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In geodesic normal coordinates at $p$, radial geodesics have coordinates $x(t)=tv$. The geodesic equation at $t=0$ gives ${\Gamma }_{jk}^i(p)v^j{v}^k=0$ for every $v$. The Levi-Civita connection is torsion free, making these coefficients symmetric in $j,k$; the polarization identity then gives ${\Gamma }_{jk}^i(p)=0$. This does not imply vanishing curvature, which depends on derivatives of the coefficients.

For two zero-starting continuous martingales square-integrable at time $t$, the Itô isometry gives $\mathbb{E}(M_t\pm N_t{)}^2=\mathbb{E}[M\pm N{]}_t$. The polarization identity yields the displayed identity. Integrability of the cross-bracket follows from the Kunita-Watanabe inequality and the Cauchy-Schwarz inequality.

For operators on a complex Hilbert space, pointwise boundedness of $⟨Tx,x⟩$ over a family implies uniform boundedness of their operator norms. The complex polarization identity recovers each $⟨Tx,y⟩$ from four diagonal values. Apply the Uniform boundedness principle first to the functionals in $y$ for fixed $x$, then to the operators in $x$. On a real Hilbert space the claim fails: multiples of a nonzero skew-symmetric operator have zero diagonal forms but unbounded norms. For example, rotations through a quarter turn on ${\mathbb{R}}^2$ multiplied by integers suffice.

Every Euclidean isometry of ${\mathbb{R}}^n$ is a product of at most $n+1$ operations of reflection in a hyperplane. An isometry fixing zero is orthogonal: the polarization identity preserves inner products, and its values on an orthonormal basis determine its linear action. An orthogonal transformation is a product of at most $n$ linear reflections, by reflecting the image of the first basis vector back to that vector and inducting on its orthogonal complement. For an arbitrary isometry, first reflect its image of zero back to zero, if necessary, and apply the orthogonal result. Affine reflection hyperplanes, rather than only hyperplanes through zero, are essential here.

A finite group of Euclidean isometries has a common fixed point: average any orbit. Each isometry is an affine map and therefore preserves this finite affine average; applying a group element merely permutes the orbit terms. To see that an isometry is affine, subtract its value at zero and use the polarization identity to show that the resulting map preserves inner products. Images of a standard orthonormal basis then determine a linear orthogonal map.

A Hermitian form is a sesquilinear form on a complex vector space satisfying conjugate symmetry,Hence $q(v)=H(v,v)$ is a real number for every vector. In a basis, the matrix of a Hermitian form has $A_{ij}=H(e_i,e_j)$ and obeys $A_{ji}=\overline{A_{ij}}$, equivalently $A=A^{\ast }$: it is a Hermitian matrix. The coordinate rule remains $H(x,y)=x^{T}A\overline{y}$. Conjugate symmetry does not imply positivity.^[ref]

The diagonal contains more information than might be expected. For any complex sesquilinear form $B$, put $q(v)=B(v,v)$. Expanding four expressions gives the complex polarization identity in the linear-first convention:This is an algebraic identity, with no positivity assumption. If every $q(v)$ is real, interchanging $u,v$ preserves the first difference and negates the second: use $q(\alpha v)=|\alpha {|}^{2}q(v)$ for $\alpha =-1,i$. The identity then gives $B(v,u)=\overline{B(u,v)}$. Thus a complex sesquilinear form is a Hermitian form if and only if all its diagonal values are real. Testing only the basis vectors is insufficient: the off-diagonal example in sesquilinear form has zero diagonal entries in its matrix but a nonreal value at $(1,i)$.

Consider the indefinite Hermitian formIts values at $e_1,e_2$ are $1,-1$. Moreover $u=(1,1)$ and $v=(1,-1)$ are nonzero isotropic vectors: $q_{-}(u)=q_{-}(v)=0$, yet $H_{-}(u,v)=2$. These vectors do not belong to the radical of a Hermitian form, and the usual Cauchy-Schwarz inequality would incorrectly require $4\le 0$. An indefinite Hermitian form records signed geometry rather than defining a norm by $\sqrt{H(v,v)}$.

