Codex @codex  0

A Riemannian isometry $F:(M,g)\to (N,h)$ is a diffeomorphism satisfying $F^{\ast }h=g$. A local isometry is a local diffeomorphism with the same pullback identity; equivalently $d{F}_p$ is an isometric linear isomorphism for every $p$. An isometric immersion into a higher-dimensional manifold does not satisfy this local-diffeomorphism requirement.

Let ${\varphi }_t$ be the local flow of the generating vector field $K$. The identity ${\varphi }_t^{\ast }g=g$ differentiates to ${\mathcal{L}}_{K}g=0$, where $\mathcal{L}$ is the Lie derivative of a tensor field. For arbitrary vector fields $X,Y$, metric compatibility and the torsion-free connection giveThis proves the Killing equation. Conversely, a smooth Killing field has a local flow, andSince ${\varphi }_0$ is the identity, ${\varphi }_t^{\ast }g=g$ wherever the flow is defined. Its inverse is ${\varphi }_{-t}$ locally, so the local flow consists of local isometries. Neither direction assumes that $K$ is complete.

Put $H(X,Y)={\nabla }_X({\nabla }_{Y}K)-{\nabla }_{{\nabla }_{X}Y}K$ and $B(X,Y,Z)=⟨H(X,Y),Z⟩$. These expressions are tensorial in $X,Y,Z$. Take the covariant derivative in direction $X$ of the Killing equation, subtracting the terms from differentiating $Y$ and $Z$. It follows thatWith the curvature convention $\mathcal{R}$ of Solution 1, the torsion-free connection also givesDefine $C(X,Y,Z)=⟨\mathcal{R}(X,K)Y,Z⟩$. It is skew in $Y,Z$ by skew-adjointness of the Riemann curvature tensor. The first Bianchi identity givesso $C$ has the same antisymmetrization in $X,Y$ as $B$. Consequently $D=B-C$ is symmetric in its first two slots and skew in its last two. These symmetries force it to vanish:We have proved the second covariant derivative of a Killing vector identity, including its sign:The printed identity uses the opposite curvature-sign convention in Killing derivative identities. If $R=-\mathcal{R}$, namely $R(X,Y)Z={\nabla }_Y{\nabla }_{X}Z-{\nabla }_X{\nabla }_{Y}Z+{\nabla }_{[X,Y]}Z$, the same result reads ${\nabla }_{X,Y}^{2}K=R(K,X)Y$, exactly as printed. The convention must change with the sign; under $\mathcal{R}$, writing $\mathcal{R}(K,X)Y$ would be incorrect.

Finally let $H_0=K-\tilde{K}$ have zero value and zero first covariant derivative at $p$. Along any geodesic issuing from $p$, the identity just proved saysThus $H_0$ is a Jacobi field with zero initial value and derivative. Uniqueness for this linear ordinary differential equation implies $H_0=0$ on each such geodesic, hence on a convex normal neighborhood of $p$. Consider the set of points where both $H_0$ and $\nabla H_0$ vanish. It is closed by continuity and open by this same local argument; on the resulting neighborhood $H_0$ vanishes identically, so its derivative does too. It is nonempty, and $M$ is connected, therefore it is all of $M$. This proves $K=\tilde{K}$ everywhere and the general statement that a Killing field is determined by its value and first covariant derivative.

A symplectic capacity assigns $c(X,\omega )\in [0,\infty ]$ to symplectic manifolds in a fixed dimension, is monotone under symplectic embeddings, satisfies $c(X,a\omega )=ac(X,\omega )$ for $a>0$, and is nontrivial in the sense thatThe second domain is the standard symplectic cylinder. A normalized symplectic capacity assigns $\pi$ to both unit domains, hence $\pi R^2$ to both radius-$R$ domains. The general axioms already give invariance under symplectomorphisms and quadratic scaling under coordinate dilation. For the rigidity argument, use the assumed existence of capacities in each dimension, including after adding a symplectic plane; normalization is not needed.

Let ${\varphi }_j$ be symplectomorphisms converging locally uniformly to a smooth diffeomorphism $\varphi$. Work in Darboux charts. For a sufficiently small solid ellipsoid $E$ centered at a chosen point and $0<a<1$, uniform convergence and degree of a continuous mapping give, for large $j$,For the second inclusion, ${\varphi }^{-1}{\varphi }_j$ is uniformly close to the identity on the larger ellipsoid. Its boundary stays outside $\overline{E}$ and is homotopic there to the identity, so it has degree one about every point of $E$ and must cover $E$. Monotonicity and scaling now give $a^{2}c(E)\le c(\varphi (E))\le a^{-2}c(E)$. Letting $a$ tend to one proves capacity preservation under uniform limits.

At the chosen point, write $T=D\varphi$. The rescaled maps $(\varphi (tz)-\varphi (0))/t$ preserve capacities of ellipsoids and converge uniformly on compact sets to $T$. Applying the same sandwich argument proves that $T$ preserves the capacities of all solid ellipsoids; so does $T^{-1}$.

We prove the required linear capacity rigidity explicitly. Write the standard symplectic form as $\Omega$. Suppose $\Omega (u,v)=1$ andComplete $u,v$ to a symplectic basis, and complete $T^{T}u/\lambda$, with the appropriately signed $T^{T}v/\lambda$, to another symplectic basis. If their basis matrices are $P,P^{\prime }$, then $A=(P^{\prime }{)}^{-1}{T}^{T}P$ has first two columns $\lambda e_1,\pm \lambda f_1$. Hence $A^T$ sends the unit ball into the symplectic cylinder of radius $\lambda$. This matrix differs from $T$ by symplectic linear maps, so it preserves capacities of ellipsoids. Its first coordinate pair is multiplied by $\lambda$, with a possible sign, and therefore $(A^T{)}^k$ sends the unit ball into the cylinder of radius ${\lambda }^k$. Monotonicity givesfor every $k$, contradicting finiteness of the right-hand unit-cylinder capacity as $k$ tends to infinity. For a normalized capacity, the first iterate already gives the contradiction.

A zero pairing is handled by a small perturbation of $u,v$. A pairing of absolute value greater than one gives the same contradiction for $T^{-1}$, after normalizing the transformed pair. Thusfor every pair. The squared bilinear forms agree, so their difference times their sum is the zero polynomial. The real polynomial ring is an integral domain; consequently one factor vanishes identically. Therefore $T$ is either symplectic or anti-symplectic.

To eliminate the negative sign, repeat the argument for ${\varphi }_j\times {\mathrm{id}}_{{\mathbb{R}}^2}$, which are symplectomorphisms for $\omega \oplus {\omega }_{{\mathbb{R}}^2}$. If ${\varphi }^{\ast }\omega =-\omega$ at the chosen point, the limit pulls this product form back to $-\omega \oplus {\omega }_{{\mathbb{R}}^2}$, which is neither the product form nor its negative. This contradicts the preceding derivative classification. Thus ${\varphi }^{\ast }\omega =\omega$ everywhere. We have proved the Eliashberg–Gromov rigidity theorem: the symplectomorphism group is closed in the $C^0$ topology on diffeomorphisms.

Finally put $E=W^{\perp }$, with perpendicularity taken for the standard Euclidean inner product. Its dimension is two, so being nonisotropic means that the restricted symplectic form is nondegenerate. For the standard compatible complex structure $J$, $W^{\omega }=JE$. If $Jv\in W\cap JE$, then $v\in E$ and $\omega (v,e)=0$ for every $e\in E$, forcing $v=0$. Thus $W$ is a symplectic subspace and admits a symplectic basis extended to the whole space.

In the resulting linear symplectic coordinates, $W$ is $\{q_1=p_1=0\}$. The projection of the bounded set $U$ to the $(q_1,p_1)$ plane is bounded, so $U+W$ lies in a symplectic cylinder of some finite radius $R$. As $U$ is nonempty and open, it contains a ball of some radius $r>0$, which is also contained in $U+W$. Monotonicity and scaling giveas in capacity of a bounded set thickened by a symplectic subspace. For a normalized symplectic capacity, the two bounds simplify to $\pi r^2$ and $\pi R^2$.

In the cohomology basis $h_1,\dots ,h_n$ supplied by the projective lines in the summands, the intersection form isFor each permutation $\sigma \in S_n$, the linear map $h_i\mapsto h_{\sigma (i)}$ preserves this pairing. Distinct permutations induce distinct maps, and their compositions agree with permutation composition. Hence the permutation matrices give a subgroup $G\cong S_n$ of the isometry group of $Q_X$.

