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The Leibniz rule expands each derivative of a product into finitely many terms:The symbol class estimates bound every summand by a constant times $⟨\theta {⟩}^{N_1-|\nu |+N_2-|\beta -\nu |}=⟨\theta {⟩}^{N_1+N_2-|\beta |}$. Hence symbol orders add under multiplication:

We now address the remaining unheaded requests. To define the oscillatory integral as a functional on the space of test functions, choose $\eta \in C_c^{\infty }({\mathbb{R}}^k)$ equal to one near zero and setThe limit is not an assertion of absolute convergence of the original frequency integral. Split off a bounded-frequency part, which is smooth in $x$. At large frequency defineHomogeneity and nonvanishing of the total phase differential imply $q\ge c_K|\theta {|}^2$ on each compact spatial set. Also $L{e}^{i\Phi }=e^{i\Phi }$. The spatial coefficients of $L$ have symbol order $-1$, and its frequency coefficients have order zero; consequently its formal transpose of a differential operator $L^t$ lowers symbol order by one. After $r>N+k$ integrations by parts, the high-frequency pairing has an absolutely integrable amplitude $(L^t{)}^r[(1-\chi (\theta ))a(x,\theta )f(x)]$, where $\chi$ is a fixed cutoff near zero. Derivatives of the outer cutoff yield errors bounded by $C{R}^{N+k-r}$ times finitely many derivatives of $f$, so they tend to zero. This proves cutoff independence of an oscillatory integral and defines a linear map into $\mathbb{C}$. Its distributional continuity is permitted as an assumption, and is also visible from this finite-derivative estimate.

The singular support of a distribution is the complement of the largest open subset on which it equals a smooth regular distribution. The stationary-direction bound for singular support needs conic support of an oscillatory amplitude. WriteIts radial lift is the closed conic enlargement of the ordinary support. The generally valid bound isIf the amplitude support is conic, this is exactly the printed bound. It is also the standard interpretation when support in the frequency directions is understood conically.

To prove the bound, take a point outside its right-hand side. Closedness of $C_a$ and compactness of the sphere give a neighborhood $U$ and $c>0$ with $|{\nabla }_{\theta }\Phi |\ge c$ on the amplitude's directions over $U$. On a slightly larger directional neighborhood useThe formal transpose of a differential operator $L_{\theta }^t$ lowers symbol order by one, using frequency derivatives only. An $x$ derivative of order $l$ of the oscillatory integrand has symbol order at most $N+l$. Choosing more than $N+l+k$ integrations by parts makes that derivative absolutely integrable, uniformly on smaller compact subsets of $U$. This works for every $l$, so $I_{\Phi }(a)$ is smooth on $U$.

The ordinary amplitude support can miss a singular-support limit phenomenon requires a qualification here: with unrestricted ordinary support the printed inclusion is false. An explicit counterexample uses $X={\mathbb{R}}^2$, $x=(s,t)$, one frequency variable and $\Phi (s,t,\theta )=s\theta$. Choose $t_j=1/j$, $w_j=1/(10j^2)$ and $\chi \in C_c^{\infty }((-1/4,1/4))$ with $\chi (0)=1$. Its translates ${\chi }_j(t)=\chi ((t-t_j)/w_j)$ have disjoint supports. Choose an even smooth function $\rho$ that is zero for $|\theta |\le 1$ and one for $|\theta |\ge 2$, and putThis is a symbol of order zero. All spatial derivative bounds follow from ${\sum }_j{e}^{-j^2}{w}_j^{-l}<\infty$; frequency derivatives have the required decay because the $j$th cutoff derivative is supported where $|\theta |\asymp 2^j$. Near every point $(s,0,\theta )$ with finite $\theta$, all large-$j$ terms vanish through the frequency cutoff and all remaining terms vanish in a small $t$ neighborhood. Thus no point $(0,0,\theta )$ lies in its ordinary support.

Nevertheless its oscillatory integral iswhere $G$ is smooth. Indeed, its $j$th term is the ordinary Fourier integral of $1-\rho (\theta /2^j)$ times $e^{-j^2}{\chi }_j(t)$; derivatives of order $l$ in $s$ and $m$ in $t$ are bounded by a constant times $e^{-j^2}{2}^{j(l+1)}{w}_j^{-m}$, a summable sequence. Since $B(t_j)\ne 0$, each $(0,t_j)$ is singular. Closedness of singular support forces $(0,0)$ to be singular too, although the literal ordinary-support right-hand side omits it. The closed conic support includes this limiting point and resolves the defect.

