nLab exponential map

Exponential maps

For the concept in category theory see at exponential object.





Exponential maps


The exponential function of classical analysis given by the series,

(1)expx i=0 x ii!, \exp x \coloneqq \sum_{i = 0}^{\infty} \frac{x^i}{i!} ,

is the solution of the differential equation

f=f f' = f

with initial value f(0)=1f(0) = 1.

It is also the limit of the following sequence of polynomial functions

f n(x)(1+xn) nf_n(x) \coloneqq \left(1 + \frac{x}{n}\right)^n
expxlim nf n(x)\exp x \coloneqq \lim_{n \to \infty} f_n(x)

This classical function is defined on the real line (or the complex plane). To generalise it to other manifolds, we need two things:

  • by its nature, the argument of the function should be a tangent vector; so in the classical function \mathbb{R} \to \mathbb{R}, the source \mathbb{R} is really the tangent space to the target \mathbb{R} at the point exp0=1\exp 0 = 1.
  • We need a covariant derivative to tell us what ff' means.

So in the end we have, for any point pp on a differentiable manifold MM with an affine connection \Del, a map exp p:T pMM\exp_p\colon T_p M \to M, which is defined at least on a neighbourhood of 00 in the tangent space T pMT_p M.

Note that pp here comes from the initial value exp p0=p\exp_p 0 = p; we usually take p=1p = 1 when we work in a Lie group, but otherwise we are really generalising the classical exponential function xpexpxx \mapsto p \exp x; every solution to f=ff' = f takes this form.

Classically, there are some other functions called ‘exponential’; given any nonzero real (or complex) number bb, the map xb xx \mapsto b^x (or even xpb xx \mapsto p\, b^x) is also an exponential map. Using the natural logarithm, we can define b xb^x in terms of the natural exponential map exp\exp:

b xexp(xlnb). b^x \coloneqq \exp (x \ln b) .

So while bb is traditionally called the ‘base’, it is really the number lnb\ln b that matters, or even better the operation of multiplication by lnb\ln b. This operation is an endomorphism of the real line (or complex plane), and every such endomorphism takes this form for some nonzero bb (and some branch of the natural logarithm, in the complex case). So we see that this generalised exponential map is simply the composite of the natural exponential map after a linear endomorphism.


See also at flow of a vector field.

With respect to an affine connection

Let MM be a differentiable manifold, let \Del be an affine connection on MM, and let pp be a point in MM. Then by the general theory of differential equations, there is a unique maximally defined partial function exp p\exp_p from the tangent space T pMT_p M to MM such that:

  • exp p=exp p\Del \exp_p = \exp_p and
  • exp p(0)=1\exp_p(0) = 1.

This function is the natural exponential map on MM at pp relative to \Del. We have exp p:UM\exp_p\colon U \to M, where UU is some neighbourhood of 00 in T pMT_p M. If MM is complete? (relative to \Del), then UU will be all of T pMT_p M.

Let MM be a Riemannian manifold (or a pseudo-Riemannian manifold) and let pp be a point in MM. Then MM may be equipped with the Levi-Civita connection lc\Del_{lc}, so we define the natural Riemannian exponential map on MM at pp to be the natural exponential map on MM at pp relative to lc\Del_{lc}.

Given any endomorphism ϕ:T pMT pM\phi\colon T_p M \to T_p M, we can also consider the exponential map on MM at pp relative to \Del with logarithmic base ϕ\phi, which is simply xexp pϕ(x)x \mapsto \exp_p \phi(x). We say ‘logarithmic base’ since a classical exponential function with base bb corresponds to an exponential function whose logarithmic base is multiplication by lnb\ln b.

