$(G,\circ)$ is an abelian group, where $x\circ y=\frac{x+y+a(1+xy)}{1+xy+a(x+y)}$

Question

Let $G=(-1,1)$ and $a\in G$ be fixed. Prove that $(G,\circ)$ is an abelian group, where $$x\circ y=\frac{x+y+a(1+xy)}{1+xy+a(x+y)}, \forall x,y\in G.$$

To me, it seems extremely tedious to prove the axioms of the group in this case. Proving associativity is horrendous and I don't believe that any other of the axioms (apart from commutativity) is provable without extremely long computations.

In order to avoid this, I tried to use the so-called structure transport i.e. finding a bijective function from $G$ to some well-known group. I couldn't come up with any function, so I don't know how to actually solve this question. I doubt that it can be solved by proving each of the group axioms, but if anyone finds a way to do this I would be both amazed and grateful.

There is a restriction: $a \in G$. But I'm having real trouble proving that there's an identity element under this operation. I suspect that identifying the identity element might give some insight into what's going on. — Robert Shore, Oct 16 '19 at 16:03
The identity satisfies $e \circ e =e$. This gives $e=-a$. The formula then reminds me of automorphisms of the unit disk. These may be your structure transport. — lhf, Oct 16 '19 at 16:05
Then, i don't think it's correct.Take, $x=y=2-\sqrt{3},a=\frac{1}{\sqrt{3}}$ — , Oct 16 '19 at 16:27
@астонвіллаолофмэллбэрг - me too: here's a wolframalpha link https://bit.ly/35FDDac — HallaSurvivor, Oct 16 '19 at 16:37
Setting $a=0$ to simplify makes this MSE question (https://math.stackexchange.com/questions/1328260/prove-that-x-y-fracxy1xy-is-a-stable-part-of-g-1-1) look useful — HallaSurvivor, Oct 16 '19 at 16:42
Oops!I applied a wrong formula, instead $tanh(x+y)$, i took $tan(x+y)$ — , Oct 16 '19 at 16:55

Will Jagy · Accepted Answer · 2019-10-17T03:06:21.927

17

the structure transport is the composition of two functions, first is hyperbolic tangent, as $$ \tanh (x+y) = \frac{\tanh x + \tanh y}{1 + \tanh x \tanh y}. $$ Second is the Mobius transformation $$ m(z) = \frac{z+a}{az+1} $$

Note that $$ \frac{p+aq}{q+ap} = m\left(\frac{p}{q} \right) $$

Works very nicely. Note that the group identity is $(-a).$ All we are doing is mapping the interval to itself by Mobius transformation, in such a way that $-a$ maps to $0.$ Then we use the known operation $ \frac{u+v}{1+uv} \; , \; $ then back by the inverse of the Mobius transformation.

edited Oct 17 '19 at 03:06

answered Oct 16 '19 at 16:47

Will Jagy

139,541

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Readers unfamiliar with "structure transport" can see here (and its links) for examples of transporting algebraic structures along a bijection. – Bill Dubuque Oct 17 '19 at 03:04
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See also Northshield's delightful paper on the associativity of the secant methd linked here. – Bill Dubuque Oct 17 '19 at 03:09

$(G,\circ)$ is an abelian group, where $x\circ y=\frac{x+y+a(1+xy)}{1+xy+a(x+y)}$

1 Answers1