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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 the Riemannian exponential map ${\exp }_p(v)={\gamma }_v(1)$, uniqueness of the geodesic equation gives ${\exp }_p(tv)={\gamma }_v(t)$. Differentiation at $t=0$ yields $(d{\exp }_p{)}_{0}v=v$. Smooth ODE dependence and the inverse function theorem therefore make ${\exp }_p$ a diffeomorphism on a neighborhood of zero, producing geodesic normal coordinates. This is a local assertion and requires no geodesic completeness.

Getzler rescaling combines the change $x\mapsto \sqrt{t}x$ in geodesic normal coordinates with the Clifford algebra filtration, assigning degree one to a derivative and to a Clifford generator and degree minus one to a coordinate. The associated graded Clifford algebra is the exterior algebra. This is useful because the spinor supertrace annihilates every Clifford monomial except the top-degree one. Applied to the Lichnerowicz formula for a twisted Dirac operator, it retains a quadratic curvature model, whose heat kernel is computed by a Gaussian ansatz; all terms of smaller rescaling degree disappear from the local index limit. Thus the cancellations in a supersymmetric heat trace can be computed before solving the complete curved heat equation.

At the centre of geodesic normal coordinates, expand the square of the Twisted Dirac operator. The diagonal terms give the rough Laplacian; the off-diagonal terms are ${\sum }_{i<j}c(e^i)c(e^j)[{\nabla }_i,{\nabla }_j]$. The tensor-product vector-bundle curvature is the sum of spin curvature and $F^W$. Substituting spin curvature $\frac{1}{4}{\sum }_{a,b}⟨R(e_i,e_j)e_a,e_b⟩c(e^a)c(e^b)$ and using the Clifford algebra relations yields the scalar term $\mathrm{Scal}/4$: the four-generator contribution vanishes by the algebraic Bianchi identity and the contracted terms are the scalar curvature. This proves the displayed identity at every point.

For $L=-{\partial }_j(g^{ij}{\partial }_i)$ in geodesic normal coordinates, the Gauss lemma gives $g^{ij}{x}_i=x_j$. Substituting $F=(4\pi t{)}^{-n/2}{e}^{-|x{|}^2/(4t)}{\sum }_{k\ge 0}{t}^k{B}_k$ into $({\partial }_t+L)F=0$ therefore yields $(x\cdot \nabla +k)B_k=-L{B}_{k-1}$, with $B_0=1$. Integration along radial rays gives the displayed formula. Every coefficient is smooth at the centre. Any smooth solution of $(x\cdot \nabla +k)h=0$ is zero for $k>0$, since $s^{k}h(sx)$ is constant and tends to zero at $s=0$. This proves uniqueness. In this special divergence-form case $L1=0$, so every $B_k$ for $k>0$ is zero: the Gaussian itself solves the equation exactly on the coordinate ball. The positive Laplace-Beltrami operator instead has a metric volume-density term and a nonconstant leading amplitude.

At any point of a Riemannian manifold, choose geodesic normal coordinates and orthonormalize their coordinate fields smoothly. At the point, the metric is the identity and its first derivatives vanish, so the orthonormalization coefficients have zero first derivatives. The Christoffel symbols also vanish there. This gives a local Riemannian orthonormal frame with the displayed property, useful for invariant identities evaluated at one point.

Use the positive Laplace-Beltrami operator $\Delta =-\mathrm{div}\nabla$. For the global discrete spectral statements, take $M$ to be a closed manifold with a smooth Riemannian metric. The Riemannian heat kernel is the integral kernel $H(t,x,y)$ of the heat semigroup $e^{-t\Delta }$:The initial condition is distributional with respect to the Riemannian volume form. Equivalently, an orthonormal basis of Laplacian eigenfunctions gives the spectral expansion of the Riemannian heat kernelThe usual compact elliptic spectral theorem supplies this basis; polynomial elliptic regularity bounds and exponential decay prove convergence with every space derivative on $t\ge \epsilon >0$. These analytic results may be used as clearly stated external ingredients. We will also construct the kernel locally and prove its short-time expansion.

