Prove addition law from duplication formula (for power series associated to elliptic genus)

Solution 1:

Well this got a non-zero amount of attention, so for the first time I'm answering my own question. Apologies for the length. A power series $f(x)\in\Bbb Q[\![x]\!]$ is admissible if $f(x)$ is odd and $f'(0)=1$.

Claim 1: for a fixed $\delta,\varepsilon$, there is a unique admissible power series $f_1(x)$ satisfying condition $(1)$. The analogous statement holds for condition $(3)$ (and of course for $(2)$, but we can avoid proving this directly).

Claim 2: $(3)\implies(2)\implies(1)$

Claim 3: $(1)\implies(3)$.

Proof of claim $3$: let $f_1$ be an admissible power series satisfying $(1)$. There is a unique admissible power series $f_3$ with the same parameter $\varepsilon$ satisfying $(3)$, and by claim $2$, $f_3$ satisfies $(1)$. By uniqueness, $f_1=f_3$, so $(1)\implies(3)$.

Proof of claim 1: to show uniqueness, by claim $2$, it suffices to show uniqueness for condition (1). We have that $f(x)f'(x)=\frac12f(2x)(1-\varepsilon f(x)^4)$. Write out $f(x)=x+a_3x^3+a_5x^5+\cdots$. Compare coefficients of $x^{2k+1}$ to see $$(2k+2)a_{2k+1}+\{\text{expression in }a_3,\ldots, a_{2k-1}\}=2^{2k}a_{2k+1}+\{\text{expression in }a_3,\ldots, a_{2k-1}\}.$$

It follows that $a_{2k+1}$ can be written in terms of $a_3,\ldots, a_{2k-1}$ and $\varepsilon$ provided that $(2k+2)\neq 2^{2k}$. The only issue is when $k=1$, but $a_3$ is already determined by $\delta$.

This also establishes existence for condition $(1)$. To show existence for condition $(3)$, we use basic calculus. Since we can write $f'(x)=\sqrt{1-2\delta f(x)^2+\varepsilon f(x)^4}$, the chain rule shows that $$f^{-1}(x)=\int_0^x\frac{dt}{\sqrt{1-2\delta t^2+\varepsilon t^4}}.$$ Recall that a formal power series $f$ has an inverse if $f(0)=0$ and $f'(0)=1$, so the previous statement makes sense and determines $f$.

Proof of claim 2: $(2)\implies(1)$ is obvious, so the bulk of the work is showing $(3)\implies(2)$. To do so, we again use single variable calculus. Note that it is enough to prove equality on an open interval, so this entails no loss of generality. Make the following definitions: \begin{align*} u&=f(x)\\ v&=f(y)\\ U&=f'(x)=\sqrt{1-2\delta u^2+\varepsilon u^4}\\ V&=f'(y)=\sqrt{1-2\delta v^2+\varepsilon v^4}\\ r&=\frac{uV+Uv}{1-\varepsilon u^2v^2} \end{align*}

Using basic calculus, observe that $g(f(x))=x$ implies $g'(f(x))f'(x)=1$, so $g'(y)=1/f'(x)$. Then with a slight abuse of notation, $$\int_0^k \frac{1}{f'(x)}dy=\int_0^kg'(y)dy=g(k)-g(0)=g(k). $$

If, in addition, $f'(x)$ can be written in terms of $y=f(x)$, then we can recover the inverse function $g$ through this process. For example, if $y=\sin x$, then $y'=\cos x=\sqrt{1-\sin^2 x}=\sqrt{1-y^2}$, and thus $\int_0^k\frac{1}{\sqrt{1-y^2}}dy=\arcsin k$ (at least on some open interval of values for $x,y$, and $k$). In our case, this means $\int_0^u\frac{du}{U}=g(u)=x$. We therefore will prove $\frac{du}{U}+\frac{dv}{V}=0$. We will consider $v$ as a function of $u$ in such a way that $r$ is a constant. Note that $u=0$ gives $v=r$, so we can integrate $\frac{du}{U}+\frac{dv}{V}=0$ and use a change of variables to see

$$\int_0^u\frac{du}{U}+\int_r^v\frac{dv}{V}=0 $$

which implies

$$\int_0^u\frac{du}{U}+\int_0^v\frac{dv}{V}=\int_0^r\frac{dv}{V}.$$

From our previous comments, this means $g(u)+g(v)=g(r)$, and thus $x+y=g(r)$, or $f(x+y)=r$. This is exactly the statement of $(3)$, which establishes our reduction.

To finish the proof of $(3)\implies(2)$, it remains to see $\frac{du}{U}+\frac{dv}{V}=0$. To this end, we now regard $v$ as a function of $u$ which makes $r$ constant. It suffices to let $u$ take values in some narrow open interval of positive real numbers, and we thus avoid the issue of whether $U$ and $V$ are defined. Since $r$ is constant, we differentiate both sides of $r=\frac{uV+Uv}{1-\varepsilon u^2v^2}$ with respect to $u$ to find

$$0=\frac{\left(V+\frac{2\varepsilon uv^3-2\delta uv}{V}\frac{dv}{du}+\frac{2\varepsilon u^3v-2\delta uv}{U}+U\frac{dv}{du}\right)(1-\varepsilon u^2v^2)+(uV+Uv)(2\varepsilon uv^2+2\varepsilon u^2v)\frac{dv}{du}}{(1-\varepsilon u^2v^2)^2}.$$ Substituting in $r$, simplifying, and rearranging gives $$\frac{du}{U}\left(UV+2\varepsilon u^3v-2\delta uv+2\varepsilon uv^2rU\right)+\frac{dv}{V}\left(UV+2\varepsilon uv^3-2\delta uv+2\varepsilon u^2vrV\right)=0.$$ The expressions in parenthesis are both nonzero for small values of $u,v$, so it now suffices to show they are equal. After cancelling, it suffices to show $$u^2+vrU=v^2+urV.$$

We compute \begin{align*} u^2+vrU-v^2-urV&=u^2+\frac{uvUV+U^2v^2}{1-\varepsilon u^2v^2}-v^2-\frac{uvUV+u^2V^2}{1-\varepsilon u^2v^2}\\ &=u^2-v^2+\frac{U^2v^2-u^2V^2}{1-\varepsilon u^2v^2}\\ &=u^2-v^2+\frac{(1-2\delta u^2+\varepsilon u^4)v^2-u^2(1-2\delta v^2+\varepsilon v^4)}{1-\varepsilon u^2v^2}\\ &=u^2-v^2+\frac{(1+\varepsilon u^4)v^2-u^2(1+\varepsilon v^4)}{1-\varepsilon u^2v^2}\\ &=\frac{u^2(1-\varepsilon u^2v^2)-v^2(1-\varepsilon u^2v^2)+v^2(1+\varepsilon u^4)-u^2(1+\varepsilon v^4)}{1-\varepsilon u^2v^2}\\ &=0, \end{align*} which establishes $(3)\implies (2)$.