For the concept in category theory see at exponential object.
transfinite arithmetic, cardinal arithmetic, ordinal arithmetic
prime field, p-adic integer, p-adic rational number, p-adic complex number
arithmetic geometry, function field analogy
The exponential function of classical analysis given by the series,
is the solution of the differential equation
with initial value $f(0) = 1$.
This classical function is defined on the real line (or the complex plane). To generalise it to other manifolds, we need two things:
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 $\exp$:
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.
See also at flow of a vector field.
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:
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 $\ln b$.
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
pretend that this is a differential equation for a function $\gamma\colon \mathbb{R} \to \mathbb{R}$, and take the solution
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 $t$.
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:
The function $\exp \colon \mathbb{C} \to \mathbb{C}^*$ is a homomorphism taking addition to multiplication:
A number of proofs may be given. One rests on the combinatorial binomial identity
(which crucially depends on the fact that multiplication is commutative), whereupon
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:
(to be expanded on)
A logarithm is a local section of an exponential map.
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
Specifically for Lie groups, a different detailed treatment of the exponential map is in
Some nice historical notes are in
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 $c = x + y$.” ↩