Past exam of the mathematics course of the University of Cambridge / 2006 / iii / Paper 21 (Cobordism)

Put $L=L_1\cap L_2$, and let $j:L\hookrightarrow M$ be inclusion. The transverse intersection theorem makes $L$ a smooth submanifold without boundary, withIt is compact because it is a closed subset of either compact input submanifold.

Write ${\nu }_r=TM{|}_{L_r}/T{L}_r$. On $L$, consider the bundle mapIts kernel is $TL$, and it is onto because $T{L}_1+T{L}_2=TM$ at every intersection point. Thus the normal bundle of a transverse intersection has the actual bundle isomorphismTransfer the direct-sum complex structure $J_1\oplus J_2$ through this isomorphism. This supplies a complex structure on the normal bundle itself, not merely after stabilization.

Geometrically, multiply the two classes by forming their external product in $M\times M$ and pulling it back along the diagonal $\Delta :M\to M\times M$. The transversality hypothesis is precisely what makes the inverse image of $L_1\times L_2$ under $\Delta$ the submanifold $L$. The normal bundle of this inverse image is the direct sum just calculated. Multiplication of the corresponding Thom classes agrees with the Thom class of this direct sum, so the complex cobordism product represented by transverse intersections isIf the inputs have complex normal ranks $a,b$, the product has degree $2(a+b)$. If they are disjoint, the representative is empty and the product is zero.

Regard ${\mathrm{Mat}}_n(\mathbb{C})$ as a real vector space of dimension $2n^2$, and let ${\mathrm{Herm}}_n(\mathbb{C})$ be the real vector space of Hermitian matrices, of dimension $n^2$. The smooth Gram maphas $F^{-1}(I)=U(n)$ and differential $D{F}_A(K)=A^{\ast }K+K^{\ast }A$. At a unitary $A$, any Hermitian matrix $H$ is obtained by taking $K=AH/2$. Indeed $A^{\ast }K=H/2$ and $K^{\ast }A=H/2$. Hence $I$ is a regular value, and the regular level set theorem proves the unitary group as a regular level set result:

Similarly let $G(A)=A^{T}A$ from ${\mathrm{Mat}}_n(\mathbb{R})$ to the symmetric real matrices, whose dimension is $n(n+1)/2$. At an orthogonal $A$ its differential is $D{G}_A(K)=A^{T}K+K^{T}A$, again onto by choosing $K=AH/2$ for a symmetric $H$. Therefore the orthogonal group as a regular level set hasMatrix multiplication is smooth on both groups, and inversion is conjugate transpose or transpose, respectively, so these manifold structures also make them Lie groups. Both groups are closed and bounded in their finite-dimensional ambient matrix spaces, hence compact.

Realification identifies a unitary complex matrix with an orthogonal transformation of ${\mathbb{R}}^{2n}$. This is a smooth injective group homomorphism, with closed image, so $U(n)$ is the stated Lie subgroup of $O(2n)$. The assumed maximal-rank quotient map has fibers diffeomorphic to $U(n)$. Consequently the compact homogeneous space, also known as the space of orthogonal complex structures, has dimensionThe original PDF's sphere is $S^{n^2}$, with a square on the exponent; its dimension exceeds $d$ by $n$.

For completeness, maps to a sphere above the dimension of a compact smooth manifold are null-homotopic. Let $X$ be a compact smooth manifold of dimension $d<q$, with $q\ge 1$, and let $f:X\to S^q\subset {\mathbb{R}}^{q+1}$ be continuous. Choose a finite open cover on which $f$ is uniformly close to its value at a chosen point of that open set, and take a smooth partition of unity subordinate to the cover. The weighted sum of those chosen sphere values is a smooth map $g_0:X\to {\mathbb{R}}^{q+1}$ uniformly within $ϵ<1$ of $f$. It never vanishes. Normalize it to $g=g_0/\Vert g_0\Vert$. Normalizing $(1-s)f+s{g}_0$ gives a homotopy from $f$ to $g$, since that vector remains within $ϵ$ of the unit vector $f$.

Every value of $g$ is critical because its differential has rank at most $d<q$. The Sard theorem makes its image have measure zero in $S^q$, so it misses some point $p$. The punctured sphere $S^q\setminus \{p\}$ is homeomorphic to ${\mathbb{R}}^q$ and contractible. Thus $g$, and hence $f$, is null-homotopic. This reasoning does not require $X$ to be connected, and all constant maps are homotopic because $S^q$ is path connected.

