Chern–Weil Theory and Characteristic Forms: Examples

The tautological line is the smooth subbundle of ${\mathbb{C}\mathbb{P}}^1\times {\mathbb{C}}^2$ whose fiber at a line $\mathrm{\ell }$ is $\mathrm{\ell }$; on $U=\{[1:z]\}$ and $V=\{[w:1]\}$ its nowhere-zero frames are $s_U([1:z])=(1,z)$ and $s_V([w:1])=(w,1)$. These smooth local trivializations make it a smooth complex line, and the induced Hermitian metric and [F3] provide a Hermitian connection.

In a local unitary frame of any Hermitian line, the metric derivative identity in [F5] gives $\omega +\bar{\omega }=0$. The line curvature structure equation has no quadratic term because a scalar one-form wedges with itself to zero, so [F6] gives ${\mathrm{\Omega }}_{\mathrm{\nabla }}=d\omega$; [F7] then gives the normalized Chern-form formula, which is real-valued by [F8]. The same scalar structure equation applies in any smooth complex frame, even when that frame is not unitary.

On $U\cap V$, $z=1\mathrm{/}w$ and $s_U=z{s}_V$; applying the connection Leibniz rule in [F4] yields ${\omega }_U-{\omega }_V=z^{-1}dz=dz\mathrm{/}z$. Set $D_U=\{[1:z]:\mathrm{\mid }z\mathrm{\mid }\le 1\}$ and $D_V=\{[w:1]:\mathrm{\mid }w\mathrm{\mid }\le 1\}$. These disks cover ${\mathbb{C}\mathbb{P}}^1$; their boundary orientations are opposite, and $z=e^{it}$ positively parametrizes $\mathrm{\partial }{D}_U$. By [F9], ${\int }_{{\mathbb{C}\mathbb{P}}^1}{\mathrm{\Omega }}_{\mathrm{\nabla }}={\int }_{D_U}d{\omega }_U+{\int }_{D_V}d{\omega }_V={\int }_{\mathrm{\partial }{D}_U}({\omega }_U-{\omega }_V)={\int }_{S^1}\frac{dz}{z}=2\pi i\mathrm{.}$ The determinant normalization [F7] therefore gives ${\int }_{{\mathbb{C}\mathbb{P}}^1}{c}_1(\mathrm{\nabla })=-1$. In particular, this curvature cannot vanish identically.

By [F10], the real de Rham class of $c_1(\mathrm{\nabla })$ is the image of $c_1(\gamma )$. The de Rham isomorphism [F11], the smooth-chain comparison [F12], the complex-oriented fundamental class [F13], and the pairing [F14] identify the calculated integral with $⟨\rho (c_1(\gamma )),[{\mathbb{C}\mathbb{P}}^1]⟩$. This proves the stated topological normalization, with the sign fixed by the complex orientation and the library's convention.

If a local unitary frame changes by $s^{\mathrm{\prime }}=gs$ for a smooth $g:U\to U(1)$, the Leibniz rule gives ${\omega }^{\mathrm{\prime }}=\omega +g^{-1}dg$. This added one-form is imaginary and closed: $\mathrm{\mid }g\mathrm{\mid }=1$ gives $d\bar{g}=-g^{-2}dg$, so $\overline{g^{-1}dg}=-g^{-1}dg$, and $d(g^{-1}dg)=-g^{-2}dg\wedge dg=0$. It is locally exact by writing $g=e^{i\theta }$ locally. It need not be globally exact: for $g(z)=z$ on $S^1$, ${\int }_{S^1}{g}^{-1}dg=2\pi i$, whereas Stokes [F9] makes the integral of an exact one-form on $S^1$ zero, applying Stokes to the real and imaginary parts. Thus curvature and Chern form are unchanged under every unitary frame change, without asserting a global primitive. The disk computation uses explicit charts and adds no choices once $\mathrm{\nabla }$ is supplied; full AC enters through the projective-line, compatible-connection, and first-Chern comparison suppliers. [A1, F3, F4, F6, F7, F9, algebra] $\mathrm{\square }$