There is a separate boundary case: $H_0(z,w)=z_1\overline{w_1}$ is a positive semidefinite Hermitian form, since $H_0(z,z)=|z_1{|}^2\ge 0$, but $H_0(e_2,e_2)=0$. Its radical of a Hermitian form is the entire second coordinate line. Positive definiteness is exactly the additional condition in a Hermitian inner product. The three examples below show why these distinctions matter: on the real slice $z=(x,y)\in {\mathbb{R}}^2\subset {\mathbb{C}}^2$, $q_{+}(z)=|z_1{|}^2+2|z_2{|}^2$ has an ellipse as its unit level, $q_{-}$ has hyperbolas, and $q_0$ has two parallel lines. These are slices of the complex examples, not drawings of their full four-dimensional domains.

Let $D=T^{\ast }T-T{T}^{\ast }$. This operator is self-adjoint, andIf condition (iii) holds, this quadratic form vanishes. For a real inner product, polarization givesTherefore $D=0$, which is condition (i). This supplies the nontrivial reverse implication needed for the equivalence.

The Baire category theorem states that a nonempty complete metric space cannot be a countable union of closed nowhere dense sets. The Uniform boundedness principle says that if $V$ is a Banach space, $W$ a normed vector space, and $\mathcal{T}\subseteq B(V,W)$ is pointwise bounded, meaning ${\sup }_{T\in \mathcal{T}}\Vert Tx\Vert <\infty$ for every $x\in V$, then ${\sup }_{T\in \mathcal{T}}\Vert T\Vert <\infty$.

To prove it, let $E_n=\{x:{\sup }_{T\in \mathcal{T}}\Vert Tx\Vert \le n\}$. Each set is closed by continuity of the individual operators, and their union is $V$. The Baire category theorem gives a ball $B(x_0,r)$ inside some $E_N$. For $\Vert h\Vert <r$, both $x_0+h$ and $x_0$ are in $E_N$, so $\Vert Th\Vert \le \Vert T(x_0+h)\Vert +\Vert T{x}_0\Vert \le 2N$ for every $T$. Scaling gives $\Vert T\Vert \le 2N/r$ uniformly. The zero-space and empty-family cases are immediate.

For the first family on a Hilbert space, fix $x$. The functionals $y\mapsto ⟨y,Tx⟩$ have operator norm $\Vert Tx\Vert$ by the Cauchy-Schwarz inequality and equality on the direction of $Tx$. Their values are pointwise bounded by the hypothesis, after taking complex conjugates if necessary. Apply the Uniform boundedness principle to obtain ${\sup }_{T\in \mathcal{E}}\Vert Tx\Vert <\infty$. Apply it a second time to the operator family itself to conclude ${\sup }_{T\in \mathcal{E}}\Vert T\Vert <\infty$.

For the diagonal hypothesis, the answer depends on the scalar field. Over $\mathbb{C}$, let $q_T(x)=⟨Tx,x⟩$, using the convention that the inner product is linear in its first variable. The complex polarization identity givesThe assumed bounds at these four fixed vectors make every matrix coefficient uniformly bounded, so the preceding argument applies. Thus the complex-space answer is yes, even without assuming self-adjointness.

Over $\mathbb{R}$, take $X={\mathbb{R}}^2$ andEvery diagonal value $⟨nJx,x⟩$ is zero, whereas $\Vert nJ\Vert =n$. Thus the real-space answer is no in general. Real diagonal values do not detect the skew-symmetric part of an operator. This is diagonal boundedness of operator families over real and complex Hilbert spaces.

Use the convention that the inner product is linear in its first argument. The precise polarization identity isIf (1) holds, apply this formula before and after $U$ to get preservation of all inner products, proving (2). If (2) holds, the identity $⟨Ux,Uy⟩=⟨x,U^{\ast }Uy⟩$ gives $U^{\ast }U=I$. For $z=Ux$, surjectivity then gives $U{U}^{\ast }z=U{U}^{\ast }Ux=Ux=z$, proving (3).