Apply it with $X=O(2n)/U(n)$ and $q=n^2$, for positive integer $n$. We obtainThe notation means homotopy classes of maps; no homotopy equivalence between the two spaces is asserted.

Let $\xi$ have real rank $r$, choose a bundle metric, and write $D(\xi ),S(\xi )$ for its disk and sphere bundles. In a multiplicative generalized cohomology theory, a Thom class in a generalized cohomology theory is a classwhose restriction to each fiber pair $(D^r,S^{r-1})$ is the suspension of the coefficient unit $1\in h^0(\mathrm{pt})$ under the chosen oriented identification. Equivalently, it is a fiberwise generator over the coefficient ring compatible with the $h$-orientation. Cup product with this class gives the Thom isomorphism theoremwhere $X$ is the base.

Let $z:X\to D(\xi )$ be the zero section and let $ȷ:h^r(D(\xi ),S(\xi ))\to h^r(D(\xi ))$ forget relative supports. The Euler class in a generalized cohomology theory isEquivalently, identify relative cohomology with reduced cohomology of the Thom space and pull its Thom class back along the zero section followed by the quotient map.

Suppose $s$ is a nowhere-zero section. Normalize it to $\hat{s}(x)=s(x)/\Vert s(x)\Vert \in S(\xi {)}_x$. The mapsgive a homotopy within $D(\xi )$ from $z$ to the sphere-valued section $\hat{s}$. The image $ȷ(u_h(\xi ))$ restricts to zero on $S(\xi )$, by the exact sequence of the pair. Homotopy invariance therefore yieldsThus a nowhere-zero section annihilates generalized Euler classes:The argument uses the actual nowhere-zero section and is valid for any chosen Thom orientation in the multiplicative theory.

For a complex bundle $E$, write $c_t(E)={\sum }_{j\ge 0}{c}_j^{MU}(E)t^j$, with $c_0(E)=1$ and $c_j(E)=0$ for $j>{\mathrm{rank}}_{\mathbb{C}}E$. The Whitney sum formula for Chern classes in complex cobordism isProducts are in the complex-cobordism cohomology ring; these classes have even degrees $2j$. The splitting principle for complex vector bundles explains the formula: on a splitting space the Chern classes are elementary symmetric functions of the first Chern classes of the line summands, and joining the two lists multiplies their total Chern polynomials.

Use the line convention for the projective bundle:Its relative tautological line bundle isIt is a complex rank-one subbundle of $p^{\ast }\eta$. Choose a Hermitian metric and let $E=\eta (1{)}^{\perp }$, of rank $n-1$, so $p^{\ast }\eta =\eta (1)\oplus E$.

Throughout, $x=c_1^{MU}(\eta (1))$, the tautological line's class, as specified before the final equation. Put $a_k=p^{\ast }{c}_k^{MU}(\eta )$ and $b_k=c_k^{MU}(E)$, with $a_0=b_0=1$. Since $c_t(\eta (1))=1+xt$, Whitney multiplication givesComparison of coefficients gives $a_k=b_k+x{b}_{k-1}$, hence recursively $b_k=a_k-x{b}_{k-1}$. Therefore the Chern classes of a tautological-line complement areand the classes for $k\ge n$ vanish by the rank.

At degree $n$ the same recursion has $b_n=0$, givingMultiplying by $(-1{)}^n$ proves the projective bundle relation in complex cobordismThese alternating signs come from the inverse formal power series $(1+xt{)}^{-1}$. They do not require identifying the First Chern class of a dual line with $-x$, which is generally incorrect in complex cobordism.

The final sentence's designation $x=c_1(\eta )$ must mean the earlier $c_1(\eta (1))$. If instead one substitutes $p^{\ast }{c}_1(\eta )$ literally, the claimed relation is false even in ordinary cohomology. For example, take $\eta =L\oplus L$ over ${\mathbb{CP}}^2$, with $a=c_1(L)$ a generator. Its projective bundle is ${\mathbb{CP}}^2\times {\mathbb{CP}}^1$, and $p^{\ast }{a}^2\ne 0$. With the incorrect substitution $x=2p^{\ast }a$, the rank-two polynomial evaluates toThe tautological-line interpretation above is the one consistent with the bundle decomposition and the stated relation.

