# Formal group law and Koenigs function conjecture?

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 $$k(x)$$.

So

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

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.

Generalization

Let $$g(2x) = A(g(x)) $$

and

$$g(x+1) = B(g(x)) $$.

(  note : $$+$$ 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 $$$$.

B) $$K_2(x)$$ satisfies $$$$.

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 .

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

——

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

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 $$\,A(x):$$ $$c_1=\frac12a_1,\; c_2=\frac13a_2 +\frac16a_1^2,\; c_3=\frac14a_3 +\frac13a_1a_2 +\frac1{24}a_1^3,\;\dots. \tag{13}$$

• 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. – mick Mar 6 at 12:18
• 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 ! – mick Mar 6 at 12:21
• $g(x)$ 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 ? – mick Mar 7 at 12:27
• 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. – Somos Mar 7 at 13:52
• 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. – mick Mar 7 at 21:39