nLab exponential map

For the concept in category theory see at exponential object.

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Idea

The exponential function of classical analysis given by the series,

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

is the solution of the differential equation

$$f' = f$$

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

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

$$f_n(x) \coloneqq \left(1 + \frac{x}{n}\right)^n$$

$$\exp x \coloneqq \lim_{n \to \infty} f_n(x)$$

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 $\exp 0 = 1$.
• We need a covariant derivative to tell us what $f'$ means.

So in the end we have, for any point $p$ on a differentiable manifold $M$ with an affine connection $\Del$, a map $\exp_p\colon T_p M \to M$, which is defined at least on a neighbourhood of $0$ in the tangent space $T_p M$.

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

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

$$b^x \coloneqq \exp (x \ln b) .$$

So while $b$ is traditionally called the ‘base’, it is really the number $\ln b$ that matters, or even better the operation of multiplication by $\ln b$. This operation is an endomorphism of the real line (or complex plane), and every such endomorphism takes this form for some nonzero $b$ (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.

Definition

See also at flow of a vector field.

With respect to an affine connection

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

• $\Del \exp_p = \exp_p$ and
• $\exp_p(0) = 1$.

This function is the natural exponential map on $M$ at $p$ relative to $\Del$. We have $\exp_p\colon U \to M$, where $U$ is some neighbourhood of $0$ in $T_p M$. If $M$ is complete? (relative to $\Del$), then $U$ will be all of $T_p M$.

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

Given any endomorphism $\phi\colon T_p M \to T_p M$, we can also consider the exponential map on $M$ at $p$ relative to $\Del$ with logarithmic base $\phi$, which is simply $x \mapsto \exp_p \phi(x)$. We say ‘logarithmic base’ since a classical exponential function with base $b$ corresponds to an exponential function whose logarithmic base is multiplication by $\ln b$.

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

$$\gamma' = v ,$$

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

$$\gamma(t) = p \exp t x ,$$

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

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

In real algebras

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

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

or

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

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

 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 $F$ to ensure that the exponential partial function is actually a total function:

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

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

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

In constructive mathematics

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

$$\left(1 + \frac{x}{n}\right)^{n} \quad \mathrm{or} \quad \sum_{i = 0}^{n} \frac{x^i}{i!}$$

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

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

In Lie groups

Note: this section is under repair.

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

Proposition

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

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

Proof

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

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

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

$$\array{ \sum_{n \geq 0} \frac{(x+y)^n}{n!} & = & \sum_{n \geq 0} \sum_{j + k = n} \frac1{j!} \frac1{k!} x^j y^k \\ & = & (\sum_{j \geq 0} \frac{x^j}{j!}) \cdot (\sum_{k \geq 0} \frac{y^k}{k!}) \\ & = & \exp(x) \cdot \exp(y) }$$

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

Let $M$ be Lie group and let $\mathfrak{g}$ be its Lie algebra $T_1 M$, the tangent space to the identity element $1$. Then $M$ may be equipped with the canonical left-invariant connection $\Del_l$ or the canonical right-invariant connection $\Del_r$. It turns out that the natural Riemannian exponential maps on $M$ at $1$ relative to $\Del_l$ and $\Del_r$ are the same; we define this to be the natural Lie exponential map on $M$ at the identity, denoted simply $\exp$. Several nice properties follow:

• $\exp$ is defined on all of $\mathfrak{g}$.
• $\exp \colon \mathfrak{g} \to G$ is a smooth map.
• If $\rho \colon \mathbb{R} \to \mathfrak{g}$ is a smooth homomorphism from the additive group $\mathbb{R}$ (i.e., if $\rho$ is an $\mathbb{R}$-linear map, uniquely determined by specifying $X = \rho(1)$), then $\exp \circ \rho_X \colon \mathbb{R} \to G$ is a smooth homomorphism.
• For $X, Y \in \mathfrak{g}$, if $[X, Y] = 0$, then the restriction of $\exp \colon \mathfrak{g} \to G$ to the subspace spanned by $X$ and $Y$ is a smooth homomorphism to $G$. In particular, $\exp \colon \mathfrak{g} \to G$ is a homomorphism if $\mathfrak{g}$ is abelian (e.g., if $G$ is a commutative Lie group).
• $\exp$ is surjective (a regular epimorphism) if $G$ is connected and compact (and also in some other situations, such as the classical cases where $G$ is $]0,\infty[$ or $\mathbb{C} \setminus \{0\}$). 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 $G$ 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 $G = SL_2(\mathbb{R})$ or $SL_2(\mathbb{C})$, both of which are connected; see here for instance.
• If $G$ 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 $1$.
• If $G$ is a matrix Lie group, then $\exp$ is given by the classical series formula (1).

(to be expanded on)

Logarithms

A logarithm is a local section of an exponential map.

Related concepts

• pane wave?

• flow of a vector field

• Euler number, e

• logarithm

• Euler's formula

• exponential exact sequence

• Hamiltonian flow

• exponential modality

• exponential ring

• one-parameter semigroup

References

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

• Errett Bishop, §7 in: Foundations of Constructive Analysis, McGraw-Hill (1967)

• Errett Bishop, Douglas Bridges §7 in: Constructive Analysis, Grundlehren der mathematischen Wissenschaften 279, Springer (1985) [doi:10.1007/978-3-642-61667-9]

See also:

• Wikipedia, exponential map (Lie theory), derivative of the exponential map, exponential map (Riemannian geometry))

• Springer eom: exponential mapping

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

• M M Postnikov, Lie groups and Lie algebras, Geometry V

Some nice historical notes are in

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

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