Past exam of the mathematics course of the University of Cambridge / 2009 / iii / Paper 17 / 1

Work with real vector bundles; for complex bundles replace the inner products below by positive-definite Hermitian forms. A smooth rank-$r$ vector bundle is a smooth total space $E$ with a smooth projection $\pi :E\to M$, a real vector space structure on each fiber $E_p$, and an open cover with vector bundle trivializationsEach ${\Phi }_i$ is a diffeomorphism over $U_i$ and is linear on fibers. The overlap maps have the form $(p,v)\mapsto (p,A_{ji}(p)v)$, where $A_{ji}:U_i\cap U_j\to G{L}_r(\mathbb{R})$ is smooth. The rank of a vector bundle may be specified separately on each connected component.

A smooth fiber metric is a positive-definite symmetric bilinear form $h_p$ on each fiber, with smoothly varying coefficients in every local frame. Equivalently it is a smooth section of ${\mathrm{Sym}}^2{E}^{\ast }$ which is positive definite at every point. Each trivialization supplies a local fiber metric $h_i$ by pulling back the Euclidean inner product.

A smooth manifold is Hausdorff and second-countable, hence paracompact. Choose a smooth, locally finite partition of unity $\{{\varphi }_i\}$ subordinate to a trivializing cover, with ${\varphi }_i\ge 0$, $\mathrm{supp}{\varphi }_i\subset U_i$, and ${\sum }_i{\varphi }_i=1$. DefineExtend each weighted term by zero outside its trivializing set; its support lies inside that set, so the extension is smooth. Local finiteness makes the sum smooth. At a point $p$ and for $0\ne v\in E_p$, every nonzero term ${\varphi }_i(p)h_i(v,v)$ is positive and at least one such term occurs. Thus $h_p(v,v)>0$. This proves that every smooth vector bundle admits a smooth fiber metric.

Let $r=\mathrm{rank}F$ and $s=\mathrm{rank}E$. Around any $p$, choose a local frame $f_1,\dots ,f_r$ of the vector subbundle $F$. Choose additional smooth sections $e_{r+1},\dots ,e_s$ of $E$ which complement these vectors at $p$. They can be obtained by constant-coordinate sections in a vector bundle trivialization. The determinant of the resulting frame is nonzero at $p$, hence on a sufficiently small neighborhood. Thusis an adapted local frame of $E$ near $p$.

On the fiberwise quotient, the classes $[e_{r+1}],\dots ,[e_s]$ form a basis. Give the quotient vector bundle its local coordinates byAn overlap between adapted frames has a smooth transition function of a vector bundle represented by an invertible block upper triangular matrix,The zero lower-left block expresses that both sets of first $r$ frame vectors span $F$. Its lower-right block $D$ is smooth and invertible. Passing to quotient coordinates removes the upper block, leaving precisely the transition $D$. These lower-right blocks satisfy the cocycle identity because the original transition matrices do. Consequently they define a compatible smooth bundle atlas of rank $s-r$. This proves $E/F$ is a smooth vector bundle, and its fiberwise quotient map $q:E\to E/F$ is a smooth vector bundle morphism. The construction is independent of the adapted frames: any two choices have the same type of compatible transition. The usual quotient topology agrees with this atlas, since in adapted coordinates $q$ is an open linear projection on each fiber.

Choose the smooth fiber metric supplied by part (a), and let $P_p:E_p\to F_p$ be orthogonal projection. These projections vary smoothly. Indeed, in a local frame of $E$ write its metric as a positive-definite matrix $G(p)$, and let the columns of $A(p)$ be a frame of $F$. ThenThe matrix $A^{t}GA$ is positive definite, so its inverse is smooth. This formula proves smoothness of the intrinsic projections in every chart. It also gives $P^2=P$ and $\mathrm{im}P=F$.

The orthogonal complement $F^{\perp }=\ker P$ is a smooth vector subbundle: applying $I-P$ to a complement frame gives a frame of its fibers, of constant rank $s-r$, after shrinking the chart. Fiberwise,and the restriction $q{|}_{F^{\perp }}:F^{\perp }\to E/F$ is a smooth fiberwise bijective vector bundle morphism. In local frames its matrix is invertible, with smooth inverse, so it is a vector bundle isomorphism.

More explicitly, the desired orthogonal splitting of a vector subbundle isIts inverse is $(f,[v])\mapsto f+(I-P)v$. Replacing $v$ by $v+w$ with $w\in F_p$ leaves $(I-P)v$ unchanged, so the inverse is well-defined. Both maps are smooth and linear on fibers, and their compositions are identities. The splitting exists globally but depends on the chosen fiber metric.