Past exam of the mathematics course of the University of Cambridge / 2005 / iii / Paper 18 / 5

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.