Via geodesics

Recall that a geodesic is a curve on a manifold whose velocity is constant (as measured along that curve relative to a given affine connection). Working naïvely, we may write

γ=v, \gamma' = v ,

pretend that this is a differential equation for a function γ:\gamma\colon \mathbb{R} \to \mathbb{R}, and take the solution

γ(t)=pexptx, \gamma(t) = p \exp t x ,

where pp is given by the initial value γ(0)=p\gamma(0) = p. We recognise this as being, morally, exp ptx\exp_p t x. This suggests (although we need more work for a proof) the following result:

Let MM be a differentiable manifold, let \Del be an affine connection on MM, and let pp be a point in MM. Given a tangent vector xx at pp, there is a unique maximal geodesic γ\gamma on MM tangent to xx at pp. If γ(1)\gamma(1) is defined (which it will be whenever MM is complete? and may be in any case), we have exp px=γ(1)exp_p x = \gamma(1). In any case, we have exp p(tx)=γ(t)\exp_p (t x) = \gamma(t) for sufficiently small tt.

In real algebras

Let \mathbb{R} be any sequentially Cauchy complete Archimedean ordered field, and let F:AlgVectF:\mathbb{R}\mathrm{Alg} \to \mathbb{R}\mathrm{Vect} be the forgetful functor which takes an \mathbb{R}-algebra AA to its underlying \mathbb{R}-vector space F(A)F(A). Given any \mathbb{R}-algebra AA whose underlying \mathbb{R}-vector space F(A)F(A) is finite-dimensional, each element of AA could be expressed as a linear combination, a finite sum of basis vectors. Then one could define the exponential function exp:AA\exp:A \to A as either

exp(x)lim n(1+xn) n\exp(x) \coloneqq \lim_{n \to \infty} \left(1 + \frac{x}{n}\right)^n


exp(x)lim n i=0 nx ii!\exp(x) \coloneqq \lim_{n \to \infty} \sum_{i = 0}^{n} \frac{x^i}{i!}

since according to the algebraic limit theorem, limits distribute over finite sums.

In particular, this is how exponential functions are defined in Clifford algebras and matrix algebras. However, the exponential functions in non-commutative algebras are not abelian group homomorphisms, because multiplication is not commutative while addition is commutative.

In the dual numbers

The dual number real algebra 𝔻[ϵ]/ϵ 2\mathbb{D} \coloneqq \mathbb{R}[\epsilon]/\epsilon^2 has a notion of exponential function, which is the solution to the functional equation exp(x+ϵ)=exp(x)(1+ϵ)\exp(x + \epsilon) = \exp(x) (1 + \epsilon) with exp(0)=1\exp(0) = 1.

 In arbitrary Archimedean ordered fields

In general, Archimedean ordered fields which are not sequentially Cauchy complete do not have an exponential function. Nevertheless, the exponential map is still guaranteed to be a partial function, because every Archimedean ordered field is a Hausdorff space and thus a sequentially Hausdorff space. Thus, an axiom could be added to an Archimedean ordered field FF to ensure that the exponential partial function is actually a total function:

Axiom of exponential function: For all elements xFx \in F, there exists a unique element exp(x)F\exp(x) \in F such that for all positive elements ϵF +\epsilon \in F_+, there exists a natural number NN \in \mathbb{N} such that for all natural numbers nn \in \mathbb{N}, if nNn \geq N, then ϵ<(1+xn) nexp(x)<ϵ-\epsilon \lt \left(1 + \frac{x}{n}\right)^{n} - \exp(x) \lt \epsilon (or equivalently, ϵ<( i=0 nx ii!)exp(x)<ϵ-\epsilon \lt \left(\sum_{i = 0}^{n} \frac{x^i}{i!}\right) - \exp(x) \lt \epsilon).

There is another axiom which uses the fact that derivatives of functions are well defined in the ordered local Artin F F -algebra F[ϵ]/ϵ 2F[\epsilon]/\epsilon^2 by the equation f(x+ϵ)=f(x)+f(x)ϵf(x + \epsilon) = f(x) + f'(x) \epsilon:

Axiom of exponential function: Let F[ϵ]/ϵ 2F[\epsilon]/\epsilon^2 be the ordered local Artin FF-algebra, with non-zero non-positive non-negative nilpotent element ϵF[ϵ]/ϵ 2\epsilon \in F[\epsilon]/\epsilon^2 where ϵ 2=0\epsilon^2 = 0 and canonical FF-algebra homomorphism h:FF[x]/x 2h:F \to F[x]/x^2. There exists a unique function exp:FF\exp:F \to F and a function exp:F[ϵ]/ϵ 2F[ϵ]/ϵ 2\exp':F[\epsilon]/\epsilon^2 \to F[\epsilon]/\epsilon^2 such that for every element xFx \in F, h(exp(x))=exp(h(x))h(\exp(x)) = \exp'(h(x)), exp(x+ϵ)=exp(x)+exp(x)ϵ\exp'(x + \epsilon) = \exp'(x) + \exp'(x) \epsilon, and exp(0)=1\exp'(0) = 1.

