[proofplan] First we check that the Chern-Simons value is independent of the smooth singular fundamental cycle used to evaluate the differential character; the boundary formula for differential characters gives a curvature integral over a $2k$-chain, and this vanishes because $M$ has dimension $2k-1$. The gauge invariance is then a direct naturality argument. A gauge transformation is a principal bundle isomorphism covering the identity map on $M$, and our convention $A^u=u^*A$ says precisely that it identifies the connection $A^u$ on the source with the connection $A$ on the target. Naturality of the Cheeger-Simons refinement identifies the two differential characters on $M$, and evaluation on the fundamental class gives equality in $\mathbb R/\mathbb Z$; the final integer statement is the definition of equality in the [quotient group](/theorems/790) $\mathbb R/\mathbb Z$. [/proofplan]

[step:Show that evaluation on the fundamental class is well-defined]Let $z_M$ and $z'_M$ be smooth singular $(2k-1)$-cycles representing the fundamental class \begin{align*} [M]\in H_{2k-1}(M;\mathbb Z). \end{align*} Since they represent the same homology class, there is a smooth singular $2k$-chain $C$ in $M$ such that \begin{align*} z_M-z'_M=\partial C. \end{align*} Write the smooth singular $2k$-chain as \begin{align*} C=\sum_{j=1}^N n_j\sigma_j, \end{align*} where $n_j\in\mathbb Z$ and each $\sigma_j:\Delta^{2k}\to M$ is a smooth singular simplex. Apply the boundary formula for the degree-$2k$ differential character $\widehat c_P(E,A)$. Its curvature is $P(F_A)$, so \begin{align*} \widehat c_P(E,A)(z_M)-\widehat c_P(E,A)(z'_M)=\widehat c_P(E,A)(\partial C)=\sum_{j=1}^N n_j\int_{\Delta^{2k}}\sigma_j^*P(F_A)\,d\mathcal L^{2k}\mod \mathbb Z. \end{align*} Because $M$ has dimension $2k-1$, every smooth $2k$-form on $M$ is zero. Hence $P(F_A)=0$ as a $2k$-form on $M$, and the curvature integral over $C$ is $0$ in $\mathbb R/\mathbb Z$. Therefore \begin{align*} \widehat c_P(E,A)(z_M)=\widehat c_P(E,A)(z'_M). \end{align*} Thus $\operatorname{CS}_{\widehat c_P}(A)$ is independent of the chosen smooth singular representative of $[M]$.[/step]

[guided]A differential character is evaluated on cycles, not directly on a homology class. Therefore we must check that choosing a different smooth singular representative of the fundamental class gives the same value. Let $z_M$ and $z'_M$ be smooth singular $(2k-1)$-cycles with \begin{align*} [z_M]=[z'_M]=[M]\in H_{2k-1}(M;\mathbb Z). \end{align*} Since the two cycles define the same homology class, their difference is a boundary: there exists a smooth singular $2k$-chain $C$ in $M$ such that \begin{align*} z_M-z'_M=\partial C. \end{align*} Write the smooth singular $2k$-chain as \begin{align*} C=\sum_{j=1}^N n_j\sigma_j, \end{align*} where $n_j\in\mathbb Z$ and each $\sigma_j:\Delta^{2k}\to M$ is a smooth singular simplex. The boundary axiom in the theorem statement says that a degree-$2k$ differential character evaluates on a boundary by summing the pullback integrals of its curvature over these standard simplices, modulo $\mathbb Z$. For the differential character $\widehat c_P(E,A)$, the curvature is $P(F_A)$. Therefore \begin{align*} \widehat c_P(E,A)(z_M)-\widehat c_P(E,A)(z'_M)=\widehat c_P(E,A)(\partial C)=\sum_{j=1}^N n_j\int_{\Delta^{2k}}\sigma_j^*P(F_A)\,d\mathcal L^{2k}\mod \mathbb Z. \end{align*} Now the dimension hypothesis enters. The manifold $M$ has dimension $2k-1$, so there are no nonzero smooth differential forms of degree $2k$ on $M$. Hence the curvature form $P(F_A)$ is the zero $2k$-form on $M$, and its integral over the smooth singular $2k$-chain $C$ is zero modulo $\mathbb Z$. Thus \begin{align*} \widehat c_P(E,A)(z_M)=\widehat c_P(E,A)(z'_M) \end{align*} in $\mathbb R/\mathbb Z$. This proves that the notation $\widehat c_P(E,A)([M])$, implemented by choosing a fundamental cycle, is well-defined.[/guided]