Flat complex connections. Let ${\mathrm{\nabla }}_0$ be the supplied flat complex-linear connection on $E$. By full AC and [F2], choose a Hermitian metric and a compatible Hermitian connection ${\mathrm{\nabla }}_1$. Both preserve the same underlying ${\mathrm{GL}}_r(\mathbb{C})$ reduction. For each $j>0$, [F1] and ${\mathrm{\Omega }}_{{\mathrm{\nabla }}_0}=0$ give $c_j({\mathrm{\nabla }}_0)=0$ pointwise. The exact transgression in [F3] gives $c_j({\mathrm{\nabla }}_1)=d{T}_j$ for a possibly complex-valued $(2j-1)$-form $T_j$. The Hermitian form $c_j({\mathrm{\nabla }}_1)$ is real-valued by [F1], so $c_j({\mathrm{\nabla }}_1)=d\mathrm{Re}(T_j)$ is exact in the real form complex. The comparison theorem [F4], applied to ${\mathrm{\nabla }}_1$, now gives ${\rho }_M(c_j(E))=0$. This step handles a flat connection that is not itself compatible with any selected Hermitian metric; it compares its zero forms to the Hermitian ones rather than applying the comparison theorem outside its hypotheses.

Real Pontryagin classes and even Chern classes. Let $D$ be a flat real connection on $V$. Every positive-degree Pontryagin form is a homogeneous positive-degree polynomial in ${\mathrm{\Omega }}_D$, so [F1] gives $p_j(D)=0$ for $j>0$. The arbitrary-real-connection clause of [F4] yields ${\rho }_M(p_j(V))=0$. The induced connection on $V_{\mathbb{C}}$ is flat. For $k=2j>0$, [F5] gives ${\rho }_M(c_{2j}(V_{\mathbb{C}}))=(-1{)}^j{\rho }_M(p_j(V))=0$; indices above the rank vanish by the stated rank conventions.

Odd Chern classes of a real complexification. Fix a connected component $B$ of $M$. By [F6], choose a homotopy equivalence $h:X\to B$ from a path-connected CW complex. This CW complex is paracompact by [F7], and $h^{\ast }(V_{\mathbb{C}}{\mathrm{\mid }}_B)$ is numerable because $V$ is numerable and numerability pulls back. The bundle is canonically isomorphic to its conjugate. Thus [F8] gives, for odd $k>0$, $c_k(h^{\ast }{V}_{\mathbb{C}})=c_k(\overline{h^{\ast }{V}_{\mathbb{C}}})=-c_k(h^{\ast }{V}_{\mathbb{C}}),$ so $2c_k(h^{\ast }{V}_{\mathbb{C}})=0$. The relation [F10] identifies this class with $h^{\ast }{c}_k(V_{\mathbb{C}}{\mathrm{\mid }}_B)$. Since $h^{\ast }$ is an isomorphism by [F9], $2c_k(V_{\mathbb{C}}{\mathrm{\mid }}_B)=0$. The torsion statement [F11] then gives ${\rho }_B(c_k(V_{\mathbb{C}}{\mathrm{\mid }}_B))=0$. This holds on every component; singular chains split as the direct sum over components, so cochains and cohomology split as products and the global real class is zero.