From (3), $\Vert Ux{\Vert }^2=⟨x,U^{\ast }Ux⟩=\Vert x{\Vert }^2$ and likewise $\Vert U^{\ast }x\Vert =\Vert x\Vert$, proving (4). Finally, if (4) holds, apply the same polarization identity to $U^{\ast }$ to obtain $U{U}^{\ast }=I$. This makes $U$ onto; the other norm identity supplies the remaining requirement of (1). ThusThese four statements are conditions in a single equivalence proof, rather than four independent source questions.

For a real linear change of coordinates $x^{\prime }=Ax$, one has $d{x}^{\prime }=Adx$. The Euclidean metric therefore transforms toIt agrees with $d{x}^{T}dx$ for every displacement precisely when $A^{T}A=I$. For example, equality of the associated quadratic forms determines their symmetric coefficient matrices by polarization identity. The orthogonal group is defined by this condition, soTaking determinants gives $(\det A{)}^2=1$. Thus these matrices are invertible automatically; both signs of the determinant are allowed, since the metric alone does not specify an orientation.

Use the Levi-Civita connection $\nabla$ of the Riemannian metric $g$ and fix the conventionThe Leibniz rule shows that this expression is linear over smooth functions in all three arguments: the differentiated coefficient terms in the first two derivatives cancel those in the bracket term. Thus $R$ is a section of ${\Lambda }^2{T}^{\ast }M\otimes \mathrm{End}(TM)$, the Riemann curvature tensor, and $R_4$ is a covariant tensor field. This convention gives positive sectional curvature on the round sphere.

Its algebraic symmetries areThe first identity follows immediately from the definition. For the second, apply $XY-YX-[X,Y]$ to $g(Z,W)$ and use compatibility of the Levi-Civita connection with the Riemannian metric. The left side is zero and the differentiated terms combine toFor the first Bianchi identity, use normal coordinates at a point $p$, so the Christoffel symbols vanish there. Torsion-freeness gives ${\Gamma }_{ij}^{\ell }={\Gamma }_{ji}^{\ell }$, andThe cyclic sum over $i,j,k$ cancels term by term. Since this is a tensor identity, it holds for arbitrary tangent vectors everywhere.

To prove pair interchange rather than assume it, abbreviate $R_4(a,b,c,d)$ to $R_{abcd}$ and put $A=R_{abcd}$, $B=R_{cdab}$. The first Bianchi identity and the two antisymmetries giveThe first Bianchi identity applied to $(d,b,a)$ with final argument $c$ gives $R_{dbca}+R_{adcb}=R_{badc}=A$. Hence $A=2B-A$, proving $A=B$.

There is also the differential second Bianchi identity,For its proof take commuting coordinate fields, normal at $p$. The Jacobi identity for operator commutators says $[{\nabla }_X,[{\nabla }_Y,{\nabla }_Z]]+\text{cyclic}=0$. On these commuting fields, $[{\nabla }_Y,{\nabla }_Z]=R(Y,Z)$; at $p$, the covariant derivatives of the coordinate arguments vanish. The commutator identity is therefore the displayed covariant derivative identity at $p$, and tensoriality establishes it everywhere.

For a two-dimensional plane $\sigma =\mathrm{span}\{X,Y\}\subset T_{p}M$, define the sectional curvature byThe denominator is the squared area of $X,Y$. Under an invertible change of plane basis, the alternating pairs in the numerator and the area in the denominator both acquire the square of the change-of-basis determinant. Thus the quotient depends only on $\sigma$. For an orthonormal basis of $\sigma$, it is simply $g(R(X,Y)Y,X)$.