Finally use Fourier inversion distributionally. The constant amplitude gives $(2\pi {)}^{-n}\int e^{ix\cdot \theta }d\theta ={\delta }_0$, and differentiation of the exponential supplies a factor $i\theta$. Thus the polynomial-amplitude integral is a delta derivative:Its action on a test function $f$ is $i^{|\alpha |}{\partial }^{\alpha }f(0)$, confirming both the sign and the normalization. This is the delta derivatives from polynomial oscillatory amplitudes identity.

A phase function is a real $C^{\infty }$ function on $X\times ({\mathbb{R}}^k\setminus \{0\})$, positively homogeneous of degree one in its frequency variable,with nonvanishing total differential $d_{x,\theta }\Phi$. Vanishing of its frequency gradient alone is allowed; those critical directions are relevant to singularities.

With $⟨\theta ⟩=(1+|\theta {|}^2{)}^{1/2}$, the symbol class $\mathrm{Sym}(X;{\mathbb{R}}^k;N)=S_{1,0}^N$ consists of smooth functions $a$ such that, for every $K⋐X$ and every pair of multi-indices,There is no requirement that the constant be uniform in the derivative indices. We use $D=(1/i)\partial$; the factors of $i$ do not affect these estimates.

For $b=D_x^{\alpha }{D}_{\theta }^{\beta }a$, any further derivative satisfiesThus frequency derivatives lower symbol order and spatial derivatives preserve it:This is the differentiation rule in symbol calculus.

A smooth vector field is a smooth section of a vector bundle $X:M\to TM$ of the tangent bundle, so $X_p\in T_{p}M$. In a manifold chart, $X=X^i{\partial }_i$ with smooth coefficients. It acts on a smooth function by $Xf=X^i{\partial }_{i}f$. For a diffeomorphism $\alpha :M\to N$, the pushforward of a vector field isThe Lie bracket of vector fields is the commutator of their actions on smooth functions:The second derivatives of $f$ cancel, so this is again a vector field. The corresponding Jacobi identity is $[X,[Y,Z]]+[Y,[Z,X]]+[Z,[X,Y]]=0$.

A local flow of $X$ is a smooth map $(t,p)\mapsto {\varphi }_t(p)$ on an open neighbourhood of $\{0\}\times M$ satisfying ${\varphi }_0(p)=p$ and ${\partial }_t{\varphi }_t(p)=X_{{\varphi }_t(p)}$. On a sufficiently small neighbourhood and time interval, each ${\varphi }_t$ is a diffeomorphism onto its image, with inverse ${\varphi }_{-t}$. Uniqueness of integral curves of a vector field gives ${\varphi }_{s+t}={\varphi }_s\circ {\varphi }_t$ wherever both sides are defined. These are local statements; no assumption of a complete vector field is required.

A smooth tensor field of type $(r,s)$ is a smooth section of $(TM{)}^{\otimes r}\otimes (T^{\ast }M{)}^{\otimes s}$, where the factors are the tangent bundle and cotangent bundle. The flow definition of the Lie derivative of a tensor field isHere pullback applies $d{\varphi }_t^{-1}$ to each vector factor and the dual of $d{\varphi }_t$ to each covector factor, evaluating $T$ at ${\varphi }_t(p)$. Thus all tensors being differentiated lie in the same fibre over $p$.

For a smooth function, ${\varphi }_t^{\ast }f=f\circ {\varphi }_t$, and the chain rule givesFor a vector field $Y$, in local coordinates the expansions ${\varphi }_t(x)=x+tX(x)+O(t^2)$ and $d{\varphi }_t(x)=I+td{X}_x+O(t^2)$ giveConsequentlyPullback preserves tensor products and tensor contractions, so differentiation makes this definition a tensor derivation. It therefore agrees on every tensor field with the Lie derivative of a tensor field determined by these two formulas.

