[proofplan] We regard the radial contraction $H(t,x)=a+t(x-a)$ as a smooth homotopy from the constant map $x\mapsto a$ to the identity map on $U$. Pulling $\omega$ back to $[0,1]\times U$, we decompose the [exterior derivative](/theorems/1525) into the $t$-direction and the $U$-directions and prove the needed Cartan-type identity for the vector field $\partial_t$ directly in coordinates. Integrating this identity in $t$ gives the difference between the endpoint pullbacks $H_1^*\omega$ and $H_0^*\omega$. The endpoint at $t=1$ is $\omega$, while the endpoint at $t=0$ vanishes because $\omega$ has positive degree and the map $H_0$ is constant. [/proofplan]

[step:Define the radial homotopy and identify the operator obtained by contraction]Define the smooth map \begin{align*} H:[0,1]\times U &\to U \\ (t,x) &\mapsto a+t(x-a). \end{align*} This map is well-defined because $U$ is star-shaped with respect to $a$. For each $t \in [0,1]$, define \begin{align*} H_t:U &\to U \\ x &\mapsto H(t,x)=a+t(x-a). \end{align*} Then $H_1=\operatorname{id}_U$ and $H_0$ is the constant map $x\mapsto a$. Let $\partial_t$ denote the coordinate vector field on $[0,1]\times U$ in the interval direction. For $\omega \in \Omega^k(U)$, define the $(k-1)$-form \begin{align*} \widetilde K\omega \in \Omega^{k-1}(U) \end{align*} by \begin{align*} (\widetilde K\omega)_x(v_1,\dots,v_{k-1}) = \int_0^1 (\iota_{\partial_t}H^*\omega)_{(t,x)} (v_1,\dots,v_{k-1}) \,d\mathcal{L}^1(t), \end{align*} where $v_1,\dots,v_{k-1}\in \mathbb{R}^n$ are viewed as tangent vectors in the $U$-directions at $(t,x)$. For every $j \in \{1,\dots,k-1\}$, \begin{align*} dH_{(t,x)}(v_j)=t v_j, \end{align*} and \begin{align*} dH_{(t,x)}(\partial_t)=x-a. \end{align*} Therefore \begin{align*} (\iota_{\partial_t}H^*\omega)_{(t,x)}(v_1,\dots,v_{k-1}) &= (H^*\omega)_{(t,x)}(\partial_t,v_1,\dots,v_{k-1})\\ &= \omega_{a+t(x-a)}(x-a,tv_1,\dots,tv_{k-1})\\ &= t^{k-1}\omega_{a+t(x-a)}(x-a,v_1,\dots,v_{k-1}). \end{align*} Thus $\widetilde K=K$ by the defining formula in the statement.[/step]

[guided]The point of this step is to connect the explicit radial formula for $K$ with the homotopy expression obtained by contracting $H^*\omega$ against the interval direction. We define \begin{align*} H:[0,1]\times U &\to U \\ (t,x) &\mapsto a+t(x-a). \end{align*} The star-shaped hypothesis says exactly that $a+t(x-a)\in U$ for every $x\in U$ and $t\in[0,1]$, so $H$ maps into $U$. For fixed $t$, define \begin{align*} H_t:U &\to U \\ x &\mapsto a+t(x-a). \end{align*} Then $H_1(x)=x$, so $H_1=\operatorname{id}_U$, and $H_0(x)=a$, so $H_0$ is constant. Let $\partial_t$ be the coordinate vector field in the $[0,1]$ direction. The contraction $\iota_{\partial_t}H^*\omega$ is a $(k-1)$-form on $[0,1]\times U$. Integrating it in $t$ gives a $(k-1)$-form on $U$: \begin{align*} (\widetilde K\omega)_x(v_1,\dots,v_{k-1}) = \int_0^1 (\iota_{\partial_t}H^*\omega)_{(t,x)} (v_1,\dots,v_{k-1}) \,d\mathcal{L}^1(t). \end{align*} Now compute the differential of $H$. If $v\in\mathbb{R}^n$ is a tangent vector in the $U$ direction, then changing $x$ by $sv$ changes $H(t,x)$ by $stv$, hence \begin{align*} dH_{(t,x)}(v)=tv. \end{align*} Changing $t$ by $s$ changes $H(t,x)$ by $s(x-a)$, hence \begin{align*} dH_{(t,x)}(\partial_t)=x-a. \end{align*} Therefore \begin{align*} (\iota_{\partial_t}H^*\omega)_{(t,x)}(v_1,\dots,v_{k-1}) &= (H^*\omega)_{(t,x)}(\partial_t,v_1,\dots,v_{k-1})\\ &= \omega_{H(t,x)}(dH_{(t,x)}(\partial_t),dH_{(t,x)}(v_1),\dots,dH_{(t,x)}(v_{k-1}))\\ &= \omega_{a+t(x-a)}(x-a,tv_1,\dots,tv_{k-1})\\ &= t^{k-1}\omega_{a+t(x-a)}(x-a,v_1,\dots,v_{k-1}). \end{align*} This is precisely the formula defining $K\omega$, so the contraction-and-integration construction agrees with the stated radial operator.[/guided]