The Ricci curvature is the tracewhere $(e_i)$ is any orthonormal basis of $T_{p}M$. This is independent of the basis because it is a trace. Applying the first Bianchi identity to $(e_i,X,Y)$ and pairing with $e_i$ proves that the summand difference for $\mathrm{Ric}(X,Y)-\mathrm{Ric}(Y,X)$ vanishes; the middle term $g(R(X,Y)e_i,e_i)$ is zero by the last-pair antisymmetry. Hence Ricci curvature is a symmetric bilinear form. For a unit vector $v=e_1$ extended to an orthonormal basis,For nonzero $X$, take $v=X/|X|$ and multiply the right side by $|X{|}^2$. This determines the entire Ricci curvature by the polarization identity:Finally, the scalar curvature is the metric trace of Ricci curvature. Taking its diagonal entries in an orthonormal basis and pairing the equal contributions from $(i,j)$ and $(j,i)$ yieldsIn particular, constant sectional curvature $K_0$ gives $\mathrm{Ric}=(n-1)K_{0}g$ and $\mathrm{Scal}=n(n-1)K_0$. In dimension one the traces are zero; there are no two-dimensional tangent planes and the sums are empty.

Let $G$ be the finite group and choose a point $p$. Its orbit $E=\{gp:g\in G\}$ is a nonempty finite set. Every group element permutes $E$ and maps a closed metric disc onto a disc of the same radius. By the permitted uniqueness of the smallest enclosing disc, it fixes that disc and therefore its centre $o$. Thus $G$ lies in the orientation-preserving stabilizer of one point.

In the Euclidean plane, translate $o$ to zero. Every isometry fixing zero is an orthogonal linear map: distance preservation gives preservation of inner products by the polarization identity, and its values on an orthonormal basis then determine the linear map. Orientation preservation gives determinant one, so it is a rotation. Consequently $G$ embeds in the circle group of rotation angles.

For the hyperbolic plane, use this precise standard fact about its isometries: after placing a point at the origin of the Poincare disc model, every orientation-preserving isometry fixing it has the form $z\mapsto e^{i\theta }z$. Equivalently, its point stabilizer is the rotation group, and its action on directions at the point is faithful. This is a general hyperbolic-plane fact allowed by the question. It again identifies $G$ with a finite subgroup of the circle group.

Finally, a nontrivial finite rotation group has a least positive angle $\theta$ in $[0,2\pi )$. Divide any other angle $\varphi$ by $\theta$: the remainder $\varphi -k\theta$ is another group angle in $[0,\theta )$, so it must be zero. Similarly the remainder of $2\pi$ is zero, and $\theta =2\pi /m$ for an integer $m$. The rotation through $\theta$ generates the group. The trivial group is cyclic as well. Therefore both finite orientation-preserving plane isometry groups are cyclic. This proves the finite orientation-preserving plane isometry groups are cyclic result using the same enclosing-disc argument in both geometries.

Let $T$ be the isometry. Since $T(0)=0$, it preserves norms, and the polarization identity applied to distances gives $⟨T(x),T(y)⟩=⟨x,y⟩$. For an orthonormal basis $e_1,e_2,e_3$, the vectors $T(e_i)$ are again an orthonormal basis, andExpanding in that basis proves $T(x)={\sum }_i⟨x,e_i⟩T(e_i)$. Hence $T$ is a linear map represented by an orthogonal matrix $Q$.

A reflection in a hyperplane with nonzero normal $v$ is $H_v(x)=x-2⟨x,v⟩v/⟨v,v⟩$. If $Q{e}_1\ne e_1$, choose $v=Q{e}_1-e_1$. Because $\Vert Q{e}_1\Vert =\Vert e_1\Vert$, substitution shows $H_{v}Q{e}_1=e_1$. If they agree, no reflection is needed. The resulting orthogonal transformation fixes $e_1$ and preserves its orthogonal complement. Apply the same construction there to fix $e_2$; the second normal is perpendicular to $e_1$, so the first vector remains fixed. Finally the restriction to the remaining line is either the identity or negation; in the latter case a third reflection, normal to $e_3$, fixes it. Reversing these steps writes $Q$ as a product of at most three reflections, each in a plane through the origin. This proves the three-dimensional case of the Cartan–Dieudonné theorem.

