[proofplan] We choose the moving frame so that the metric on $E$ is fixed in time, which allows all curvature components to be compared in one fixed Euclidean vector bundle. We then pull the covariant curvature evolution equation for the Ricci flow back from $TU$ to $E$. The time derivative of the moving frame produces precisely the Ricci-linear terms needed to cancel the linear terms in the fully covariant curvature evolution. After cancellation, the remaining terms are the pulled-back rough Laplacian and, with the curvature-operator normalization used below, Hamilton's fibrewise quadratic expression $2(\mathcal R^2+\mathcal R^{\#})$. [/proofplan]

[step:Verify that the moving frame fixes the bundle metric]Let $U\subset M$ be the spatial [open set](/page/Open%20Set) and let $J\subset I$ be the time interval on which the moving frame is defined. The Uhlenbeck moving-frame equation used here is \begin{align*} \partial_t(\iota_t a)=\operatorname{Ric}_{g(t)}^{\sharp}(\iota_t a) \end{align*} for every time-independent smooth local section $a$ of $E$, where $\operatorname{Ric}_{g(t)}^{\sharp}:TU\to TU$ is the Ricci endomorphism determined by $g(t)(\operatorname{Ric}_{g(t)}^{\sharp}X,Y)=\operatorname{Ric}_{g(t)}(X,Y)$. Fix $t_0\in J$ and define the reference bundle metric $h$ on $E$ by $h:=\iota_{t_0}^{*}g(t_0)$. Let $h_{\Lambda^2}$ denote the metric induced by $h$ on $\Lambda^2E$. Let $a,b$ be local smooth sections of $E$. Define the time-dependent scalar function $H_{a,b}:U \times J \to \mathbb{R}$ by \begin{align*} H_{a,b}(x,t)=g(t)_x(\iota_t a(x),\iota_t b(x)). \end{align*} Using $\partial_t g(t)=-2\operatorname{Ric}_{g(t)}$ and the moving-frame equation, we compute \begin{align*} \partial_t H_{a,b}=-2\operatorname{Ric}_{g(t)}(\iota_t a,\iota_t b)+g(t)(\operatorname{Ric}_{g(t)}^{\sharp}\iota_t a,\iota_t b)+g(t)(\iota_t a,\operatorname{Ric}_{g(t)}^{\sharp}\iota_t b). \end{align*} By the defining relation for the Ricci endomorphism, this becomes \begin{align*} \partial_t H_{a,b}=-2\operatorname{Ric}_{g(t)}(\iota_t a,\iota_t b)+\operatorname{Ric}_{g(t)}(\iota_t a,\iota_t b)+\operatorname{Ric}_{g(t)}(\iota_t a,\iota_t b)=0. \end{align*} Thus $\iota_t^*g(t)$ is independent of $t$, and by the definition of $h$ at $t_0$ we have $\iota_t^*g(t)=h$ for all $t\in J$. All contractions on $E$ and $\Lambda^2E$ are therefore performed with the fixed metrics $h$ and $h_{\Lambda^2}$.[/step]

[guided]Let $U\subset M$ be the spatial open set and let $J\subset I$ be the time interval on which the moving frame is defined. Fix $t_0\in J$ and set $h:=\iota_{t_0}^{*}g(t_0)$. Let $h_{\Lambda^2}$ be the metric induced by $h$ on $\Lambda^2E$. The purpose of the Uhlenbeck frame is to prove that this metric does not depend on the time at which it is computed. For local sections $a,b$ of $E$, define $H_{a,b}:U \times J \to \mathbb{R}$ by \begin{align*} H_{a,b}(x,t)=g(t)_x(\iota_t