LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedjudge pass (gpt-6.1-sol)

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

The transformation law of the weight-two Eisenstein series E_2

Statement

The weight-two Eisenstein series $E_{2} (τ) = 1 - 24 \sum_{n \geq 1} σ_{1} (n) q^{n}$ satisfies $E_{2} (τ + 1) = E_{2} (τ)$ and, for every $γ = (\begin{matrix} a & b \\ c & d \end{matrix}) \in S L_{2} (Z)$ , $E_{2} ⁣ (\frac{a τ + b}{c τ + d}) = (c τ + d)^{2} E_{2} (τ) - \frac{6 i c}{π} (c τ + d) .$ In particular $E_{2} (- 1 / τ) = τ^{2} E_{2} (τ) - \frac{6 i}{π} τ$ , so $E_{2}$ is not a modular form of weight $2$ .

Facts & Assumptions

Given: The series $E_{2} = 1 - 24 \sum_{n \geq 1} σ_{1} (n) q^{n}$ , $q = e^{2 π i τ}$ , and its half-normalisation $H : = (π^{2} / 6) E_{2}$ , all on $H$ (The level-one Eisenstein series E_k and the weight-two series E_2, The unit disc, the upper half-plane, and Blaschke factors).

[F1]

$π \cot (π z) = \frac{1}{z} + \sum_{n \geq 1} \frac{2 z}{z^{2} - n^{2}}$ with local uniform convergence on $C ∖ Z$ (The Mittag-Leffler expansion of pi cotangent), and $\sin w = w - \frac{w^{3}}{6} + O (w^{5})$ , $\cos w = 1 - \frac{w^{2}}{2} + O (w^{4})$ (The exponential definitions of complex sine, cosine, hyperbolic sine, and hyperbolic cosine equal their entire power series).

[F2]

Lipschitz: for $τ \in H$ , $\sum_{n \in Z} (τ + n)^{- 2} = (- 2 π i)^{2} \sum_{r \geq 1} r q^{r} = - 4 π^{2} \sum_{r \geq 1} r q^{r}$ , absolutely on the left (The Lipschitz formula for the reciprocal-power sums).

[F3]

On every compact $K \subseteq H$ there is $c_{K} > 0$ with $∣ m z + n ∣ \geq c_{K} \max (∣ m ∣, ∣ n ∣)$ for all $(m, n) \neq (0, 0)$ and $z \in K$ This estimate is derived locally: if $ℑ z \geq y_{0} > 0$ , $∣ ℜ z ∣ \leq X$ , then $∣ m ∣ \leq ∣ m z + n ∣ / y_{0}$ and $∣ n ∣ \leq ∣ m z + n ∣ + X ∣ m ∣ \leq (1 + X / y_{0}) ∣ m z + n ∣$ , so one may take $c_{K} = \min (y_{0}, (1 + X / y_{0})^{- 1})$ .

[F4]

The $M$ -test gives uniform convergence of a series dominated by a summable real majorant, which may depend on a parameter (Weierstrass M-test for complex-valued function series); comparison with a convergent $p$ -series and bounded monotone partial sums give convergence of positive series (If $0 \leq a_{k} \leq b_{k}$ eventually, convergence of $\sum b_{k}$ gives convergence of $\sum a_{k}$ , and divergence of $\sum a_{k}$ gives divergence of $\sum b_{k}$ , For rational $p > 0$ , $\sum 1 / k^{p}$ converges iff $p > 1$ , A series of nonnegative terms converges iff its partial sums are bounded, and then the sum is their supremum); an absolutely summable double family may be regrouped and reindexed (Fubini for double series: if $\sum_{i} \sum_{j} ∣ a_{i j} ∣$ converges then both iterated sums and the sum along every bijection $N \to N \times N$ converge to one and the same value, If $\sum ∣ a_{k} ∣$ converges then $\sum a_{k}$ converges).