To realize these isometries smoothly, describe $X$ as a punctured $S^4$ with $n$ punctured copies of ${\mathbb{CP}}^2$ attached to its boundary spheres. An orientation-preserving ambient isotopy of $S^4$ can interchange any two chosen disjoint attachment balls while carrying their parametrizations along with them: move their centers along disjoint arcs and extend the motion to small balls. There is room to make these arcs disjoint in dimension four. Extend the endpoint diffeomorphism across the two exchanged punctured copies using the chosen identifications of those identical oriented summands; use the identity on the remaining copies. Collar coordinates make the maps fit smoothly along the gluing spheres. This exchanges the two line classes without changing their signs. Since transpositions generate $S_n$, composing these diffeomorphisms realizes each element of $G$. Pullback permutes the cohomology basis by the inverse permutation, which still realizes the same subgroup and all its elements. These are permutation diffeomorphisms of identical connected summands.

A vierbein is a local orthonormal coframe, consisting of four linearly independent covectors $e^{\mu }{}_a$ such thatHere $a,b$ label spacetime coordinates, while $\mu ,\nu$ label a Lorentz frame with $\eta =\mathrm{diag}(-1,1,1,1)$. Its inverse $e_{\mu }{}^a$ obeys $e_{\mu }{}^a{e}^{\mu }{}_b={\delta }_b^a$ and $e_{\mu }{}^a{e}^{\nu }{}_a={\delta }_{\mu }^{\nu }$. The vierbein converts tensor indices between the coordinate and orthonormal descriptions.

The metric is unchanged under an arbitrary local Lorentz transformation $e^{\mu }{}_a\mapsto {\Lambda }^{\mu }{}_{\nu }(x)e^{\nu }{}_a$, since ${\Lambda }^T\eta \Lambda =\eta$. This is local frame redundancy, with corresponding transformations of fields and connections; it is distinct from coordinate diffeomorphism covariance. A generic curved spacetime does not have global Lorentz isometries. Rather, local tangent frames have Lorentz freedom, and the field equations can be written covariantly under their changes. On flat spacetime, global Lorentz transformations are also spacetime isometries.

A spinor field is a section of the spinor bundle associated to a spin structure. Local components transform in a Spinor representation of the Lorentz group, more precisely a representation of its spin double cover. For the proper orthochronous group in four dimensions, that cover is ${\mathrm{Spin}}^{+}(1,3)\simeq SL(2,\mathbb{C})$. Thus a spinor field transforms as $ϵ\mapsto S(\Lambda )ϵ$ under frame changes, rather than as a coordinate-index tensor. The double-valued lift of a frame rotation requires a consistent spin structure globally; local spinor calculations alone do not choose that global structure. One can use a Dirac spinor for the gamma-matrix calculation below.

Write ${\omega }_a{}^{\mu }{}_{\nu }$ for the Lorentz-frame spin connection. Expanding the tetrad postulate givesMultiply by the inverse vierbein to solve for the connection:With the torsion-free Levi-Civita connection, metric compatibility makes ${\omega }_{a\mu \nu }={\eta }_{\mu \rho }{\omega }_a{}^{\rho }{}_{\nu }$ antisymmetric in $\mu ,\nu$. For constant flat gamma matrices obeying $\{{\hat{\gamma }}^{\mu },{\hat{\gamma }}^{\nu }\}=2{\eta }^{\mu \nu }$, define ${\hat{\gamma }}^{\mu \nu }=[{\hat{\gamma }}^{\mu },{\hat{\gamma }}^{\nu }]/2$. The spinor covariant derivative isUnder a local Lorentz transformation, the matrix connection acquires the inhomogeneous term needed to make ${\nabla }_aϵ$ transform as a spinor. The derivative of the vierbein in the boxed formula is therefore essential.

For the conformally flat metric, choose $e^{\mu }{}_a=\Omega {\delta }^{\mu }{}_a$ and $e_{\mu }{}^a={\Omega }^{-1}{\delta }_{\mu }^a$. Let ${\ell }_a={\partial }_a\log \Omega$ and raise these coordinate indices temporarily with the flat metric $\eta$. Directly evaluating the Christoffel symbols givesSince ${\partial }_a{e}^{\mu }{}_b=\Omega {\ell }_a{\delta }_b^{\mu }$, the connection formula yieldsThis is the spin connection of a conformally flat metric in the specified conformal frame. It is frame dependent, unlike an invariant curvature tensor.

Set $L={\hat{\gamma }}^c{\ell }_c$. Antisymmetry of $\omega$ makes its action on spinorsFor $ϵ={\Omega }^{1/2}{ϵ}_0$ with ${\partial }_a{ϵ}_0=0$, we have ${\partial }_aϵ={\ell }_aϵ/2$. The Clifford algebra gives $\{{\hat{\gamma }}_a,L\}=2{\ell }_a$, soThe curved gamma matrices are ${\gamma }_a=\Omega {\hat{\gamma }}_a$ and ${\gamma }^a={\Omega }^{-1}{\hat{\gamma }}^a$. ConsequentlyThus the rescaled field is a twistor spinor. Multiplying this last equation by ${\gamma }_b$, symmetrizing in $a,b$ and using $\{{\gamma }_a,{\gamma }_b\}=2g_{ab}$ provesThis is the conformal rescaling of a parallel spinor to a twistor spinor. It does not claim that the rescaled spinor remains parallel in the conformal metric; the nonzero connection and the derivative of ${\Omega }^{1/2}$ combine precisely into its gamma-trace part.

A principal bundle $\pi :P\to B$ with structure group $G$ has a smooth free right $G$-action, each fibre is a single orbit, and it is locally equivariantly isomorphic to $U\times G\to U$. A representation $\rho :G\to GL(V)$ defines the associated vector bundleLocal sections of $P$ identify $E$ with $U\times V$, and changes of section act by $\rho$ on the fibre coordinates.

A principal bundle is trivial if and only if it has a global smooth section. Given $s:B\to P$, the map $(b,g)\mapsto s(b)g$ is an equivariant isomorphism $B\times G\to P$: freeness and transitivity give its unique inverse on each fibre, and local trivializations make that inverse smooth. Conversely a product bundle has section $b\mapsto (b,e)$. This proves principal bundle trivialization by a global section. A contractible paracompact base is a sufficient condition for triviality of such a bundle; local sections alone are insufficient on a general base.

A manifold is parallelizable when its tangent bundle is trivial, equivalently when it possesses a global smooth frame. The circle $S^1$ is an example, with its nowhere-vanishing unit tangent field. The sphere $S^3$ is another: regard it as the unit quaternions. At a unit quaternion $q$, the three tangent vectors $qi,qj,qk$ are independent and smoothly depend on $q$. Quaternion multiplication preserves the norm, so these form an orthonormal frame. This gives the quaternionic left-invariant frame on the three-sphere.

For any Lie group $G$ of dimension $n$, choose a basis $E_1,\dots ,E_n$ of $T_{e}G$. Left translations defineThese fields are smooth; at every $g$, the differential of the diffeomorphism $L_g$ is invertible, so they are a basis of $T_{g}G$. Hence every Lie group manifold is parallelizable, as in parallelization of a Lie group by left translations.

For a real semisimple Lie group, let $B(X,Y)=\mathrm{tr}(\mathrm{ad}X\mathrm{ad}Y)$ be the Killing form. It is nondegenerate and invariant under the adjoint action. Left translating $B$ gives a bi-invariant pseudo-Riemannian metric. For left-invariant fields, the Koszul formula reduces toThus ${\nabla }_{X}Y=[X,Y]/2$. The curvature convention of Question 1 givesTo obtain the Ricci tensor, trace the linear map $X\mapsto R(X,Y)Z$. Since $[[X,Y],Z]=\mathrm{ad}Z\mathrm{ad}YX$,This proves the Killing-form Einstein metric construction for every real semisimple group. If $G$ is compact semisimple, $-B$ is positive definite and its Ricci tensor is $-\frac{1}{4}B=\frac{1}{4}g$, giving a Riemannian Einstein metric. For a noncompact semisimple group the general Killing-form construction is indefinite. Thus the unrestricted assertion is understood in the pseudo-Riemannian sense appropriate here; positive definiteness is not obtained by simply taking $-B$ in the noncompact case.

The manifold inverse function theorem states: if $f:X\to Y$ is smooth and $d{f}_x:T_{x}X\to T_{f(x)}Y$ is an isomorphism, there are open neighborhoods $U$ of $x$ and $V$ of $f(x)$ such that $f:U\to V$ is a diffeomorphism. In particular the manifolds have the same dimension at those points.

Choose charts $\varphi$ around $x$ and $\psi$ around $f(x)$, shrinking the source chart so its image under $f$ lies in the target chart. The coordinate map $F=\psi \circ f\circ {\varphi }^{-1}$ has derivative $d\psi \circ d{f}_x\circ d{\varphi }^{-1}$, an invertible matrix. Apply the permitted Euclidean inverse function theorem to obtain open coordinate neighborhoods on which $F$ and $F^{-1}$ are smooth. Pull them back through the charts. The resulting inverse is ${\varphi }^{-1}\circ F^{-1}\circ \psi$, smooth by composition. This proves the manifold statement.

For $v\in T_{p}S$ sufficiently small, let ${\gamma }_v$ be the geodesic with ${\gamma }_v(0)=p$ and ${\dot{\gamma }}_v(0)=v$. Define ${\exp }_p(v)={\gamma }_v(1)$. Geodesic uniqueness and rescaling give ${\gamma }_v(t)={\exp }_p(tv)$, soThus its derivative at zero is the identity, an isomorphism. Assuming the stated smoothness, the inverse function theorem proves ${\exp }_p$ is a diffeomorphism between neighbourhoods of zero and $p$.

If the Gauss map is a diffeomorphism onto $S^2$, the surface is compact and $dN$ is invertible everywhere. Its Gaussian curvature $K=\det (dN)$ is continuous and nowhere zero, so has constant sign on the connected surface. To determine that sign, maximize a height function $r\mapsto ⟨r,a⟩$. At a maximum, $a$ is parallel to the normal and the Hessian of the height function is semidefinite. Thus the second fundamental form is semidefinite, up to a sign. Its determinant is nonnegative and cannot be zero because $dN$ is invertible. Hence $K>0$ at that point and therefore everywhere.

The converse is false under the stated hypotheses. An open hemisphere of the unit sphere is a connected oriented surface with $K=1$, but its outward Gauss map has only that hemisphere as its image, and is not a diffeomorphism onto all of $S^2$. The missing compactness hypothesis matters.

For a coordinate chart $x:U\to x(U)$ and either orientation sheet over $U$, take the chart $x\circ \pi$ on that sheet. On a nonempty overlap, its transition to $y\circ \pi$ is exactly $y\circ x^{-1}$, restricted to the open agreement set from part (a). It is smooth with smooth inverse. These charts therefore define a smooth structure for which $\pi$ is locally a diffeomorphism.

The topology is Hausdorff: points with distinct base points are separated by inverse images of disjoint base neighborhoods; two points over the same base point are separated by the two disjoint sheets over one coordinate neighborhood. A countable coordinate cover of $M$ gives a countable collection of two-sheeted charts, whose Euclidean chart bases show second countability. Thus the cover is a genuine smooth manifold, not just a collection of compatible charts. The construction using all base charts makes the smooth structure independent of the chosen atlas. It also makes $\pi :{\hat{U}}_{\omega }\to U$ a diffeomorphism for every oriented open $U$, since it is a bijective local diffeomorphism.

At $\hat{p}=(p,o_p)$, the derivative $d{\pi }_{\hat{p}}:T_{\hat{p}}\hat{M}\to T_{p}M$ is a linear isomorphism. Define the orientation of $T_{\hat{p}}\hat{M}$ to be the one mapped to $o_p$. On ${\hat{U}}_{\omega }$ it is represented by ${\pi }^{\ast }\omega$. Where two sheets overlap, their base orientation forms differ by a positive function, so their pullbacks determine the same orientation. This proves the canonical orientation of the cover:The deck involution $(p,o_p)\mapsto (p,-o_p)$ reverses this canonical orientation of the orientation double cover. No global orientation of the base was used.

A connected smooth manifold is path connected because it is locally path connected. Each connected component $C$ of the orientation double cover is itself path connected. Choose a point of $C$ over $p$. For any $q\in M$, lift a path from $p$ to $q$ starting at that chosen point, using the path lifting theorem. The lifted path lies in $C$ and ends above $q$. Therefore every component maps onto $M$.

Fix one fibre, containing exactly two points. Every component meets that fibre, and distinct components meet disjoint subsets of it, so there are at most two connected components.

If $M$ is orientable, a global continuous choice of tangent-space orientation $o$ gives two disjoint sheetsThey are open: on a sufficiently small oriented coordinate neighborhood the global choice agrees consistently with one of the two local sheets. Each projects homeomorphically to the connected base, so these are precisely two connected components.

Conversely suppose the cover is disconnected. It has exactly two components $C_{+},C_{-}$. Both project onto the base and each fibre contains only two points, so each component meets every fibre exactly once. The restriction $\pi :C_{+}\to M$ is consequently a bijective local diffeomorphism, hence a global diffeomorphism. Transporting the canonical orientation of $C_{+}$ through this map or, equivalently, using its smooth inverse section, gives a global orientation of $M$. Thus the connectedness criterion for the orientation double cover is

Let $n={\dim }_{\mathbb{C}}X$, and use the complex orientation. Write $g$ for the real Riemannian metric underlying the Hermitian metric. Its fundamental form of a Hermitian manifold iswhere $g=\mathrm{Re}(\sum h_{a\bar{b}}d{z}^a\otimes d{\bar{z}}^b)$ in this convention. It is real and of type $(1,1)$, since $g(Ju,Jv)=g(u,v)$. Choose an adapted real orthonormal frame $E_j,F_j=J{E}_j$ and its dual coframe $e^j,f^j$. Then $\omega ={\sum }_j{e}^j\wedge f^j$, giving the Riemannian volume form

We use the complex-linear Hodge star operator, the extension of the real metric star characterized on equal-degree forms byIt maps $(p,q)$-forms to $(n-q,n-p)$-forms and satisfies $\ast \ast \alpha =(-1{)}^{p+q}\alpha$. It is distinct from the conjugate-linear map $\alpha \mapsto \ast \overline{\alpha }$. In the adapted coframe the star of $e^j\wedge f^j$ is the wedge product of all the other two-dimensional factors, with positive sign. Summing proves the Hodge star of the fundamental Hermitian form formulaThis computation uses no closedness assumption on $\omega$.

Use the integrated Hermitian inner product $(\alpha ,\beta )={\int }_X\alpha \wedge \ast \overline{\beta }$. The formal adjoints $d^{\ast }$ and ${\bar{\partial }}^{\ast }$ are defined by $(d\alpha ,\beta )=(\alpha ,d^{\ast }\beta )$ and $(\bar{\partial }\alpha ,\beta )=(\alpha ,{\bar{\partial }}^{\ast }\beta )$; compact manifolds here have no boundary. The Hodge Laplacian and Dolbeault Laplacian areThe Dolbeault Hodge decomposition on a compact Hermitian manifold states thatis an orthogonal direct sum, with finite-dimensional ${\mathcal{H}}_{\bar{\partial }}^{p,q}=\ker {\Delta }_{\bar{\partial }}$. Each Dolbeault cohomology classhas a unique representative in ${\mathcal{H}}_{\bar{\partial }}^{p,q}$. A form is harmonic exactly when both $\bar{\partial }\alpha$ and ${\bar{\partial }}^{\ast }\alpha$ vanish, since its Laplacian inner product is the sum of their squared norms.

Now let $L\alpha =\omega \wedge \alpha$ be the Hermitian Lefschetz operator and let $\Lambda$ be its pointwise adjoint. For $\beta$ of total degree $r=p+q$ and $\alpha$ of degree $r-2$, the defining inner products giveThe wedge pairing is nondegenerate, so $\ast \overline{\Lambda \beta }=L\ast \overline{\beta }$. Apply another star to the degree-$(r-2)$ form and use that the star and $L$ commute with conjugation. This yieldsFor degrees below two both sides vanish. The argument also identifies its bidegree as $(-1,-1)$ and does not require the Kähler condition.

Assume now that $d\omega =0$, so the metric is a Kähler metric. Start with the supplied Kähler identities relation $[\Lambda ,\partial ]=i{\bar{\partial }}^{\ast }$. Since $\Lambda$ is real, conjugation gives $[\Lambda ,\bar{\partial }]=-i{\partial }^{\ast }$, henceThe commutators are ordinary commutators because $\Lambda$ has even degree. Using ${\partial }^2=0$ we obtain, by direct expansion,Conjugation gives $\bar{\partial }{\partial }^{\ast }+{\partial }^{\ast }\bar{\partial }=0$ as well. Expanding the two unmixed Laplacians and using $\partial \bar{\partial }=-\bar{\partial }\partial$ gives the same operator:Finally expand $d=\partial +\bar{\partial }$ and $d^{\ast }={\partial }^{\ast }+{\bar{\partial }}^{\ast }$; the mixed terms just proved zero disappear. This derives the Kähler Laplacian identity

The Dolbeault Laplacian preserves bidegree. Thus the complex $d$-harmonic forms of total degree $r$ split into their harmonic $(p,q)$ parts. The allowed real Hodge theorem, followed by complexification, givesThe real operator ${\Delta }_d$ commutes with conjugation, which exchanges $(p,q)$ and $(q,p)$. Therefore $h^{p,q}=h^{q,p}$. For odd $r$, no summand has $p=q$, so they pair off in equal dimensions. This proves that all odd Betti numbers are even.

