Riemannian manifold

If ${\displaystyle (M,g)}$ and ${\displaystyle (N,h)}$ are two Riemannian manifolds, with ${\displaystyle f:M\to N}$ a diffeomorphism, then ${\displaystyle f}$ is called an isometry if ${\displaystyle g=f^{\ast }h,}$ i.e. if

${\displaystyle g_{p}(u,v)=h_{f(p)}(df_{p}(u),df_{p}(v))}$

for all ${\displaystyle p\in M}$ and ${\displaystyle u,v\in T_{p}M.}$

One says that a map ${\displaystyle f:M\to N,}$ not assumed to be a diffeomorphism, is a local isometry if every ${\displaystyle p\in M}$ has an open neighborhood ${\displaystyle U}$ such that ${\displaystyle f:U\to f(U)}$ is an isometry (and thus a diffeomorphism).

One says that the Riemannian metric ${\displaystyle g}$ is continuous if ${\displaystyle g_{ij}:U\to \mathbb {R} }$ are continuous when given any smooth coordinate chart ${\displaystyle (U,x).}$ One says that ${\displaystyle g}$ is smooth if these functions are smooth when given any smooth coordinate chart. One could also consider many other types of Riemannian metrics in this spirit.

In most expository accounts of Riemannian geometry, the metrics are always taken to be smooth. However, there can be important reasons to consider metrics which are less smooth. Riemannian metrics produced by methods of geometric analysis, in particular, can be less than smooth. See for instance (Gromov 1999) and (Shi and Tam 2002).

Let ${\displaystyle x^{1},\ldots ,x^{n}}$ denote the standard coordinates on ${\displaystyle \mathbb {R} ^{n}.}$ Then define ${\displaystyle g_{p}^{\mathrm {can} }:T_{p}\mathbb {R} ^{n}\times T_{p}\mathbb {R} ^{n}\to \mathbb {R} }$ by

${\displaystyle \left(\sum _{i}a_{i}{\frac {\partial }{\partial x^{i}}},\sum _{j}b_{j}{\frac {\partial }{\partial x^{j}}}\right)\longmapsto \sum _{i}a_{i}b_{i}.}$

Phrased differently: relative to the standard coordinates, the local representation ${\displaystyle g_{ij}:U\to \mathbb {R} }$ is given by the constant value ${\displaystyle \delta _{ij}.}$

This is clearly a Riemannian metric, and is called the standard Riemannian structure on ${\displaystyle \mathbb {R} ^{n}.}$ It is also referred to as Euclidean space of dimension n and g_ij^can is also called the (canonical) Euclidean metric.

Let ${\displaystyle (M,g)}$ be a Riemannian manifold and let ${\displaystyle N\subset M}$ be an embedded submanifold of ${\displaystyle M,}$ which is at least ${\displaystyle C^{1}.}$ Then the restriction of g to vectors tangent along N defines a Riemannian metric over N.

• For example, consider ${\displaystyle S^{n-1}=\{x\in \mathbb {R} ^{n}:(x^{1})^{2}+\cdots +(x^{n})^{2}=1.\},}$ which is a smooth embedded submanifold of the Euclidean space with its standard metric. The Riemannian metric this induces on ${\displaystyle S^{n-1}}$ is called the standard metric or canonical metric on ${\displaystyle S^{n-1}.}$
• There are many similar examples. For example, every ellipsoid in ${\displaystyle \mathbb {R} ^{3}}$ has a natural Riemannian metric. The graph of a smooth function ${\displaystyle f:\mathbb {R} ^{3}\to \mathbb {R} }$ is an embedded submanifold, and so has a natural Riemannian metric as well.