[F5]

If $G$ is differentiable on $[n, n + 1]$ with integrable $G^{'} = f$ , then $\int_{n}^{n + 1} f = G (n + 1) - G (n)$ ; consequently $∣ f (n) - \int_{n}^{n + 1} f ∣ \leq 2 \sup_{[n, n + 1]} ∣ f^{'} ∣$ (The second fundamental theorem: if $G$ is differentiable on $[a, b]$ with $G^{'} = f$ and $f$ is integrable, then $\int_{a}^{b} f = G (b) - G (a)$ ).

[F6]

If $φ (t, ε)$ and $\partial_{ε} φ (t, ε)$ are continuous on $R \times I$ , one slice is absolutely integrable, and $\partial_{ε} φ$ is dominated on each compact interval of $ε$ by a $t$ -integrable function, then $Φ (ε) = \int_{R} φ (t, ε) d t$ is differentiable with $Φ^{'} (ε) = \int_{R} \partial_{ε} φ (t, ε) d t$ (Differentiation under an improper multiple integral under an integrable derivative bound).

Proof

1.1F1F4givenalgebra

We first record $ζ (2) = π^{2} / 6$ . By [F1], $π \cot (π z) = \frac{1}{z} - 2 z \sum_{n \geq 1} \frac{1}{n^{2} - z^{2}}$ for $z \notin Z$ ; for $∣ z ∣ \leq 1 / 2$ one has $\frac{1}{n^{2} - z^{2}} = \frac{1}{n^{2}} + \frac{z^{2}}{n^{2} (n^{2} - z^{2})}$ with $\frac{∣ z ∣^{2}}{n^{2} ∣ n^{2} - z^{2} ∣} \leq \frac{2 ∣ z ∣^{2}}{n^{4}}$ , so $\sum_{n \geq 1} \frac{2 z}{z^{2} - n^{2}} = - 2 z \sum_{n \geq 1} \frac{1}{n^{2}} + O (z^{3})$ by [F4] and $π \cot (π z) = \frac{1}{z} - 2 z \sum_{n \geq 1} n^{- 2} + O (z^{3})$ . On the other hand [F1] gives $π \cot (π z) = π \frac{\cos π z}{\sin π z} = π \frac{1 - (π z)^{2} / 2 + O (z^{4})}{π z - (π z)^{3} / 6 + O (z^{5})} = \frac{1}{z} - \frac{π^{2}}{3} z + O (z^{3})$ . Comparing coefficients of $z$ and using $\sum n^{- 2} > 0$ gives $\sum_{n \geq 1} n^{- 2} = π^{2} / 6$ .

2.1F2F3F4step 1.1givenalgebra

With $q = e^{2 π i τ}$ and $∣ q ∣ < 1$ , [F2] gives $\sum_{n \in Z} (m τ + n)^{- 2} = - 4 π^{2} \sum_{r \geq 1} r q^{m r}$ for each $m \geq 1$ , so $\sum_{m \geq 1} \sum_{n \in Z} (m τ + n)^{- 2} = - 4 π^{2} \sum_{m, r \geq 1} r q^{m r} = - 4 π^{2} \sum_{n \geq 1} σ_{1} (n) q^{n}$ , the last step regrouping the absolutely summable family $(r q^{m r})_{m, r \geq 1}$ by $n = m r$ [F4]. Hence, by 1.1, $H (τ) = ζ (2) + \sum_{m \geq 1} \sum_{n \in Z} (m τ + n)^{- 2} .$ For $ε > 0$ define the regularised series $H_{ε} (z) : = \frac{1}{2} \sum_{(m, n) \neq (0, 0)}^{'} (m z + n)^{- 2} ∣ m z + n ∣^{- 2 ε}$ ; pairing $(m, n)$ with $(- m, - n)$ and writing $f_{m, ε} (t) : = (m z + t)^{- 2} ∣ m z + t ∣^{- 2 ε}$ , we have $H_{ε} (z) = ζ (2 + 2 ε) + \sum_{m \geq 1} \sum_{n \in Z} f_{m, ε} (n)$ , and this family is absolutely summable for $ε > 0$ : on a compact $K$ with $c = c_{K}$ , the shell of pairs with $\max (∣ m ∣, ∣ n ∣) = j$ has $8 j$ elements each bounded by $(c j)^{- 2 - 2 ε}$ , so the shell sum is $8 c^{- 2 - 2 ε} j^{- 1 - 2 ε}$ , summable: compare with $j^{- 1 - r}$ for a rational $0 < r < 2 ε$ and apply [F4].