For a real or complex vector bundle $E$, choose a fiber metric. Its Thom space is the pointed quotientcollapsing the sphere bundle to one basepoint. When the base is compact, this is the one-point compactification of the total space: fiberwise radial compression identifies the total space with the interior of $D(E)$, and all directions approaching its boundary become the single point at infinity.

Decompose ${\mathbb{C}}^{n+m+2}=V\oplus W$ with $V={\mathbb{C}}^{n+1}$, $W={\mathbb{C}}^{m+1}$, so the subspace being collapsed is $\mathbb{P}(V)={\mathbb{CP}}^n$. Every line outside it has a representative $(v,w)$ with $w\ne 0$. Its projection to $W$ gives the line $\ell =\mathbb{C}w\in \mathbb{P}(W)={\mathbb{CP}}^m$, and there is a unique complex-linear map $A:\ell \to V$ such that $A(w)=v$. The line in $V\oplus W$ is precisely the graph of $A$. This construction is unaffected by replacing $(v,w)$ with $(\lambda v,\lambda w)$.

Let $\gamma ={\eta }_1$ be the tautological line over $\mathbb{P}(W)$. The graph construction and its inverse are continuous, indeed smooth in the usual affine charts, and identifyA Hermitian metric gives a complex-linear bundle isomorphism $\bar{\gamma }\cong {\gamma }^{\ast }$. With the Hermitian form linear in its first argument, it sends $\bar{v}$ to the functional $w\mapsto ⟨w,v⟩$. Linearity in the conjugate vector is exactly what makes this a complex, rather than merely real, isomorphism. Thus the bundle is $({\bar{\eta }}_1{)}^{\oplus (n+1)}$, with the conjugation bars present in the original PDF.

Collapsing the closed subspace ${\mathbb{CP}}^n$ in compact Hausdorff space ${\mathbb{CP}}^{n+m+1}$ gives the one-point compactification of its open complement: neighborhoods of the collapsed point have compact complements, and conversely every compact subset of the complement is closed in the ambient space. The graph homeomorphism therefore extends over the added point. This proves the projective-space quotient as a Thom space of conjugate tautological lines:

The graph coordinates also determine the normal complex structures requested in the second part. The copy $\mathbb{P}(W)$ corresponds to the zero section of the graph bundle, whose normal bundle is that bundle itself. For a coordinate embedding ${\mathbb{CP}}^k\subset {\mathbb{CP}}^n$, use $W={\mathbb{C}}^{k+1}$ and its complementary coordinate space $V={\mathbb{C}}^{n-k}$. Then the normal bundle of a linear complex-projective embedding isThe complex structure is the linear structure in these graph coordinates. Equivalently, applying the preceding homeomorphism with its parameters $n-k-1$ and $k$ identifies the Thom space of this normal bundle with ${\mathbb{CP}}^n/{\mathbb{CP}}^{n-k-1}$, after reordering coordinates. Define ${\nu }_m$ in the identical way. The Hermitian-metric identification with the conjugate line does not change the complex normal orientation.

The original coordinate embeddings may be nested, so first move one into transverse position. Keep ${\mathbb{CP}}^k=\mathbb{P}(A)$ for the first $k+1$ coordinates, and move ${\mathbb{CP}}^m$ to $\mathbb{P}(B)$ for the last $m+1$ coordinates. A unitary coordinate permutation accomplishes this. Every unitary matrix can be joined to the identity by diagonalizing it and varying its eigenvalue phases continuously, so this move is an ambient isotopy. Its trace gives a cobordism carrying the transported normal complex structure, and hence leaves the input cobordism class unchanged.

If $k+m\ge n$, thenTheir projectivizations are transverse and meet in ${\mathbb{CP}}^r$, where $r=k+m-n$. The direct-sum normal structure from question 1 restricts toThe quotient map defining the transverse normal-bundle isomorphism is complex-linear here, so it agrees with the standard normal complex structure and contributes no orientation sign. If $k+m<n$, the first and last coordinate subspaces have no nonzero vector in common, so their projectivizations are disjoint.

Consequently the product of linear projective subspaces in complex cobordism isThe product degree is $2(2n-k-m)$, equal to the real codimension of the stated intersection when it exists. In the equality case $k+m=n$, the representative is one point with its complex-oriented normal space.