Past exam of the mathematics course of the University of Cambridge / 2007 / iii / Paper 19 (Cobordism)

Write $M{U}^{\ast }={\Omega }_U^{\ast }$ for complex cobordism. A complex orientation of a smooth map is chosen stable complex normal data, not a complex structure on either manifold separately. More concretely, replace the continuous map by a homotopic smooth map $f_0$, choose an embeddingand give its normal bundle ${\nu }_j$ a stable complex structure on a real vector bundle. The auxiliary dimension can be chosen so the normal rank is even; further stabilization makes the normal bundle actually complex. Equivalence means stabilization by trivial complex summands and homotopy of the oriented embedding data. The stable identityidentifies this as orientation of the relative stable normal bundle. For odd relative codimension one uses an odd auxiliary suspension coordinate; the degree formula below is unchanged. This is the geometric factorization model for complex-oriented maps used in Algebraic and Geometric Topology 18, geometric cobordism discussion.

Let $d=\dim Y-\dim X$ and $r={\mathrm{rank}}_{\mathbb{R}}{\nu }_j=d+N$. The chosen normal orientation gives a Thom class $u_{{\nu }_j}$ of degree $r$ and the Thom isomorphism in complex cobordism. For compact manifolds the Pontryagin-Thom collapse of a tubular neighborhood to its Thom space isFor $a\in M{U}^q(X)$, define the Gysin map in complex cobordism by the compositeThus $f_{!}$ raises degree by the real relative codimension $d$. This definition depends on the specified orientation and is invariant under the stated equivalences of its normal data. The plus sign on a space denotes adjoining a disjoint basepoint.

For the embedding $i:L\hookrightarrow M$ with normal rank $2n$, let $c:M_{+}\to \mathrm{Th}(\nu )$ be its tubular collapse. The Thom construction gives $i_{!}(1)=c^{\ast }{u}_{\nu }$. Restricted to $L$, this collapse is exactly the zero section $z:L\to \mathrm{Th}(\nu )$. ThereforeThis is the self-intersection formula in complex cobordism. For an actual complex rank-$n$ normal bundle its top Chern class in complex cobordism is the zero-section pullback of its Thom class, so the right side is $c_n^{MU}(\nu )$. Equivalently, under the splitting principle for complex vector bundles both are the product of the first Chern classes of the line summands. The same argument gives $i^{\ast }{i}_{!}(a)=a{e}_{MU}(\nu )$ for every $a$.

There is a real distinction if the printed word “stable” is interpreted literally. The Euler class of a stably complex real bundle need not equal $c_n^{MU}$ of its stable complex class. For the diagonal embedding $S^4\hookrightarrow S^4\times S^4$, the normal bundle is $T{S}^4$. It has a stable complex structure withso its stable complex rank-two class is trivial and has second Chern class zero. However the ordinary integral image of $i^{\ast }{i}_{!}(1)$ is $e(T{S}^4)$, which evaluates to the Euler characteristic $2$. Thus the always-valid answer is the cobordism Euler class. The requested top-Chern identification is valid for an actual complex normal bundle, or if “top Chern class” here is explicitly being used as the Euler-class terminology for the chosen normal orientation.

Let $e_1(x),\dots ,e_n(x)$ be the prescribed ordered orthonormal framing. The unordered Stiefel bundle is the quotient of the orthonormal $k$-frame bundle by the action of the symmetric group permuting its frame vectors. Inside it take only the unordered subsets of the distinguished framing:Projection onto $x$ defines the subset cover of a framed vector bundle $p_k:X_k\to X$. Each sheet is globally labeled by a $k$-element index subset, soand this is a $\left(\genfrac{}{}{0ex}{}{n}{k}\right)$-sheeted cover. The whole unordered Stiefel fiber is generally continuous, not this finite cover; it is the specified framing that selects these sheets. Even if the given framing is only continuous, the product description gives the covering its smooth structure.

A covering is a local diffeomorphism, with $T{X}_k\cong p_k^{\ast }TX$, so its relative normal data has the canonical rank-zero stable framing of a vector bundle. Its degree-zero framed cobordism transfer definesThese are the exotic subset-cover classes of a framed bundle. Set $l_0=1$, using the identity cover, and $l_k=0$ outside $0\le k\le n$.

For framed bundles $\eta ,\zeta$ of ranks $n,m$, use the concatenated framing of their direct sum. A $k$-element subset uniquely contains $i$ vectors from the first frame and $j$ from the second, where $i+j=k$. This gives an isomorphism of framed covers over $X$,Transfer is additive under disjoint union, since the tubular collapse is the sum of the collapses of the disjoint sheets. It is multiplicative under fiber product: external products of the two collapse and Thom constructions give the transfer of the product cover over $X\times X$, and pullback by the diagonal gives the fiber-product cover over $X$. Its product framing is exactly the canonical relative framing used here. Consequently

With the literal globally ordered framing in the question, these covers are all trivial and each sheet is the identity map with its canonical relative framing. Thus their classes simplify further toThe displayed sum formula then also follows from the binomial convolution. Nontrivial covering monodromy would require only an unordered frame reduction, rather than the global ordered frame assumed here. In particular no extra nonconstant characteristic classes are produced by this literal construction. The original PDF places these classes in degree zero; the converted TeX's degree $-1$ and omitted letter $l$ are transcription errors.