[step:Differentiate the pulled-back form in the interval direction]Let $\eta:=H^*\omega\in\Omega^k([0,1]\times U)$. We prove the identity \begin{align*} \partial_t\eta=d(\iota_{\partial_t}\eta)+\iota_{\partial_t}(d\eta), \end{align*} where $\partial_t\eta$ denotes coefficientwise differentiation of $\eta$ in the $t$ variable. Write local coordinates on $[0,1]\times U$ as $(t,x_1,\dots,x_n)$. Every $k$-form $\eta$ decomposes uniquely as \begin{align*} \eta=dt\wedge \alpha+\beta, \end{align*} where $\alpha\in\Omega^{k-1}([0,1]\times U)$ and $\beta\in\Omega^k([0,1]\times U)$ contain only the differentials $dx_1,\dots,dx_n$. Since $\iota_{\partial_t}\eta=\alpha$, and since the exterior derivative decomposes as \begin{align*} d=dt\wedge \partial_t+d_x, \end{align*} where $d_x$ is the exterior derivative in the $x$ variables, we compute \begin{align*} d\eta &= d(dt\wedge\alpha+\beta)\\ &= -dt\wedge d\alpha+dt\wedge\partial_t\beta+d_x\beta\\ &= -dt\wedge d_x\alpha+dt\wedge\partial_t\beta+d_x\beta. \end{align*} Contracting with $\partial_t$ gives \begin{align*} \iota_{\partial_t}(d\eta) = -d_x\alpha+\partial_t\beta. \end{align*} Also \begin{align*} d(\iota_{\partial_t}\eta) = d\alpha = dt\wedge\partial_t\alpha+d_x\alpha. \end{align*} Adding these two identities yields \begin{align*} d(\iota_{\partial_t}\eta)+\iota_{\partial_t}(d\eta) = dt\wedge\partial_t\alpha+\partial_t\beta = \partial_t(dt\wedge\alpha+\beta) = \partial_t\eta. \end{align*}[/step]

