logarithm

Classically, a logarithm is a partially-defined smooth homomorphism from a multiplicative group of numbers to an additive group of numbers. As such, it is a local section of an exponential map. As exponential maps can be generalised to Lie groups, so can logarithms.

Consider the field of real numbers; these numbers form a Lie group under addition (which we will call simply $\mathbb{R}$), while the nonzero numbers form a Lie group under multiplication (which we will call $\mathbb{R}^*$). The multiplicative group has two connected components; we will focus attention on the identity component (which we will call $\mathbb{R}^+$), consisting of the positive numbers.

The Lie groups $\mathbb{R}$ and $\mathbb{R}^+$ are in fact isomorphic. In fact, there is one isomorphism for each positive real number $b$ other than $1$; this number $b$ is called the **base**. Fixing a base, the map from $\mathbb{R}^+$ to $\mathbb{R}$ is called the **real logarithm with base $b$**, written $x \mapsto \log_b x$; the map from $\mathbb{R}$ to $\mathbb{R}^+$ is the **real exponential map with base $b$**, written $x \mapsto b^x$.

The real logarithms are handily defined using the Riemann integral of the reciprocal as follows:

(1)$\array {
\ln x & \coloneqq \int_1^x \frac{1}{t} \,\mathrm{d}t ;\\
\log_b x & \coloneqq \frac{\ln x}{\ln b} .\\
}$

Note that $\ln$ is itself a logarithm, the **natural logarithm**, whose base is $\mathrm{e} = 2.71828182845\ldots$. (The exponential map may similarly be defined as an infinite series, but I'll leave that for its own article.)

Now consider the field of complex numbers; these also form a Lie group under addition (which we call $\mathbb{C}$), while the nonzero numbers form a Lie group under multiplication (which we call $\mathbb{C}^*$). Now the multiplicative group is connected, so we would like to use all of it.

However, $\mathbb{C}$ and $\mathbb{C}^*$ are *not* isomorphic. Indeed, the multiplication map

$\mathbb{R}^* \times S^1 \to \mathbb{C}^*$

exhibits $\mathbb{C}^*$ as a biproduct of $\mathbb{R}^*$ and the circle group $S^1$, so that homomorphisms $\mathbb{C}^* \to \mathbb{C}$ are given by pairs of homomorphisms $f \colon \mathbb{R}^* \to \mathbb{C}$, $g \colon S^1 \to \mathbb{C}$. But every homomorphisms $g \colon S^1 \to \mathbb{C}$ is trivial: the restriction of $g$ to the torsion subgroup of $S^1$ is trivial since $\mathbb{C}$ is torsionfree, and since the torsion subgroup is dense in $S^1$, any Lie group homomorphism $S^1 \to \mathbb{C}$ must also be trivial. Therefore, every homomorphism $h \colon \mathbb{C}^* \to \mathbb{C}$ factors through the projection $\mathbb{C}^* \to \mathbb{R}^*$. It quickly follows that no such $h$ can be injective, nor can such $h$ be surjective.

Taking advantage of biproduct representations $\mathbb{C} \cong \mathbb{R} \oplus \mathbb{R}$ and $\mathbb{C}^* \cong \mathbb{R}^* \oplus S^1$, we can classify homomorphisms from $\mathbb{C}$ to $\mathbb{C}^*$. Each is given by a 4-tuple of real numbers $(a, b, c, d)$:

$\phi_{a, b, c, d}(x + i y) = e^{a x} e^{i b x} e^{c y} e^{i d y}.$

The cases where $a = d$, $b = -c$ correspond to those homomorphisms that are holomorphic functions (i.e., that satisfy the Cauchy-Riemann equations). Putting $w = a + b i$, we have

$\phi_{a, b, -b, a}(z) = e^{w z}$

with one such homomorphism for each complex number $w$, and these homomorphisms are surjections whenever $w \ne 0$. (N.B.: these homomorphisms are not uniquely determined by their values at $z = 1$, since we have $e^w = e^{w'}$ whenever $w - w'$ is an integer multiple of $2 \pi i$, and yet the homomorphisms $z \mapsto e^{w z}$ and $z \mapsto e^{w' z}$ will be different unless $w = w'$.)

So we have these surjections (the **complex exponential map** $z \mapsto e^{w z}$, for $w \ne 0$), which are regular epimorphisms but not split epimorphisms. However, while they have no sections (being not split), they have quite a few local sections, and the domains of the maximal local sections are precisely the connected simply connected open dense subspaces $R$ of $\mathbb{C}^*$. A **complex logarithm with exponential base $w$ on $R$** is this $R$-defined section of the complex exponential map $z \mapsto e^{w z}$. Supposing $R$ given, we denote this by $\log_{[w]}$ (but please note that in the context of real logarithms, this would ordinarily be denoted $\log_b$ where $b = e^w$).