3.1F3F4step 2.1givenalgebra

The regularised series satisfies $H_{ε} (γ τ) = (c τ + d)^{2} ∣ c τ + d ∣^{2 ε} H_{ε} (τ)$ for every $γ = (\begin{matrix} a & b \\ c & d \end{matrix}) \in S L_{2} (Z)$ . Indeed $m γ τ + n = \frac{m^{'} τ + n^{'}}{c τ + d}$ with $(m^{'}, n^{'}) = (m a + n c, m b + n d)$ , and $(m, n) \mapsto (m^{'}, n^{'})$ is a bijection of $Z^{2}$ because its matrix $(\begin{matrix} a & c \\ b & d \end{matrix})$ has determinant $a d - b c = 1$ ; since $(m γ τ + n)^{- 2} ∣ m γ τ + n ∣^{- 2 ε} = (c τ + d)^{2} ∣ c τ + d ∣^{2 ε} (m^{'} τ + n^{'})^{- 2} ∣ m^{'} τ + n^{'} ∣^{- 2 ε}$ , the absolutely summable family of 2.1 may be reindexed and regrouped by [F4], giving the displayed law.

4.1F3F4F5step 3.1givenalgebra

Fix $τ \in H$ , $y : = Im τ > 0$ , and $m \geq 1$ . Set $I_{ε} (m τ) : = \int_{R} f_{m, ε} (t) d t$ and $A_{m} (ε) : = \sum_{n \in Z} [f_{m, ε} (n) - \int_{n}^{n + 1} f_{m, ε} (t) d t]$ . The improper integral $I_{ε} (m τ)$ is the sum over $n \in Z$ of $\int_{n}^{n + 1} f_{m, ε}$ , so $\sum_{n} f_{m, ε} (n) = I_{ε} (m τ) + A_{m} (ε)$ , and summing over $m$ gives the exact identity $H_{ε} (τ) = ζ (2 + 2 ε) + \sum_{m \geq 1} A_{m} (ε) + \sum_{m \geq 1} I_{ε} (m τ)$ for $ε > 0$ . The substitution $m τ + t = m y (u + i)$ shows $I_{ε} (m τ) = (m y)^{- 1 - 2 ε} I (ε)$ with $I (ε) : = \int_{R} (u + i)^{- 2} (1 + u^{2})^{- ε} d u$ . Finally $\sum_{m \geq 1} A_{m} (ε)$ converges uniformly for $ε \in [- 1 / 4, 1 / 4]$ : by [F5], $∣ f_{m, ε} (n) - \int_{n}^{n + 1} f_{m, ε} ∣ \leq 2 \sup_{[n, n + 1]} ∣ f_{m, ε}^{'} ∣$ , and differentiating the product gives $∣ f_{m, ε}^{'} (t) ∣ \leq (2 + 2 ∣ ε ∣) ∣ m τ + t ∣^{- 3 - 2 ε}$ ; on $[n, n + 1]$ the quantity $∣ m τ + t ∣$ is comparable to $∣ m τ + n ∣$ (their difference has modulus at most $1$ , and $∣ m τ + t ∣ \geq Im m τ = m y$ ; for $∣ m τ + n ∣ > 2$ this gives comparison, and only finitely many $n$ fail it for a given $m$ , none at all once $m y > 2$ ), so the absolute sum of all row differences is dominated uniformly in $ε$ by a constant times $\sum_{j \geq 1} 8 j j^{- 5 / 2} < \infty$ using [F3] and [F4]. The double series is thus uniformly convergent by the M-test; each row difference is continuous in $ε$ , and regrouping into $A_{m}$ preserves the limit.