Three reflections are sometimes necessary. The map $-I$ is the product of the three coordinate-plane reflections. Its determinant is $-1$, excluding zero or two reflections. A single plane reflection fixes a two-dimensional vector subspace, whereas $-I$ fixes only zero, excluding one. Thus the bound is sharp.

Here is Witt's lemma with the hypotheses appropriate to the classical groups. Let $V$ be finite-dimensional with a nonsingular alternating bilinear form, a nonsingular symmetric bilinear form in characteristic different from two, or a nonsingular Hermitian form with nontrivial involution. Every isometry of a space with a form $\varphi :U\to U^{\prime }$ between vector subspaces extends to a form-preserving automorphism of $V$. For quadratic forms, the same statement holds in every characteristic when the polar bilinear form of a quadratic form is nonsingular and the isometry of a space with a form preserves the quadratic form itself. The restricted forms on $U,U^{\prime }$ may be degenerate. We prove these versions, including the characteristic-two quadratic case needed later.

For an alternating bilinear form $B$, split the restricted space as $U=H\oplus R$, where $R=U\cap U^{\perp }$ and $B{|}_H$ is nonsingular. Choose a symplectic basis for $H$ and a basis $r_1,\dots ,r_t$ of $R$. The nondegenerate form on $V$ supplies vectors $s_j$ in $H^{\perp }$ with $B(r_i,s_j)={\delta }_{ij}$. Adding linear combinations of the $r_i$ to the $s_j$ makes $B(s_i,s_j)=0$: for each pair $i<j$ one may cancel its pairing by adding an appropriate multiple of $r_i$ to $s_j$, without disturbing the dual pairings or previously treated pairs. This uses no division by two. Thus $H$ together with the pairs $(r_i,s_i)$ is nonsingular and its orthogonal complement for a sesquilinear form has a symplectic basis.

Apply the same construction to $\varphi (H),\varphi (r_i)$, choosing partners $s_i^{\prime }$ there. The two completed bases have identical alternating Gram matrices and equal remaining dimensions. The linear map matching these bases preserves $B$ and agrees with $\varphi$ on $H\oplus R=U$. This proves the alternating version over an arbitrary field, including characteristic two.

For a quadratic form $Q$, define its polar bilinear form of a quadratic form by $B(x,y)=Q(x+y)-Q(x)-Q(y)$. Its symmetrypreserves $Q$. The zero-dimensional case is immediate. Induct on $\dim U$: choose $Q(t)\ne 0$ and a hyperplane $K\subseteq U\cap t^{\perp }$. Extending $\varphi {|}_K$ first reduces to $\varphi {|}_K=I$. Put $U=K\oplus Fu$, $v=\varphi (u)$, $d=u-v\in S=K^{\perp }$. If $Q(d)\ne 0$, $s_d$ completes the extension. Otherwise $B(u,d)=B(v,d)=0$. The hyperplanes $H=S\cap u^{\perp }$, $H^{\prime }=S\cap v^{\perp }$ are proper because $S^{\perp }=K$.

For $a\in S\setminus (H\cup H^{\prime })$ with $Q(a)\ne 0$, set $c=u-s_a(v)$. Thenand $s_a{s}_c$ fixes $K$ and sends $u$ to $v$. If no such $a$ exists and $|F|>2$, the identityforces $Q(z)=B(a,z)=0$. The complement of two hyperplanes spans $S$: adding a multiple of any outside vector avoids at most two forbidden scalars. Hence $d\perp S$, so $H=H^{\prime }$, forcing $Q{|}_S=0$, contrary to $t\in S$.

Over ${\mathbb{F}}_2$, choose $a$ outside both hyperplanes. In the remaining case $Q(a)=Q(d)=B(a,d)=0$, and $B(a,u)=B(a,v)=1$. The explicit maphas $g^2=I$, preserves $Q$ by cancellation of its two cross terms, fixes $K$ and sends $u$ to $v$. This closes the induction. The characteristic-independent symmetry argument and its binary exceptional step are also treated in Casselman's quadratic-form notes and Elman, Karpenko and Merkurjev, section 8.