Let ${\displaystyle (M,g)}$ be a Riemannian manifold and let ${\displaystyle f:\Sigma \to M}$ be a differentiable map. Then one may consider the pullback of ${\displaystyle g}$ via ${\displaystyle f}$, which is a symmetric 2-tensor on ${\displaystyle \Sigma }$ defined by

${\displaystyle (f^{\ast }g)_{p}(v,w)=g_{f(p)}{\big (}df_{p}(v),df_{p}(w){\big )},}$

where ${\displaystyle df_{p}(v)}$ is the pushforward of ${\displaystyle v}$ by ${\displaystyle f.}$

In this setting, generally ${\displaystyle f^{\ast }g}$ will not be a Riemannian metric on ${\displaystyle \Sigma ,}$ since it is not positive-definite. For instance, if ${\displaystyle f}$ is constant, then ${\displaystyle f^{\ast }g}$ is zero. In fact, ${\displaystyle f^{\ast }g}$ is a Riemannian metric if and only if ${\displaystyle f}$ is an immersion, meaning that the linear map ${\displaystyle df_{p}:T_{p}\Sigma \to T_{f(p)}M}$ is injective for each ${\displaystyle p\in \Sigma .}$

• An important example occurs when ${\displaystyle (M,g)}$ is not simply-connected, so that there is a covering map ${\displaystyle {\widetilde {M}}\to M.}$ This is an immersion, and so the universal cover of any Riemannian manifold automatically inherits a Riemannian metric. More generally, but by the same principle, any covering space of a Riemannian manifold inherits a Riemannian metric.
• Also, an immersed submanifold of a Riemannian manifold inherits a Riemannian metric.

Let ${\displaystyle (M,g)}$ and ${\displaystyle (N,h)}$ be two Riemannian manifolds, and consider the cartesian product ${\displaystyle M\times N}$ with the usual product smooth structure. The Riemannian metrics ${\displaystyle g}$ and ${\displaystyle h}$ naturally put a Riemannian metric ${\displaystyle {\widetilde {g}}}$ on ${\displaystyle M\times N,}$ which can be described in a few ways.

• Considering the decomposition ${\displaystyle T_{(p,q)}(M\times N)\cong T_{p}M\oplus T_{q}N,}$ one may define

${\displaystyle {\widetilde {g}}_{p,q}(u\oplus x,v\oplus y)=g_{p}(u,v)+h_{q}(x,y).}$

• Let ${\displaystyle (U,x)}$ be a smooth coordinate chart on ${\displaystyle M}$ and let ${\displaystyle (V,y)}$ be a smooth coordinate chart on ${\displaystyle N.}$ Then ${\displaystyle (U\times V,(x,y))}$ is a smooth coordinate chart on ${\displaystyle M\times N.}$ For convenience let ${\displaystyle \operatorname {Sym} _{n\times n}^{+}}$ denote the collection of positive-definite symmetric ${\displaystyle n\times n}$ real matrices. Denote the coordinate representation of ${\displaystyle g}$ relative to ${\displaystyle (U,x)}$ by ${\displaystyle g_{U}:U\to \operatorname {Sym} _{m\times m}^{+}}$ and denote the coordinate representation of ${\displaystyle h}$ relative to ${\displaystyle (V,y)}$ by ${\displaystyle h_{V}:V\to \operatorname {Sym} _{n\times n}^{+}.}$ Then the local coordinate representation of ${\displaystyle {\widetilde {g}}}$ relative to ${\displaystyle (U\times V,(x,y))}$ is ${\displaystyle {\widetilde {g}}_{U\times V}:U\times V\to \operatorname {Sym} _{(m+n)\times (m+n)}^{+}}$ given by

${\displaystyle (p,q)\mapsto {\begin{pmatrix}g_{U}(p)&0\\0&h_{V}(q)\end{pmatrix}}.}$

A standard example is to consider the n-torus ${\displaystyle T^{n},}$ define as the n-fold product ${\displaystyle S^{1}\times \cdots \times S^{1}.}$ If one gives each copy of ${\displaystyle S^{1}}$ its standard Riemannian metric, considering ${\displaystyle S^{1}\subset \mathbb {R} ^{2}}$ as an embedded submanifold (as above), then one can consider the product Riemannian metric on ${\displaystyle T^{n}.}$ It is called a flat torus.