On a sufficiently small neighbourhood of the diagonal, let $r=d(x,y)$ and write $x={\exp }_y\xi$ in geodesic normal coordinates, with $d{V}_x=J(y,\xi )d\xi$. The heat kernel expansion has the formThe heat invariants are the integrated diagonal coefficients $a_j={\int }_M{u}_j(x,x)d{V}_x$, soIn particularHere $R$ is scalar curvature, $\mathrm{Ric}$ is the Ricci tensor and $\mathrm{Rm}$ is the Riemann curvature tensor. The last expression is the integrated second scalar heat coefficient; a local divergence term integrates to zero on a closed manifold. The exponent of the leading heat-trace term determines dimension, and these coefficients determine volume, total scalar curvature, and the displayed integrated curvature combination. On a surface $R=2K$, where $K$ is Gaussian curvature, so $a_1=\frac{1}{3}\int KdA=\frac{2\pi }{3}\chi (M)$ by the Gauss-Bonnet theorem, and $a_2=\frac{1}{15}\int K^{2}dA$. Thus the Euler characteristic, and the genus of a connected orientable closed surface, are determined. The first three coefficients also detect constant Gaussian curvature, since they determine $\int (K-\overline{K}{)}^{2}dA$. These conclusions do not claim that the pointwise curvature function or the metric is determined.

Here is the local construction. Put $g_t=(4\pi t{)}^{-m/2}{e}^{-r^2/(4t)}$. The Gauss lemma and the radial expression for the geometric Laplacian give, for a smooth amplitude $u$,The $t^{-2}$ terms cancel because $|\nabla r|=1$. Define $D=r{\partial }_r+(r/2){\partial }_r\log J$. The singular term disappears with $u_0=J^{-1/2}$, normalized by $u_0(y,y)=1$. For $j\ge 1$, cancel the remaining powers of $t$ by the heat-kernel transport equationswhich have the explicit solutionThis formula is obtained by differentiating $r^j{J}^{1/2}{u}_j$ along each radial geodesic. It proves smoothness jointly in $x,y$, including the diagonal, by induction: the integrand is smooth on a fixed compact integration interval. A homogeneous solution of $(D+j)u=0$ has $r^j{J}^{1/2}u$ constant along a ray, and smoothness at $r=0$ forces this constant to vanish for $j>0$. Thus the smooth transport coefficients are unique. In particular $u_j(y,y)=-({\Delta }_x{u}_{j-1})(y,y)/j$. Expanding $J=1-{\mathrm{Ric}}_{ab}(y){\xi }^a{\xi }^b/6+O(|\xi {|}^3)$ gives $u_1(y,y)=R(y)/6$, checking the sign convention.

Choose a smooth cutoff $\eta (x,y)$ that is one near the diagonal and supported where these normal coordinates exist, and form the heat parametrixIt tends to the identity kernel as $t\downarrow 0$: in normal coordinates the Gaussian is an approximate identity, and $u_{0}J=J^{1/2}\to 1$ at the centre. Its residual $R_N=({\partial }_t+{\Delta }_x)P_N$ is $\eta g_t{t}^N{\Delta }_x{u}_N$ plus cutoff-derivative terms supported a positive distance from the diagonal. Those latter terms are smaller than every power of $t$. Hence $R_N=O(t^{N-m/2})$ uniformly. Every prescribed finite number of derivatives has the same kind of estimate after increasing $N$; differentiating a Gaussian costs only finitely many powers of $t$.

