Let $f(x,y)$ be a symmetric real function and a formal group law

$$G(x + y) = f(G(x),G(y)). \tag{1}$$

Consider the equation

$$ h(2x) = f(h(x),h(x)) = A(h(x)). \tag{2}$$

This equation has many solutions.

Compute a solution to that equation with the fixed point at $0$ and its Koenigs function, and call the solution $k(x)$.


$$k(2x) = A(k(x)). \tag{3}$$

Then it seems it is always true that

Conjecture $T$:

$$ G(x) = k(x), \tag{4}$$ $$ G(x+y) = k(x+y) = f(k(x),k(y)). \tag{5}$$

Of course we can not use The Koenigs function if its conditions are not met, and there is no way around it. In other words The fixed point of $A(x)$ (at $0$) should not be parabolic and must be strictly positive.

I searched the Internet for Koenigs function and formal group law but did not find them combined.

Is conjecture $T$ true? How is it proved?


After hesitation I considered posting A generalisation that might be helpful.


Let $ g(2x) = A(g(x)) [1]$


$ g(x+1) = B(g(x)) [2] $.

( [3] note : $[1]+[2]$ define $g$ uniquely and equivalently they define $f(x,y)$ uniquely )

Let $ K_1(x) $ be the solution to $ K_1(2x) = A(K_1(x)) $ obtained by using the associated koenigs function.

Likewise Let $ K_2(x) $ be the solution to $ K_2(x+1) = B(K_2(x)) $ obtained by using the associated koenigs function.

Conjectures :

A) $K_1(x) $ satisfies $[2]$.

B) $K_2(x) $ satisfies $[1]$.

C) $K_1 = K_2$

D) $ K_1 = g $

E) $ K_2 = g $

Ofcourse these conjectures relate to each other and proving a few is equivalent to proving them all because of note [3].

Ps: it might be intresting to consider ( or generalize the conjecture ) replacing [2] with $g(x + q) = B_q(g(x)) $ for Some ( or any ) values of $q$ say eg 2 or e.


  • 4
    $\begingroup$ What is $G$? What is $A$? $\endgroup$ Mar 3, 2019 at 19:31
  • 3
    $\begingroup$ Well, $G$ would seem to be an exponential map (inverse of logarithm) from the additive formal group to what might be the formal group $f$. $\endgroup$
    – Lubin
    Mar 3, 2019 at 23:24
  • $\begingroup$ Pleae give a non-trivial example of your generalization.[1] and [2]. $\endgroup$
    – Somos
    Mar 9, 2019 at 16:27
  • $\begingroup$ I think the velocity addition formula ( tanh = g ) serves as an example. @Somos. Not to easy and not too hard. $\endgroup$
    – mick
    Mar 9, 2019 at 19:38

1 Answer 1


We are given a formal group law defined by $$ f(x,y) = f(y,x) = \sum_{n=0}^\infty P_n(x,y) \tag{1} $$ which satisfies the associativity equation $$ f(x,f(y,z)) = f(f(x,y),z) \tag{2} $$ and where the symmetric polynomials $\,P_n(x,y)\,$ are $$ P_0 \!=\! x\! + \!y,\; P_1 \!=\! a_1 x\,y,\; P_2 \!=\! a_2x\,y\,(x\!+\!y),\,\\ P_3 \!=\! a_3x\,y\,(x^2\!+\!y^2) \!+\! \frac12(2a_1a_2\!+\!3a_3)x^2y^2,\, \dots\,. \tag{3}$$ We seek a homomorphism function $\,g(x)\,$ that satisfies $$ g(x\!+\!y) \!=\! f(g(x),g(y)) \;\textrm{where}\; g(x) \!=\! x \!+\! \sum_{n=1}^\infty c_n x^n. \tag{4} $$ Notice that if we define $$ A(x) := 1 \!+\! \sum_{n=1}^\infty a_n x^n = f^{(0,1)}(x,0) \tag{5} $$ and differentiate equation $(4)$ w.r.t. $\,y\,$ at $\,y=0\,$ we get $$ g'(x) = A(g(x)) \tag{6} $$ and leads to an iterative method to find $\,g(x).\,$ Define the recurrence relation sequence $$ g_{n+1}(x) :=\! \int_0^x \! A(g_n(y)) \,dy \;\;\textrm{with}\;\; g_0(x) := O(x). \tag{7} $$ Since $\, g_n(x) \to g(x)\,$ as $\,n \to \infty,\,$ $\,g(x)\,$ is uniquely determined.

Now we use another approach. Using equation (4) define the function $$ B(x) := f(x,x) \;\;\textrm{with}\;\; g(2x) = B(g(x)) \tag{8}$$ and construct the recurrence relation sequence $$ g_{n+1}(x) \!:=\! g_n(x) \!-\! \frac{g_n(2x) \!-\! B(g_n(x)) }{ (2^{n+1}\!-\!2) }\!+\! O(x^{n+2}) \tag{9} $$ with $\, g_1(x) \!:=\! x \!+\! O(x^2). \,$ Unlike equation $(6)$ which leads immediately to a recursion which adds one more term of the power series for each iteration, equation $(8)$ does not do so. However, a standard trick is to add an extra term with an unknown coefficient and use equation $(8)$ to solve for the unknown coefficient which leads to equation $(9)$. The sequence $\,g_n(x)\,$ is the same as in equation $(7)$ and that proves conjecture $T$.