Let ${\displaystyle g_{0}}$ and ${\displaystyle g_{1}}$ be two Riemannian metrics on ${\displaystyle M.}$ Then, for any number ${\displaystyle \lambda \in [0,1],}$

${\displaystyle {\tilde {g}}:=\lambda g_{0}+(1-\lambda )g_{1}}$

is also a Riemannian metric on ${\displaystyle M.}$ More generally, if ${\displaystyle a}$ and ${\displaystyle b}$ are any two positive numbers, then ${\displaystyle ag_{0}+bg_{1}}$ is another Riemannian metric.

If ${\displaystyle \gamma$ :[a,b]\to M} is differentiable, then it assigns to each ${\displaystyle t\in (a,b)}$ a vector ${\displaystyle \gamma '(t)}$ in the vector space ${\displaystyle T_{\gamma (t)}M,}$ the size of which can be measured by the norm ${\displaystyle |\cdot |_{\gamma (t)}.}$ So ${\displaystyle t\mapsto |\gamma '(t)|_{\gamma (t)}}$ defines a nonnegative function on the interval ${\displaystyle (a,b).}$ The length is defined as the integral of this function; however, as presented here, there is no reason to expect this function to be integrable. It is typical to suppose g to be continuous and ${\displaystyle \gamma }$ to be continuously differentiable, so that the function to be integrated is nonnegative and continuous, and hence the length of ${\displaystyle \gamma ,}$

${\displaystyle L(\gamma )=\int _{a}^{b}|\gamma '(t)|_{\gamma (t)}\,dt,}$

is well-defined. This definition can easily be extended to define the length of any piecewise-continuously differentiable curve.

In many instances, such as in defining the Riemann curvature tensor, it is necessary to require that g has more regularity than mere continuity; this will be discussed elsewhere. For now, continuity of g will be enough to use the length defined above in order to endow M with the structure of a metric space, provided that it is connected.

Precisely, define ${\displaystyle d_{g}:M\times M\to [0,\infty )}$ by

${\displaystyle d_{g}(p,q)=\inf\{L(\gamma ):\gamma {\text{ a piecewise continuously differentiable curve from }}p{\text{ to }}q\}.}$

It is mostly straightforward to check the well-definedness of the function ${\displaystyle d_{g},}$ its symmetry property ${\displaystyle d_{g}(p,q)=d_{g}(q,p),}$ its reflexivity property ${\displaystyle d_{g}(p,p)=0,}$ and the triangle inequality ${\displaystyle d_{g}(p,q)+d_{g}(q,r)\geq d_{g}(p,r),}$ although there are some minor technical complications (such as verifying that any two points can be connected by a piecewise-differentiable path). It is more fundamental to understand that ${\displaystyle p\neq q}$ ensures ${\displaystyle d_{g}(p,q)>0,}$ and hence that ${\displaystyle d_{g}}$ satisfies all of the axioms of a metric.

(Sketched) Proof that ${\displaystyle p\neq q}$ implies ${\displaystyle d_{g}(p,q)>0.}$

Briefly: there must be some precompact open set around p which every curve from p to q must escape. By selecting this open set to be contained in a coordinate chart, one can reduce the claim to the well-known fact that, in Euclidean geometry, the shortest curve between two points is a line. In particular, as seen by the Euclidean geometry of a coordinate chart around p, any curve from p to q must first pass though a certain "inner radius." The assumed continuity of the Riemannian metric g only allows this "coordinate chart geometry" to distort the "true geometry" by some bounded factor. To be precise, let ${\displaystyle (U,x)}$ be a smooth coordinate chart with ${\displaystyle x(p)=0}$ and ${\displaystyle q\notin U.}$ Let ${\displaystyle V\ni x}$ be an open subset of ${\displaystyle U}$ with ${\displaystyle {\overline {V}}\subset U.}$ By continuity of ${\displaystyle g}$ and compactness of ${\displaystyle {\overline {V}},}$ there is a positive number ${\displaystyle \lambda }$ such that ${\displaystyle g(X,X)\geq \lambda \|X\|^{2}}$ for any ${\displaystyle r\in V}$ and any ${\displaystyle X\in T_{r}M,}$ where ${\displaystyle \|\cdot \|}$ denotes the Euclidean norm induced by the local coordinates. Let R denote ${\displaystyle \sup\{r>0:B_{r}(0)\subset x(V)\},}$ to be used at the final step of the proof. Now, given any piecewise continuously-differentiable path ${\displaystyle \gamma$ :[a,b]\to M} from p to q, there must be some minimal ${\displaystyle \delta >0}$ such that ${\displaystyle \gamma (a+\delta )\notin V;}$ clearly ${\displaystyle \gamma (a+\delta )\in \partial V.}$ The length of ${\displaystyle \gamma }$ is at least as large as the restriction of ${\displaystyle \gamma }$ to ${\displaystyle [a,a+\delta ].}$ So ${\displaystyle L(\gamma )\geq {\sqrt {\lambda }}\int _{a}^{a+\delta }\|\gamma '(t)\|\,dt.}$ The integral which appears here represents the Euclidean length of a curve from 0 to ${\displaystyle x(\partial V)\subset \mathbb {R} ^{n}}$, and so it is greater than or equal to R. So we conclude ${\displaystyle L(\gamma )\geq {\sqrt {\lambda }}R.}$

