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}


\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.