Past exam of the mathematics course of the University of Cambridge / 2007 / iii / Paper 15 / 3 / b / Solution

Define the curvature form of a connection by the operatorIt is alternating in $X,Y$. It is also tensorial. For instance, replacing $s$ by $fs$ makes the coefficient of $s$ from derivatives of $f$ equal to $X(Yf)-Y(Xf)-[X,Y]f=0$; the remaining mixed terms $(Xf){\nabla }_{Y}s$ and $(Yf){\nabla }_{X}s$ cancel between the two second derivatives. Thus $F(X,Y)(fs)=fF(X,Y)s$. Replacing $X$ by $fX$, the unwanted term $-(Yf){\nabla }_{X}s$ cancels the opposite term from $[fX,Y]=f[X,Y]-(Yf)X$. Hence $F(fX,Y)=fF(X,Y)$, and the other input follows by alternation.

Consequently $F_p$ depends only on the tangent vectors $X_p,Y_p$ and on $s_p$, giving a smooth global element of ${\Omega }^2(M;\mathrm{End}E)$. In the coefficient-column convention of part (a), applying $d_A$ twice to an arbitrary $E$-valued $r$-form givesFor sections this agrees with the operator definition, so the local Cartan curvature matrix equation isHere $F=\frac{1}{2}{\sum }_{i,j}{F}_{ij}d{x}^i\wedge d{x}^j$, with $F_{ij}=F({\partial }_i,{\partial }_j)$. The same calculation for all degrees provesUnder $e^{\prime }=eg$, applying the covariance of $d_A$ twice gives $F^{\prime }=g^{-1}Fg$. This is exactly the transition rule of the endomorphism bundle, independently confirming that $F$ is a well-defined bundle-valued two-form.