The observation that underlies the above proof, about comparison between lengths measured by g and Euclidean lengths measured in a smooth coordinate chart, also verifies that the metric space topology of ${\displaystyle (M,d_{g})}$ coincides with the original topological space structure of ${\displaystyle M.}$

Although the length of a curve is given by an explicit formula, it is generally impossible to write out the distance function ${\displaystyle d_{g}}$ by any explicit means. In fact, if ${\displaystyle M}$ is compact then, even when g is smooth, there always exist points where ${\displaystyle d_{g}:M\times M\to \mathbb {R} }$ is non-differentiable, and it can be remarkably difficult to even determine the location or nature of these points, even in seemingly simple cases such as when ${\displaystyle (M,g)}$ is an ellipsoid.

As in the previous section, let ${\displaystyle (M,g)}$ be a connected and continuous Riemannian manifold; consider the associated metric space ${\displaystyle (M,d_{g}).}$ Relative to this metric space structure, one says that a path ${\displaystyle c:[a,b]\to M}$ is a unit-speed geodesic if for every ${\displaystyle t_{0}\in [a,b]}$ there exists an interval ${\displaystyle J\subset [a,b]}$ which contains ${\displaystyle t_{0}}$ and such that

${\displaystyle d_{g}(c(s),c(t))=|s-t|\qquad \forall s,t\in J.}$

Informally, one may say that one is asking for ${\displaystyle c}$ to locally 'stretch itself out' as much as it can, subject to the (informally considered) unit-speed constraint. The idea is that if ${\displaystyle c:[a,b]\to M}$ is (piecewise) continuously differentiable and ${\displaystyle |c'(t)|_{c(t)}=1}$ for all ${\displaystyle t,}$ then one automatically has ${\displaystyle d_{g}(c(s),c(t))\leq |s-t|}$ by applying the triangle inequality to a Riemann sum approximation of the integral defining the length of ${\displaystyle c.}$ So the unit-speed geodesic condition as given above is requiring ${\displaystyle c(s)}$ and ${\displaystyle c(t)}$ to be as far from one another as possible. The fact that we are only looking for curves to locally stretch themselves out is reflected by the first two examples given below; the global shape of ${\displaystyle (M,g)}$ may force even the most innocuous geodesics to bend back and intersect themselves.

• Consider the case that ${\displaystyle (M,g)}$ is the circle ${\displaystyle S^{1}}$ with its standard Riemannian metric, and ${\displaystyle c:\mathbb {R} \to S^{1}}$ is given by ${\displaystyle t\mapsto (\cos t,\sin t).}$ Recall that ${\displaystyle d_{g}}$ is measured by the lengths of curves along ${\displaystyle S^{1}}$, not by the straight-line paths in the plane. This example also exhibits the necessity of selecting out the subinterval ${\displaystyle J,}$ since the curve ${\displaystyle c}$ repeats back on itself in a particularly natural way.
• Likewise, if ${\displaystyle (M,g)}$ is the round sphere ${\displaystyle S^{2}}$ with its standard Riemannian metric, then a unit-speed path along an equatorial circle will be a geodesic. A unit-speed path along the other latitudinal circles will not be geodesic.
• Consider the case that ${\displaystyle (M,g)}$ is ${\displaystyle \mathbb {R} ^{2}}$ with its standard Riemannian metric. Then a unit-speed line such as ${\displaystyle t\mapsto (2^{-1/2}t,2^{-1/2}t)}$ is a geodesic but the curve ${\displaystyle c}$ from the first example above is not.