P.S. About Koenigs function notice that the definition is that given function $\,F(z)\,$ the function $\,k(z)\,$ is the Koenigs function for $\,F\,$ iff $\, k(F(z)) = F'(0)k(z).\,$ First, using the inverse function $\,k^{-1}(z)\,$ we get the alternative form $\,k^{-1}(F'(0)z) = F(k^{-1}(z)).\,$ Second, this is for analytic functions but it naturally extends to formal power series and, in that case, the restriction to $\,|k'(0)|<1\,$ does not apply. If we combine these two together then your equations $(2)$ and $(4)$ state that $\,k^{-1}(x)\,$ is the Koenigs function for $\,A(x).\,$ Now my equation $(8)$ states that $\,g^{-1}(x)\,$ is the Koenings function for $\,B(x)\,$. They say essentially the same thing.

Some details summarized are:

Suppose we have two formal power series $\,f(x),\,g(x)\,$ that satisfy equation $(4)$. The function $\,g(x)\,$ has an inverse function $\,g^{-1}(x)\,$ and applying it to equation $(4)$ gives $$ f(x,y)=f\left(g(g^{-1}(x)),\; g(g^{-1}(y))\right)=g(g^{-1}(x)+g^{-1}(y)).\tag{10} $$ Addition being commutative, associative and with a zero immediately implies $$ f(x,y) = f(y,x), \; f(x,f(y,z)) = f(f(x,y),z), \; f(x,0) = x, \tag{11} $$ and also, that the function $\,f(x,y)\,$ is uniquely determined by the function $\,g(x),\,$ implying that the $\,a_n\,$ are polynomials in $\,c_n:$ $$ a_1=+2c_1,\; a_2=-2c_1^2+3c_2,\; a_3=+4c_1^3-8c_1c_2+4c_3,\;\dots \tag{12} $$ and, of course, $\,g(x)\,$ is uniquely determined by the function $\,A(x):$ $$ c_1=\frac12a_1,\; c_2=\frac13a_2 +\frac16a_1^2,\; c_3=\frac14a_3 +\frac13a_1a_2 +\frac1{24}a_1^3,\;\dots. \tag{13} $$

  • $\begingroup$ Dear Somos. Unfortunately my math skills are still relatively weak and I am new to it. In other words I assume your answer is correct but I have trouble understanding. I think you are going a bit to fast for me. Maybe a longer answer will illuminate me. This is The first time I think about formal groups laws , so Im a beginner. More concrete : The polynomials you define are not completely clear to me , nor how The $a_i$ are related. But maybe only The first polynomial matters , and you merely point out that we have a taylor. Im not sure how A and B relate , I assume that matters. $\endgroup$
    – mick
    Mar 6, 2019 at 12:18
  • $\begingroup$ Also I assume your last equation is suppose to be The koenigs function. You actually nowhere mention The koenigs function ! I assume you have 7) and 9) both convergent and equal to each other by a “ telescoping argument “ but I fail to see it. Sorry for my confusion ! $\endgroup$
    – mick
    Mar 6, 2019 at 12:21
  • $\begingroup$ $g(x)$ is The functional inverse of koenigs function : $ 2^n c^n(x) $ where $c$ is The functional inverse $D(x) $ defined by $ g(2x) = D(g(x)) $. How is that equal to (9) WITH proof ? Is (9) derived from a kind of Newton iterations to find the inverse function and then plugged in The koenigs ? The thing is(9) comes out of nowhere without explaination. And how (7) = (9) is also not explained. I believe it ... but I need to see it. My skills may be limited sorry , but I think a bit more details would do miracles. So , Is Newton iteration used ? Or similar ? Am I even thinking in The right way ? $\endgroup$
    – mick
    Mar 7, 2019 at 12:27
  • $\begingroup$ You are correct about the inverse Koenings function which I had overlooked before. I added a sentence of explanation of the standard trick in equation (9). As for (7)=(9), do the math. $\endgroup$
    – Somos
    Mar 7, 2019 at 13:52
  • $\begingroup$ I understand that (7) is a Nice way tot compute the unique solution $g$ and therefore also (9) assuming equality. Convergeance also assumed. I have not thought deep about convergeance of radius of convergeance but because of the uniqueness of both g and Taylor series in general this seems unneccessary. But the Point =I can post methods to compute g too.Computing g wasnt the problem of the OP. Showing the equivalence with koenigs method was.You claim (7)= (9) = inverse koenigs ... Without proof ? This sounds like repeating the question.It really needs to be shown beyond doubt.Observing ≠ proof. $\endgroup$
    – mick
    Mar 7, 2019 at 21:39

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