# Eisenstein series identities, directly

Let ${\Omega \subseteq \mathbf C}$ be a lattice. The Weierstrass ${\wp}$-function is defined as

$\displaystyle \wp(z) = \frac{1}{z^2} + \sum_{\omega \in \Omega^*} \left(\frac{1}{(z-\omega)^{2}} - \frac{1}{\omega^{2}}\right).$

It is ${\Omega}$-invariant, meromorphic, and has a double pole at each lattice point and no other poles. Its Laurent expansion at the origin is

$\displaystyle \wp(z) = \frac{1}{z^2} + c_2z^2 + c_4z^4 + c_6z^6 + \dots,$

where ${c_{2m} = (2m+1)\sum \frac{1}{\omega^{(2m+2)}}}$. Its derivative is

$\displaystyle \wp'(z) = \sum_{\omega \in \Omega} \frac{1}{(z-\omega)^3} = \frac{-2}{z^3} + 2c_2z + 4c_4z^3 + 6c_6z^5 +\dots.$

The functions ${\wp}$ and ${\wp'}$ satisfy

$\displaystyle \wp'(z)^2 = 4\wp(z)^2 - g_2\wp(z) - g_3$

in terms of the quantities ${g_2 = 20c_2}$ and ${g_3 = 28c_4}$. Now if we take ${\Lambda = \left<1, \tau\right>}$, where ${\tau}$ is in the upper half-plane, then ${\sum \frac{1}{\omega^{2m}} = G_{2k}(\tau)}$, where ${G_{2k}}$ is the weight ${2k}$ Eisenstein series. These Eisenstein series, or rather the normalized Eisenstein series ${E_{2k} =\frac{G_{2k}}{G_{2k}(i\infty)}= \frac{G_{2k}}{2\zeta(2k)}}$, satisfy certain relations, such as: (wikipedia)

$\displaystyle \begin{array}{rcl} E_{8} &=& E_4^2 \\ E_{10} &=& E_4\cdot E_6 \\ 691 \cdot E_{12} &=& 441\cdot E_4^3+ 250\cdot E_6^2 \\ E_{14} &=& E_4^2\cdot E_6 \\ 3617\cdot E_{16} &=& 1617\cdot E_4^4+ 2000\cdot E_4 \cdot E_6^2 \\ 43867 \cdot E_{18} &=& 38367\cdot E_4^3\cdot E_6+5500\cdot E_6^3 \\ 174611 \cdot E_{20} &=& 53361\cdot E_4^5+ 121250\cdot E_4^2\cdot E_6^2 \\ 77683 \cdot E_{22} &=& 57183\cdot E_4^4\cdot E_6+20500\cdot E_4\cdot E_6^3 \\ 236364091 \cdot E_{24} &=& 49679091\cdot E_4^6+ 176400000\cdot E_4^3\cdot E_6^2 + 10285000\cdot E_6^4. \end{array}$

In most basic texts on modular forms, these identities are derived by proving and exploiting the fact that the space ${M_{2k}}$ of modular forms of weight ${2k}$ is finite-dimensional. For instance, the fact that ${E_8}$ and ${E_4^2}$ have the same value at ${i\infty}$, combined with ${\dim M_{8} = 1}$, implies ${E_8=E_4^2}$. However, there is another, more “hands on” way to derive these identities.

Let us prove that ${E_8 = E_4^2}$. Substituting Laurent series at the origin in the equation ${\wp'(z)^2 - 4\wp(z)^2 + g_2\wp(z) + g_3=0}$, we see after some rearranging that the function

$\displaystyle (12c_2^2 - 36c_6)z^2 + (12c_2c_4 - 44c_8)z^4 + (-4c_2^3 + 4c_4^2 + 20c_2c_6 - 52c_10)z^6 + ...$

is identically ${0}$, and therefore

$\displaystyle \begin{array}{rcl} 0 &= &12c_2^2 - 36c_6\\ 0 &=& 12c_2c_4 - 44c_8 \\ 0 &=& -4c_2^3 + 4c_4^2 + 20c_2c_6 - 52c_{10}\\ & \dots & \end{array}$

Now

$\displaystyle c_{2m} = (2m+1)G_{2m+2} = (2m+1)2\zeta(2m+2)E_{2m+2} = (-1)^{m+1}(2m+1)\frac{(2\pi)^{2m+2}}{(2m+2)!}B_{2m+2}.$

Thus, for instance, ${c_6 = 14 \zeta(8)E_8 = \frac{14}{9450}\pi^8E_8}$. Making these substitutions in the first equation and factoring out ${\pi^8}$, we see that

$\displaystyle 0= \frac{12 \cdot 36}{90^2}E_4^2 - \frac{14 \cdot 36}{9450}E_8.$

Since ${\frac{12 \cdot 36}{90^2} = \frac{14 \cdot 36}{9450}}$, we have ${E_4^2 = E_8}$.