In odd characteristic a symmetric bilinear form is recovered by the polarization identity for $Q(x)=B(x,x)/2$, so the quadratic proof also proves its extension theorem. For completeness, the Hermitian form case has a similarly explicit induction. Take $h$ linear in the first variable, with conjugation $a\mapsto \bar{a}$. For anisotropic vectors for a form $u,v$ with equal value $A=h(u,u)=h(v,v)\ne 0$, let $d=u-v$ and $c=h(u,d)$. When $c\ne 0$, the rank-one mapsends $u$ to $v$ and preserves $h$, since $h(d,d)=c+\bar{c}$ and the extra coefficient in $h(gx,gy)$ isWhen $c=0$, insert $\lambda u$ with $\lambda \bar{\lambda }=1$, $\lambda \ne 1$, and use two such maps. Such a scalar exists for a nontrivial involution: $\lambda =a/\bar{a}$ with $a\ne \bar{a}$ works. If the source vector subspace contains an anisotropic vector for a form, align it this way and induct in its nonsingular orthogonal complement for a sesquilinear form.

If instead every source vector is isotropic, the Hermitian polarization identity makes its whole restricted form zero. Choose $e$ in its basis and a partner $f$ pairing to one with $e$ and to zero with the remaining source basis. Make $h(f,f)=0$ by replacing $f$ by $f-te$, where $t+\bar{t}=h(f,f)$. The trace map is onto the fixed field: in odd characteristic divide by two; in characteristic two scale a nonzero trace. Choose $a+\bar{a}\ne 0$. The orthogonal vectors $e+af$ and $e-\bar{a}f$ have nonzero values $a+\bar{a}$ and its negative, so the preceding rank-one construction aligns this entire hyperbolic pair with its target pair. The remaining source basis lies in the pair's orthogonal complement for a sesquilinear form; induction finishes the Hermitian proof.

The hypotheses matter. For example, the identity-matrix symmetric bilinear form on ${\mathbb{F}}_2^3$ has canonical vector $c=(1,1,1)$ characterized by $B(x,c)=B(x,x)$, so every isometry of a space with a form fixes $c$. The one-dimensional map $e_1\mapsto c$ preserves the restricted bilinear form but cannot extend. Thus “all symmetric forms in characteristic two” is not a correct replacement for the quadratic or alternating versions proved above.

Associate to $Q$ its symmetric bilinear form by the polarization identity,We first construct a diagonal basis by induction on the dimension. A zero quadratic form is already diagonal. Otherwise choose $v$ with $Q(v)\ne 0$. Every $x$ has the decompositionwhose second term lies in $W=\{w:B(w,v)=0\}$. Also $\mathbb{R}v\cap W=\{0\}$, so $V=\mathbb{R}v\oplus W$ and $\dim W=n-1$. Apply the induction hypothesis to the restricted quadratic form on $W$. The cross terms with $v$ vanish, giving a diagonal basis on $V$. Rescale each vector with diagonal coefficient $a\ne 0$ by $1/\sqrt{|a|}$, and reorder positive, negative and zero coefficients. This yieldsNo nondegeneracy assumption was used.

For uniqueness, let $V_{+}$ be the positive coordinate subspace and $V_{\le 0}$ the span of all negative and zero coordinate vectors. Their dimensions are $p$ and $n-p$. Any subspace $W$ on which $Q$ has positive definiteness intersects $V_{\le 0}$ only in zero. The dimension formula for a sum of subspaces therefore gives $\dim W\le p$. Conversely $V_{+}$ itself has dimension $p$ and positive restriction. Thus $p$ is the largest possible dimension of a positive subspace, an intrinsic description independent of the basis. Applying the same argument to $-Q$ shows that $q$ is the largest dimension of a negative subspace. This proves Sylvester's law of inertia, including uniqueness of the zero count $n-p-q$.

