Past exam of the mathematics course of the University of Cambridge / 2005 / iii / Paper 18 (Fibre Bundles)

Sending a matrix to its last column gives the special orthogonal sphere fibrationThe action is transitive and the stabilizer of that column is $SO(n-1)$. Local orthonormal frames, obtained by Gram-Schmidt process near a chosen unit vector, give local sections and hence local bundle trivializations. This allows the long exact sequence of homotopy groups of a fibration to be applied.

At the bottom of the induction, $SO(1)$ is a point and $SO(2)$ is the rotation circle $S^1$. For $SO(3)$, identify ${\mathbb{R}}^3$ with imaginary quaternions. A unit quaternion acts by $v\mapsto qv{q}^{-1}$, preserving length and orientation. Every rotation through angle $\theta$ about a unit imaginary axis $u$ comes from $q=\cos (\theta /2)+u\sin (\theta /2)$. The kernel is precisely $\{1,-1\}$, since a quaternion commuting with all imaginary quaternions is real. Thus this gives the twofold covering space $S^3\to SO(3)$. Its simply connected total space is the universal cover, yielding ${\pi }_{1}SO(3)=\mathbb{Z}/2$. Coverings induce isomorphisms on higher homotopy groups, so ${\pi }_{2}SO(3)={\pi }_2{S}^3=0$ and ${\pi }_{3}SO(3)={\pi }_3{S}^3=\mathbb{Z}$.

For $n\ge 4$, both ${\pi }_1{S}^{n-1}$ and ${\pi }_2{S}^{n-1}$ vanish. The Serre fibration sequence therefore identifies ${\pi }_{1}SO(n-1)$ with ${\pi }_{1}SO(n)$ and propagates vanishing of ${\pi }_2$. Path-connectedness also propagates from connected fiber and base. Consequently the full low-degree answer isHere ${\pi }_0$ records the single connected component.

For the remaining third group use $SO(3)\to SO(4)\to S^3$. The relevant exact segment, using the supplied fourth homotopy group, isThe first homomorphism is zero since a torsion group cannot map nontrivially to $\mathbb{Z}$. Thus it reduces to a short exact sequence $0\to \mathbb{Z}\to {\pi }_{3}SO(4)\to \mathbb{Z}\to 0$. These groups are abelian and the final free group has a lift of its generator, so the sequence splits. ThereforeThese are the low homotopy groups of special orthogonal groups.

An immersion $M^4\to {\mathbb{R}}^5$ would inject $TM$ into the trivial rank-five bundle. Taking its orthogonal complement gives a normal real line bundle $\nu$ with $TM\oplus \nu \cong {\epsilon }^5$. The Whitney product formula for Stiefel–Whitney classes requiresThe real line bundle need not be orientable, but it has no Stiefel–Whitney classes above degree one.

Let $a,b$ be the pulled-back degree-one generators from the two real projective planes. Their mod-two cohomology ring, by the Künneth theorem, is ${\mathbb{F}}_2[a,b]/(a^3,b^3)$. To compute the tangent classes, at a line $L$ the projective tangent space is $\mathrm{Hom}(L,L^{\perp })$. Adding $\mathrm{Hom}(L,L)\cong \mathbb{R}$ gives $\mathrm{Hom}(L,{\mathbb{R}}^3)\cong 3L$, with the real line identified with its dual using a metric. Thus the stabilized tangent bundle of real projective space gives $T{\mathbb{RP}}^2\oplus {\epsilon }^1\cong 3\gamma$ andThe tangent bundle of the product is the direct sum of the pulled-back tangent bundles. Since $(1+a+a^2)(1+a)=1$ in this truncated ring, and similarly for $b$,The degree-two class $ab$ is nonzero by the product-ring basis. It cannot be a class of the rank-one bundle $\nu$. Equivalently $w_1(TM{)}^2+w_2(TM)=ab\ne 0$, violating the codimension-one Stiefel–Whitney immersion obstruction. Hence ${\mathbb{RP}}^2\times {\mathbb{RP}}^2$ admits no immersion in ${\mathbb{R}}^5$.

Apply the splitting principle for complex vector bundles simultaneously to $E$ and $F$. This can be implemented by successive projective bundle pullbacks; the projective bundle theorem makes their composite pullback injective on integral cohomology. After pullback write $E=L_1\oplus L_2$ and $F=M_1\oplus M_2$, with Chern roots $x_1,x_2$ and $y_1,y_2$. LetSince the First Chern class of a tensor product of complex line bundles is the sum of their first Chern classes, the four roots of $E\otimes F$ are $x_i+y_j$, $1\le i,j\le 2$.