For $N$ large enough that the residual is continuous and bounded up to time zero, use the Volterra parametrix correction. For kernels defineSet $Q=-R_N+R_N\ast R_N-R_N\ast R_N\ast R_N+\cdots$. A bounded residual with bound $C$ givesso this series converges and $Q+R_N+R_N\ast Q=0$. Because $P_N$ has identity initial kernel, differentiating the convolution gives $({\partial }_t+{\Delta }_x)(P_N\ast Q)=Q+R_N\ast Q$. Thusis an exact kernel with the correct initial condition. Taking $N$ sufficiently large gives any required finite regularity; uniqueness of the heat equation makes the kernels obtained from different $N$ equal. For uniqueness, the difference applied to smooth initial data has zero initial value, and the energy identity $\frac{d}{dt}\Vert v{\Vert }_2^2=-2\Vert \nabla v{\Vert }_2^2$ forces $v=0$. Parabolic regularity then gives smoothness for positive time.

The Gaussian part of $P_N$ has uniformly bounded spatial $L^1$ norm, so the correction is $O(t^{N+1-m/2})$. This proves, uniformly near the diagonal,with corresponding derivative estimates on increasing the truncation order. In particular on every parabolic neighbourhood $r\le B\sqrt{t}$ it is the stated Gaussian expansion with relative remainder $O(t^{N+1})$, and on the diagonal it yields the complete heat trace expansion. The standard off-diagonal Gaussian version on compact subsets of a sufficiently small convex normal neighbourhood follows from the same convolution estimates with the Gaussian factor retained: the inequalitylocalizes the convolution to the unique joining geodesic; its transverse Hessian is positive, and Gaussian integration gives the successive transport orders. The additive estimate above already specifies an appropriate uniform near-diagonal remainder without asserting a relative estimate far from the diagonal on the basis of a crude bound.

If $M$ is noncompact, use the minimal Friedrichs extension heat semigroup; if there is a boundary, specify boundary conditions. The same construction is local in the interior. One may extend a relatively compact coordinate neighbourhood to a closed auxiliary manifold; localization errors are supported a positive distance away and are exponentially small near the smaller diagonal neighbourhood. Thus the local coefficients remain the same. Finite integrated heat invariants and a discrete spectrum require the global hypotheses used above.

Fix any point $p$ and use a normal orthonormal frame there. Such frames exist: in geodesic normal coordinates, $g_{ij}(p)={\delta }_{ij}$, ${\partial }_k{g}_{ij}(p)=0$ and the Christoffel symbols vanish at $p$. Applying the Gram-Schmidt process smoothly to the coordinate fields gives identity coefficients at $p$ and zero first derivatives there, so it produces the required frame. Reverse one frame vector if necessary to match the orientation.

Because the Levi-Civita connection is torsion-free, $[E_j,E_k](p)=({\nabla }_{E_j}{E}_k-{\nabla }_{E_k}{E}_j)(p)=0$. The supplied formula for the exterior derivative of a one-form givesThus $d{\omega }_i(p)=0$, and the graded Leibniz rule implies $d{\theta }_i(p)=0$. Differentiate the formula in part (b):Since $d{f}_i={\sum }_j{E}_j(f_i){\omega }_j$, every summand with $j\ne i$ vanishes because ${\omega }_j$ already occurs in ${\theta }_i$. The remaining term has ${\omega }_i\wedge {\theta }_i=(-1{)}^{i-1}{\omega }_g$, so both signs cancel. HenceThe point was arbitrary, proving the global exterior derivative of contracted volume form identity

Use the nonnegative convention for the Laplace-Beltrami operator. For a smooth function $f$ on a Riemannian manifold $(M,g)$, three intrinsic definitions areHere $d$ is the exterior derivative, $d^{\ast }$ its formal $L^2$ adjoint for the Riemannian volume density, $\mathrm{grad}f$ the Riemannian gradient, and $\nabla$ the Levi-Civita connection. The bilinear form $\nabla df$ is the Riemannian Hessian. The equally common nonpositive convention negates all three definitions simultaneously.