Now put ${\psi }_t=\alpha \circ {\varphi }_t\circ {\alpha }^{-1}$. The chain rule givesThus ${\psi }_t$ is the local flow of ${\alpha }_{\ast }X$. In the case $M=N$, if ${\alpha }_{\ast }X=X$, uniqueness of integral curves of a vector field makes ${\psi }_t={\varphi }_t$ on their common domains, or $\alpha \circ {\varphi }_t={\varphi }_t\circ \alpha$. Conversely, differentiating this commuting identity at zero gives $d{\alpha }_p{X}_p=X_{\alpha (p)}$, hence ${\alpha }_{\ast }X=X$. This proves that diffeomorphism invariance of a vector field is equivalent to commuting with its local flow, with every identity understood on the domain where its compositions exist.

In a manifold chart, write a smooth differential form as $\eta ={\sum }_I{a}_{I}d{x}^{i_1}\wedge \cdots \wedge d{x}^{i_p}$, with $I$ increasing. Define its exterior derivative byTo check that this definition is global, change coordinates to $y$. The chain rule expresses $d{x}^i$ as ${\sum }_j({\partial }_j{x}^i)d{y}^j$. Differentiating it gives zero because the second partial derivatives of $x^i$ are symmetric and $d{y}^k\wedge d{y}^j$ is antisymmetric. The coordinate definition satisfies the graded Leibniz rule, so differentiating the transformed expression for $\eta$ leaves exactly ${\sum }_{I}d{a}_I\wedge d{x}^{i_1}\wedge \cdots \wedge d{x}^{i_p}$. This agrees with the original formula and proves independence of the manifold chart. The same cancellation of mixed derivatives gives $d^2=0$.

The de Rham cohomology is thereforethe vector space of closed differential forms modulo exact differential forms.

Here is a Green-theorem proof of exactness on a disc. Identify the disc with a round disc centred at zero, and let $\eta =Pdx+Qdy$ be a closed differential form. Set $f(x)={\int }_{[0,x]}\eta$. For $x$ and $x+h$ in the disc, Green theorem on the oriented triangle with vertices $0,x,x+h$ gives zero boundary integral, since ${\partial }_{x}Q-{\partial }_{y}P=0$. HenceTaking $h=\epsilon v$ and dividing by $\epsilon$ yields $d{f}_x(v)={\eta }_x(v)$. The radial integral defining $f$ is smooth, so $\eta =df$ andDegenerate triangles follow by continuity. The same proof works on any star-shaped plane domain, including the whole plane.

Integration requires an orientation: take $M$ to be an oriented compact $n$-dimensional smooth manifold without boundary. The Generalized Stokes theorem says ${\int }_{M}d\beta =0$ for every smooth $(n-1)$-form $\beta$. Every $n$-form is closed by dimension, and changing it by an exact differential form leaves its integral unchanged. Thusis a well-defined linear map $H_{\mathrm{dR}}^n(M;\mathbb{R})\to \mathbb{R}$. Compactness alone does not supply the orientation omitted from the printed integration statement.

For an $n$-dimensional Lie group, choose a nonzero $\lambda \in {\Lambda }^n{T}_e^{\ast }G$ and define the invariant volume form on a Lie group byEach $d{L}_{g^{-1}}$ is an isomorphism, so $\omega$ is nowhere zero and defines an orientation. The composition identity $L_{(hg{)}^{-1}}\circ L_h=L_{g^{-1}}$ gives $L_h^{\ast }\omega =\omega$. Moreover, every left-invariant differential form of degree $n$ is determined by its value at $e$, so the space of such forms is one-dimensional.

Under the supplied invariant-representative assumption, all top-degree de Rham cohomology classes are multiples of $[\omega ]$. For compact $G$, orient it so $\omega$ is positive; then ${\int }_G\omega >0$, whereas the Generalized Stokes theorem makes every exact top form have integral zero. Therefore $[\omega ]\ne 0$ andunder that assumption. For actual compact Lie groups, this argument applies when $G$ is connected. The supplied assumption fails for disconnected groups: on $S^1\times \mathbb{Z}/2$, the two components have independent degree-one classes, while globally left-invariant top forms form only a one-dimensional space. In general the top-degree de Rham cohomology of a compact Lie group is ${\mathbb{R}}^c$, where $c$ is its number of connected components.