Note that unit-speed geodesics, as defined here, are by necessity continuous, and in fact Lipschitz, but they are not necessarily differentiable or piecewise differentiable.

As above, let ${\displaystyle (M,g)}$ be a connected and continuous Riemannian manifold. The Hopf–Rinow theorem, in this setting, says that (Gromov 1999)

• if the metric space ${\displaystyle (M,d_{g})}$ is complete (i.e. every ${\displaystyle d_{g}}$-Cauchy sequence converges) then
  • every closed and bounded subset of ${\displaystyle M}$ is compact.
  • given any ${\displaystyle p,q\in M}$ there is a unit-speed geodesic ${\displaystyle c:[a,b]\to M}$ from ${\displaystyle p}$ to ${\displaystyle q}$ such that ${\displaystyle d_{g}(c(s),c(t))=|s-t|}$ for all ${\displaystyle s,t\in [a,b].}$

The essence of the proof is that once the first half is established, one may directly apply the Arzelà–Ascoli theorem, in the context of the compact metric space ${\displaystyle {\overline {B_{2d_{g}(p,q)}(p)}},}$ to a sequence of piecewise continuously-differentiable unit-speed curves from ${\displaystyle p}$ to ${\displaystyle q}$ whose lengths approximate ${\displaystyle d_{g}(p,q).}$ The resulting subsequential limit is the desired geodesic.

The assumed completeness of ${\displaystyle (M,d_{g})}$ is important. For example, consider the case that ${\displaystyle (M,g)}$ is the punctured plane ${\displaystyle \mathbb {R} ^{2}\smallsetminus \{(0,0)\}}$ with its standard Riemannian metric, and one takes ${\displaystyle p=(1,0)}$ and ${\displaystyle q=(-1,0).}$ There is no unit-speed geodesic from one to the other.

Let ${\displaystyle (M,g)}$ be a connected and continuous Riemannian manifold. As with any metric space, one can define the diameter of ${\displaystyle (M,d_{g})}$ to be

${\displaystyle \operatorname {diam} (M,d_{g})=\sup\{d_{g}(p,q):p,q\in M\}.}$

The Hopf–Rinow theorem shows that if ${\displaystyle (M,d_{g})}$ is complete and has finite diameter, then it is compact. Conversely, if ${\displaystyle (M,d_{g})}$ is compact, then the function ${\displaystyle d_{g}:M\times M\to \mathbb {R} }$ has a maximum, since it is a continuous function on a compact metric space. This proves the following statement:

• If ${\displaystyle (M,d_{g})}$ is complete, then it is compact if and only if it has finite diameter.

This is not the case without the completeness assumption; for counterexamples one could consider any open bounded subset of a Euclidean space with the standard Riemannian metric.

Note that, more generally, and with the same one-line proof, every compact metric space has finite diameter. However the following statement is false: "If a metric space is complete and has finite diameter, then it is compact." For an example of a complete and non-compact metric space of finite diameter, consider

${\displaystyle M={\Big \{}{\text{continuous functions }}f:[0,1]\to \mathbb {R} {\text{ with }}\sup _{x\in [0,1]}|f(x)|\leq 1{\Big \}}}$

with the uniform metric

${\displaystyle d(f,g)=\sup _{x\in [0,1]}|f(x)-g(x)|.}$

So, although all of the terms in the above corollary of the Hopf–Rinow theorem involve only the metric space structure of ${\displaystyle (M,g),}$ it is important that the metric is induced from a Riemannian structure.