In local coordinates let $g^{ij}$ be the inverse metric matrix and $\rho =\sqrt{\det (g_{ij})}$. By the defining relation $g(\mathrm{grad}f,V)=df(V)$,The divergence of a Riemannian vector field can be defined by ${\mathcal{L}}_V(d{V}_g)=(\mathrm{div}V)d{V}_g$. Differentiating the coordinate volume density gives $\mathrm{div}V={\rho }^{-1}{\partial }_i(\rho V^i)$. Thus the second definition becomesFor compactly supported smooth $h$, coordinate integration by parts showsSince this holds for every such $h$, the first and second definitions agree. This argument also works on a nonorientable Riemannian manifold, using its volume density; no global orientation is needed.

For the third definition,The Christoffel symbols satisfy ${\Gamma }_{ik}^i={\partial }_k\log \rho$. This follows by tracing their formula and using ${\partial }_k\log \det g=\mathrm{tr}(g^{-1}{\partial }_{k}g)$. Metric compatibility givesSubstitute both identities into the coordinate divergence expression to obtainThis proves all three definitions equivalent. In geodesic normal coordinates at a point $p$, the common value is $-{\sum }_i{\partial }_i^{2}f(p)$. Finally, on a closed manifold,confirming the chosen positive Laplace-Beltrami operator sign convention.

Let $M$ be a closed manifold of even dimension $n=2m$, with a Riemannian metric and a spin structure. Let $S=S^{+}\oplus S^{-}$ be its spinor bundle, and let $W$ be a Hermitian vector bundle with a unitary connection of vector-bundle curvature $F^W=({\nabla }^W{)}^2$. We use Clifford multiplication satisfying $c(v)c(w)+c(w)c(v)=-2⟨v,w⟩$, so the Twisted Dirac operatoris self-adjoint and $D_W^{-}=(D_W^{+}{)}^{\ast }$. Here $(e_i)$ is an oriented local orthonormal basis and $(e^i)$ its dual. The Atiyah-Singer index theorem in this setting isThe second factor is the Chern character; the first is the A-hat formThe square root has constant term one. All functions of the vector-bundle curvature matrices mean power series truncated above the dimension of $M$. The brackets select the top-degree differential form. The Bianchi identity shows that these forms are closed, and Chern-Weil connection transgression shows that their de Rham cohomology classes do not depend on the chosen connections on a vector bundle.

The global step is supersymmetric quantum mechanics. On the Hilbert space $L^2(S\otimes W)$, take the grading operator $\Gamma$, the odd supercharge $D_W$, and the nonnegative Hamiltonian $H=D_W^2$. They obey $\Gamma D_W=-D_W\Gamma$ and $D_W^2=H$. The principal symbol of $D_W$ is invertible away from the zero covector, so elliptic regularity and Rellich-Kondrachov compactness theorem give a smooth discrete orthonormal eigenbasis for $H$, finite-dimensional eigenspaces, and a trace-class operator $e^{-tH}$ for $t>0$. Define its supertrace by $\mathrm{Str}A=\mathrm{Tr}(\Gamma A)$.

For every positive eigenvalue $\lambda$ of $H$, the map ${\lambda }^{-1/2}{D}_W$ is an isomorphism between its positive and negative grading eigenspaces, with inverse itself. Their contributions to the supertrace cancel. The zero eigenspaces are $\ker D_W^{+}$ and $\ker D_W^{-}$. ConsequentlyThis also proves independence of $t$ without differentiating an unbounded operator inside a trace. If $K_t$ is the heat kernel, the heat kernel trace formula givesThe remaining task is to compute this local integrand as $t\downarrow 0$.