A closed genus-two hyperbolic surface. In ${\mathbb{H}}^2$ take a regular octagon whose interior angle is $\pi \mathrm{/}4$. Such an octagon exists: if $\mathrm{\ell }$ is its side length and $\theta$ its angle, the right triangle from the center to a vertex and a side midpoint gives $\sin (\theta \mathrm{/}2)=\frac{\cos (\pi \mathrm{/}8)}{\cosh (\mathrm{\ell }\mathrm{/}2)}\mathrm{.}$ Thus $\theta =\pi \mathrm{/}4$ when $\cosh (\mathrm{\ell }\mathrm{/}2)=\cot (\pi \mathrm{/}8)=1+\sqrt{2}$. Identify its sides by the oriented genus-two word $a_1{b}_1{a}_1^{-1}{b}_1^{-1}{a}_2{b}_2{a}_2^{-1}{b}_2^{-1}\mathrm{.}$ The quotient is compact, oriented, and has genus two; all eight vertices become one point and their angles sum to $8(\pi \mathrm{/}4)=2\pi$, so the hyperbolic metric extends smoothly across that point. Call this closed surface $\mathrm{\Sigma }$. By [F16], its universal cover is ${\mathbb{H}}^2$. Its orientation-preserving deck transformations are fractional-linear transformations, giving a homomorphism $\rho :\mathrm{\Pi }={\pi }_1(\mathrm{\Sigma })\to \mathrm{PSL}(2,\mathbb{R})$.

The deck representation's projective-line bundle. Let $\eta =({\mathbb{H}}^2\times {\mathbb{R}\mathbb{P}}^1)\mathrm{/}\mathrm{\Pi }⟶\mathrm{\Sigma },$ where $\mathrm{\Pi }$ acts diagonally through $\rho$ and its projective action. A nonzero tangent direction at $z\in {\mathbb{H}}^2$ determines the endpoint of the unique orthogonal geodesic ray to the ideal boundary ${\mathbb{R}\mathbb{P}}^1$; this map is deck-equivariant. Hence $\eta$ is the unit direction circle bundle of $T\mathrm{\Sigma }$. The tangent Euler number is the Euler characteristic (Milnor–Stasheff; the printed locator is recorded in the source notes), and the genus-two polygon has one 0-cell, four 1-cells, and one 2-cell. Therefore $⟨e(\eta ),[\mathrm{\Sigma }]⟩=\chi (\mathrm{\Sigma })=1-4+1=-2.$ The sign is fixed by the chosen orientation; only its evenness and nonvanishing are used below.

Lift to a flat oriented plane bundle. Write $D(\eta )$ and $S(\eta )$ for the associated disk and circle bundles. The mod-two Thom class $u\in H^2(D(\eta ),S(\eta );{\mathbb{F}}_2)$ maps to the reduction of $e(\eta )$, which is zero because its integer evaluation is $-2$ on the connected oriented surface. By exactness of the pair sequence in [F12], $u$ is the connecting image of a class $a\in H^1(S(\eta );{\mathbb{F}}_2)$. On each fiber, the connecting map $H^1(S^1;{\mathbb{F}}_2)\to H^2(D^2,S^1;{\mathbb{F}}_2)$ takes its nonzero class to the Thom generator; hence $a$ restricts nontrivially to the circle fiber. This degree-one class defines a double cover $\widehat{\eta }\to \eta$ whose restriction to every fiber is the connected degree-two circle cover.

The flat connection and its curvature. Form the associated real two-plane bundle $\widehat{V}=({\mathbb{H}}^2\times {\mathbb{R}}^2)\mathrm{/}\mathrm{\Pi }⟶\mathrm{\Sigma },\gamma \cdot (z,v)=(\gamma z,\widehat{\rho }(\gamma )v)\mathrm{.}$ Its transition matrices lie in $\mathrm{SL}(2,\mathbb{R})$, so the bundle is oriented. In an evenly covered chart, use the standard basis of ${\mathbb{R}}^2$ as a frame. On each overlap the two chosen lifts differ by one deck transformation, so the transition matrix is the constant matrix $\widehat{\rho }(\gamma )$. These constant cocycles define the smooth bundle by [F13]. Put ${\omega }_{\alpha }=0$ in each such frame. Since every transition matrix $A_{\alpha \beta }$ is constant, $A_{\alpha \beta }^{-1}{\omega }_{\alpha }{A}_{\alpha \beta }+A_{\alpha \beta }^{-1}d{A}_{\alpha \beta }=0={\omega }_{\beta }\mathrm{.}$ Thus [F14] glues the zero matrices to a global connection $D$, and [F15] gives ${\mathrm{\Omega }}_D=0$. The unit-circle bundle of $\widehat{V}$ is $\widehat{\eta }$, so its Thom-normalized Euler number is $-1$ by step 1.6. Changing the chosen orientation could reverse that sign, but not its nonvanishing.