Compute their six pairwise products integrally. The two pairs within a fixed $x_i$ row contributeThe four pairs between the two rows contributeTheir sum is the second elementary symmetric polynomial of the tensor roots. Injectivity of the pullback returns the resulting identity to $X$, givingProducts here are cup products in integral cohomology. This second Chern class of a tensor product of rank-two bundles formula uses no division by two and therefore holds without assuming the integral cohomology of $X$ is torsion-free.

Let $p:E\to S^r$ be the Serre fibration. For $r\ge 1$, cover the base by $U=S^r\setminus \{s\}$ and a small open ball $V$ about $s$. Write $A=p^{-1}U$ and $B=p^{-1}V$. The contractible-base assumption gives $A\simeq F$ and identifies the pair $(B,A\cap B)$, up to fiberwise homotopy, with $(F\times V,F\times (V\setminus \{s\}))$.

Excision for the open cover $E=A\cup B$ gives $H_i(E,A)\cong H_i(B,A\cap B)$. The relative base pair hasby the deformation retraction of the punctured ball onto $S^{r-1}$ and its relative homology sequence. The relative Künneth theorem therefore givesThere is no Tor term because the only nonzero base group is free. The long exact sequence of a pair, replacing $A$ by its homotopy-equivalent fiber, now becomesThe last arrow is the relative boundary transported through these identifications; it can encode twisting, so it has not been set to zero. This proves the relative homology of a fibration over a sphere sequence. In the dimension-zero case, the total space is the disjoint union of the two fibers and the same formula is the split sequence for that union.

Use the based path space and the path-loop fibration $\Omega S^r\to P{S}^r\to S^r$, where paths begin at a fixed basepoint and the projection evaluates their final endpoint. The total space is contractible: the homotopy $\gamma (t)\mapsto \gamma (st)$, $s\downarrow 0$, contracts every path to the constant path. The fiber is path-connected because $S^r$ is simply connected. Hence $H_0(\Omega S^r)=\mathbb{Z}$ and all positive homology of $P{S}^r$ vanishes.

The exact sequence from part (i), for $i\ge 2$, gives $H_{i-r}(\Omega S^r)\cong H_{i-1}(\Omega S^r)$. Put $d=r-1$. Taking negative-degree homology to be zero, this says $H_j\cong H_{j-d}$ for every $j\ge 1$. Starting with $H_0=\mathbb{Z}$ determines every group:There are no torsion groups hidden in the recurrence, since each step is an actual integral isomorphism. This is the integral homology of the loop space of a sphere.

Choose the unit map $\eta :S^{r-1}\to \Omega S^r$ adjoint to the identity under $\Sigma S^{r-1}\cong S^r$. Loop adjunction gives ${\pi }_j(\Omega S^r)\cong {\pi }_{j+1}(S^r)$, so the loop space is $(r-2)$-connected and ${\pi }_{r-1}(\Omega S^r)\cong \mathbb{Z}$. The adjoint of $\eta$ is the degree-one identity, so $\eta$ sends the generator of ${\pi }_{r-1}(S^{r-1})$ to that generator. The Hurewicz theorem for both $(r-2)$-connected spaces, and its naturality, then show that ${\eta }_{\ast }$ is an isomorphism on $H_{r-1}$. It also is on $H_0$, since both spaces are connected. All other positive degrees below $2r-2$ vanish in both spaces. ThusThis proves the required existence with a specific map, without using the suspension theorem in advance.

Use the unit $\eta$ just constructed and replace it by the inclusion $A=S^{r-1}\hookrightarrow Y=\mathrm{Cyl}(\eta )$ into its mapping cylinder; $Y\simeq \Omega S^r$. Put $m=2r-3$. The homology isomorphisms from question 4(ii) and the relative exact sequence give $H_j(Y,A;\mathbb{Z})=0$ for $j\le m$.

Both $A$ and $Y$ are simply connected for $r\ge 3$. For the degree-two start, the absolute Hurewicz theorem identifies their ${\pi }_2$ with $H_2$; naturality and the homology isomorphism show ${\eta }_{\ast }$ is an isomorphism on ${\pi }_2$, so ${\pi }_2(Y,A)=0$. The relative fundamental group also vanishes. If there were a first nonzero relative homotopy group in some degree $3\le k\le m$, the Relative Hurewicz theorem for a $(k-1)$-connected simply connected pair would identify it with $H_k(Y,A)$, which is zero. Induction therefore gives ${\pi }_j(Y,A)=0$ for all $j\le m$.