a(x),\iota_t b(x)). \end{align*} This function is the metric on $E$ obtained by pulling $g(t)$ back through $\iota_t$. Differentiate $H_{a,b}$ with respect to $t$. There are three contributions: the time derivative of $g(t)$, the time derivative of the first frame vector, and the time derivative of the second frame vector. Since $g(t)$ evolves by Ricci flow and $\iota_t$ satisfies the Uhlenbeck frame equation, we get \begin{align*} \partial_t H_{a,b}=(\partial_t g(t))(\iota_t a,\iota_t b)+g(t)(\partial_t(\iota_t a),\iota_t b)+g(t)(\iota_t a,\partial_t(\iota_t b)). \end{align*} Using the Ricci flow equation and the moving-frame equation gives \begin{align*} \partial_t H_{a,b}=-2\operatorname{Ric}_{g(t)}(\iota_t a,\iota_t b)+g(t)(\operatorname{Ric}_{g(t)}^{\sharp}\iota_t a,\iota_t b)+g(t)(\iota_t a,\operatorname{Ric}_{g(t)}^{\sharp}\iota_t b). \end{align*} By definition of the Ricci endomorphism $\operatorname{Ric}_{g(t)}^{\sharp}$, the last two terms are both equal to $\operatorname{Ric}_{g(t)}(\iota_t a,\iota_t b)$. Hence \begin{align*} \partial_t H_{a,b}=-2\operatorname{Ric}_{g(t)}(\iota_t a,\iota_t b)+\operatorname{Ric}_{g(t)}(\iota_t a,\iota_t b)+\operatorname{Ric}_{g(t)}(\iota_t a,\iota_t b)=0. \end{align*} So the pulled-back metric is constant in time. This is the key structural reason for using the moving frame: curvature can now be viewed as an endomorphism of the fixed vector bundle $\Lambda^2E$, rather than as an object living in a vector bundle whose metric is changing with time.[/guided]

[step:Pull the curvature tensor and the connection back to $E$] Let $\mathfrak X(U)$ denote the space of smooth vector fields on the spatial open set $U$, and let $\Gamma(E)$ denote the space of smooth local sections of $E$ over $U$. For each $t \in J$, define the pullback connection $\nabla^E(t):\mathfrak X(U) \times \Gamma(E) \to \Gamma(E)$ by requiring \begin{align*} \iota_t(\nabla^E_X(t)a)=\nabla^{g(t)}_X(\iota_t a) \end{align*} for every vector field $X \in \mathfrak X(U)$ and every local section $a$ of $E$. This induces a connection on $\Lambda^2E$ and hence on $\operatorname{End}(\Lambda^2E)$. Let $\Delta_{\nabla^E(t)}$ denote the rough Laplacian induced by this connection. In a local $g(t)$-orthonormal frame $(e_1,\dots,e_n)$ of $TU$, with pulled-back frame $(a_1,\dots,a_n)$ of $E$ determined by $\iota_t a_i=e_i$, this operator is \begin{align*} \Delta_{\nabla^E(t)}\mathcal R=\sum_{i=1}^n\left(\nabla^E_{e_i}(t)\nabla^E_{e_i}(t)\mathcal R-\nabla^E_{\nabla^{g(t)}_{e_i}e_i}(t)\mathcal R\right). \end{align*} Let $\Delta_{g(t)}$ denote the rough Laplacian on covariant curvature tensors induced by the Levi-Civita connection $\nabla^{g(t)}$. In the same local $g(t)$-orthonormal frame, it is defined by \begin{align*} \Delta_{g(t)}R_{g(t)}=\sum_{i=1}^n\left(\nabla^{g(t)}_{e_i}\nabla^{g(t)}_{e_i}R_{g(t)}-\nabla^{g(t)}_{\nabla^{g(t)}_{e_i}e_i}R_{g(t)}\right). \end{align*} Because $\iota_t$ intertwines $\nabla^E(t)$ with $\nabla^{g(t)}$, it also intertwines the induced connections on $\Lambda^2E$ and $\Lambda^2TU$; tracing with the corresponding $h$-orthonormal and $g(t)$-orthonormal frames shows that the rough Laplacian of the pulled-back curvature operator is the pullback of the rough Laplacian of the curvature operator on $\Lambda^2TU$. [/step]