[guided]We need the usual Cartan identity only for the special vector field $\partial_t$, so we prove that special case directly. This avoids hiding the main computation inside a named theorem. Let \begin{align*} \eta:=H^*\omega\in\Omega^k([0,1]\times U). \end{align*} In the product coordinates $(t,x_1,\dots,x_n)$, every form splits into the part containing $dt$ and the part containing no $dt$. Thus there are unique forms $\alpha\in\Omega^{k-1}([0,1]\times U)$ and $\beta\in\Omega^k([0,1]\times U)$, both involving only $dx_1,\dots,dx_n$, such that \begin{align*} \eta=dt\wedge\alpha+\beta. \end{align*} Because contraction with $\partial_t$ removes the leading $dt$ and kills forms with no $dt$, we have \begin{align*} \iota_{\partial_t}\eta=\alpha. \end{align*} The exterior derivative on the product decomposes into the $t$ part and the $x$ part: \begin{align*} d=dt\wedge\partial_t+d_x, \end{align*} where $d_x$ differentiates only with respect to the variables $x_1,\dots,x_n$. Applying this to $\eta=dt\wedge\alpha+\beta$ gives \begin{align*} d\eta &= d(dt\wedge\alpha)+d\beta\\ &= -dt\wedge d\alpha+dt\wedge\partial_t\beta+d_x\beta\\ &= -dt\wedge d_x\alpha+dt\wedge\partial_t\beta+d_x\beta. \end{align*} The term $dt\wedge dt\wedge\partial_t\alpha$ vanishes because $dt\wedge dt=0$. Now contract this identity with $\partial_t$. Since $\alpha,\beta,d_x\alpha,d_x\beta$ contain no $dt$, we obtain \begin{align*} \iota_{\partial_t}(d\eta) = -d_x\alpha+\partial_t\beta. \end{align*} On the other hand, \begin{align*} d(\iota_{\partial_t}\eta) = d\alpha = dt\wedge\partial_t\alpha+d_x\alpha. \end{align*} Adding the two expressions cancels the $d_x\alpha$ terms: \begin{align*} d(\iota_{\partial_t}\eta)+\iota_{\partial_t}(d\eta) = dt\wedge\partial_t\alpha+\partial_t\beta. \end{align*} But coefficientwise differentiation of $\eta=dt\wedge\alpha+\beta$ in the $t$ direction is \begin{align*} \partial_t\eta = dt\wedge\partial_t\alpha+\partial_t\beta. \end{align*} Therefore \begin{align*} \partial_t\eta=d(\iota_{\partial_t}\eta)+\iota_{\partial_t}(d\eta). \end{align*}[/guided]

[step:Integrate the interval derivative to obtain the endpoint difference]For $v_1,\dots,v_k\in\mathbb{R}^n$, evaluating the identity from the previous step on the $U$-direction tangent vectors $v_1,\dots,v_k$ gives \begin{align*} \frac{\partial}{\partial t} \bigl[(H_t^*\omega)_x(v_1,\dots,v_k)\bigr] = \bigl(d(\iota_{\partial_t}H^*\omega)+\iota_{\partial_t}d(H^*\omega)\bigr)_{(t,x)}(v_1,\dots,v_k). \end{align*} Since $H$ and $\omega$ are smooth, the integrand is continuous in $t$. Integrating with respect to $\mathcal{L}^1$ on $[0,1]$ and applying the [fundamental theorem of calculus](/theorems/632) for the real-valued function \begin{align*} t\mapsto (H_t^*\omega)_x(v_1,\dots,v_k) \end{align*} yields \begin{align*} (H_1^*\omega)_x(v_1,\dots,v_k)-(H_0^*\omega)_x(v_1,\dots,v_k) = \int_0^1 \bigl(d(\iota_{\partial_t}H^*\omega)+\iota_{\partial_t}d(H^*\omega)\bigr)_{(t,x)}(v_1,\dots,v_k) \,d\mathcal{L}^1(t). \end{align*} The exterior derivative commutes with pullback, so \begin{align*} d(H^*\omega)=H^*(d\omega). \end{align*} Also, differentiation in the $x$ variables commutes with integration over the compact interval $[0,1]$, because the coefficients of $\iota_{\partial_t}H^*\omega$ are smooth in $(t,x)$. Hence \begin{align*} \int_0^1 d(\iota_{\partial_t}H^*\omega)_{(t,x)}(v_1,\dots,v_k) \,d\mathcal{L}^1(t) = (dK\omega)_x(v_1,\dots,v_k), \end{align*} and \begin{align*} \int_0^1 (\iota_{\partial_t}H^*(d\omega))_{(t,x)}(v_1,\dots,v_k) \,d\mathcal{L}^1(t) = (Kd\omega)_x(v_1,\dots,v_k). \end{align*} Therefore \begin{align*} H_1^*\omega-H_0^*\omega=dK\omega+Kd\omega. \end{align*}[/step]