If $1 \in R$, then a complex natural logarithm on $R$ may be defined using the contour integral with the same formula (1) as for the real natural logarithm. We merely insist that the integral be done along a contour within the region $R$. (Since $R$ is connected, there is such a contour; since $R$ is simply connected and $t \mapsto 1/t$ is holomorphic, the result is unique.) Note that if $x \in \mathbb{R}^+ \subseteq R$, then the real and complex natural logarithms of $x$ will be equal.

The natural exponential map is periodic? (with period $2 \pi \mathrm{i}$), and it is possible to add any multiple of this period to the natural logarithm of any $x \ne 1$ by suitably changing the region $R$. We then obtain the most general notion of maximally-defined complex logarithm with any base by using the formulas

$\array {
\ln x & \coloneqq C + \int_{\mathrm{e}^C}^x \frac{1}{t} \,\mathrm{d}t,\\
\log_{[w]} x & \coloneqq \frac{\ln x}{w} .\\
}$

In the classical examples, the multiplicative groups $\mathbb{R}^+$ and $\mathbb{C}^*$ both Lie groups. The additive groups $\mathbb{R}$ and $\mathbb{C}$ are also Lie groups, but they are more than this: they are Lie algebras. (The additive group of a Lie algebra is always a Lie group. Actually, since these are abelian Lie algebras, their Lie-algebra structure is easy to miss, but of course they are vector spaces.) And what's more, each additive group is *the* Lie algebra of the corresponding Lie group.

This generalises. Given any Lie group $G$, let $\mathfrak{g}$ be its Lie algebra. Then we have an exponential map $\exp\colon \mathfrak{g} \to G$, which is surjective under certain conditions (most famously when $G$ is connected and compact, but also in the classical cases, even though $G$ is not compact). More generally, given any automorphism $\phi$ of $\mathfrak{g}$, we have a map $x \mapsto \exp(\phi(x))$, which is a homomorphism of Lie groups. Any local section of this map may be called a **logarithm base $\phi$** on $G$ (denoted $\log_{[\phi]}$ with the bracket as in the previous section); any local section of $\exp$ itself may be called a **natural logarithm** on $G$.

**(Mercator series)**

The Taylor series of the natural logarithm around $1 \in \mathbb{R}$ is the following series:

(2)$\begin{aligned}
\underoverset{n = 0}{\infty}{\sum}
\tfrac{1}{n!}
\left(
\frac{d^n}{ d x^n} ln(1 + x)
\right)_{\vert x = 0}
x^n
&
\;\;
=
\;\;
\underoverset{n = 1}{\infty}{\sum}
\frac
{(-1)^{n+1}}
{n}
x^n
\\
&
\;\;
=
\;\;
x
- \tfrac{1}{2} x^2
+ \tfrac{1}{3} x^3
- \tfrac{1}{4} x^4
+ \cdots
\,.
\end{aligned}$

For the first two terms notice that

$ln(1 + x)
\;\xrightarrow{x \to 0}\;
ln(1)
\,=\,
0$

and that the derivative of the natural logarithm is:

$\frac{d}{d x} \ln(1 + x)
\;=\;
\tfrac{1}{1+x}
\;\xrightarrow{ x \to 0 }\;
1
\,.$

From here on, noticing for $k \in \mathbb{N}_+$ that:

$\frac{d}{d x}
\left(
\frac{1}{(1 + x)^k}
\right)
\;=\;
- k
\frac{1}{(1 + x)^{k+1}}
\;\xrightarrow{x \to 0}\;
- k$

we get, for $n \in \mathbb{N}_+$:

$\begin{aligned}
\frac{d^n}{d x^n}
ln(1 + x)
&
\;=\;
\frac{d^{n-1}}{d x^{n-1}}
\left(
\frac{1}{1 + x}
\right)
\\
&
\;=\;
(n-1)! \cdot (-1)^{n-1}
\frac{1}{(1 + x)^{n-1}}
\\
&
\;\xrightarrow{ x \to 0 }\;
(n-1)! \cdot (-1)^{n+1}
\end{aligned}
\,.$

Plugging this into the defining equation on the left of (2) and using

$\frac{(n-1)!}{n!} = \frac{1}{n}$

yields the claim.

Historical textbooks:

- Nicholas Mercator,
*Logarithmotechnia: Sive Methodus constuendi Logrithmos*, London 1667 (GoogleBooks)

(…)

Last revised on July 25, 2021 at 03:08:33. See the history of this page for a list of all contributions to it.