Failure without metric compatibility. The real image ${\rho }_{\mathrm{\Sigma }}(e(\widehat{V}))$ evaluates to $-1$ on the real fundamental class, so it is nonzero. If the flat connection $D$ were compatible with any Euclidean metric, the Euler clause of [F4] would identify that real class with the de Rham class of its metric-compatible Pfaffian form. Since ${\mathrm{\Omega }}_D=0$, that form is identically zero by [F1], contradicting the evaluation $-1$. Therefore $D$ is flat but compatible with no Euclidean metric, and the unqualified Euler conclusion is false.

Boundary and rank cases. If $M$ is empty all cohomology groups in clauses 1–3 are zero. For clause 3 in rank $2m>0$, flatness gives $\mathrm{\Omega }=0$, and the degree-$m$ Pfaffian is zero; [F4] therefore gives ${\rho }_M(e(W))=0$. At rank zero the empty Pfaffian and the Euler class in the canonical unit orientation are both $1$, so the real Euler class is the degree-zero unit, not a positive-degree vanishing class. Rank-zero bundles have no positive-degree Chern or Pontryagin classes; for rank one, the determinant/transgression argument of step 1.1 still applies, while positive Pontryagin classes vanish by the rank cutoff. Indices above the complex rank or the real Pontryagin cutoff vanish by definition; degree zero is excluded from the vanishing assertions and is recorded separately in clause 3. The flatness calculations and comparison theorem apply on boundary charts under their stated scopes. The counterexample is a fixed closed surface and has rank two. Full AC is used in [A1] through [F2], [F4], [F6], [F8], and [F12], including the componentwise CW models; after the connections, representation, and frames are supplied, the local matrix and quotient calculations make no further choices. There is no endpoint parameter and no biconditional in this example.

The second determinant coefficient. In a real frame $e_1,\dots ,e_r$ the complexified connection restricts to $\mathrm{\nabla }$ on $E\otimes 1$ and is complex-linear in the scalar factor, so it has the same connection matrix and hence the same curvature matrix $\mathrm{\Omega }$ over $\mathbb{C}$. Put $A=-\mathrm{\Omega }\mathrm{/}(2\pi i)$. Its entries are $2$-forms and therefore commute, so the usual expansion $\det (I+A)=1+e_1(A)+e_2(A)+\cdots$ in principal minors is valid, with $e_1(A)=\mathrm{tr}A$ and $e_2(A)=\frac{1}{2}((\mathrm{tr}A{)}^2-\mathrm{tr}(A\wedge A))$ by Newton's identity. Since $(2\pi i{)}^2=-4{\pi }^2$, $c_2({\mathrm{\nabla }}_{\mathbb{C}})=\frac{\mathrm{tr}(\mathrm{\Omega }\wedge \mathrm{\Omega })-\mathrm{tr}\mathrm{\Omega }\wedge \mathrm{tr}\mathrm{\Omega }}{8{\pi }^2},$ and [F2] turns this into $p_1(\mathrm{\nabla })=-c_2({\mathrm{\nabla }}_{\mathbb{C}})=[(\mathrm{tr}\mathrm{\Omega }{)}^2-\mathrm{tr}(\mathrm{\Omega }\wedge \mathrm{\Omega })]\mathrm{/}(8{\pi }^2)$. For $r\le 1$ the rank cutoff gives $c_2=0=p_1$.

