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

A connection on a vector bundle is a linear operator $\nabla :\Gamma (E)\to {\Omega }^1(M;E)$ satisfying $\nabla (fs)=df\otimes s+f\nabla s$. Choose coordinates $x^i$ and a local frame $e_1,\dots ,e_m$, and define the connection matrix by $\nabla e_b={\sum }_a{A}^a{}_b\otimes e_a$, with $A^a{}_b={\sum }_i{A}^a{}_{bi}d{x}^i$. For a section $s={\sum }_a{s}^a{e}_a$, the coefficient column satisfiesThe symbol $d$ on the right means the ordinary exterior derivative of the coefficient functions.

Extend uniquely to $E$-valued differential forms by the graded Leibniz ruleIn the same local trivialization, an $E$-valued form is a column $\sigma =({\sigma }^a)$ and the covariant exterior derivative isMoving the one-form $A$ past the scalar $r$-form produces exactly the $(-1{)}^r$ in the graded rule. This explains the sign rather than replacing the rule by an ungraded product formula.

The endomorphism bundle has the induced endomorphism bundle connection characterized on sections by ${\nabla }_X^{\mathrm{End}}(T)s={\nabla }_X(Ts)-T{\nabla }_{X}s$. Its coefficient formula is $dT+AT-TA$. On an endomorphism-valued $r$-form $T$, extend as a graded derivation of the endomorphism-valued exterior product, givingExplicitly its $(a,b)$ entry is $d{T}^a{}_b+{\sum }_c{A}^a{}_c\wedge T^c{}_b-(-1{)}^r{\sum }_c{T}^a{}_c\wedge A^c{}_b$. This version is characterized byThe endomorphism-valued version also satisfies $d_A(T\wedge S)=d_{A}T\wedge S+(-1{)}^{r}T\wedge d_{A}S$. Products here combine composition of endomorphisms with wedge products of coefficient forms; matrix factors must retain their order.

These local formulas are independent of the local frame. For $e^{\prime }=eg$, the change of frame of a vector-bundle connection gives $A^{\prime }=g^{-1}Ag+g^{-1}dg$, ${\sigma }^{\prime }=g^{-1}\sigma$ and $T^{\prime }=g^{-1}Tg$. Using $d(g^{-1})=-g^{-1}(dg)g^{-1}$ gives $d_{A^{\prime }}{\sigma }^{\prime }=g^{-1}{d}_A\sigma$ and $d_{A^{\prime }}{T}^{\prime }=g^{-1}(d_{A}T)g$. Thus both extended operators are globally defined.

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.

Use the covariant exterior derivative on the endomorphism bundle and remember that $F$ has degree two. The Bianchi identity is $d_{A}F=dF+A\wedge F-F\wedge A=0$. The local formula from part (b) givesThe two lines cancel, establishingBecause this is an endomorphism-valued equation and the operators transform covariantly under a change of local frame, this local calculation proves the global identity.

For a line bundle over $k=\mathbb{R}$ or $\mathbb{C}$, every fibre endomorphism is multiplication by a scalar. Thus $\mathrm{End}E$ is canonically the trivial scalar bundle, even when $E$ itself is nontrivial. In a local frame $A$ is a scalar one-form, so $A\wedge A=0$ and $F=dA$ locally. Also $A\wedge F=F\wedge A$, since their degrees are one and two. The Bianchi identity therefore reduces to the ordinary closedness conditionThis does not assert that the local scalar one-forms $A$ combine into a global one-form.

For two connections on a vector bundle $\nabla ,{\nabla }^{\prime }$, the difference is $C^{\infty }(M)$-linear in the section: the two $df\otimes s$ terms cancel. It is consequently a global endomorphism-valued one-form $a$. In rank one, $a$ is a global ordinary scalar one-form. Locally $A^{\prime }=A+a$, and the scalar cross terms and $a\wedge a$ vanish, givingThis is a global equality, so the two closed differential forms differ by an exact differential form. It proves the connection-independent curvature class of a line bundle: