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Put $r^2=y_1^2+y_2^2$ and write the Riemannian metric as a conformal rescaling of a Riemannian metric:The base metric is flat, so its Ricci scalar is zero and its Laplacian is $\Delta ={\partial }_1^2+{\partial }_2^2$. The formula for scalar curvature under conformal rescaling simplifies in dimension two toDifferentiate explicitly:Since $e^{-2\omega }=1+r^2$, the required Ricci scalar is

The Poincare disc model isThis Riemannian metric has constant Gaussian curvature $-1$.

For $a\in D$ and $|\eta |=1$, the Möbius transformationis an isometry of the disc, and every orientation-preserving disc isometry has this form. In particular, $T_{z_1,1}$ sends $z_1$ to zero and $z_2$ to a point of modulusRotations are also isometries, so it remains only to integrate the metric along a radial geodesic from $0$ to $\rho$:Thus the Hyperbolic distance in the Poincare disc is

The Levi-Civita connection of a Riemannian manifold $(M,g)$ is the unique connection on a vector bundle on $TM$ that is torsion-free and compatible with the Riemannian metric. Any such connection must satisfy the Koszul formulaNondegeneracy of $g$ determines ${\nabla }_{X}Y$ uniquely from the right-hand side. Conversely, define $\nabla$ by this formula. Direct substitution shows that it is $C^{\infty }$-linear in $X$, satisfies the Leibniz rule in $Y$, preserves $g$, and obeys ${\nabla }_{X}Y-{\nabla }_{Y}X=[X,Y]$. It is therefore a torsion-free metric connection. This proves the Existence and uniqueness of the Levi-Civita connection.

For the metric vector bundle $E$, choose the stated orthonormal local frame $(e_1,\dots ,e_m)$ and write $d_A{e}_j=A^i{}_j{e}_i$. Since $⟨e_i,e_j⟩={\delta }_{ij}$, metric compatibility givesHence the connection matrix in an orthonormal frame is skew-symmetric:

On an oriented Riemannian $d$-manifold, the Hodge star operator is defined byIt is an orthogonal map $\ast :{\Lambda }^r{T}_x^{\ast }M\to {\Lambda }^{d-r}{T}_x^{\ast }M$ and satisfies ${\ast }^2=(-1{)}^{r(d-r)}$. In dimension $d=2n$ on middle-degree forms, the Adjoint of the Hodge star on middle-degree forms isThus it is self-adjoint when $n$ is even, but it is not always self-adjoint. For $n=1$ on the oriented Euclidean plane,so its matrix in the orthonormal basis $(dx,dy)$ is skew-adjoint.

The Laplace-Beltrami operator on differential forms is the Hodge Laplacianwhere the codifferential $\delta$ is the formal $L^2$ adjoint of the exterior derivative. On a compact manifold without boundary, if $\Delta \omega =\lambda \omega$ for a nonzero differential form $\omega$, then integration by parts givesTherefore the nonnegativity of the Hodge Laplacian yields

Expand $X^a=x^a+\theta {\psi }^a$, $D{X}^a={\psi }^a-i\theta {\dot{x}}^a$, and $g_{ab}(X)=g_{ab}(x)+\theta {\psi }^c{\partial }_c{g}_{ab}$. Performing the Berezin integral givesAfter a fermionic integration by parts, this is the manifestly covariant expressionUnder a diffeomorphism, $x$ is a point of $M$, $\psi \in T_{x}M$ is a tangent vector, $D_t\psi$ is its covariant derivative along a curve, and every index is contracted with the Riemannian metric; the action is therefore invariant.

An almost complex structure $J$ on $(M,{\omega }_M)$ is a compatible almost complex structure when $J^2=-I$, ${\omega }_M(Ju,Jv)={\omega }_M(u,v)$, andis a positive-definite inner product. To prove existence, choose any Riemannian metric $h$ and define $A$ by ${\omega }_M(u,v)=h(Au,v)$. The metric construction of a compatible almost complex structureis smooth and compatible, so the space is nonempty.