For $S^2$, use $U=S^2\setminus \{N\}$ and $V=S^2\setminus \{S\}$. Stereographic projection identifies each with the plane. The disc argument gives $\eta {|}_U=d{f}_U$ and $\eta {|}_V=d{f}_V$ for any closed differential one-form $\eta$. The overlap $U\cap V$ is connected, so $d(f_U-f_V)=0$ makes the difference a constant. Adjust one potential by that constant; they glue to a global smooth function $f$ with $df=\eta$. ThusFinally, $S^3$ is a compact connected Lie group: the identificationidentifies it smoothly with the special unitary group $SU(2)$. Matrix multiplication and inversion supply its smooth group operations. The preceding invariant-volume argument therefore gives

Let $i:M\hookrightarrow N$ be the inclusion. The restriction of a connection to an embedded submanifold is the pullback connection on $i^{\ast }TN=TN{|}_M$. Concretely, for a tangent vector field $X$ on $M$ and a local section $S$ of $TN{|}_M$, extend $S$ smoothly off $M$ and define ${\nabla }_{X}S$ by differentiating the extension in direction $di(X)$. Two extensions differ by a field whose coefficients vanish on $M$. Their derivatives along every tangent curve in $M$ vanish too, and the connection's coefficient terms are multiplied by the zero field. The answer is therefore independent of the extension. Dependence only on the value of the first argument follows from the connection's linearity over smooth functions.

The induced Riemannian metric is positive definite, so the fibrewise orthogonal projection $\pi :TN{|}_M\to TM$ is smooth. For tangent fields define the projected ambient connection $D_{X}Y=\pi ({\nabla }_{X}Y)$. It is real-linear and satisfiesbecause $\pi Y=Y$. Thus $D$ is a Koszul connection on $M$.

The Levi-Civita connection is characterized by the torsion-free connection and metric connection conditionsThese determine it uniquely. For the ambient Levi-Civita connection, the Gauss formula is ${\nabla }_{X}Y=D_{X}Y+II(X,Y)$, withLinearity over smooth functions in $X$ is immediate. In $Y$, the additional $X(f)Y$ term is tangent and is killed by $I-\pi$. Thus tensoriality makes $II$ a bilinear map $TM{\times }_{M}TM\to \nu M$ into the normal bundle. Its antisymmetric part isbecause the Lie bracket of vector fields tangent to $M$ is tangent. This proves the symmetry of the second fundamental form in an arbitrary Riemannian ambient manifold.

Write the curvature operators as $R^N(v,w)y={\nabla }_v{\nabla }_{w}y-{\nabla }_w{\nabla }_{v}y-{\nabla }_{[v,w]}y$ and $R^M$ with $D$ in place of $\nabla$. To match the four-slot notation in the requested identity, useThis convention is stated because permuting the slots can change the displayed signs.

If $\eta$ is a normal field and $x$ is tangent, metric compatibility applied to $⟨\eta ,x⟩=0$ gives the tangential derivative of a normal field identityUsing the Gauss formula and this identity,Here the normal part of ${\nabla }_v(D_{w}y)$ pairs to zero with $x$. Interchange $v,w$ and subtract; the bracket term obeys $⟨{\nabla }_{[v,w]}y,x⟩=⟨D_{[v,w]}y,x⟩$. It follows thatRearranging proves the Gauss equation in a curved ambient manifold:All expressions are tensorial, so choosing local extensions of the four tangent vectors proves the pointwise identity at $P$. With flat Euclidean ambient space, $R^N=0$ and this reduces to the usual Gauss equation.

On smooth functions, the Lie derivative of a function satisfiesOn an arbitrary vector field $W$, the Lie derivative of a vector field givesThe middle equality follows from the Jacobi identity for the Lie bracket of vector fields. One can verify that identity without assuming the result being proved: view vector fields as derivations acting on smooth functions, expand their commutators, and cancel the six compositions. Equality as derivations implies equality as vector fields.

For the second identity, let $A,B,C$ be any operators on either smooth functions or vector fields. Associativity of composition givesEach term cancels in the sum. Applying this to $A={\mathcal{L}}_X$, $B={\mathcal{L}}_Y$, $C={\mathcal{L}}_Z$ proves both requested identities on both kinds of arguments:The commutator identity for Lie derivatives expresses that the Lie derivative of a tensor field represents the Lie bracket of vector fields.