In fact, we have a bijection between ${\{(\wp_\Omega(z), \wp_\Omega'(z)) : \Omega \subseteq C\}}$ and the set of infinituples ${(c_2, c_4, c_6, \dots)}$ of complex numbers satisfying the infinite system of equations ${I}$:

$\displaystyle \begin{array}{rcl} 0 &= &12c_2^2 - 36c_6\\ 0 &=& 12c_2c_4 - 44c_8 \\ 0 &=& -4c_2^3 + 4c_4^2 + 20c_2c_6 - 52c_{10}\\ & \dots & \end{array}$

Luckily, the values of ${c_2}$ and ${c_4}$ determine all of the others, and the ring ${\mathbf C[c_2, c_4, c_6, \dots]/I}$ is generated by ${c_2}$ and ${c_4}$, and is in fact isomorphic to ${\mathbf C[c_2, c_3] = \mathbf C[G_4, G_6]}$. This means precisely that ${\{(\wp_\Omega(z), \wp_\Omega'(z)) : \Omega \subseteq C\}}$ is in bijection with the closed points of ${\text{Proj} (\mathbf C[G_4, G_6]) = \mathbf P^1_{\mathbf C}}$.

# A divisibility identity for Euler’s totient function

In this note I will give a Galois-theoretic proof that for a prime ${p}$ and positive integer ${n}$,

$\displaystyle n \mid \frac{\varphi(p^n-1)}{\varphi(p-1)}.$

I’d love to see a more elementary proof if you can come up with one.
First we need the following:

Lemma 1 Let ${Z_n}$ be the cyclic group with ${n}$ elements. Let ${m}$ be a positive divisor of ${n}$, and consider ${Z_m}$ as a subgroup of ${Z_n}$. Then the number of automorphisms of ${Z_n}$ which fix ${Z_m}$ pointwise is equal to ${\varphi(n)/\varphi(m)}$ (which, in particular, is an integer).

Proof of the Lemma: Note that any automorphism of ${Z_n}$ fixes ${Z_m}$, though not necessarily pointwise: indeed ${Z_n}$ has a unique subgroup of order ${m}$, and thus any automorphism of ${Z_n}$ must take this subgroup to itself. Thus we have a group homomorphism ${\text{Aut}(Z_n) \rightarrow \text{Aut}(Z_m)}$ which is easily seen to be surjective; its kernel is precisely the subgroup consisting of those automorphisms of ${Z_n}$ which fix ${Z_m}$ pointwise. The statement follows by comparing orders. ${\square}$

Now to prove the initial claim, consider the field extension ${\mathbf{F}_{p^n}/\mathbf{F}_p}$. Basic Galois theory tells that this is a Galois extension of degree ${n}$. Consider the canonical homomorphism

$\psi: \displaystyle \text{Gal}(\mathbf{F}_{p^n}/\mathbf{F}_p) \rightarrow \text{Aut}(\mathbf{F}_{p^n}^\times)$

which restricts an ${\mathbf{F}_p}$-automorphism ${\sigma}$ to the group of units of ${\mathbf{F}_{p^n}}$. Clearly it is an injective homomorphism since ${\sigma}$ is completely determined by where it sends the units. Moreover for any ${\sigma \in \text{Gal}(\mathbf{F}_{p^n}/\mathbf{F}_p)}$, $\psi(\sigma)$ lies in the subgroup of ${\mathbf{F}_{p^n}^\times}$ of those automorphisms fixing pointwise the cyclic subgroup ${\mathbf{F}_{p}^\times}$ of order ${p-1}$, because the Galois group consists of $\mathbf{F}_p$-homomorphisms. By the lemma the subgroup of these automorphisms has order ${\frac{\varphi(p^n-1)}{\varphi(p-1)}}$, whereas ${\text{Gal}(\mathbf{F}_{p^n}/\mathbf{F}_p)}$ has order ${n}$. This does it.