The extremal subspaces themselves need not be unique. For $Q(x,y)=x^2-y^2$, the coordinate axes give one positive and one negative line. For any $t\ne 0$, putThen $Q(u)=1$, $Q(v)=-1$, $B(u,v)=0$, and the determinant of the change of basis is one. The distinct lines $\mathbb{R}u$ and $\mathbb{R}v$ give another positive/negative decomposition, with the same intrinsic counts $p=q=1$.

The polarization identity for quadratic covariation, together with $\mathbb{E}[M\pm N{]}_s=\mathbb{E}(M_s\pm N_s{)}^2$, yieldsThe bracket identity for each zero-starting square-integrable martingale is the Itô isometry for integration of the constant integrand against that martingale. Polarizing both identities gives the displayed covariance identity for continuous square-integrable martingales. The bracket is integrable: its total variation is at most $\sqrt{[M{]}_s[N{]}_s}$ by the Kunita-Watanabe inequality, whose expectation is finite by the Cauchy-Schwarz inequality. Therefore condition (iii) is equivalent to (ii).

Let $T$ be an isometry, and set $S(x)=T(x)-T(0)$. Then $S(0)=0$, and preservation of distances gives $|S(x)|=|x|$. The polarization identity yieldsFor a standard orthonormal basis $e_1,e_2,e_3$, put $v_i=S(e_i)$. These three vectors are an orthonormal basis. Since $⟨S(x),v_i⟩=x_i$, we have $S(x)={\sum }_i{x}_i{v}_i=Qx$ for an orthogonal matrix $Q$. Thusan affine map.

For a finite isometry group $G$, take $c=|G{|}^{-1}{\sum }_{g\in G}g(0)$. An affine map preserves an affine average, so for any $h\in G$,The orbit barycentre is a common fixed point. This proves the fixed point of a finite Euclidean isometry group result.

Use the given Pestov identity for a geodesic flow for the function $\phi$ constructed above. Since $VX\phi =-2H\phi$, its left side is $-4\Vert H\phi {\Vert }_{L^2(\mu )}^2$. Moving this to the right givesAll terms are nonnegative, and $-K$ is uniformly positive by compactness. The smooth derivatives therefore vanish pointwise:Since $X,H,V$ span the tangent bundle, $\phi =c$ is constant, component by component if necessary.

On each circle fibre, $V^{2}u+u=c$ is the elementary periodic differential equation $u_{\vartheta \vartheta }+u=c$. Its solutions areThe first harmonics transform as a covector when the local oriented orthonormal frame changes. Their coefficients are smooth, either directly from the smooth fibre equation or from their sine/cosine integral formulas. Thus they define a smooth one-form $\eta$ on $M$ with $u(x,v)=c+{\eta }_x(v)$.

Along a geodesic, the velocity is parallel, so differentiating givesSet $Z={\eta }^{♯}/2$. ThenEquality initially holds on unit vectors, then on every vector by homogeneity. The polarization identity yields the tensor identityThis proves that $\beta$ is a potential symmetric 2-tensor, with the precise factor of two in the requested convention.

First the common boundary distance function determines the induced tangential Riemannian metric. For a boundary curve $c(t)$ with $c(0)=x$ and $c^{\prime }(0)=w\in T_x\partial M$, the local first-order distance expansion givesFor clarity, the upper bound follows from the length of the boundary curve. In a small coordinate neighbourhood the Riemannian metric is arbitrarily close to its value at $x$, so the length of any sufficiently short joining path is bounded below by that constant-metric displacement norm with an arbitrarily small relative error. A minimizing short path cannot leave that neighbourhood. These two bounds prove the limit. Equality of the boundary distance functions then gives $|w{|}_{g_1}=|w{|}_{g_2}$. The polarization identity yieldsThis alone fixes no choice of normal coordinates, so a boundary-fixing gauge is needed to compare the full tangent-space Riemannian metrics.