Fix $p\in M$, use geodesic normal coordinates $x$, and trivialize $S\otimes W$ by radial parallel transport. Expanding $D_W^2$ at $p$ gives the Lichnerowicz formula for a twisted Dirac operator:Indeed the terms with identical indices give the rough Laplacian. The other terms are ${\sum }_{i<j}c(e^i)c(e^j)[{\nabla }_i,{\nabla }_j]$. Decompose this vector-bundle curvature into spin curvature and $F^W$; substituting the spin curvature into the Clifford algebra relations and the algebraic Bianchi identity leaves $\mathrm{Scal}/4$. Thus the calculation retains the curvature of the twisting bundle explicitly.

There is an especially strong cancellation in the spinor supertrace. With $\Gamma =i^{m}c(e^1)\cdots c(e^{2m})$, every ordered product of distinct Clifford generators of degree less than $2m$ has zero supertrace, whileFor an even product of lower degree, conjugate $\Gamma$ times that product by a missing Clifford generator: it changes sign, so its matrix trace is zero. An odd product exchanges the grading summands and also has zero supertrace. For the full product, $(c(e^1)\cdots c(e^{2m}){)}^2=(-1{)}^m$, and $\dim S=2^m$, giving the displayed constant.

To exploit this, identify the Clifford algebra as a filtered vector space with the exterior algebra by sending an ordered Clifford product to the corresponding wedge product of differential forms. Implement Getzler rescaling: put $x=\epsilon y$, $t={\epsilon }^{2}s$, multiply the degree-$k$ exterior component of the kernel by ${\epsilon }^{-k}$, and multiply the whole kernel by ${\epsilon }^{2m}$. A derivative and a Clifford generator have rescaling degree one; a coordinate has degree minus one. Clifford contractions lower the degree by two relative to the wedge product of differential forms, so the leading product becomes exterior multiplication.

In the radially parallel frame, the linear Taylor term of the connection is determined by its vector-bundle curvature: if $A_i$ is its matrix, then $A_i(x)=\frac{1}{2}{x}^j{F}_{ji}(p)+O(|x{|}^2)$. To check the coefficient, radial gauge means $x^i{A}_i=0$, and integration of $x^j{F}_{ji}$ along the radial segment gives $A_i(x)={\int }_0^{1}s{x}^j{F}_{ji}(sx)ds$. Combining this formula with the spin representation, the symmetries of Riemann curvature tensor, and the Lichnerowicz formula for a twisted Dirac operator, the rescaled square has leading modelHere $\mathcal{R}$ is the skew matrix of tangent-bundle curvature two-forms at $p$; changing the curvature-matrix sign convention changes the sign inside the covariant derivatives but not the determinant below. The coefficients act in $\Lambda T_p^{\ast }M\otimes \mathrm{End}(W_p)$. The scalar-curvature term has degree zero and disappears after multiplication by ${\epsilon }^2$; the linear spin-connection term and the Clifford action of $F^W$ have degree two and survive. Higher Taylor terms have smaller degree and disappear.

For clarity, the analytic meaning of this limiting calculation is the following. Construct the local heat kernel expansion to any fixed order and insert the Taylor expansion of the rescaled coefficients into its transport equations. At each exterior degree, the leading equation is the heat equation of ${\mathcal{H}}_p$; the remaining terms acquire positive powers of $\epsilon$. Equivalently, successive substitutions ingive the same expansion. Polynomial spatial factors are integrable against the local Gaussian kernels, and only finitely many exterior degrees exist. A cutoff outside a fixed normal neighbourhood contributes $O(e^{-c/t})$. Choosing the initial expansion order arbitrarily large bounds its differentiated remainder to any required order. This proves convergence of the rescaled diagonal kernel to the model diagonal kernel, uniformly in $p$ on the compact manifold. Its degree-$2m$ component has no net power of $\epsilon$, precisely the component detected by the spinor supertrace.