In constructive mathematics

In classical mathematics, one could prove that the modulated Cantor real numbers C\mathbb{R}_C are sequentially Cauchy complete and equivalent to the HoTT book real numbers H\mathbb{R}_H. However, in constructive mathematics, the above cannot be proven; while the HoTT book real numbers H\mathbb{R}_H are still sequentially Cauchy complete, the modulated Cantor real numbers C\mathbb{R}_C in general cannot be proven to be sequentially Cauchy complete. In particular, this means that the sequences

(1+xn) nor i=0 nx ii!\left(1 + \frac{x}{n}\right)^{n} \quad \mathrm{or} \quad \sum_{i = 0}^{n} \frac{x^i}{i!}

do not have a limit for all modulated Cantor real numbers x Cx \in \mathbb{R}_C. However, the sequences, by definition of C\mathbb{R}_C, do have a limit for all rational numbers xx \in \mathbb{Q}; this means that one could restrict the domain of the exponential function to the rational numbers exp: C\exp:\mathbb{Q} \to \mathbb{R}_C, and define it in the usual manner:

  • For all rational numbers xx \in \mathbb{Q}, there exists a unique modulated Cantor real number exp(x) C\exp(x) \in \mathbb{R}_C such that for all positive rational numbers ϵ +\epsilon \in \mathbb{Q}_+, there exists a natural number NN \in \mathbb{N} such that for all natural numbers nn \in \mathbb{N}, if nNn \geq N, then ϵ<(1+xn) nexp(x)<ϵ-\epsilon \lt \left(1 + \frac{x}{n}\right)^{n} - \exp(x) \lt \epsilon (or equivalently, ϵ<( i=0 nx ii!)exp(x)<ϵ-\epsilon \lt \left(\sum_{i = 0}^{n} \frac{x^i}{i!}\right) - \exp(x) \lt \epsilon).

In Lie groups

Note: this section is under repair.

The classical exponential function exp: *\exp \colon \mathbb{R} \to \mathbb{R}^* or exp: *\exp \colon \mathbb{C} \to \mathbb{C}^* satisfies the fundamental property:


The function exp: *\exp \colon \mathbb{C} \to \mathbb{C}^* is a homomorphism taking addition to multiplication:

exp(x+y)=exp(x)exp(y)\exp(x + y) = \exp(x) \cdot \exp(y)

A number of proofs may be given. One rests on the combinatorial binomial identity

(x+y) n= j+k=nn!j!k!x jy k(x + y)^n = \sum_{j + k = n} \frac{n!}{j! k!} x^j y^k

(which crucially depends on the fact that multiplication is commutative), whereupon

n0(x+y) nn! = n0 j+k=n1j!1k!x jy k = ( j0x jj!)( k0y kk!) = exp(x)exp(y)\array{ \sum_{n \geq 0} \frac{(x+y)^n}{n!} & = & \sum_{n \geq 0} \sum_{j + k = n} \frac1{j!} \frac1{k!} x^j y^k \\ & = & (\sum_{j \geq 0} \frac{x^j}{j!}) \cdot (\sum_{k \geq 0} \frac{y^k}{k!}) \\ & = & \exp(x) \cdot \exp(y) }

An alternative proof begins with the observation that f=expf = \exp is the solution to the system f=ff' = f, f(0)=1f(0) = 1. For each yy, the function g 1:xf(x)f(y)g_1 \colon x \mapsto f(x) f(y) is a solution to the system g=gg' = g, g(0)=f(y)g(0) = f(y), as is the function g 2:xf(x+y)g_2 \colon x \mapsto f(x + y). Then by uniqueness of solutions to ordinary differential equations (over connected domains; see, e.g., here), g 1=g 2g_1 = g_2, i.e., f(x+y)=f(x)f(y)f(x + y) = f(x)f(y) for all x,yx, y.1