Use the metric on $M$ induced by its given Euclidean embedding. For two smooth maps define the C1 metric on a smooth mapping space bywhere the differential has domain $T_{x}M$ with the induced inner product. Compactness makes both suprema finite. The norm properties give symmetry and the triangle inequality, and the first term makes distance zero imply equality of maps. Thus convergence in $d_1$ is uniform convergence of both maps and differentials.

Because $i$ is an immersion, $D{i}_x$ is injective at every point. The continuous function $|D{i}_{x}v|$ on the compact unit tangent bundle has a positive minimum $a$. ThereforeFor all sufficiently large $n$, ${\sup }_x\Vert D{f}_{n,x}-D{i}_x\Vert <a/2$, and thenConsequently every sufficiently late $f_n$ is an immersion. If $M$ has dimension zero it is a finite set, the immersion assertion is automatic, and eventual injectivity follows directly from separation of its finitely many image points; we henceforth treat positive dimension.

For injectivity, choose finitely many coordinate maps ${\varphi }_b$ on convex balls, with smaller coordinate neighborhoods covering $M$ and closures lying inside the larger balls. Write $I_b=i\circ {\varphi }_b$ and $F_{n,b}=f_n\circ {\varphi }_b$. On the finitely many relevant compact coordinate closures, the differentials of the $I_b$ have a common positive lower bound $s$, and convergence in $d_1$ givesfor all late $n$. The assumed uniform second-derivative bound gives a common bound $L$ for $D^2{F}_{n,b}$ on those larger coordinate neighborhoods. If the bound is expressed using an intrinsic Hessian, the coordinate chain rule also involves first derivatives and derivatives of the fixed charts; their bounds are uniform by C1 convergence and compactness. Thus $L$ can be larger than the printed constant $C$, but is independent of $n$.

For $u,v$ whose segment lies in one such coordinate ball, Taylor formula with integral remainder givesChoose $\delta >0$ with $L\delta \le s/2$. Then, whenever $0<|u-v|<\delta$,The Lebesgue number lemma applied to the finite smaller-chart cover, together with uniform continuity of its coordinate inverses on the relevant compact sets, gives $\rho >0$ such that every pair $x,y\in M$ with $0<|x-y|<\rho$ lies in one such chart with coordinate distance below $\delta$. The displayed estimate therefore excludes all sufficiently close pairs of equal images, uniformly in late $n$.

The remaining pairs form the compact setIf it is nonempty, injectivity of the smooth embedding $i$ gives a positive minimum $b$ of $|i(x)-i(y)|$ on it. Uniform convergence makes ${\sup }_x|f_n(x)-i(x)|<b/3$ for all sufficiently large $n$, and hence on $K_{\rho }$,If $K_{\rho }$ is empty there is nothing left to check. Combining the near-pair and distant-pair bounds makes $f_n$ injective for every sufficiently late $n$. It is already an immersion, and a continuous injection from compact $M$ into a Hausdorff space is a homeomorphism onto its image. By compact injective immersion is an embedding,This proves the requested stability of compact smooth embeddings using the stated second-derivative assumption. That assumption supplies a convenient Taylor constant; the stronger C1 openness result can replace it by the uniform continuity of the limiting differential on finitely many small coordinate balls.

The exactness axiom of a generalized cohomology theory gives a natural long exact sequence for each CW pair:For $(X,X)$, the restriction $h^n(X)\to h^n(X)$ is the identity. Its preceding relative-to-absolute map has zero image. The previous connecting map is zero too, since it follows the surjective identity in degree $n-1$. Exactness at the relative group therefore makes its map both injective and zero, givingin particular $h^n(x_0,x_0)=0$. No dimension axiom is being used; a generalized theory can have nonzero $h^n(x_0)$ in many degrees.

The wedge-product notation for pairs here means the smash product of CW pairs,Its quotient is the based smash product $(X/A)\wedge (Y/B)$. A multiplicative theory has an external productIf the two spaces are the same, pullback by the diagonal gives the relative cup product in $h^{k+m}(X,A\cup B)$. This is different from the wedge sum of two based spaces.

First read the covering subcomplexes as pointed subcomplexes, so $x_0\in A_j$. Contractibility makes the inclusion $\{x_0\}\hookrightarrow A_j$ a homotopy equivalence, and thus $h^r(A_j,x_0)=0$ in every degree, by exactness and homotopy invariance of cohomology. The relative exact sequence for the triple $\{x_0\}\subseteq A_j\subseteq X$ givesConsequently each homogeneous $w_j$ lifts to $v_j\in h^{r_j}(X,A_j)$. Repeated relative multiplication givesNaturality maps this product to the product of the $w_j$ in $h^{\ast }(X,x_0)$. ThereforeFor nonhomogeneous classes the same conclusion follows by distributivity. This is nilpotence from a contractible subcomplex cover.