A Riemannian manifold M is geodesically complete if for all p ∈ M, the exponential map exp_p is defined for all v ∈ T_pM, i.e. if any geodesic γ(t) starting from p is defined for all values of the parameter t ∈ R. The Hopf–Rinow theorem asserts that M is geodesically complete if and only if it is complete as a metric space.

If M is complete, then M is non-extendable in the sense that it is not isometric to an open proper submanifold of any other Riemannian manifold. The converse is not true, however: there exist non-extendable manifolds that are not complete.

Riemannian metrics are defined in a way similar to the finite-dimensional case. However there is a distinction between two types of Riemannian metrics:

• A weak Riemannian metric on ${\displaystyle M}$ is a smooth function ${\displaystyle g:TM\times TM\to \mathbb {R} ,}$ such that for any ${\displaystyle x\in M}$ the restriction ${\displaystyle g_{x}:T_{x}M\times T_{x}M\to \mathbb {R} }$ is an inner product on ${\displaystyle T_{x}M.}$
• A strong Riemannian metric on ${\displaystyle M}$ is a weak Riemannian metric, such that ${\displaystyle g_{x}}$ induces the topology on ${\displaystyle T_{x}M.}$ Note that if ${\displaystyle M}$ is not a Hilbert manifold then ${\displaystyle g}$ cannot be a strong metric.

• If ${\displaystyle (H,\langle \,\cdot ,\cdot \,\rangle )}$ is a Hilbert space, then for any ${\displaystyle x\in H,}$ one can identify ${\displaystyle H}$ with ${\displaystyle T_{x}H.}$ By setting for all ${\displaystyle x,u,v\in H}$ ${\displaystyle g_{x}(u,v)=\langle u,v\rangle }$ one obtains a strong Riemannian metric.
• Let ${\displaystyle (M,g)}$ be a compact Riemannian manifold and denote by ${\displaystyle \operatorname {Diff} (M)}$ its diffeomorphism group. It is a smooth manifold (see here) and in fact, a Lie group. Its tangent bundle at the identity is the set of smooth vector fields on ${\displaystyle M.}$ Let ${\displaystyle \mu }$ be a volume form on ${\displaystyle M.}$ Then one can define ${\displaystyle G,}$ the ${\displaystyle L^{2}}$ weak Riemannian metric, on ${\displaystyle \operatorname {Diff} (M).}$ Let ${\displaystyle f\in \operatorname {Diff} (M),}$ ${\displaystyle u,v\in T_{f}\operatorname {Diff} (M).}$ Then for ${\displaystyle x\in M,u(x)\in T_{f(x)}M}$ and define ${\displaystyle G_{f}(u,v)=\int _{M}g_{f(x)}(u(x),v(x))d\mu (x).}$ The ${\displaystyle L^{2}}$ weak Riemannian metric on ${\displaystyle \operatorname {Diff} (M)}$ induces vanishing geodesic distance, see Michor and Mumford (2005).

Length of curves is defined in a way similar to the finite-dimensional case. The function ${\displaystyle d_{g}:M\times M\to [0,\infty )}$ is defined in the same manner and is called the geodesic distance. In the finite-dimensional case, the proof that this function is a metric uses the existence of a pre-compact open set around any point. In the infinite case, open sets are no longer pre-compact and so this statement may fail.

• If ${\displaystyle g}$ is a strong Riemannian metric on ${\displaystyle M}$, then ${\displaystyle d_{g}}$ separates points (hence is a metric) and induces the original topology.
• If ${\displaystyle g}$ is a weak Riemannian metric but not strong, ${\displaystyle d_{g}}$ may fail to separate points or even be degenerate.

For an example of the latter, see Valentino and Daniele (2019).

In the case of strong Riemannian metrics, a part of the finite-dimensional Hopf–Rinow still works.

Theorem: Let ${\displaystyle (M,g)}$ be a strong Riemannian manifold. Then metric completeness (in the metric ${\displaystyle d_{g}}$) implies geodesic completeness (geodesics exist for all time). Proof can be found in (Lang 1999, Chapter VII, Section 6). The other statements of the finite-dimensional case may fail. An example can be found here.

If ${\displaystyle g}$ is a weak Riemannian metric, then no notion of completeness implies the other in general.