Identify the compact symplectic manifold $N$ with its image under the symplectic embedding. Along $N$ there is a symplectic splittingChoose compatible almost complex structures on both summands and take their direct sum. Its associated metric makes the two summands orthogonal. Extend this metric from the closed submanifold $N$ to all of $M$ using a partition of unity, and apply the metric construction again. Along $N$ it recovers the prescribed direct sum, so the resulting global compatible $J$ satisfies $J(TN)=TN$. This is the relative extension of a compatible almost complex structure.

For the two compact complex curves, use the supplied holomorphic coordinates at their transverse intersection. There $C_1\cup C_2$ is $\{xy=0\}$. Replace it in a small ball by the complex annulus $\{xy=ϵ\}$ and use a cutoff in a surrounding annulus to rejoin the unchanged curves. For sufficiently small nonzero $ϵ$, the result is an embedded symplectic surface $\Sigma$. This local replacement is a bordism between the old and new cycles, soThis is the symplectic smoothing of a positive transverse node.

At critical coupling, an $N$-vortex solution of the Abelian Higgs model satisfies the Bogomolny vortex equations. The Abelian Higgs vortex moduli space ${\mathcal{M}}_N$ consists of these solutions modulo gauge transformations. The unordered zeros of the Higgs field specify the $N$ vortex positions, soas a complex manifold. A tangent vector is represented by a linearized solution orthogonal to infinitesimal gauge transformations. Substitution into the field-theory kinetic energy giveswhich defines the natural $L^2$ Riemannian metric on the moduli space. Its geodesics give the slow-motion approximation.

For $N=1$, translation invariance makes ${\mathcal{M}}_1$ a flat plane. For $N=2$, the center-of-mass plane factors from a rotationally symmetric relative moduli space. Far from coincidence, exchanging the vortices identifies the relative separation $z$ with $-z$, so the relative space is asymptotic to a cone of angle $\pi$. The exact metric smoothly rounds its apparent tip at coincident vortices; the regular local coordinate is proportional to $z^2$.

A head-on relative geodesic passes smoothly through this coincidence point. In terms of the asymptotic separation coordinate, it emerges on the perpendicular axis, producing Right-angle scattering of Abelian Higgs vortices. For signed impact parameter $b$, the deflection is odd in $b$, approaches zero as $|b|\to \infty$, and approaches the two signed values $\pm \pi /2$ as $b\to 0^{\pm }$. Within the geodesic approximation the trajectory, and hence this graph, is independent of incoming speed; speed only changes its parametrization in time.

For identical point particles in a bounded repulsive central potential, the relative coordinate moves in an ordinary plane rather than the rounded exchange cone. At energy below the finite central barrier, a head-on encounter is reflected and gives backscattering through $\pi$, so the available deflections range from zero to nearly $\pi$. At energy above the barrier, a head-on trajectory passes through the center and has zero deflection; the magnitude reaches a maximum at nonzero $|b|$ and then returns to zero. Increasing speed reduces this maximum. Thus point particles can exhibit backscattering at low speed and near-transparent head-on passage at high speed, whereas slow identical vortices have the speed-independent right-angle limit forced by the geometry of ${\mathcal{M}}_2$.

A unit-speed curve on a smooth embedded surface is a geodesic exactly when its acceleration is normal to the surface, equivalently when its tangential acceleration vanishes.

Every unit-speed geodesic on the cylinder through $(1,0,0)$ has the formIndeed, unrolling the cylinder to its universal cover ${\mathbb{R}}^2$ turns these curves into straight lines. Directly,which is parallel to the cylinder's radial normal vector, verifying the geodesic characterization. Such a geodesic is closed exactly when $b=0$; its image is then the horizontal circle $z=0$.