There is an alternative sufficient convention: if $X$ is path connected, the $A_j$ need not contain $x_0$. The absolute image $\bar{w}$ of a reduced class restricts to zero at $x_0$, hence at any other point since the point inclusions are homotopic along a path. On a nonempty contractible $A_j$ it therefore restricts to zero, so exactness lifts ${\bar{w}}_j$ to $h^{\ast }(X,A_j)$. Their relative product vanishes as above. The map $h^{\ast }(X,x_0)\to h^{\ast }(X)$ is injective: restriction to the basepoint is split surjective by the map $X\to \mathrm{pt}$, making the previous connecting map zero. Thus vanishing of the absolute product gives vanishing of the reduced product too.

Without either the pointed-subcomplex or connectedness convention, the printed conclusion is false. Take $X=\{x_0,x_1\}$ covered by its two singleton subcomplexes and $h^{\ast }=H^{\ast }(-;\mathbb{Z})$. Both subcomplexes are contractible, but $H^0(X,x_0)$ contains the function $w$ with values $0,1$. Pointwise multiplication gives $w^2=w\ne 0$. This states precisely the needed interpretation rather than assuming a contractible component lies in the basepoint component.

Let $R=M{U}^{\ast }(\mathrm{pt})$ be the coefficient ring of complex cobordism. We use its standard coefficient theorem: $R\cong \mathbb{Z}[b_1,b_2,\dots ]$ with $|b_j|=-2j$. In particular $R$ is concentrated in nonpositive even degrees. Let $H$ be the quaternionic tautological line bundle on ${\mathbb{HP}}^2$, regarded as a complex rank-two bundle by right multiplication by a fixed imaginary quaternion. SetWe will proveThe argument must establish the multiplication as well as the additive groups, since collapse of a spectral sequence alone does not settle ring extensions.

The quaternionic projective space has one cell in each dimension $0,4,8$, giving integral cohomology $\mathbb{Z}$ in exactly these three degrees. The sphere bundle of $H$ is the quaternionic Hopf fibrationWrite $u=e(H_{\mathbb{R}})=c_2(H)$ for its ordinary degree-four class, fixing its sign by this equality. The Gysin sequence of a sphere bundle gives surjections, in fact isomorphisms,Indeed the corresponding neighboring cohomology groups of $S^{11}$ vanish. Both source and target groups are $\mathbb{Z}$, so these maps make $u$ and $u^2$ generators. There are no higher cohomology groups. This proves the cohomology ring of quaternionic projective space in the form $H^{\ast }({\mathbb{HP}}^2;\mathbb{Z})=\mathbb{Z}[u]/(u^3)$.

Apply the multiplicative Atiyah-Hirzebruch spectral sequence for complex cobordism:The skeletal construction obtains this sequence from the long exact sequences of successive skeleta; the quotients are wedges of spheres, and the first differential is the ordinary cellular differential with coefficient groups $R^t$. Its multiplication on the associated graded group is the ordinary cup product with these coefficients. The finite eight-dimensional skeleton makes the filtration finite, so there is no convergence ambiguity.

Only columns $s=0,4,8$ and even rows $t$ are nonzero. If $r$ is odd, the target column $s+r$ is odd and zero. If $r$ is even, the target coefficient degree $t-r+1$ is odd and zero. The same observations exclude incoming differentials. Consequently $E_{\infty }=E_2$, with associated graded $R$-module freely generated by $1,u,u^2$ in filtrations $0,4,8$.

The canonical map from complex cobordism to integral cohomology preserves the chosen complex orientation and hence sends $q$ to $u$. The class $q$ vanishes on the zero-dimensional skeleton since $R^4=0$, and its leading skeletal term is $u$ in filtration four. Multiplicativity makes the leading term of $q^2$ the generator $u^2$ in filtration eight. Thus $1,q,q^2$ lift the associated graded basis.

This resolves all module extensions explicitly. For a class of total degree $d$, subtract its coefficient $a_0\in R^d$ in filtration zero. From the remainder subtract $a_{1}q$, where $a_1\in R^{d-4}$ is its leading coefficient in filtration four, and then subtract $a_2{q}^2$ with $a_2\in R^{d-8}$. The remaining filtration is beyond dimension eight and is zero. Conversely a relation $a_0+a_{1}q+a_2{q}^2=0$ has zero leading coefficient in each of the three successive filtrations, forcing all three $a_j$ to vanish. ThereforeFinally $q^3$ has degree twelve. Every associated graded contribution in that total degree would have coefficient degree $12-s>0$ for $s=0,4,8$, and all such coefficients vanish. Thus $M{U}^{12}({\mathbb{HP}}^2)=0$ and $q^3=0$. The polynomial map $R[t]/(t^3)\to M{U}^{\ast }({\mathbb{HP}}^2)$ taking $t$ to $q$ is surjective by the basis argument and injective by its uniqueness. This proves the complex cobordism ring of the quaternionic projective plane as a graded ring over the stated coefficient ring, including the absence of any additional coefficient-dependent relation.