[step:Differentiate the pulled-back curvature components] Fix a point $(x,t)\in U \times J$. Choose local sections $a,b,c,d$ of $E$ near $x$. Define vector fields along $U \times J$ by \begin{align*} A=\iota_t a,\qquad B=\iota_t b,\qquad C=\iota_t c,\qquad D=\iota_t d. \end{align*} The pulled-back curvature tensor is \begin{align*} \mathcal R_{a b c d} = R_{g(t)}(A,B,C,D). \end{align*} Differentiating in time gives \begin{align*} \partial_t \mathcal R_{a b c d}=(\partial_t R_{g(t)})(A,B,C,D)+R_{g(t)}(\partial_t A,B,C,D)+R_{g(t)}(A,\partial_t B,C,D)+R_{g(t)}(A,B,\partial_t C,D)+R_{g(t)}(A,B,C,\partial_t D). \end{align*} Using $\partial_t A=\operatorname{Ric}_{g(t)}^{\sharp}A$ and the analogous identities for $B,C,D$, the frame-derivative contribution is \begin{align*} R_{g(t)}(\operatorname{Ric}_{g(t)}^{\sharp}A,B,C,D)+R_{g(t)}(A,\operatorname{Ric}_{g(t)}^{\sharp}B,C,D)+R_{g(t)}(A,B,\operatorname{Ric}_{g(t)}^{\sharp}C,D)+R_{g(t)}(A,B,C,\operatorname{Ric}_{g(t)}^{\sharp}D). \end{align*} [/step]

[step:Cancel the Ricci-linear terms in the covariant curvature evolution]We use the curvature sign convention \begin{align*} R_{g(t)}(X,Y,Z,W)=g(t)((\nabla^{g(t)}_X\nabla^{g(t)}_Y-\nabla^{g(t)}_Y\nabla^{g(t)}_X-\nabla^{g(t)}_{[X,Y]})Z,W) \end{align*} for smooth vector fields $X,Y,Z,W \in \mathfrak X(U)$. Since $g(t)$ is a smooth Ricci flow on $U \times J$, the curvature tensor, the Ricci tensor, and the covariant derivatives appearing below are smooth. Thus the hypotheses of Hamilton's covariant curvature evolution identity for a smooth Ricci flow are satisfied; this is the standard covariant tensor evolution formula from Hamilton's Ricci-flow curvature computation. In a local $g(t)$-orthonormal frame $(e_1,\dots,e_n)$, write $R_{i j k l}=R_{g(t)}(e_i,e_j,e_k,e_l)$ and define the auxiliary tensor $B_{g(t)}$ by \begin{align*} B_{i j k l}=\sum_{p,q=1}^n R_{p i q j}R_{p k q l}. \end{align*} Hamilton's quadratic tensor $Q_{g(t)}$ is the algebraic tensor with components \begin{align*} Q_{i j k l}=2(B_{i j k l}-B_{i j l k}-B_{i l j k}+B_{i k j l}). \end{align*} This is Hamilton's covariant curvature evolution identity written with the curvature sign convention displayed above and with the curvature operator defined by \begin{align*} h_{\Lambda^2}(\mathcal R(u\wedge v),w\wedge z)=R_{g(t)}(\iota_tu,\iota_tv,\iota_tw,\iota_tz) \end{align*} for $u,v,w,z\in E_x$. The factor $2$ in the definition of $Q_{g(t)}$ is part of this normalization; it is the factor that converts the covariant tensor