The Lie derivative of a function is differentiation along the flow, while the Lie derivative of a vector field compares a vector with its pullback by that flow. Their index-free expressions areHere $X(f)$ is the directional derivative of the smooth function, and $[X,Y]$ is the Lie bracket of vector fields, characterized by $[X,Y](f)=X(Y(f))-Y(X(f))$. To see the second expression directly, if ${\varphi }_t$ is the local flow of $X$, differentiating $(d{\varphi }_t{)}^{-1}Y\circ {\varphi }_t$ at $t=0$ gives $dY(X)-dX(Y)=[X,Y]$. This uses the flow definition of the Lie derivative of a tensor field.

The test-function space is $\mathcal{D}({\mathbb{R}}^n)=C_c^{\infty }({\mathbb{R}}^n)$. Its topology is the strict inductive limit topology of the spaces ${\mathcal{D}}_K$ of smooth functions supported in a fixed compact set $K$, each with the seminorms $p_{K,j}(\phi )={\max }_{|\alpha |\le j}{\sup }_K|{\partial }^{\alpha }\phi |$. In particular, a sequence converges in $\mathcal{D}$ precisely when its supports lie in one compact set and all its derivatives converge uniformly. The indices $\alpha$ use multi-index notation.

The distribution space ${\mathcal{D}}^{\prime }({\mathbb{R}}^n)$ consists of continuous complex-linear forms on $\mathcal{D}$. Equivalently, for every compact $K$ there are $C_K$ and an integer $q_K\ge 0$ such thatWe use distributional convergence: $u_j\to u$ means $⟨u_j,\phi ⟩\to ⟨u,\phi ⟩$ for every $\phi \in \mathcal{D}$. The pairings are bilinear, with no complex conjugation. A test function $f$ is identified with its regular distribution $⟨f,\phi ⟩=\int f\phi$.

For a distribution $u$ and a test function $\psi$, their smoothing convolution with a test function isIt is a smooth function, with ${\partial }_x^{\alpha }(u\ast \psi )(x)=⟨u_y,{\partial }^{\alpha }\psi (x-y)⟩$. Indeed, for $x$ in a compact neighborhood all translated tests have support in one compact set, and their difference quotients converge in the space of test functions. Notice that $u\ast \psi$ need not be compactly supported.

Choose a nonnegative mollifier $\rho \in C_c^{\infty }(B_1)$ with $\int \rho =1$, and write ${\rho }_{\epsilon }(x)={\epsilon }^{-n}\rho (x/\epsilon )$. If $\tilde{\rho }(x)=\rho (-x)$, thenThe right-hand test tends to $\phi$ in $\mathcal{D}$: its supports lie in $\mathrm{supp}\phi +{\overline{B}}_1$ for $\epsilon \le 1$, and every derivative converges uniformly by the approximate-identity argument. Thus the smooth regularizations converge to $u$ as distributions.

To obtain actual compactly supported approximants, choose a smooth cutoff function $\chi$ equal to one on $B_1$ and supported in $B_2$, and setFor each fixed test function $\phi$, the cutoff is identically one on its support once $j$ is large. Hence $⟨f_j,\phi ⟩=⟨u,{\tilde{\rho }}_{1/j}\ast \phi ⟩\to ⟨u,\phi ⟩$. This proves $\mathcal{D}$ is dense in ${\mathcal{D}}^{\prime }$, and in fact establishes the density of test functions in distributions and gives a convergent approximating sequence for each distribution. The expanding cutoff is essential when $u$ has noncompact support of a distribution.

For the radial limit, take a test function $\phi$ and introduce $t=r^2-1$ in polar coordinates. The Jacobian $rdr=dt/2$ givesWhen $\phi$ is supported away from the origin, $A$ vanishes near $t=-1$ and extends to a compactly supported smooth function on the whole line. The folded sine approximation to a Dirac delta is exposed by setting $B(s)=A(s)+A(-s)$: integration by parts givesby the Riemann-Lebesgue lemma. ThereforeThis surface delta distribution is arclength measure on the unit circle. By the level-set normalization of a surface delta, it is $2\delta (|x{|}^2-1)$: the factor two cancels the gradient magnitude $|\nabla (|x{|}^2-1)|=2$ on the circle. It is not twice arclength measure.