We can compute the model kernel rather than invoke a local index formula. Set $F=F_p^W$ and seek its kernel from the origin in the formThe scalar even exterior coefficients of $\mathcal{R}$ commute. On this ansatz the rotational first-order term vanishes because $Q$ is a function of ${\mathcal{R}}^2$. Expanding the model operator gives $-\sum {\partial }_i^2+\frac{1}{2}(\mathcal{R}y)\cdot \nabla +\frac{1}{16}{y}^T{\mathcal{R}}^{2}y+F$, so substitution into the heat equation givesThe Euclidean initial condition $Q(s)\sim s^{-1}I$, $a(s)\sim (4\pi s{)}^{-m}$ selectsThese expressions have well-defined power series at $\mathcal{R}=0$. This is the curvature version of the Gaussian calculation underlying the Mehler formula for Hermite functions.

At $s=1$, $y=0$, multiply the top exterior coefficient by the spinor supertrace constant. Since $(4\pi {)}^{-m}(-2i{)}^m=(2\pi i{)}^{-m}$, and every curvature factor is a two-form, the result isIntegration and the constant supertrace established above prove the boxed Atiyah-Singer index theorem. The mechanism is that odd symmetry pairs the nonzero spectrum globally, while the Clifford filtration makes the surviving local contribution an explicitly solvable curvature oscillator. Odd-dimensional spin Dirac operators have no canonical chiral index of this type; their self-adjoint Fredholm index is zero. For comparison of normalization conventions, Raphaël Ponge's local index calculation writes the same unnormalized curvature model and spinor supertrace constant.

The Gauss lemma says that, for $v,w\in T_{p}M$ for which the exponential map is defined,In particular radial and angular directions are orthogonal, and the radial direction retains its length.

Take the variation through geodesics $\alpha (t,s)={\exp }_p(t(v+sw))$, and put $T={\partial }_t\alpha$, $J={\partial }_s\alpha$. The Levi-Civita connection is metric compatible and torsion-free, so ${\nabla }_{t}J={\nabla }_{s}T$. Each $t$-curve is a geodesic, hence ${\nabla }_{t}T=0$, and its constant squared speed is $|v+sw{|}_p^2$. ThereforeAt $t=0$ the variation starts at the fixed point $p$, so $J(0,s)=0$. Integrating from $0$ to $1$ and setting $s=0$ gives the displayed identity, because $T(1,0)=(d{\exp }_p{)}_{v}v$ and $J(1,0)=(d{\exp }_p{)}_{v}w$.

In geodesic normal coordinates based on an orthonormal basis at $p$, choose $v=x$ and $w$ to be the $j$th coordinate vector. The lemma givesThese identities also hold at $x=0$ by continuity. They are the radial identities required in part (b).

Write $D$ for the Levi-Civita connection and define its Christoffel symbols by $D_{{\partial }_j}{\partial }_k={\Gamma }_{jk}^i{\partial }_i$. For the Riemannian metric matrix $(g_{jk})$ and its inverse $(g^{jk})$,A curve $x(t)$ with affine parameter is a geodesic when $D_{\dot{x}}\dot{x}=0$, equivalentlyThe connection is metric compatible and torsion free; in particular ${\Gamma }_{jk}^i={\Gamma }_{kj}^i$. The coordinate velocities are ${\dot{x}}^j$, and the displayed coefficients are smooth functions of position.

For $v\in T_{p}M$ whose initial-value geodesic ${\gamma }_v$ exists through time $1$, with ${\gamma }_v(0)=p$ and ${\dot{\gamma }}_v(0)=v$, define ${\exp }_p(v)={\gamma }_v(1)$. The domain is open and contains a neighborhood of zero. Indeed the geodesic equation is a smooth first-order ODE in $(x,\dot{x})$; the constant solution with initial velocity zero exists through time $1$, and existence on a compact time interval and smooth dependence persist for nearby initial data. Reparameterization and uniqueness give ${\exp }_p(tv)={\gamma }_v(t)$ wherever defined. Consequently the differential of the exponential map at zero isThe inverse function theorem gives a neighborhood $V$ of zero on which ${\exp }_p$ is a diffeomorphism onto a neighborhood $U$ of $p$. Shrink $V$ to a ball in $T_{p}M$ and choose an orthonormal basis there. Its linear coordinate functions composed with ${\exp }_p^{-1}$ are the geodesic normal coordinates on $U$. This establishes well-defined local coordinates without assuming geodesic completeness.

