An observation more than a question.

Take any function $f$ (additional hypotheses may follow) and evaluate it on any two points $x_1, x_2\in\mathcal{D}_f$. Define then the sum of these two points as the value the function takes on the sum of the inverse values, namely $$ x_1 +_f x_2 = f\left(f^{-1}(x_1)+f^{-1}x_2\right)\equiv F\left(f^{-1}(x_1),f^{-1}(x_2)\right) $$ where essentially the idea is to rewrite the right hand side in terms of each of the singular values $f^{-1}(x_i)$. This has been triggered by the following explicit example with $f=\tanh$, which gives back the relativistic composition of velocities $$ x_1+_f x_2=\tanh\left(\tanh^{-1}(x_1)+\tanh^{-1}(x_2)\right)=\frac{x_1+x_2}{1+x_1\cdot x_2} $$ which reads nothing but ($c=1$) $$ u+_fv=\frac{u+v}{1+uv} $$ Question is: have you ever come across any other example where the above can be useful and generalised?

EDIT: I have adjusted the notation to make it more precise following the comments below.

  • 1
    $\begingroup$ As an aside the paper "Nothing But Relativity" linked in the earlier thread, shows that the only other physically meaningful composition rule for velocities in the boring Galilean one. $\endgroup$ Jul 7 '15 at 22:08
  • $\begingroup$ Two unclear things: $f$ is a function of one argument, what is the notation $f(f(\cdot),f(\cdot))$? In addition, $x_1 +_f x_2$ does not imply the formula for $u$, $\nu$ below, apart from objects $z$ such that $z=\tanh z$ (I am not sure there are). $\endgroup$
    – yuggib
    Jul 7 '15 at 22:14
  • $\begingroup$ @yuggib Ooops, that was a typo, I edited it. I want to express the rhs in terms of the $f(x_1),f(x_2)$ only. The formula is implied once you consider $u,v$ not as $x_1,x_2$ but rather as their values $f(x_1),f(x_2)$: basically the idea is to sum the values of the function instead of just summing the variables and see what the composition law is. $\endgroup$
    – gented
    Jul 7 '15 at 22:21
  • $\begingroup$ The formula $u+_f v$ continues to seem wrong to me (but maybe it is just the late hour): if in the formula above you substitute on the right hand side $\tanh x_1$ by $u$ and the other by $v$, you get the right hand side below; however the left hand side above then reads $\tanh^{-1} u +_f \tanh^{-1} v$ and not $u +_f v$. And unrelated, in general I think it is not always possible to write $f(x+y)= F(f(x_1),f(x_2))$. $\endgroup$
    – yuggib
    Jul 7 '15 at 22:34
  • $\begingroup$ You're right: I have to adjust the notation. And yes, it's not always possible to write $f(x+y)=F(f(x),f(y))$ which causes more obstructions. $\endgroup$
    – gented
    Jul 7 '15 at 22:49

Another important example: the Pythagorean Sum defined by $f(x) = \sqrt{x}$.

As Doetoe says, you're simply transporting the binary operation + to the domain of $f$ if $f$ is e.g. bijective. If your domain is a real interval, your question actually describes the following:

Which operation is calculated by the addition of affine co-ordinates implemented by the slide action of a slide rule with a scale defined by $f$?

In this context the function $f(x)$ defines what number sits at a distance $x$ along the scale and $f^{-1}(y)$ defines the distance along the scale from the scale's beginning to the point where $y$ lies on the scale. So in the basic slide rule scales, we have $f = \exp$ and then the main (ABCD scales) are a physical embodiment of the isomorphism between the multiplicative Lie group of positive reals and the additive real Lie group: $f$ is here the isomorphism $\exp:\mathbb{R}\to\mathbb{R}^+$ so that $x\,y = x +_{\exp} y = \exp(\exp^{-1}(x) + \exp^{-1}(y))$. Doetoe's example of the Campbell Baker Hausdorff formula generalizes the $\exp$ notion to noncommutative Lie groups.