The relative homotopy sequence now makes ${\eta }_{\ast }:{\pi }_i(S^{r-1})\to {\pi }_i(\Omega S^r)$ an isomorphism when $i+1\le m$, and a surjection at $i=m$. Under loop-suspension adjunction, the adjoint of $\eta \circ a$ is $\Sigma a$, so this particular isomorphism is the suspension homomorphism. Consequently the requested Freudenthal suspension theorem isFor $i=0$ this is read as the identification of connected-component sets; all degree-one groups here also vanish. The proof is the homology comparison proof of Freudenthal suspension.

To compute the specified unstable groups, $S^1$ has contractible universal cover, so ${\pi }_2(S^1)=0$. The Hopf fibration $S^1\to S^3\to S^2$ has exact segment $0={\pi }_3{S}^1\to {\pi }_3{S}^3\to {\pi }_3{S}^2\to {\pi }_2{S}^1=0$, giving ${\pi }_3{S}^2=\mathbb{Z}$. The supplied starting value is ${\pi }_4{S}^3=\mathbb{Z}/2$. To pass from the group over $S^{n-1}$ to that over $S^n$, use $r=n,i=n$; the condition $n\le 2n-4$ holds exactly when $n\ge 4$. Thus every later step is an isomorphism, andThis is the first homotopy group above the dimension of a sphere. The stable range deliberately excludes the step from $n=2$ to $n=3$, so the different groups there do not conflict with the theorem.

The orthogonal group acts transitively on unoriented lines in ${\mathbb{R}}^n$. The stabilizer of the line spanned by the first coordinate isThus $G/H={\mathbb{RP}}^{n-1}$; using only $O(n-1)$ would stabilize a vector and give a sphere instead. For the universal contractible free $G$-space $EG$, the associated bundle $EG/H\to EG/G$ gives the Serre fibration $G/H\to BH\to BG$. Its total space identifies up to homotopy with $BO(1)\times BO(n-1)$.

If $n$ is odd, $n-1$ is even and the rational cohomology of ${\mathbb{RP}}^{n-1}$ is just $\mathbb{Q}$ in degree zero. This follows from the integral projective cellular differentials, alternating zero and multiplication by two, with the even-dimensional top class killed rationally. The cohomological Serre spectral sequence therefore has only the rowwith all other rows zero. The local system in degree zero is trivial because the fiber is connected. There can be no differential or extension, and the multiplicative edge map $H^{\ast }(BO(n);\mathbb{Q})\to H^{\ast }(BH;\mathbb{Q})$ is an isomorphism of rings.

Also $BO(1)={\mathbb{RP}}^{\infty }$ has rational cohomology $\mathbb{Q}$ in degree zero only, by the same cellular calculation. The Künneth theorem identifies $H^{\ast }(BH;\mathbb{Q})$ with $H^{\ast }(BO(n-1);\mathbb{Q})$. The inclusion fixing the first coordinate factors through $H$, setting its $O(1)$ coordinate to one; hence the resulting isomorphism is the actual restriction along this inclusion:This includes $n=1$, with $BO(0)$ a point, and proves rational odd-rank stabilization of orthogonal classifying spaces.

For even $n$, the fiber is odd-dimensional real projective space and has a further rational class in degree $n-1$. The spectral sequence has a possible additional row, now with determinant-sign local coefficients: an orientation-reversing orthogonal matrix acts with degree $-1$ on the covering sphere and hence on the odd-dimensional projective space. Therefore the one-row collapse argument is unavailable; this top row must not be treated as automatically untwisted.

There is an actual failure of injectivity in every even rank. Write $n=2m$. The universal class $p_m\in H^{4m}(BO(2m);\mathbb{Q})$ is nonzero: pull the universal bundle back to the sum of the real underlying bundles of $m$ complex lines over $({\mathbb{CP}}^{\infty }{)}^m$. If their first Chern classes are $x_j$, the Pontryagin formula $p_m(E)=(-1{)}^m{c}_{2m}(E\otimes \mathbb{C})$ gives $p_m={\prod }_j{x}_j^2\ne 0$. On $BO(2m-1)$ the restricted bundle is ${\epsilon }^1\oplus {\gamma }_{2m-1}$; its top complex Chern class $c_{2m}$ vanishes by the rank bound and the trivial summand. Thus $p_m$ restricts to zero. The natural restriction is not injective for even $n$, as the extra fiber cohomology already warns.