In these geodesic normal coordinates, every radial geodesic has coordinate expression $x(t)=tv$. Its equation at $t=0$ reads ${\Gamma }_{jk}^i(p)v^j{v}^k=0$ for every $v$. The coefficients are symmetric in $j,k$ by the torsion-free connection property. Evaluating at the coordinate unit vectors and their pairwise sums, or using the polarization identity, therefore givesThis is the fact that Christoffel symbols vanish at the center of normal coordinates. Also $g_{ij}(p)={\delta }_{ij}$, since $(d{\exp }_p{)}_0$ is the identity in an orthonormal basis.

For sufficiently small $r>0$, the geodesic sphere is $S_r(p)={\exp }_p\{v:|v{|}_{g_p}=r\}$. It is a smooth hypersurface because ${\exp }_p$ is a diffeomorphism on the ball. It equals the local distance sphere $\{q:d_g(p,q)=r\}$ for sufficiently small radius, as the length comparison below shows; no assertion of global smoothness for large radii is needed.

The full Gauss lemma isfor $v$ in a star-shaped domain of ${\exp }_p$ and $w\in T_{p}M$. To prove it, set $F(t,s)={\exp }_p(t(v+sw))$, $T={\partial }_{t}F$ and $S={\partial }_{s}F$, for $s$ near zero. Every $t$-curve is a geodesic and has squared speed $|v+sw{|}_{g_p}^2$. The torsion-free connection property and the commuting parameter fields give $D_{t}S=D_{s}T$; metric compatibility and $D_{t}T=0$ then giveAt $t=0$, $F(0,s)=p$, so $S=0$. Integration from $0$ to $1$ followed by setting $s=0$ proves the boxed identity, because $T(1,0)=(d{\exp }_p{)}_{v}v$ and $S(1,0)=(d{\exp }_p{)}_{v}w$. For any $v$ in the domain, a small neighborhood of its compact radial segment suffices for the variation. The zero vector is covered directly.

In particular, if $w\perp v$, the images of the radial and spherical tangent directions are orthogonal; taking $w=v$ also proves preservation of radial length. Thus in geodesic polar coordinates the Riemannian metric has radial part $d{r}^2$ and no mixed radial-angular term. Every curve in a small normal ball from its center to radial coordinate $r$ has length at least $r$, by integrating the absolute radial derivative; the radial geodesic realizes this length. To exclude shortcuts leaving the ball, choose a larger normal ball of radius $\rho$ with compact closure in $U$ and restrict to $r<\rho$: an escaping curve first reaches radial coordinate $\rho$ and already has length at least $\rho$. This proves the asserted small-radius distance interpretation and orthogonality of radial geodesics to geodesic spheres.

For $v$ in the domain of the exponential map at $p$, let ${\gamma }_v$ be the unique geodesic with initial data $(p,v)$ and define ${\exp }_p(v)={\gamma }_v(1)$. The domain consists of initial velocities whose geodesics exist on $[0,1]$; it is open, contains zero, and need not be the whole tangent space. The ordinary differential equation theorem used here says that a smooth vector field has unique maximal integral curves, with an open flow domain and smooth dependence on initial point and time. Applied to the geodesic equation on $TM$, it gives smoothness of the exponential map. Rescaling the parameter gives ${\exp }_p(tv)={\gamma }_v(t)$ and hence $(d{\exp }_p{)}_0(v)=v$. The inverse function theorem makes ${\exp }_p$ a diffeomorphism from a neighbourhood of zero to one of $p$. Coordinates in an orthonormal tangent basis transported by this map are geodesic normal coordinates.