Yes. In polar coordinates $(r,\theta )$ on ${\mathbb{R}}^2\setminus \{0\}$, use the Riemannian metricThe coordinate $u=\log r$ identifies this surface isometrically with the flat cylinder $\mathbb{R}\times S^1$. Through each point, the circle $u=\text{constant}$ is a closed geodesic, and every other geodesic has nonzero linear motion in $u$ and is not closed. Thus every point lies on a unique closed geodesic.

Write $g=\mathrm{Re}h$ for the underlying Riemannian metric of the Hermitian manifold. Its fundamental form of a Hermitian manifold isIt is real and skew-symmetric. In a unitary coframe it is $\omega ={\sum }_{j=1}^n{e}^j\wedge J{e}^j$, which also shows that it has type $(1,1)$ and thatThe Hodge star operator is characterized, after complex-linear extension, byExpanding in the same unitary coframe gives

The Hodge Laplacian and Dolbeault Laplacian areThe Dolbeault Hodge decomposition on a compact Hermitian manifold says that every Dolbeault class has a unique ${\Delta }_{\bar{\partial }}$-harmonic representative andIf ${\Delta }_{\bar{\partial }}\eta =0$, thenso $\eta$ is $\bar{\partial }$-closed and ${\bar{\partial }}^{\ast }$-closed. If also $\eta =\bar{\partial }\xi$, then $\Vert \eta {\Vert }_2^2=⟨{\bar{\partial }}^{\ast }\eta ,\xi ⟩=0$, hence $\eta =0$.

Now suppose $X$ is compact and Kähler. With $d^c=i(\bar{\partial }-\partial )$ and $d^{\ast }={\partial }^{\ast }+{\bar{\partial }}^{\ast }$, the Kähler identities make the mixed anticommutators vanish and give ${\Delta }_{\partial }={\Delta }_{\bar{\partial }}$. Consequently

Let $\alpha =d^c\gamma$ and $d\alpha =0$. The Kähler Laplacian identity implies that the $d$-Laplacian commutes with $d^c$. Since a harmonic form is $d^{c\ast }$-closed, $\alpha$ is orthogonal to every harmonic form. If $G$ is the Green operator of the Hodge Laplacian, thenwhere the $d^{\ast }dG\alpha$ term vanishes because $dG\alpha =Gd\alpha =0$. The Green operator commutes with $d^c$, and the anticommutation identity just proved givesTherefore, for the $(k-2)$-form $\beta =-d^{\ast }G\gamma$,This is the d d c lemma in the form required here.

Write $q^1=u$, $q^2=v$, ${\sigma }_i={\partial }_i\sigma$, and let $g_{ij}={\sigma }_i\cdot {\sigma }_j$ be the first fundamental form. Sincedifferentiating ${\sigma }_u\cdot \dot{\gamma }$ and ${\sigma }_v\cdot \dot{\gamma }$, or equivalently taking the two inner products of the second formula with ${\sigma }_k$, givesThe matrix $(g_{k\ell })$ is invertible because $\sigma$ is a regular embedded surface parametrization. Consequently the two geodesic equations hold exactly when $\ddot{\gamma }$ is orthogonal to both tangent vectors ${\sigma }_u$ and ${\sigma }_v$, which says precisely that $\ddot{\gamma }$ is a normal vector. This is the ambient acceleration criterion for a surface geodesic.

Because $\dot{\gamma }$ is tangent and $\ddot{\gamma }$ is normal,Thus an affinely parametrized geodesic has constant speed, as recorded by constant speed of an affinely parametrized geodesic.

For a surface of revolution, use profile arc length $s$ and azimuth $\varphi$. Its Riemannian metric isThe azimuth is an ignorable coordinate, so the corresponding geodesic equation has the first integralIf $v=|\dot{\gamma }|$ and $\theta$ is the oriented angle with the parallel, then the component of velocity along the parallel isSince $v$ is constant,This is the Clairaut first integral for a surface of revolution.