quadratic term into $2(\mathcal R^2+\mathcal R^{\#})$ on $\Lambda^2E_x$ below. With this convention, Hamilton's covariant curvature evolution identity under Ricci flow gives \begin{align*} (\partial_t R_{g(t)})(A,B,C,D)=(\Delta_{g(t)}R_{g(t)})(A,B,C,D)+Q_{g(t)}(A,B,C,D)-R_{g(t)}(\operatorname{Ric}_{g(t)}^{\sharp}A,B,C,D)-R_{g(t)}(A,\operatorname{Ric}_{g(t)}^{\sharp}B,C,D)-R_{g(t)}(A,B,\operatorname{Ric}_{g(t)}^{\sharp}C,D)-R_{g(t)}(A,B,C,\operatorname{Ric}_{g(t)}^{\sharp}D). \end{align*} Substituting this identity into the formula for $\partial_t\mathcal R_{a b c d}$ from the previous step cancels the four Ricci-linear terms exactly. Therefore \begin{align*} \partial_t \mathcal R_{a b c d}=(\Delta_{g(t)}R_{g(t)})(A,B,C,D)+Q_{g(t)}(A,B,C,D). \end{align*} By the intertwining of the connections established above, the first term is \begin{align*} (\Delta_{g(t)}R_{g(t)})(A,B,C,D) = h_{\Lambda^2}\bigl((\Delta_{\nabla^E(t)}\mathcal R)(a\wedge b),c\wedge d\bigr). \end{align*} Thus the pulled-back evolution is \begin{align*} \partial_t\mathcal R=\Delta_{\nabla^E(t)}\mathcal R+\mathcal Q(\mathcal R), \end{align*} where $\mathcal Q(\mathcal R)$ is the endomorphism of $\Lambda^2E$ determined by the pullback of $Q_{g(t)}$ through the fixed metric $h_{\Lambda^2}$.[/step]

[guided]The delicate point is that the moving frame contributes four Ricci-linear terms, and those terms must cancel with the linear part of the covariant curvature evolution before the curvature can be regarded as satisfying a clean heat-type equation on the fixed bundle $\Lambda^2E$. We use the curvature sign convention \begin{align*} R_{g(t)}(X,Y,Z,W)=g(t)((\nabla^{g(t)}_X\nabla^{g(t)}_Y-\nabla^{g(t)}_Y\nabla^{g(t)}_X-\nabla^{g(t)}_{[X,Y]})Z,W) \end{align*} for smooth vector fields $X,Y,Z,W\in\mathfrak X(U)$. Since $g(t)$ is a smooth Ricci flow on $U\times J$, all curvature tensors and covariant derivatives in the following identity are smooth. These are exactly the regularity hypotheses required for Hamilton's covariant curvature evolution identity for a smooth Ricci flow, the standard covariant tensor evolution formula from Hamilton's Ricci-flow curvature computation. In this sign convention, the identity says that \begin{align*} (\partial_t R_{g(t)})(A,B,C,D)=(\Delta_{g(t)}R_{g(t)})(A,B,C,D)+Q_{g(t)}(A,B,C,D)-R_{g(t)}(\operatorname{Ric}_{g(t)}^{\sharp}A,B,C,D)-R_{g(t)}(A,\operatorname{Ric}_{g(t)}^{\sharp}B,C,D)-R_{g(t)}(A,B,\operatorname{Ric}_{g(t)}^{\sharp}C,D)-R_{g(t)}(A,B,C,\operatorname{Ric}_{g(t)}^{\sharp}D), \end{align*} where, in a local $g(t)$-orthonormal frame $(e_1,\dots,e_n)$, the tensor $Q_{g(t)}$ is defined by \begin{align*} B_{i j k l}=\sum_{p,q=1}^n R_{p i q j}R_{p k q l},\qquad Q_{i j k l}=2(B_{i j k l}-B_{i j l k}-B_{i l j k}+B_{i k j l}). \end{align*} The displayed factor $2$ is a normalization statement, not a cosmetic convention: with the curvature operator defined by \begin{align*} h_{\Lambda^2}(\mathcal R(u\wedge v),w\wedge z)=R_{g(t)}(\iota_tu,\iota_tv,\iota_tw,\iota_tz), \end{align*} this tensor $Q_{g(t)}$ corresponds to $2(\mathcal R^2+\mathcal R^{\#})$. Now substitute Hamilton's identity into the time derivative formula from the previous step. The four terms produced by differentiating the moving frame are \begin{align*} R_{g(t)}(\operatorname{Ric}_{g(t)}^{\sharp}A,B,C,D)+R_{g(t)}(A,\operatorname{Ric}_{g(t)}^{\sharp}B,C,D)+R_{g(t)}(A,B,\operatorname{Ric}_{g(t)}^{\sharp}C,D)+R_{g(t)}(A,B,C,\operatorname{Ric}_{g(t)}^{\sharp}D). \end{align*} These are exactly the negatives of the four Ricci-linear terms in Hamilton's covariant evolution identity, so they cancel term by term. Therefore \begin{align*} \partial_t \mathcal R_{a b c d}=(\Delta_{g(t)}R_{g(t)})(A,B,C,D)+Q_{g(t)}(A,B,C,D). \end{align*} Finally, $\iota_t$ intertwines the pulled-back connection $\nabla^E(t)$ with $\nabla^{g(t)}$, and the trace defining the rough Laplacian is taken over corresponding orthonormal frames. Hence \begin{align*} (\Delta_{g(t)}R_{g(t)})(A,B,C,D) = h_{\Lambda^2}\bigl((\Delta_{\nabla^E(t)}\mathcal R)(a\wedge b),c\wedge d\bigr). \end{align*} This gives the pulled-back evolution \begin{align*} \partial_t\mathcal R=\Delta_{\nabla^E(t)}\mathcal R+\mathcal Q(\mathcal R), \end{align*} where $\mathcal Q(\mathcal R)$ is the endomorphism of $\Lambda^2E$ determined by the pulled-back tensor $Q_{g(t)}$ through $h_{\Lambda^2}$.[/guided]

[step:Identify the remaining algebraic term as $2(\mathcal R^2+\mathcal R^{\#})$]If $n=1$, then $\Lambda^2E=0$, so $\mathcal R=0$, $\mathcal R^2=0$, and $\mathcal R^{\#}=0$; the asserted equation is therefore the zero identity. Assume from now on that $n\geq 2$. At each point $x\in U$, the metric [vector space](/page/Vector%20Space) $(E_x,h_x)$ identifies $\Lambda^2E_x$ with $\mathfrak{so}(E_x,h_x)$ by \begin{align*} u\wedge v \longmapsto \bigl(w\mapsto h_x(v,w)u-h_x(u,w)v\bigr). \end{align*} Through this identification, define $\mathcal R^2:=\mathcal R\circ\mathcal R$. Define $\mathcal R^{\#}:\Lambda^2E_x\to\Lambda^2E_x$ by the Lie bracket structure constants of $\mathfrak{so}(E_x,h_x)$ as follows: if $(\omega_\alpha)$ is any $h_{\Lambda^2}$-orthonormal local frame of $\Lambda^2E$ and \begin{align*} [\omega_\alpha,\omega_\beta]=\sum_\gamma c_{\alpha\beta\gamma}\omega_\gamma \end{align*} defines the structure constants, then \begin{align*} h_{\Lambda^2}(\mathcal R^{\#}\omega_\alpha,\omega_\beta)=\frac{1}{2}\sum_{\gamma,\delta,\eta,\theta}c_{\alpha\gamma\eta}c_{\beta\delta\theta}h_{\Lambda^2}(\mathcal R\omega_\gamma,\omega_\delta)h_{\Lambda^2}(\mathcal R\omega_\eta,\omega_\theta). \end{align*} The tensor $Q_{g(t)}$ is converted into an endomorphism $\mathcal Q(\mathcal R):\Lambda^2E_x\to\Lambda^2E_x$ by the rule \begin{align*} h_{\Lambda^2}(\mathcal Q(\mathcal R)(u\wedge v),w\wedge