The geodesic sphere of radius $r$ is $S_r(p)=\{x:d(p,x)=r\}$. For small $r>0$, it is ${\exp }_p(\{v:\Vert v\Vert =r\})$, a smooth compact hypersurface; larger distance spheres need not be smooth. The Gauss lemma statesIn particular radial and angular directions are orthogonal and the radial coordinate measures arc length.

For a piecewise smooth path $c$, put $L(c)=\int \Vert \dot{c}(t)\Vert dt$. The Riemannian distance is $d(x,y)=\inf L(c)$ over such paths from $x$ to $y$. Connectedness makes it finite. Reversal and concatenation give symmetry and the triangle inequality; positivity and compatibility with the manifold topology follow locally from geodesic normal coordinates and the Gauss lemma.

Choose $r_0>0$ so that ${\exp }_p$ is a diffeomorphism on a neighbourhood of the closed tangent ball of radius $2r_0$. For $r<r_0$, radial geodesics realize the distance from $p$ on that ball. Indeed, Gauss lemma bounds any path remaining in the normal ball below by its radial change, while a path leaving the radius-$2r_0$ ball already has length at least $2r_0$. Thus the radius-$r$ sphere is the compact exponential image described above.

Given $q\ne p$, take $0<\delta <\min \{r_0,d(p,q)\}$. The continuous function $z\mapsto d(z,q)$ attains its minimum at some $p_0\in S_{\delta }(p)$. Every path from $p$ to $q$ crosses that sphere, since the distance from $p$ is continuous. Splitting at a crossing gives length at least $\delta +{\min }_{S_{\delta }(p)}d(\cdot ,q)$. Taking the infimum over paths, and using the triangle inequality in the other direction, proves the distance splitting through a small geodesic sphere equalityThis compact local argument does not assume completeness or the existence of a globally minimizing geodesic to $q$.

For the Lie-group clause, a bi-invariant Riemannian metric has adjoint-invariant identity inner product. Differentiating this invariance gives $⟨[X,Y],Z⟩+⟨Y,[X,Z]⟩=0$. The Koszul formula for left-invariant vector fields has no metric-derivative terms, and this skew-adjointness reduces it toThis is the Levi-Civita connection of a bi-invariant metric.

Let $\sigma$ be the integral curve through the identity of the left-invariant field with identity value $v$. Uniqueness and left translation show $\sigma (s+t)=\sigma (s)\sigma (t)$ wherever the local curves are defined. The field is complete: a fixed local existence interval about the identity translates to an equally long interval at every point of the group. Restarting the curve before any proposed finite endpoint extends it beyond that endpoint. Thus $\sigma$ exists on $\mathbb{R}$, and uniqueness now gives the group law for every $s,t$.

Since ${\nabla }_{X}X=0$, this integral curve is a geodesic. Conversely any geodesic starting at the identity has the same initial data as one of these curves and agrees with it by uniqueness. Therefore the geodesics of a bi-invariant metric are one-parameter subgroups conclusion isThe zero velocity gives the constant, trivial subgroup. Here ${\exp }_G$ is the Exponential map of a Lie group; for this metric it agrees at the identity with the Riemannian exponential.

For $v$ near $0\in T_{p}M$, let ${\gamma }_v$ be the unique geodesic with ${\gamma }_v(0)=p$ and ${\dot{\gamma }}_v(0)=v$. The exponential map is ${\exp }_p(v)={\gamma }_v(1)$.

The constant initial velocity $0$ gives ${\exp }_p(0)=p$. Varying the initial velocity through $sv$ yields ${\exp }_p(sv)={\gamma }_v(s)$, soThus $(d{\exp }_p{)}_0$ is the identity. The inverse function theorem makes ${\exp }_p$ a diffeomorphism from a neighborhood of $0$ onto a neighborhood of $p$. Coordinates from an orthonormal basis of $T_{p}M$ are the geodesic normal coordinates.