In the Poincare half-plane model, the Riemannian metricassigns to a smooth curve $\gamma (t)=(x(t),y(t))$ the lengthThe metric is conformal to the Euclidean metric, so hyperbolic and Euclidean angles agree. Its area element is

The two geodesics from $e^{i\alpha }$ and $e^{i\beta }$ to infinity are the vertical lines $x=\cos \alpha$ and $x=\cos \beta$, while the third side is the unit semicircle. The triangle lies above that semicircle, so its area isIts interior angles are $\alpha$, $\pi -\beta$, and zero at the ideal vertex, and henceTriangulating a geodesic polygon with $n$ sides into $n-2$ triangles givesthe area of a hyperbolic geodesic polygon and the polygonal Gauss-Bonnet theorem.

Choose a Riemannian metric on $X$, using a partition of unity if necessary. For each $p$, letSince $F$ is a submersion, $D_{p}F{|}_{H_p}:H_p\to T_{F(p)}Y$ is an isomorphism. Its inverse depends smoothly on $p$, sodefines a smooth vector field on $X$ satisfying $D_{p}F(v(p))=w(F(p))$. The formula also gives $v(p)=0$ whenever $w(F(p))=0$.

The connection is orthogonal, or metric-compatible, when $\nabla g=0$; equivalently, its parallel transport preserves the Riemannian metric. Compatibility of the induced connections with tensor contraction givesfor all vector fields $U,V$. Taking $U={\partial }_j$ and $V={\partial }_k$ yieldsConversely, this coordinate identity makes every component of $\nabla g$ vanish, so it is equivalent to orthogonality.

The Kähler manifold structure gives a Riemannian metric, its volume form, the complex orientation, and a Hermitian inner product on complex differential forms. The complex Hodge star operator is the complex-linear map characterized byOn $r$-forms in real dimension $m=2n$, the codifferential isequivalently the formal adjoint of $d$ for the $L^2$ inner product. Similarly ${\bar{\partial }}^{\ast }$ is the formal adjoint of $\bar{\partial }$. Define the Hodge Laplacian and Dolbeault Laplacian by

Expanding $d=\partial +\bar{\partial }$, the Kähler identities make the mixed anticommutators vanish and imply ${\Delta }_{\partial }={\Delta }_{\bar{\partial }}$. Hence the Kähler Laplacian identity is

Let $L\gamma =\omega \wedge \gamma$ be the Lefschetz operator of a Kähler manifold. The Kähler identities also implyThus, if ${\Delta }_{\bar{\partial }}\alpha =0$, thenThis is the fact that the Lefschetz operator preserves harmonic forms.

The Dolbeault Hodge decomposition on a compact Hermitian manifold states thatan orthogonal direct sum, where ${\mathcal{H}}_{\bar{\partial }}^{p,q}=\ker {\Delta }_{\bar{\partial }}$.

Suppose $\eta =\bar{\partial }\gamma$ has type $(p,q)$. Apply this decomposition to $\gamma \in {\Omega }^{p,q-1}$. The harmonic and $\bar{\partial }$-exact pieces disappear after applying $\bar{\partial }$, so for some $\beta \in {\Omega }^{p,q}$,Put $\theta ={\bar{\partial }}^{\ast }\beta$. If also $\partial \eta =0$, thenThe Kähler anticommutation identity ${\bar{\partial }}^{\ast }\partial +\partial {\bar{\partial }}^{\ast }=0$ and $({\bar{\partial }}^{\ast }{)}^2=0$ giveTherefore $\partial \theta$ is ${\Delta }_{\bar{\partial }}$-harmonic. By ${\Delta }_{\partial }={\Delta }_{\bar{\partial }}$ it is also $\partial$-harmonic, but it is $\partial$-exact; orthogonality of harmonic and exact forms forcesThis proves both requested claims: $\partial {\bar{\partial }}^{\ast }\beta =0$ is harmonic, and ${\bar{\partial }}^{\ast }\beta =\theta$ is $\partial$-closed.