For a test function that can meet the origin, the endpoint cannot be discarded. Now $A(-1)=\pi \phi (0)$. Its right derivative is integrable, with $A^{\prime }(-1+s)=(\pi /4)\Delta \phi (0)+O(s)$ from differentiated Taylor expansion: the angular average cancels odd Taylor terms, so $A(t)=\pi \phi (0)+(\pi /4)(1+t)\Delta \phi (0)+O((1+t{)}^2)$ near $t=-1$. Applying integration by parts separately on the negative and positive intervals givesThe last two integrals tend to zero by the Riemann-Lebesgue lemma. Thus the radial quadratic oscillation defect in two dimensions gives the stronger oscillating point-mass defect formulaChoose $\phi$ supported in $B_{1/2}$ with $\phi (0)=1$. Its pairing is $-\pi \cos m+o(1)$, which does not converge. For completeness, if $\cos m$ had a limit $L$, the recurrence $\cos (m+1)+\cos (m-1)=2\cos 1\cos m$ would force $L=0$, whereas the even subsequence identity $\cos (2m)=2{\cos }^{2}m-1$ would then force $L=-1$. Consequently there is no limit in ${\mathcal{D}}^{\prime }({\mathbb{R}}^2)$.

For the Lie derivative of a tensor field, the scalar rule follows because a vector field $X$ differentiates products of smooth functions. On vector fields the Lie bracket of vector fields satisfiesThis follows by applying both sides to an arbitrary smooth function and expanding the two compositions of derivations. Both the scalar and vector-field operators are real-linear, so the extension theorem applies and gives the unique contraction-compatible tensor operator ${\mathcal{L}}_X$.

For a type-$(1,1)$ tensor $A$, regard it as the pointwise endomorphism of $TM$ obtained by contracting its covector slot with a vector. The endomorphism-induced tensor derivation starts withIt is real-linear and obeys $D_A(fY)=fA(Y)=f{D}_{A}Y+(D_{A}f)Y$. The scalar operator zero is a derivation in every dimension, so there is no zero-dimensional obstruction here. Consequently it also extends uniquely. On a differential one-form, the two operators areTheir actions on arbitrary tensor fields follow by the tensor-product rule; $D_A$ adds $A$ in every contravariant slot and subtracts its dual action in every covariant slot.

Functions in this calculation belong to $C^{\infty }(M)$: the printed $C^{\infty }(X)$ in Q2(c) is a typographical error. The contraction $C_2^1$ pairs the sole covector with the new vector $Y$.

The tensor denoted $X\otimes df$ acts as the rank-one endomorphism $Y\mapsto df(Y)X=Y(f)X$. In the paper's covector-first ordering it is written $df\otimes X$, using the canonical interchange of tensor factors; this is the same endomorphism.

On any smooth function $h$,On a vector field $Y$, the Lie bracket of vector fields givessoMultiplication by $f$ preserves the tensor-product Leibniz rule and commutation with tensor contractions, since the latter are linear over smooth functions. Hence the difference is a real-linear contraction-compatible tensor derivation whose scalar operator is zero. Its agreement with $D_{X\otimes df}$ on functions and vector fields determines its action on every tensor by uniqueness:For example, on a differential one-form this gives $-\omega (X)df$. Equivalently ${\mathcal{L}}_{fX}\omega =f{\mathcal{L}}_X\omega +\omega (X)df$, which checks the sign independently. This is the scaled Lie derivative defect identity.

An orientation of a smooth $n$-dimensional manifold can be specified by a nowhere-vanishing smooth $n$-form $\omega$: an ordered tangent basis is positive precisely when $\omega$ evaluates positively on it. Multiplication by a positive smooth function preserves the orientation. This is the differential-form formulation of the equivalent formulations of orientability of a smooth manifold.

Use the standard convention that the compact connected manifold is nonempty and has no boundary. Choose a positive volume form $\omega$ for its orientation. It is a closed differential form, because there are no nonzero $(n+1)$-forms. For $n>0$, compactness makes its integral finite, and positivity on a coordinate neighborhood together with nonnegativity elsewhere givesIf $\omega =d\eta$, the Generalized Stokes theorem would give ${\int }_M\omega ={\int }_{\partial M}\eta =0$, a contradiction. ThereforeFor $n=0$, a nonempty connected manifold is a point, and $H_{\mathrm{dR}}^0(M)=\mathbb{R}$ directly. The boundary hypothesis matters: a closed interval is an oriented compact connected manifold with boundary with $H_{\mathrm{dR}}^1=0$. Thus the result is about ordinary boundaryless manifolds, not an unrestricted version including boundary.