Logarithms

# Logarithms

## Idea

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.

## Definitions

### Logarithms of real numbers

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.)

### Logarithms of complex numbers

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

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

### Logarithms and Lie groups

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 $G$.

## Properties

###### Proposition

(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}

###### Proof

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:

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