Finally, $\theta \in \mathrm{im}{\bar{\partial }}^{\ast }$ is orthogonal to $\ker \bar{\partial }$, and hence to every $\bar{\partial }$-harmonic form. Since the $\partial$- and $\bar{\partial }$-harmonic spaces agree on a compact Kähler manifold, the $\partial$-closed form $\theta$ has zero harmonic component in its $\partial$-Hodge decomposition. It follows that $\theta =\partial \varphi$ for some $\varphi \in {\Omega }^{p-1,q-1}(X)$. Hencewhich is the ddbar lemma in this case.

The pointwise inequality $\hat{g}(X,X)\ge g(X,X)$ impliesEvery $d_{\hat{g}}$-Cauchy sequence $(x_j)$ is therefore $d_g$-Cauchy. Since $g$ is geodesically complete, the Hopf-Rinow theorem makes $(M,d_g)$ a complete metric space, so $x_j\to x$ in $d_g$ for some $x\in M$.

On a coordinate neighbourhood with compact closure around $x$, smooth positive-definite Riemannian metrics are uniformly equivalent. Thus there is $C>0$ such thatthere. For all sufficiently large $j$, a short $g$-geodesic from $x$ to $x_j$ stays in this neighbourhood, and henceThus $(M,d_{\hat{g}})$ is complete. Another application of Hopf-Rinow shows that $\hat{g}$ is geodesically complete.

Yes to both questions. Since ${\omega }_g$ has top degree, $d{\omega }_g=0$. Moreover $\ast {\omega }_g=1$, so the formula for the codifferential givesThereforeand the Riemannian volume form is a harmonic differential form.

The Levi-Civita connection preserves both the Riemannian metric and its chosen orientation. At any point, extend a positively oriented orthonormal basis to a local frame whose covariant derivatives vanish at that point. Differentiating ${\omega }_g(e_1,\dots ,e_n)=1$ there gives $\nabla {\omega }_g=0$. Hence ${\omega }_g$ is a parallel differential form.

For closed manifolds with the product Riemannian metric, the positive Laplace-Beltrami operator is the sum of the two factor operators. Products of their orthonormal eigenbases form a complete basis, and eigenvalues add. Hence the product eigenspace at $\lambda$ is the displayed finite orthogonal sum of tensor products. The discrete assertion requires a discrete spectral setting.

For a Riemannian metric on an ambient manifold and an embedded submanifold, let $\pi :TN{|}_M\to TM$ be orthogonal projection. Project the restriction of a connection to an embedded submanifold to obtain $D_{X}Y=\pi {\nabla }_{X}Y$. Since $\pi Y=Y$, the Leibniz rule and linearity over smooth functions hold, making $D$ an affine connection. If the ambient connection is the Levi-Civita connection, pairing with tangent vectors shows that $D$ is a metric connection; projecting the zero-torsion identity shows it is a torsion-free connection. By uniqueness, $D$ is the induced Levi-Civita connection.

For a smooth map $f:M\to N$, the pullback of a Riemannian metric is $(f^{\ast }g{)}_p(u,v)=g_{f(p)}(d{f}_{p}u,d{f}_{p}v)$. It is positive definite precisely when $d{f}_p$ is injective at every point, that is, when $f$ is an immersion. In coordinates, $(f^{\ast }g{)}_{ij}=g_{ab}(f(y)){\partial }_i{f}^a{\partial }_j{f}^b$.

A symmetric two-tensor satisfying this equation gives a quadratic geodesic first integral. Examples include the Riemannian metric, symmetric products of metric-dual Killing vector fields, and the square of a Killing-Yano two-form.

For a smooth function $f$ on a Riemannian manifold, its Riemannian gradient is the unique vector field satisfying $g({\nabla }_{g}f,V)=df(V)$ for every vector field $V$. It depends on the Riemannian metric, and its negative generates the downward gradient flow.