[guided]We now turn the infinitesimal identity in $t$ into a global identity between forms on $U$. Fix $x\in U$ and tangent vectors $v_1,\dots,v_k\in\mathbb{R}^n$. The coefficient \begin{align*} t\mapsto (H_t^*\omega)_x(v_1,\dots,v_k) \end{align*} is a smooth real-valued function on $[0,1]$, because $H$ and $\omega$ are smooth. Evaluating \begin{align*} \partial_t(H^*\omega) = d(\iota_{\partial_t}H^*\omega)+\iota_{\partial_t}d(H^*\omega) \end{align*} on the $U$-direction vectors $v_1,\dots,v_k$ gives \begin{align*} \frac{\partial}{\partial t} \bigl[(H_t^*\omega)_x(v_1,\dots,v_k)\bigr] = \bigl(d(\iota_{\partial_t}H^*\omega)+\iota_{\partial_t}d(H^*\omega)\bigr)_{(t,x)}(v_1,\dots,v_k). \end{align*} The ordinary fundamental theorem of calculus applies to this smooth real-valued function, so \begin{align*} (H_1^*\omega)_x(v_1,\dots,v_k)-(H_0^*\omega)_x(v_1,\dots,v_k) = \int_0^1 \bigl(d(\iota_{\partial_t}H^*\omega)+\iota_{\partial_t}d(H^*\omega)\bigr)_{(t,x)}(v_1,\dots,v_k) \,d\mathcal{L}^1(t). \end{align*} We now identify the two terms on the right. First, the exterior derivative commutes with pullback, hence \begin{align*} d(H^*\omega)=H^*(d\omega). \end{align*} Second, the operation $d$ in the $x$ variables commutes with integration over $[0,1]$: the coefficients of $\iota_{\partial_t}H^*\omega$ are smooth functions of $(t,x)$, and the interval of integration is compact. Therefore differentiating the coefficient of \begin{align*} \int_0^1 \iota_{\partial_t}H^*\omega\,d\mathcal{L}^1(t) \end{align*} is the same as integrating the differentiated coefficient. Since this integrated contraction is exactly $K\omega$, we get \begin{align*} \int_0^1 d(\iota_{\partial_t}H^*\omega)_{(t,x)}(v_1,\dots,v_k) \,d\mathcal{L}^1(t) = (dK\omega)_x(v_1,\dots,v_k). \end{align*} Similarly, replacing $\omega$ by $d\omega$ in the defining formula for $K$ gives \begin{align*} \int_0^1 (\iota_{\partial_t}H^*(d\omega))_{(t,x)}(v_1,\dots,v_k) \,d\mathcal{L}^1(t) = (Kd\omega)_x(v_1,\dots,v_k). \end{align*} Since the equality holds for every $x\in U$ and every $v_1,\dots,v_k\in\mathbb{R}^n$, it is an equality of $k$-forms: \begin{align*} H_1^*\omega-H_0^*\omega=dK\omega+Kd\omega. \end{align*}[/guided]

[step:Evaluate the endpoint pullbacks and conclude the formula] Since $H_1=\operatorname{id}_U$, \begin{align*} H_1^*\omega=\omega. \end{align*} Since $H_0:U\to U$ is the constant map $x\mapsto a$, its differential satisfies \begin{align*} d(H_0)_x(v)=0 \end{align*} for every $x\in U$ and every $v\in\mathbb{R}^n$. Because $k\geq 1$, for every $v_1,\dots,v_k\in\mathbb{R}^n$, \begin{align*} (H_0^*\omega)_x(v_1,\dots,v_k) = \omega_a(0,\dots,0) = 0. \end{align*} Thus $H_0^*\omega=0$. Substituting the endpoint values into \begin{align*} H_1^*\omega-H_0^*\omega=dK\omega+Kd\omega \end{align*} gives \begin{align*} \omega=dK\omega+Kd\omega. \end{align*} This is the desired homotopy formula. [/step]