z)=Q_{g(t)}(\iota_tu,\iota_tv,\iota_tw,\iota_tz) \end{align*} for all $u,v,w,z\in E_x$. With the factor $2$ included in the earlier definition of $Q_{g(t)}$, comparison with the structure-constant formula gives the single consistent normalization \begin{align*} \mathcal Q(\mathcal R)=2(\mathcal R^2+\mathcal R^{\#}). \end{align*} The formula is invariant under change of $h_{\Lambda^2}$-orthonormal frame because it is defined by the Lie bracket and the fixed metric on $\Lambda^2E$. Hence \begin{align*} \partial_t\mathcal R=\Delta_{\nabla^E(t)}\mathcal R+2(\mathcal R^2+\mathcal R^{\#}). \end{align*} This is the asserted curvature evolution equation in the Uhlenbeck moving frame.[/step]

[guided]This final step is convention-sensitive because two normalizations are being compared: the covariant four-tensor $Q_{g(t)}$ and the endomorphism expression built from the Lie algebra structure on $\Lambda^2E_x$. We first fix the algebraic model. At the point $x$, the metric $h_x$ identifies $\Lambda^2E_x$ with $\mathfrak{so}(E_x,h_x)$ by \begin{align*} u\wedge v \longmapsto \bigl(w\mapsto h_x(v,w)u-h_x(u,w)v\bigr). \end{align*} This identification supplies a Lie bracket on $\Lambda^2E_x$. Define $\mathcal R^2:=\mathcal R\circ\mathcal R$. To define $\mathcal R^{\#}$, choose an $h_{\Lambda^2}$-orthonormal frame $(\omega_\alpha)$ of $\Lambda^2E$ and define the structure constants $c_{\alpha\beta\gamma}$ by \begin{align*} [\omega_\alpha,\omega_\beta]=\sum_\gamma c_{\alpha\beta\gamma}\omega_\gamma. \end{align*} Then $\mathcal R^{\#}:\Lambda^2E_x\to\Lambda^2E_x$ is the endomorphism determined by \begin{align*} h_{\Lambda^2}(\mathcal R^{\#}\omega_\alpha,\omega_\beta)=\frac{1}{2}\sum_{\gamma,\delta,\eta,\theta}c_{\alpha\gamma\eta}c_{\beta\delta\theta}h_{\Lambda^2}(\mathcal R\omega_\gamma,\omega_\delta)h_{\Lambda^2}(\mathcal R\omega_\eta,\omega_\theta). \end{align*} The right-hand side is tensorial in the orthonormal frame because it is written using the Lie bracket and the fixed [inner product](/page/Inner%20Product) $h_{\Lambda^2}$, so the definition is independent of the chosen orthonormal frame. Now convert the covariant tensor $Q_{g(t)}$ from the previous step into an endomorphism of $\Lambda^2E_x$. By definition, $\mathcal Q(\mathcal R):\Lambda^2E_x\to\Lambda^2E_x$ is characterized by \begin{align*} h_{\Lambda^2}(\mathcal Q(\mathcal R)(u\wedge v),w\wedge z)=Q_{g(t)}(\iota_tu,\iota_tv,\iota_tw,\iota_tz) \end{align*} for all $u,v,w,z\in E_x$. The earlier definition of $Q_{g(t)}$ included the factor $2$: \begin{align*} Q_{i j k l}=2(B_{i j k l}-B_{i j l k}-B_{i l j k}+B_{i k j l}). \end{align*} That factor is exactly the normalization difference between the covariant four-tensor expression and the Lie-algebraic endomorphism $\mathcal R^2+\mathcal R^{\#}$. Therefore the endomorphism corresponding to $Q_{g(t)}$ is \begin{align*} \mathcal Q(\mathcal R)=2(\mathcal R^2+\mathcal R^{\#}). \end{align*}[/guided]