[step:Use the gauge transformation as an isomorphism of bundles with connection]Let \begin{align*} \operatorname{id}_M:M\to M \end{align*} denote the identity map. Since $u:E\to E$ is a gauge transformation, it is a smooth principal $G$-bundle automorphism covering $\operatorname{id}_M$, so \begin{align*} \pi\circ u=\operatorname{id}_M\circ \pi=\pi. \end{align*} By the convention in the statement, \begin{align*} A^u=u^*A. \end{align*} Therefore $u$ is an isomorphism from the principal bundle with connection $(E,A^u)$ to the principal bundle with connection $(E,A)$, in the precise sense that the pullback of the target connection $A$ along $u$ is the source connection $A^u$.[/step]

[guided]From this point onward, the argument uses only the naturality axiom of the chosen differential refinement; the hypotheses on $P$, $c_P$, and integrality are part of the setup that supplies the differential character whose value defines the invariant. The point of fixing the gauge convention is to make the naturality identity have a definite direction. A gauge transformation is not merely a diffeomorphism of the total space: it is a smooth principal $G$-bundle automorphism \begin{align*} u:E\to E \end{align*} covering the identity map \begin{align*} \operatorname{id}_M:M\to M. \end{align*} Thus it satisfies \begin{align*} \pi\circ u=\pi. \end{align*} Equivalently, $u$ is a bundle isomorphism over $\operatorname{id}_M$. The transformed connection is defined in this theorem by \begin{align*} A^u:=u^*A. \end{align*} This equation says exactly that, if the target copy of $E$ is equipped with the connection $A$, then pulling that connection back along $u$ gives the source connection $A^u$. Hence $u$ is an isomorphism of principal bundles with connection from $(E,A^u)$ to $(E,A)$. This is the hypothesis needed in order to apply the naturality property of the differential refinement $\widehat c_P$.[/guided]

[step:Apply the stated naturality property of the differential characteristic class]By the naturality property of the chosen Cheeger-Simons refinement stated in the theorem, applied to the isomorphism \begin{align*} u:(E,A^u)\to (E,A) \end{align*} covering $\operatorname{id}_M$, one has \begin{align*} \widehat c_P(E,A^u)=\operatorname{id}_M^*\widehat c_P(E,A) \end{align*} as degree-$2k$ differential characters on $M$. Since pullback by $\operatorname{id}_M$ is the identity operation on differential characters, this gives \begin{align*} \widehat c_P(E,A^u)=\widehat c_P(E,A). \end{align*}[/step]

[guided]The theorem statement includes the precise naturality axiom needed here: whenever a principal bundle isomorphism $\Phi:Q'\to Q$ covers a smooth map $f:X'\to X$ and the source connection is the pullback of the target connection, the corresponding differential characters satisfy \begin{align*} \widehat c_P(Q',\Phi^*B)=f^*\widehat c_P(Q,B). \end{align*} In the present situation, the source and target manifolds are both $M$, the bundle isomorphism is \begin{align*} u:(E,A^u)\to (E,A), \end{align*} and the base map is $\operatorname{id}_M:M\to M$. The connection condition is exactly $A^u=u^*A$, verified in the previous step. Therefore the naturality axiom gives \begin{align*} \widehat c_P(E,A^u)=\operatorname{id}_M^*\widehat c_P(E,A). \end{align*} Pullback by the identity map leaves every differential character on $M$ unchanged, so \begin{align*} \operatorname{id}_M^*\widehat c_P(E,A)=\widehat c_P(E,A). \end{align*} Combining the two equalities yields \begin{align*} \widehat c_P(E,A^u)=\widehat c_P(E,A). \end{align*}[/guided]