The only other operation to my knowledge that was done on real slide rules that makes use of your pattern was the Pythagorean sum, where we put $f(x) = \sqrt{x}$. The $P$ and $Q$ scales of some of the Hemmi-Sun Japanese slide rules encoded this pattern. The number $x^2$ lies at distance $x$ along the $P$. You then bring the zero on the $Q$ scale to this point, and the number $y^2$ lies at a distance $y$ along this scale. When you add the two and read back on the $P$ scale, you're looking at the number $z$ such that $z^2 = x^2 + y^2$: so the $P$ and $Q$ scales were used to work out the hypotenuse length from the two other sides of a right angled triangle. Useful for trigonometry and electrical calculations, which these particular rules were designed for. So Pythagoras's theorem is an important example of your pattern.

There is a virtual slide rule with these $P$ and $Q$ scales at:

Mark Armburst, Virtual Hemmi 153 Slide Rule at Mark's Math

To see the $P$ and $Q$ scales you need to push the "Flip to Other Side" button.

To my knowledge, no slide rule gave the calculation with $f=\tanh$, which would have been useful for special relativistic calculations! Albert Einstein was known to have used a Nestler 23R rule, which was very basic.


Let $X$ be a set with a binary operation $\star:X\times X\to X$, which could be a group structure or whatever you want. Let $f:X\to Y$ be a mapping to some other set and let $F(x,y) := f(x\star y)$ (your middle expression is this one with the operation +). If $f$ is sufficiently nice, e.g. a bijection, then your construction transports the binary operation from $X$ to $Y$.

In your example, for a given velocity $v\in(-1,1)$ we have a boost $\Lambda(v)\in O(1,1)$, giving us the bijection $f = \Lambda^{-1}$:

$$\Lambda^{-1}: O(1,1)\to(-1,1)$$

The group structure of $O(1,1)$ is transported to the interval $(-1,1)$ by this map:

$$u\star v = \Lambda^{-1}(\Lambda(u)\Lambda(v)) = \frac{u+v}{1 + uv}$$

When both already come with their binary structure so that $f$ is an isomorphism we get the most trivial case, in which you can use $f$ to choose to use the group operation in the group (or whatever it is) it is easiest in, this is often done to turn multiplication into addition through $\exp$ and $\log$.

A less trivial example of some interest in physics is given by the Baker-Campbell-Hausdorff formula, in which $f = \exp$, the exponential map $\mathfrak g\to G$ of a Lie algebra to the Lie group it is the tangent space of (at the identity). Strictly speaking this would correspond to a map $Y\to X$ in the preceding but that doesn't really change anything essential, only its inverse may not be defined on the whole of $G$.

This map has a lot of structure, but in general it is not a homomorphism. The formula defines a map $Z$ of two Lie algebra elements $X,Y\in\mathfrak g$ that map to an element $Z(X,Y)\in\mathfrak g$ whose image under $\exp$ is the product $\exp(X)\exp(Y)\in G$ where defined.

Explicitly, let $x,y\in G$, write $X = \log(x), Y = \log(y)$, then

$$xy = \exp(X)\exp(Y) = \exp Z(X,Y)$$

(so $Z$ is our binary operation and $F = \exp\circ\, Z$) where

$$\begin{split}\log(xy) = Z(X,Y) &{}= \log(\exp X\exp Y) \\ &{}= X + Y + \frac{1}{2}[X,Y] + \frac{1}{12}\left ([X,[X,Y]] +[Y,[Y,X]]\right ) \\ &{}\quad - \frac {1}{24}[Y,[X,[X,Y]]] \\ &{}\quad - \frac{1}{720}\left([Y,[Y,[Y,[Y,X]]]] + [X,[X,[X,[X,Y]]]] \right) + \cdots\end{split}$$

In this way you can obtain or approximate the product of two elements in $G$ from the Lie algebra, which often has advantages.

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    $\begingroup$ There's a highly amusing passage somewhere in Chapter 1 of the Bourbaki Volume 8 "Groupes et algèbres de Lie" where it talks about the priority of Campbell, Baker and Hausdorff and it says something like: "each found the demonstrations of his forerunners unconvincing(!)". That statement always makes me giggle and gives some comfort that I'm not the only one with about a 5% comprehension rate in reading technical literature (I reckon I need to read a paper about 20 times on average to "get" it). $\endgroup$ Aug 3 '15 at 10:09

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