Metric connections. Let $\mathrm{\nabla }$ be Euclidean-compatible and let $e_1,\dots ,e_r$ be a local orthonormal frame, so $h(e_i,e_j)={\delta }_{ij}$. Applying [F4] to these frame sections gives $0=Xh(e_i,e_j)=h({\mathrm{\nabla }}_X{e}_i,e_j)+h(e_i,{\mathrm{\nabla }}_X{e}_j)={\omega }_{ji}(X)+{\omega }_{ij}(X)$ for every vector field $X$, hence ${\omega }^{\mathsf{T}}=-\omega$. Transposing [F5] and using $(\omega \wedge \omega {)}^{\mathsf{T}}=-{\omega }^{\mathsf{T}}\wedge {\omega }^{\mathsf{T}}$ together with the anticommutativity of one-form coefficients gives ${\mathrm{\Omega }}^{\mathsf{T}}=d({\omega }^{\mathsf{T}})-{\omega }^{\mathsf{T}}\wedge {\omega }^{\mathsf{T}}=-\mathrm{\Omega }$. Thus every diagonal entry of $\mathrm{\Omega }$ vanishes and $\mathrm{tr}\mathrm{\Omega }=0$. Step 1.1 then gives $c_2({\mathrm{\nabla }}_{\mathbb{C}})=\mathrm{tr}(\mathrm{\Omega }\wedge \mathrm{\Omega })\mathrm{/}(8{\pi }^2)$ and $p_1(\mathrm{\nabla })=-\mathrm{tr}(\mathrm{\Omega }\wedge \mathrm{\Omega })\mathrm{/}(8{\pi }^2)$.

The correction term is genuine. On the trivial rank-two real bundle over ${\mathbb{R}}^4$ take the standard frame, let $\omega =\mathrm{diag}(x_{2}d{x}_1,x_{4}d{x}_3)$, and let $\mathrm{\nabla }=d+\omega$. Then $\omega \wedge \omega =0$ and $\mathrm{\Omega }=d\omega =\mathrm{diag}(d{x}_2\wedge d{x}_1,d{x}_4\wedge d{x}_3)$. Hence $\mathrm{tr}\mathrm{\Omega }=d{x}_2\wedge d{x}_1+d{x}_4\wedge d{x}_3$ is nonzero, while $\mathrm{tr}(\mathrm{\Omega }\wedge \mathrm{\Omega })=(d{x}_2\wedge d{x}_1{)}^2+(d{x}_4\wedge d{x}_3{)}^2=0$. Step 1.1 gives $c_2({\mathrm{\nabla }}_{\mathbb{C}})=-\frac{\mathrm{tr}\mathrm{\Omega }\wedge \mathrm{tr}\mathrm{\Omega }}{8{\pi }^2}=-\frac{d{x}_1\wedge d{x}_2\wedge d{x}_3\wedge d{x}_4}{4{\pi }^2}\mathrm{\ne }0,$ whereas the uncorrected expression $-\mathrm{tr}(\mathrm{\Omega }\wedge \mathrm{\Omega })\mathrm{/}(8{\pi }^2)$ of step 2.1 evaluates to $0$; the trace formula therefore requires metric compatibility.

Rank two. Let $r=2$, let the bundle be oriented, and let $\mathrm{\Omega }=\left(\begin{array}{cc}0& -F\\ F& 0\end{array}\right)$ in a positively oriented orthonormal frame. Then $\mathrm{\Omega }\wedge \mathrm{\Omega }=\left(\begin{array}{cc}-F\wedge F& 0\\ 0& -F\wedge F\end{array}\right)$, so $\mathrm{tr}(\mathrm{\Omega }\wedge \mathrm{\Omega })=-2F\wedge F$ and step 2.1 gives $p_1(\mathrm{\nabla })=F\wedge F\mathrm{/}(4{\pi }^2)$. By [F3] and the $2\times 2$ normalization, $e(\mathrm{\nabla })=\mathrm{Pf}(\mathrm{\Omega }\mathrm{/}(2\pi ))=-F\mathrm{/}(2\pi )$, hence $e(\mathrm{\nabla })\wedge e(\mathrm{\nabla })=F\wedge F\mathrm{/}(4{\pi }^2)=p_1(\mathrm{\nabla })$.