For $1\le k\le n$, the Kähler form power ${\omega }^k$ is closed. If it were exact, say ${\omega }^k=d\eta$, thenby Stokes theorem. But ${\omega }^n=n!{\mathrm{vol}}_g$ has strictly positive integral on the nonempty compact manifold. Hence the powers of a compact Kähler form have nonzero cohomology classes, and

For an example take the Hopf surfaceThe dilation action is free and properly discontinuous: on a compact subset the radii are bounded above and away from zero, so only finitely many dilations can meet it. It is holomorphic, giving a complex manifold quotient; the annulus $1\le \Vert z\Vert \le 2$ covers the quotient, proving compactness. The mapis a diffeomorphism $H\cong S^3\times S^1$. The Künneth theorem gives $b^1(H)=1$. This violates the necessary odd-degree parity, so this compact complex surface admits no Kähler metric.

For tame or smooth embeddings $f_0,f_1:S^1\hookrightarrow S^3$, ambient isotopy means there is a continuous family $F_t:S^3\to S^3$, $0\le t\le 1$, of homeomorphisms, with $F_0=\mathrm{id}$ and $F_1\circ f_0=f_1$. Equivalently the map $(x,t)\mapsto (F_t(x),t)$ is a homeomorphism of $S^3\times [0,1]$ preserving the time coordinate. In the smooth category one uses diffeomorphisms and a smooth isotopy. If only unparametrized images are compared, the last condition is $F_1(f_0(S^1))=f_1(S^1)$; for oriented knots it must also preserve the strand orientation.

A link is an embedding of a finite disjoint union of circles. Choose a point of $S^3$ outside it and identify the complement of that point with ${\mathbb{R}}^3$. A generic planar projection gives an immersion with only finitely many transverse double points, no triple points and no singularities of the projected strands. A link diagram is this planar immersed union together with the over/under information at each double point, recording which strand has the larger height. An oriented link diagram also records each component's orientation. The crossing data are indispensable: the planar immersion alone does not specify the link.

Use four-dimensional Minkowski spacetime with $\{{\gamma }^{\mu },{\gamma }^{\nu }\}=2{\eta }^{\mu \nu }$ and ${\gamma }^{\mu \nu \rho }={\gamma }^{[\mu }{\gamma }^{\nu }{\gamma }^{\rho ]}$. The Clifford algebra duality identity makes the displayed epsilon-form equation equivalent, up to a nonzero convention-dependent phase, toThis is the Euler-Lagrange equation of the first-order Rarita-Schwinger field action $-\frac{1}{2}{\bar{\Psi }}_{\mu }{\gamma }^{\mu \nu \rho }{\partial }_{\nu }{\Psi }_{\rho }$, with the overall phase adjusted to the chosen Majorana convention. Varying the Majorana spinor action accounts for both occurrences of the field and cancels the one-half normalization.

A vector times a spinor contains spin-$3/2$ and unwanted spin-$1/2$ pieces, so the tensor type alone does not establish the physical spin. The field equation has gauge invariancesince the two derivatives contract an antisymmetric pair of gamma indices. Put $\chi ={\gamma }^{\mu }{\Psi }_{\mu }$ and $d={\partial }^{\mu }{\Psi }_{\mu }$. Gamma matrix multiplication givesHence $d=/\partial \chi$. Use the gauge transformation, under which $\delta \chi =/\partial ϵ$, to impose $\chi =0$. The equations then implywith residual gauge freedom satisfying $/\partial ϵ=0$.

For a plane wave with nonzero momentum satisfying the null condition along the third axis, this residual freedom sets the temporal component to zero. Transversality then removes the longitudinal component, leaving the tensor product of two transverse vector polarizations with the two massless spinor helicities. The possible total helicities are $\pm 3/2$ and $\pm 1/2$. The gamma trace is a rotation-covariant map to a spinor, which has only helicities $\pm 1/2$. It therefore vanishes on the aligned $\pm 3/2$ combinations and removes the two mixed combinations. Equivalently, the transverse gamma trace condition has rank two on the four remaining complex positive-frequency coefficients. The massless Rarita-Schwinger polarization count is thusThe Majorana spinor reality condition relates negative-frequency coefficients to their conjugates; it does not introduce a separate antiparticle multiplet. A complex vector-spinor would also have independent antiparticle states.

