieldThe division is legitimate in the truncated cohomology ring, since $1+x$ is a unit. The degree-$2k$ component isIt must vanish because $2k>\mathrm{rank}\nu$, while $x^{2k}\ne 0$. Therefore $k$ must be odd.

Use the standard complex K-theory Thom class convention. For a rank-$r$ complex vector bundle, its K-theory Euler class is the pullback of the K-theory Thom class along the zero section, namelyHere ${\Lambda }^j$ denotes an exterior power, and a Hermitian metric identifies the dual bundle with the conjugate vector bundle. Choosing ${\lambda }_{-1}(E)$ instead changes the K-theory Euler class by the unit $(-1{)}^r[\det E]$, so its kernel and cokernel computations are unchanged.

For the disk bundle $D(E)$ and sphere bundle $S(E)$, the K-theory Thom class identifies $K^i(D(E),S(E))$ with $K^i(X)$. The exact sequence of this pair becomes the K-theory Gysin sequence of a sphere bundle:The groups are two-periodic by Bott periodicity.

View Real projective space as the circle bundle of ${\gamma }^{\otimes 2}$ over Complex projective space, where $\gamma \to {\mathbb{CP}}^n$ is the tautological bundle. Indeed, $S^{2n+1}/\{\pm 1\}={\mathbb{RP}}^{2n+1}$ maps to $S({\gamma }^{\otimes 2})$ by $[v]\mapsto v\otimes v$; in each fiber squaring identifies exactly $v$ and $-v$.

Use Complex K-theory of complex projective space in the formThe K-theory Euler class of ${\gamma }^{\otimes 2}$ is $1-[\gamma {]}^{-2}$. Multiplying it by the unit $-[\gamma {]}^2$ gives $1-[\gamma {]}^2=2t-t^2$. The K-theory Gysin sequence of a sphere bundle therefore givesUnder the projection $q:{\mathbb{RP}}^{2n+1}\to {\mathbb{CP}}^n$, the pullback vector bundle $q^{\ast }\gamma$ is the complexification of a real vector bundle $L$ of the tautological bundle of Real projective space: the fiber map sends $[v,z]$ to $zv$. Thus $t$ pulls back to $1-[L]$. Put $\mu =[L]-1$. The relation $L^{\otimes 2}\cong 1$ gives ${\mu }^2=-2\mu$, and $t^{n+1}=0$ gives $2^n\mu =0$. HenceFor completeness, these relations give precisely the abelian group $\mathbb{Z}\oplus \mathbb{Z}/2^n$. In the basis $t,\dots ,t^n$, the image of multiplication by $2t-t^2$ is generated by $2t-t^2,2t^2-t^3,\dots ,2t^n$; its relation matrix is triangular with determinant $2^n$, and the relations express every $t^j$ as $2^{j-1}t$. Thus there is exactly one cyclic torsion subgroup of order $2^n$.

If $n\ge 1$ and $a=a_0+a_{1}t+\cdots +a_n{t}^n$ is in the kernel, successive coefficients of $(2t-t^2)a$ give $2a_0=0$, $2a_1-a_0=0$, and so on, up to $2a_{n-1}-a_{n-2}=0$. Hence $a_0=\cdots =a_{n-1}=0$, with $a_n$ arbitrary. For $n=0$ the multiplier is zero on $R=\mathbb{Z}$. In all cases,

Finally let $E=L^{\oplus k}$ with $k\ge 1$, over $B={\mathbb{RP}}^{2n+1}$. Since $L^{\ast }\cong L$, multiplicativity of the K-theory Euler class givesOn $K^0(B)=\mathbb{Z}\oplus (\mathbb{Z}/2^n)\mu$, the image of multiplication by this K-theory Euler class is the subgroup generated by $2^{k-1}\mu$: multiplication by an integer produces every such multiple, while multiplication by $\mu$ only adds multiples of $2^k\mu$. ThereforeMultiplication by $\mu$ is zero on $K^{-1}(B)\cong \mathbb{Z}$, since $2^n\mu =0$ and this abelian group has no nonzero torsion elements. Thus the K-theory Gysin sequence of a sphere bundle gives a short exact sequenceIt splits as an exact sequence of abelian groups because its quotient is a free abelian group. ConsequentlyThe convention $\mathbb{Z}/1=0$ covers $n=0$ and $k=1$. For example, when $k=1$ the sphere bundle is $(S^{2n+1}\times S^1)/\{(v,z)\sim (-v,-z)\}$, a mapping torus of the antipodal map; its Topological K-theory has no nonzero torsion elements, agreeing with the formula. If $k=0$ is allowed, the rank-zero vector bundle has an empty unit sphere bundle, so $K^0(S(E))=0$ instead.

Let $q$ denote the quotient map and write $[X]$ for a point of Real projective space. For $i=1,\dots ,n+1$, take the open setIt is open in the quotient topology because its inverse image under $q$ is the open set $\{X:X_i\ne 0\}$. These sets cover, since a nonzero vector has at least one nonzero coordinate. Define the manifold chartThe hat means that the $i$th coordinate is omitted. Ratios are unchanged by multiplying $X$ by any nonzero real number, so the chart is well defined. Its inverse inserts $1$ into position $i$ and takes the resulting equivalence class. The ratios and this inverse are continuous, giving a homeomorphism $U_i\to {\mathbb{R}}^n$.

For a compact description of the smooth transition maps, write $a_i=1$ and let $a_k=X_k/X_i$ for $k\ne i$. On $U_i\cap U_j$ one has $a_j\ne 0$, andEach coordinate is a rational smooth function on the open domain $a_j\ne 0$; reversing $i,j$ gives a smooth inverse. Thus these $n+1$ charts form the standard affine atlas of real projective space.

For completeness, the underlying space is Hausdorff: normalizing $X$ identifies it with the antipodal quotient of the unit sphere, and the continuous injective map $[X]\mapsto X{X}^T/(X^{T}X)$ into the space of symmetric matrices separates any two distinct classes by disjoint open neighborhoods. It is also a second-countable space: pull back the countable bases of rational balls in these finitely many charts, and take their union. Therefore

The standard CW complex structure on Real projective space ${\mathbb{RP}}^n$ has one cell in each dimension $0,1,\dots ,n$. Its cellular chain complex has $C_q\cong \mathbb{Z}$ and differentialConsequentlyAfter tensoring the cellular complex with ${\mathbb{F}}_2=\mathbb{Z}/2$, every differential vanishes, so

Because every $C_i$ is free abelian, reduction modulo $n$ gives a short exact sequence of chain complexesIts long exact sequence in homology isIf $\bar{z}$ is a cycle modulo $n$, choose a lift $z\in C_i$. Then $dz=nw$ for some $w\in C_{i-1}$, and the connecting homomorphism is $\hat{\beta }[\bar{z}]=[w]$.

The coefficient sequencesimilarly givesAt chain level, if $z$ lifts a mod-$n$ cycle and $dz=nw$ modulo $n^2$, then the Bockstein homomorphism is $\beta [\bar{z}]=[\bar{w}]$.

There is a morphism from the first short exact sequence to the second whose three vertical maps are reduction modulo $n$, reduction modulo $n^2$, and the identity on $C\otimes \mathbb{Z}/n$. Naturality of connecting homomorphisms givesExactness of the first long exact sequence says $\hat{\beta }{\rho }_{\ast }=0$, and hence