[step:Evaluate the equal differential characters on the fundamental class]The manifold $M$ is closed and oriented of dimension $2k-1$, so it has a fundamental class \begin{align*} [M]\in H_{2k-1}(M;\mathbb Z). \end{align*} Choose a smooth singular $(2k-1)$-cycle $z_M$ representing $[M]$. Evaluating the equality of differential characters from the previous step on this cycle gives \begin{align*} \widehat c_P(E,A^u)(z_M)=\widehat c_P(E,A)(z_M) \end{align*} in $\mathbb R/\mathbb Z$. By the definition of the integral Chern-Simons invariant, the left-hand side is $\operatorname{CS}_{\widehat c_P}(A^u)$ and the right-hand side is $\operatorname{CS}_{\widehat c_P}(A)$. Therefore \begin{align*} \operatorname{CS}_{\widehat c_P}(A^u)=\operatorname{CS}_{\widehat c_P}(A) \end{align*} in $\mathbb R/\mathbb Z$.[/step]

[guided]A degree-$2k$ differential character on $M$ can be evaluated on smooth singular $(2k-1)$-cycles, with values in $\mathbb R/\mathbb Z$. Since $M$ is closed and oriented of dimension $2k-1$, its orientation determines the fundamental class \begin{align*} [M]\in H_{2k-1}(M;\mathbb Z). \end{align*} Choose a smooth singular $(2k-1)$-cycle $z_M$ representing $[M]$. The previous step proved equality of the two differential characters themselves: \begin{align*} \widehat c_P(E,A^u)=\widehat c_P(E,A). \end{align*} Equal differential characters have equal values on every admissible cycle, so evaluating both sides on $z_M$ gives \begin{align*} \widehat c_P(E,A^u)(z_M)=\widehat c_P(E,A)(z_M) \end{align*} in $\mathbb R/\mathbb Z$. The definition of the integral Chern-Simons invariant identifies these two evaluations as \begin{align*} \widehat c_P(E,A^u)(z_M)=\operatorname{CS}_{\widehat c_P}(A^u) \end{align*} and \begin{align*} \widehat c_P(E,A)(z_M)=\operatorname{CS}_{\widehat c_P}(A). \end{align*} Therefore \begin{align*} \operatorname{CS}_{\widehat c_P}(A^u)=\operatorname{CS}_{\widehat c_P}(A) \end{align*} in $\mathbb R/\mathbb Z$.[/guided]

[step:Translate equality modulo $\mathbb Z$ into the statement about representatives]Let $r,s\in \mathbb R$ be real representatives of $\operatorname{CS}_{\widehat c_P}(A^u)$ and $\operatorname{CS}_{\widehat c_P}(A)$, respectively. The equality just proved in $\mathbb R/\mathbb Z$ means precisely that $r+\mathbb Z=s+\mathbb Z$. By the definition of the quotient group $\mathbb R/\mathbb Z$, this is equivalent to \begin{align*} r-s\in \mathbb Z. \end{align*} This proves the representative formulation and completes the proof.[/step]

[guided]Choose [real numbers](/page/Real%20Numbers) $r,s\in \mathbb R$ whose residue classes in $\mathbb R/\mathbb Z$ are the two Chern-Simons invariants: \begin{align*} r+\mathbb Z=\operatorname{CS}_{\widehat c_P}(A^u) \end{align*} and \begin{align*} s+\mathbb Z=\operatorname{CS}_{\widehat c_P}(A). \end{align*} The equality proved above says that these two cosets in the quotient group $\mathbb R/\mathbb Z$ are equal: \begin{align*} r+\mathbb Z=s+\mathbb Z. \end{align*} By the definition of equality of cosets modulo the subgroup $\mathbb Z\subset\mathbb R$, this holds exactly when the difference of the representatives belongs to $\mathbb Z$: \begin{align*} r-s\in \mathbb Z. \end{align*} Thus any two real representatives of the two Chern-Simons classes differ by an integer. This is the representative formulation and completes the proof.[/guided]