Let $n_i$ be the inward unit normals and letbe sufficiently small boundary normal coordinates. Their pullbacks have the form $C_i^{\ast }{g}_i=d{r}^2+h_i(r)$, with $h_i(0)=h$. On a collar define $\psi =C_2\circ C_1^{-1}$. It fixes the boundary, has $D\psi (w)=w$ for tangential $w$, and sends $n_1$ to $n_2$. For tangent-space vectors $w+a{n}_1$ and $z+b{n}_1$ at the boundary, therefore,This proves the desired full Riemannian metric equality locally at the boundary.

We also justify extending this local identification to a global diffeomorphism. Interpolate the positive Riemannian metrics by $g_s=(1-s)g_1+s{g}_2$, for $0\le s\le 1$. Let $C_s$ be their inward unit-normal exponential collars. Compactness of the boundary and parameter interval gives a common small collar width for all $s$. The intermediate Riemannian metrics need not be simple; only these local collars are used.

On the image of $C_s$, define the time-dependent vector fieldwhere $\chi$ is one near zero and vanishes before the outer edge of the collar; extend $Y_s$ by zero elsewhere. This is a smooth vector field on $M$, vanishing at the boundary because $C_s(x,0)=x$. Its time-dependent flow ${\Psi }_s$ is a global diffeomorphism fixing the boundary. For every sufficiently small fixed $r$, the curve $s\mapsto C_s(x,r)$ solves its equation, since $\chi (r)=1$. Uniqueness gives ${\Psi }_s(C_0(x,r))=C_s(x,r)$. Consequently ${\Psi }_1$ agrees near the boundary with $C_2{C}_1^{-1}$. Taking $\psi ={\Psi }_1$ proves

We use the following existence and uniqueness statement for parallel transport: for a smooth affine connection, a smooth curve $\gamma :[0,1]\to M$, an initial time $t_0$ and a vector $v\in T_{\gamma (t_0)}M$, there is a unique smooth vector field along a map $V$ on the whole curve satisfying $D_{t}V=0$ and $V(t_0)=v$. In a frame of a vector bundle this is the linear ordinary differential equation $\dot{V}=-\Gamma (\dot{\gamma })V$. Its smooth coefficient matrix is bounded on every compact time subinterval in a trivializing neighborhood, and the linear existence theorem gives a solution throughout that subinterval. A finite subdivision of $[0,1]$ into such neighborhoods and uniqueness patch the solutions. No completeness hypothesis on $M$ is needed.

For arbitrary fields $U,V$ along a curve, the definition of the induced covariant derivative of the metric tensor givesThis can also be checked by differentiating $g_{ij}(\gamma (t))U^i(t)V^j(t)$ and substituting the formula in part (a). If $\nabla$ is a metric connection and $V$ is parallel, it follows that $d(g(V,V))/dt=0$. Positivity of the Riemannian metric makes $|V|=\sqrt{g(V,V)}$ constant. Thus parallel transport preserves lengths.

Conversely assume every parallel field along every smooth curve has constant length. Fix $p\in M$ and arbitrary $x,v\in T_{p}M$. Choose a curve through $p$ with velocity $x$ at an interior time $t_0$. Explicitly, in a coordinate chart $z$ centered at $p$, take $\gamma (t)=z^{-1}(h(t)d{z}_p(x))$, where $h(t_0)=0$, $h^{\prime }(t_0)=1$, and $h$ has sufficiently small amplitude and support near $t_0$. A smooth bump function produces such $h$, so the curve is defined on all of $[0,1]$ and remains inside the chart. By the stated parallel transport theorem there is a parallel field with $V(t_0)=v$. Evaluating the derivative of its squared length givesFor fixed $x$, the bilinear form $B_x(v,w)=({\nabla }_{x}g{)}_p(v,w)$ is symmetric. Its diagonal values vanish, so polarization identity givesSince $p,x,v,w$ were arbitrary, $\nabla g=0$. A connection is metric-compatible if and only if all parallel fields have constant length. Polarization also shows that its parallel transport preserves all inner products, not just lengths.