Past exam of the mathematics course of the University of Cambridge / 2009 / iii / Paper 17 / 5

A connection on a vector bundle is an $\mathbb{R}$-linear mapwhich satisfies $\nabla (fs)=df\otimes s+f\nabla s$ for every smooth function $f$ and section $s$. Here ${\Omega }^1(M;E)=\Gamma (T^{\ast }M\otimes E)$. Equivalently, writing ${\nabla }_{X}s=(\nabla s)(X)$, it is linear over smooth functions in $X$, linear over $\mathbb{R}$ in $s$, and obeysChoose a trivializing open cover and a local frame $e_{i1},\dots ,e_{ir}$ on each $U_i$. Differentiating the coefficient functions gives a local connection on a vector bundle:No compatibility of these local connections is assumed. Choose a smooth locally finite partition of unity $\{{\varphi }_i\}$ subordinate to the cover and defineextending each weighted term by zero outside $U_i$. Its support lies inside $U_i$, so each extension is smooth, and local finiteness makes the whole sum smooth. The operation is linear over smooth functions in $X$. For the Leibniz rule,using ${\sum }_i{\varphi }_i=1$. This proves every smooth vector bundle admits a connection by the construction of a vector bundle connection by a partition of unity.

Fix the coefficient-column convention: write the local frame as a row $e=(e_1,\dots ,e_r)$ and a section as $s=eu$ for a column $u$ of functions. Define the connection matrix $\theta =({\theta }_{ij})$ byEach entry is a smooth one-form. Define the curvature form of a connection byThe same Leibniz cancellations as for tangent-bundle curvature show that this is linear over smooth functions in $X,Y,s$. Thus it is a two-form with values in the endomorphisms of $E$. Its curvature matrix $\Theta =({\Theta }_{ij})$ is specified by $R^E(X,Y)e_j={\sum }_i{e}_i{\Theta }_{ij}(X,Y)$.

The covariant exterior derivative on coefficient forms is $d+\theta \wedge$. Squaring it on $u$ givesTherefore the Cartan curvature matrix equation isMatrix-valued wedge products here combine matrix multiplication with the wedge product of differential forms, in the stated order.

Using $d^2=0$ and the graded Leibniz rule,On the other hand, associativity of the endomorphism-valued exterior product givesThe cubic terms cancel, proving the requested Bianchi identity

Let another local frame be $e^{\prime }=eA$ with $A:U\to G{L}_r(\mathbb{R})$ smooth. Since the coefficient columns satisfy $u=A{u}^{\prime }$, the Leibniz rule gives the change of frame of a vector-bundle connection:The curvature form of a connection is linear over smooth functions in its section argument, so its matrix transforms homogeneously:ConsequentlyThe entries of $A$ are functions, of degree zero, so ordinary cyclic invariance of trace applies without a graded sign. The local forms therefore glue to a globally defined two-form $\alpha =\mathrm{tr}\Theta$, the trace of vector-bundle curvature.

Apply the Bianchi identity proved in part (b):For the first term,The middle interchange has sign $(-1{)}^{2\cdot 1}=1$, and the next equality simply renames $i,j$. ThusOne can also check closedness locally: the off-diagonal terms in $\mathrm{tr}(\theta \wedge \theta )={\sum }_{i,j}{\theta }_{ij}\wedge {\theta }_{ji}$ cancel in pairs, while the diagonal terms vanish. Hence $\alpha =d(\mathrm{tr}\theta )$ in each frame, and $d^2=0$ gives the same conclusion.