Past exam of the mathematics course of the University of Cambridge / 2010 / iii / Paper 15 / 2

A smooth rank-$r$ vector bundle over a smooth manifold $M$ consists of a smooth total space $E$, a smooth projection $\pi :E\to M$, and an $r$-dimensional real vector space structure on each fibre $E_p={\pi }^{-1}(p)$. Each point has a neighborhood $U$ with a vector bundle trivializationthat is a diffeomorphism, commutes with projection to $U$, and is a linear isomorphism on every fibre. The overlap maps have the form $(p,v)\mapsto (p,h(p)v)$ with $h:U\cap V\to G{L}_r(\mathbb{R})$ smooth. For a complex vector bundle, replace $\mathbb{R}$ by $\mathbb{C}$ and require complex-linear fibre maps.

A connection on a vector bundle is an $\mathbb{R}$-linear mapwhere $\Gamma (E)$ is the module of smooth sections, ${\Omega }^1(M;E)$ consists of vector-bundle-valued differential forms of degree one, and $f$ is a smooth function. Equivalently, evaluating on a vector field $X$ gives operators ${\nabla }_X$ with ${\nabla }_{fX}s=f{\nabla }_{X}s$ and ${\nabla }_X(fs)=X(f)s+f{\nabla }_{X}s$. A connection on a vector bundle is local: its value at a point depends on the germ of the section of a vector bundle, so the same definition applies to local sections.

Realize the pullback vector bundle asIts fibre over $p$ is canonically $E_{\varphi (p)}$, with the same vector space operations. If a vector bundle trivialization over $U\subset M$ is ${\tau }_U(v)=(\pi (v),\hat{v})$, give the pullback vector bundle the local trivializationDeclare these maps to be local homeomorphisms and use coordinate charts on $M^{\prime }$ to obtain a smooth atlas on the total space. On overlaps the smooth transition maps are $(p,w)\mapsto (p,h(\varphi (p))w)$, hence smooth and fibrewise linear. They satisfy the same cocycle identities as the original transition maps. This gives a well-defined smooth vector bundle, with smooth projection and rank $r$. It also gives the subspace topology inherited from $M^{\prime }\times E$: inside each product ${\varphi }^{-1}(U)\times {\pi }^{-1}(U)$ the defining condition is the graph of $\varphi$ in the base coordinates. Thus the total space is Hausdorff and second countable, as required for a smooth manifold.

For a smooth section of a vector bundle $s$ over $U$, its pulled-back section is more precisely $p\mapsto (p,s(\varphi (p)))$. If $s$ has component column $u:U\to {\mathbb{R}}^r$ in the original vector bundle trivialization, its component column in $T_U$ is $u\circ \varphi$, which is smooth. The pullback is a smooth rank-$r$ vector bundle, and every local section pulls back smoothly.

The expression $d\varphi (X)$ is in general a vector field along a map, rather than a vector field on $M$. Its intended meaning in the required formula is pointwise evaluation of the vector-bundle-valued differential form $\nabla s$:using the canonical identification $({\varphi }^{\ast }E{)}_p=E_{\varphi (p)}$. This interpretation works for every smooth map, including a constant map or a map with self-intersections.

Choose a frame of a vector bundle over $U$, written as a row $e=(e_1,\dots ,e_r)$. Define its connection one-form $\Omega$ byfor component columns $u$. The pulled-back frame $e^{\varphi }=({\varphi }^{\ast }{e}_1,\dots ,{\varphi }^{\ast }{e}_r)$ is a frame of a vector bundle over ${\varphi }^{-1}U$. For any component column $v$ of smooth functions on that open set, defineHere the matrix entries of ${\varphi }^{\ast }\Omega$ are ordinary pullbacks of a differential form. The displayed operator is linear and satisfies the Leibniz rule, hence is locally a connection on a vector bundle.

To prove these local operators glue, change the original frame of a vector bundle to $\tilde{e}=eh$. Expanding $\nabla (ehu)$ gives the change of frame of a vector-bundle connectionThe pullback of a differential form and the chain rule giveThis is precisely the required transformation law for the pulled-back frame ${\tilde{e}}^{\varphi }=e^{\varphi }(h\circ \varphi )$. Thus the definitions agree and give a global pullback connection.

For $s=eu$, substitute $v=u\circ \varphi$ in the defining formula and use $d(u\circ \varphi )(X)=du(d\varphi (X))$. This proves the required identity. Conversely that identity fixes ${\nabla }_X^{\prime }({\varphi }^{\ast }{e}_a)$; the Leibniz rule then fixes the derivative of every ${\sum }_a{v}^a{\varphi }^{\ast }{e}_a$, proving uniqueness. If the identity is initially imposed only on global sections of a vector bundle, multiply each local frame section by a smooth bump function equal to one near the point in question and supported in $U$, and extend by zero. These global sections agree with the frame locally, so the same uniqueness argument applies. The pullback connection exists and is unique.

The curvature form of a connection is the endomorphism-valued differential two-form defined byIt is alternating in $X,Y$. To see its tensoriality, replacing $X$ by $fX$ produces an extra $-Y(f){\nabla }_{X}s$ from the second term and an extra $+Y(f){\nabla }_{X}s$ from $[fX,Y]=f[X,Y]-Y(f)X$, so these cancel. Alternation gives linearity over smooth functions in $Y$ too. Replacing $s$ by $fs$ produces the additional coefficient $(X(Yf)-Y(Xf)-[X,Y]f)s=0$; all terms involving one derivative of $s$ also cancel. Thus $R^{\nabla }$ belongs to ${\Omega }^2(M;\mathrm{End}E)$.

In the coefficient-column convention of part (b), expanding the definition givesThe first three terms are $d\Omega (X,Y)$, and the last two are $(\Omega \wedge \Omega )(X,Y)$. This proves the Cartan curvature matrix equation $F=d\Omega +\Omega \wedge \Omega$, with matrix multiplication combined with the wedge product of differential forms.

The pullback connection has matrix ${\Omega }^{\prime }={\varphi }^{\ast }\Omega$. Since the pullback of a differential form commutes with the exterior derivative and the wedge product of differential forms, its curvature form of a connection has matrixChanges of frame of a vector bundle conjugate both sides by the same pulled-back transition matrix, so this identity is intrinsic. In fibre notation the answer isBoth sides act on the same fibre $E_{\varphi (p)}$. No claim that $d\varphi$ preserves Lie brackets of vector fields for an arbitrary map is needed.