Let MM be Lie group and let 𝔤\mathfrak{g} be its Lie algebra T 1MT_1 M, the tangent space to the identity element 11. Then MM may be equipped with the canonical left-invariant connection l\Del_l or the canonical right-invariant connection r\Del_r. It turns out that the natural Riemannian exponential maps on MM at 11 relative to l\Del_l and r\Del_r are the same; we define this to be the natural Lie exponential map on MM at the identity, denoted simply exp\exp. Several nice properties follow:

  • exp\exp is defined on all of 𝔤\mathfrak{g}.
  • exp:𝔤G\exp \colon \mathfrak{g} \to G is a smooth map.
  • If ρ:𝔤\rho \colon \mathbb{R} \to \mathfrak{g} is a smooth homomorphism from the additive group \mathbb{R} (i.e., if ρ\rho is an \mathbb{R}-linear map, uniquely determined by specifying X=ρ(1)X = \rho(1)), then expρ X:G\exp \circ \rho_X \colon \mathbb{R} \to G is a smooth homomorphism.
  • For X,Y𝔤X, Y \in \mathfrak{g}, if [X,Y]=0[X, Y] = 0, then the restriction of exp:𝔤G\exp \colon \mathfrak{g} \to G to the subspace spanned by XX and YY is a smooth homomorphism to GG. In particular, exp:𝔤G\exp \colon \mathfrak{g} \to G is a homomorphism if 𝔤\mathfrak{g} is abelian (e.g., if GG is a commutative Lie group).
  • exp\exp is surjective (a regular epimorphism) if GG is connected and compact (and also in some other situations, such as the classical cases where GG is ]0,[]0,\infty[ or {0}\mathbb{C} \setminus \{0\}). See this post by Terence Tao, Proposition 1; see also the first comment which indicates an alternative proof based on the fact that maximal tori in GG are all conjugate to one another. Note also that the exponential map might not be surjective if the compactness assumption is dropped, as in the case of G=SL 2()G = SL_2(\mathbb{R}) or SL 2()SL_2(\mathbb{C}), both of which are connected; see here for instance.
  • If GG is compact, then it may be equipped with a Riemannian metric that is both left and right invariant (see Tao’s post linked in the previous remark); then the Lie exponential map is the same as the Riemannian exponential map at 11.
  • If GG is a matrix Lie group, then exp\exp is given by the classical series formula (1).

(to be expanded on)


A logarithm is a local section of an exponential map.


Discussion in constructive analysis of the exponential function on real numbers:

See also:

An extensive treatment for the general exponential map for an affine connection, for exponential map for Riemannian manifolds and the one for Lie groups is

  • Sigurdur Helgason, Differential geometry, Lie groups and symmetric spaces

Specifically for Lie groups, a different detailed treatment of the exponential map is in

Some nice historical notes are in

  • Wilfried Schmid, Poincare and Lie groups, Bull. Amer. Math. Soc. 6:2, 1982 pdf

Discussion in point-free topology:

  1. A previous edit offered even more detail: “An alternative proof begins with the premise that each solution of the ordinary differential equation g=0g' = 0 is locally constant. Suppose cc is a complex number. As exp=exp\exp' = \exp, we find that (exp(x)exp(cx))=exp(x)exp(cx)+exp(x)(exp(cx))=0(\exp(x) \exp(c - x))' = \exp(x) \exp(c - x) + \exp(x) (-\exp(c-x)) = 0. Hence, by the premise and the connectedness of the domain of exp\exp (either {\mathbb{R}} or {\mathbb{C}}), we obtain exp(x)exp(cx)=exp(0)exp(c)\exp(x)\exp(c - x) = \exp(0)\exp(c). The initial condition exp(0)=1\exp(0) = 1 then yields exp(x)exp(cx)=exp(c)\exp(x)\exp(c - x) = \exp(c). The result follows by setting c=x+yc = x + y.”

Last revised on December 12, 2023 at 09:58:51. See the history of this page for a list of all contributions to it.