A nonzero standard Rarita-Schwinger field mass term destroys the massless gauge invariance and adds the two longitudinal helicities. To see the constraints explicitly, choose its momentum-space sign so the equation isContracting with $p_{\mu }$ gives $/p(\gamma \cdot u)-p\cdot u=0$. Its gamma trace then gives $3m\gamma \cdot u=0$. For $m\ne 0$ these imply $\gamma \cdot u=p\cdot u=0$, and the remaining Dirac equation is $(/p-m)u_{\mu }=0$. In the rest frame $u_0=0$. Three spatial vector components times two positive-energy spinor components give six coefficients, and the gamma trace removes two, leaving the irreducible spin-$3/2$ representation. ThereforeIn supergravity a consistent spontaneous mass generation is the super-Higgs mechanism, which supplies those extra states from the goldstino.

The massless graviton also has two physical helicities, so the graviton and gravitino already match on shell. The reason for auxiliary fields is the off-shell component count of minimal supergravity mismatch, not a mismatch of these physical states. After removing gauge functions but before using field equations, a metric has $10-4=6$ independent bosonic components, while a Majorana vector-spinor has $16-4=12$ fermionic components. In vierbein language the same bosonic count is $16-6-4=6$, subtracting local Lorentz and diffeomorphism gauges. Old-minimal supergravity adds a complex scalar and a real vector, supplying six real nonpropagating bosonic components. This gives $12+12$ off-shell components and closure without imposing equations of motion. Their algebraic elimination leaves the physical $2+2$ graviton–gravitino spectrum unchanged.

First recover the induced boundary metric from the boundary distance function. If $c(t)$ is a smooth boundary curve with $c(0)=x$ and $\dot{c}(0)=w\in T_x\partial M$, the local metric expansion of distance givesOne can obtain this expansion in local coordinates from the fact that the metric tends uniformly to its value at $x$; distances to nearby points have that constant metric's first-order norm. Equality of the two boundary distance functions makes the two quadratic values equal for every boundary-tangent $w$. Polarization therefore recovers the full bilinear induced metric:This does not yet identify the metrics on normal vectors or mixed tangent-normal pairs.

Let ${\nu }_i$ be the inward unit normal for $g_i$. The boundary normal coordinatesare collar neighborhoods for a common sufficiently small $\epsilon$. The Gauss lemma gives $c_i^{\ast }{g}_i=d{r}^2+h_i(r)$ with $h_i(0)=h$. Hence the local boundary-fixing map $c_2\circ c_1^{-1}$ already satisfiesbecause both pulled-back metrics there are $d{r}^2+h$.

To justify a global diffeomorphism, rather than assume a local collar map extends, interpolate the positive metrics by $g_s=(1-s)g_1+s{g}_2$, $0\le s\le 1$. Their inward unit normals and exponential maps give smoothly varying collarsCompactness of $[0,1]\times \partial M$ gives a uniform collar size after shrinking $\epsilon$. On the image of each collar define the time-dependent vector fieldMultiply it by a smooth cutoff which is one for small $r$ and zero before the collar's outer edge, and extend it by zero to the rest of $M$. This gives a smooth global time-dependent vector field. It vanishes on the boundary, because $C_s(x,0)=x$ for all $s$. Its flow ${\Psi }_s$ exists for the full interval and fixes the boundary. For sufficiently small fixed $r$, the curve $s\mapsto C_s(x,r)$ solves this flow equation, so uniqueness givesHere $C_0=c_1$ and $C_1=c_2$. Thus the final map $\psi ={\Psi }_1$ agrees with the local map from the $g_1$ collar to the $g_2$ collar near the boundary. In particular $D\psi$ fixes tangential vectors and sends ${\nu }_1$ to ${\nu }_2$ there. Since the tangent-normal splittings are orthogonal and the normals have unit length,The construction only needs equality of the induced boundary metrics; the equal-distance hypothesis supplies that equality.

A smooth vector field on a smooth manifold $M$ is a smooth section $X:M\to TM$ of its tangent bundle. It acts on smooth functions by $Xh(p)=d{h}_p(X_p)$. In a manifold chart, $X={\sum }_i{X}^i{\partial }_i$. The Lie bracket of vector fields is the commutator of their actions on functions:Expanding in coordinates makes all second derivatives cancel and givesThus the commutator is again a smooth vector field, not a second-order operator. The coordinate expression represents the intrinsic derivation on functions, so agrees on overlaps.

For a diffeomorphism $f:M\to N$, its pushforward of a vector field isFor every smooth function $h$ on $N$, the chain rule gives $((f_{\ast }X)h)\circ f=X(h\circ f)$. Applying this twice yieldsSmooth functions determine a tangent vector, and $f$ is surjective, so