Boundary cases. If $M$ is empty then every form space is zero and the identities hold trivially. If $r=0$ or $r=1$, then $c_2=p_1=0$ in clause 1 by the rank cutoff; for $r=1$ and a Euclidean-compatible connection, step 2.1 makes the $1\times 1$ curvature matrix skew-symmetric, hence zero, so $p_1=-\mathrm{tr}(\mathrm{\Omega }\wedge \mathrm{\Omega })\mathrm{/}(8{\pi }^2)=0$. The Euler form is defined only in even rank, so clause 3 has no rank-one case. If $\mathrm{\Omega }=0$, then both sides of each displayed identity in clauses 2 and 3 vanish, with $F=0$ there. All three clauses are pointwise local statements in frame coefficients, so they restrict to boundary charts unchanged; no choice principle, parameter, or limiting process occurs, clause 2 assumes metric compatibility while clause 1 does not, and no converse of clause 2 is asserted.

By [F6], $E={\mathbb{R}}^2\times \mathbb{C}$ is the product line; give its fibres the standard complex structure and $h(z,w)=z\bar{w}$. In the global frame, write ${\mathrm{\nabla }}_{a}s=ds+as$, where $a_0=0$ and $a_1=ixdy$. Each operator is complex-linear and satisfies ${\mathrm{\nabla }}_a(fs)=dfs+f{\mathrm{\nabla }}_{a}s$, so it is a connection. For either $a$, $h({\mathrm{\nabla }}_{a,X}s,t)+h(s,{\mathrm{\nabla }}_{a,X}t)=X(s\bar{t})+(a(X)+\overline{a(X)})s\bar{t}$. Here $a_0=0$ and $\overline{a_1}=-a_1$, so this equals $Xh(s,t)$; both connections are Hermitian for the stated metric, with the compatibility convention of [F3].

The structure equation [F2] gives ${\mathrm{\Omega }}_0=0$. For $a_1=ixdy$, $d{a}_1=idx\wedge dy$ and $a_1\wedge a_1=-x^{2}dy\wedge dy=0$, so ${\mathrm{\Omega }}_1=idx\wedge dy$.

Expanding the degree-one term of the determinant in the Chern-form definition [F1] gives $c_1=-\mathrm{\Omega }\mathrm{/}(2\pi i)$ for this rank-one bundle. Consequently $c_1({\mathrm{\nabla }}_0)=0$ and $c_1({\mathrm{\nabla }}_1)=-dx\wedge dy\mathrm{/}(2\pi )$. The latter is nonzero, since its value on $({\mathrm{\partial }}_x,{\mathrm{\partial }}_y)$ is $-1\mathrm{/}(2\pi )$; hence the representative forms are not equal.

Put $\eta =-xdy\mathrm{/}(2\pi )$. Direct differentiation gives $d\eta =-dx\wedge dy\mathrm{/}(2\pi )=c_1({\mathrm{\nabla }}_1)-c_1({\mathrm{\nabla }}_0)$. This is also the degree-one transgression in [F4]: its polynomial is $P_1(B)=-B\mathrm{/}(2\pi i)$ and $A=ixdy$, so $T_{P_1}({\mathrm{\nabla }}_0,{\mathrm{\nabla }}_1)=P_1(A)=\eta$. Thus the endpoint forms are unequal while [F5] identifies their de Rham classes; [F7] ensures these closed forms represent classes. The example uses only the displayed product bundle and connections; it requires no choice axiom. [F1, F4, F5, F7, step 1.2, step 2.1, given, algebra] $\mathrm{\square }$