5.1F4F5F6step 2.1step 4.1givenalgebra

We let $ε \to 0^{+}$ . First, $ζ (2 + 2 ε) \to ζ (2)$ : for $0 < ε \leq 1 / 4$ and every $N$ , $∣ \sum_{n} n^{- 2 - 2 ε} - \sum_{n} n^{- 2} ∣ \leq \sum_{n \leq N} ∣ n^{- 2 - 2 ε} - n^{- 2} ∣ + 2 \sum_{n > N} n^{- 2}$ , and the inner estimate is as small as desired for $N$ large and then $ε$ small. Second, $\sum_{m \geq 1} A_{m} (ε) \to \sum_{m \geq 1} A_{m} (0)$ because the convergence is uniform on $[- 1 / 4, 1 / 4]$ by 4.1 and each $A_{m}$ is continuous in $ε$ ; and $A_{m} (0) = \sum_{n \in Z} (m τ + n)^{- 2}$ because $I_{0} (m τ) = \int_{R} (m τ + t)^{- 2} d t = 0$ , the primitive being $- (m τ + t)^{- 1}$ . Third, $I (0) = \int_{R} (u + i)^{- 2} d u = 0$ (same primitive) and $I^{'} (0) = - \int_{R} (u + i)^{- 2} \log (1 + u^{2}) d u = - π$ : the differentiation is licensed by [F6] separately on real and imaginary parts, with base slice $ε = 0$ absolutely integrable (its modulus is $(1 + u^{2})^{- 1}$ ), since for $∣ ε ∣ \leq 1 / 4$ the $ε$ -difference quotients of $(u + i)^{- 2} (1 + u^{2})^{- ε}$ are bounded by $(1 + u^{2})^{- 3 / 4} \log (1 + u^{2})$ , which is integrable on $R$ , and a primitive of $- \log (1 + t^{2}) / (t + i)^{2}$ is $\frac{1 + \log (1 + t^{2})}{t + i} - \arctan t$ , whose endpoint difference is $- π$ . Fourth, by [F5] and [F4] the decreasing function $t \mapsto t^{- 1 - 2 ε}$ satisfies $\int_{1}^{\infty} t^{- 1 - 2 ε} d t \leq \sum_{m \geq 1} m^{- 1 - 2 ε} \leq 1 + \int_{1}^{\infty} t^{- 1 - 2 ε} d t$ , that is $\sum_{m \geq 1} m^{- 1 - 2 ε} = \frac{1}{2 ε} + O (1)$ . Hence $\sum_{m \geq 1} I_{ε} (m τ) = y^{- 1 - 2 ε} I (ε) \sum_{m \geq 1} m^{- 1 - 2 ε} \to \frac{1}{y} \cdot (- \frac{π}{2}) = - \frac{π}{2 y}$ as $ε \to 0^{+}$ , and therefore $H_{ε} (τ) \to H (τ) - \frac{π}{2 y}$ , using 2.1, 4.1 and $2 ε \to 0$ .

6.1step 3.1step 5.1givenalgebra∎

Taking $ε \to 0^{+}$ in 3.1 and using 5.1 with $∣ c τ + d ∣^{2 ε} \to 1$ , $H (γ τ) - \frac{π}{2 Im (γ τ)} = (c τ + d)^{2} (H (τ) - \frac{π}{2 y}) .$ Since $Im (γ τ) = y / ∣ c τ + d ∣^{2}$ , this rearranges to $H (γ τ) = (c τ + d)^{2} H (τ) + \frac{π}{2 y} (∣ c τ + d ∣^{2} - (c τ + d)^{2})$ . Writing $w : = c τ + d$ we have $w^{2} - ∣ w ∣^{2} = w (w - \overline{w}) = 2 i c y w$ , so the correction is $- π i c (c τ + d)$ and $H (γ τ) = (c τ + d)^{2} H (τ) - π i c (c τ + d)$ . Multiplying by $6 / π^{2}$ proves the stated transformation law; $E_{2} (τ + 1) = E_{2} (τ)$ is immediate from $E_{2} (τ + 1) = 1 - 24 \sum σ_{1} (n) q^{n} = E_{2} (τ)$ ; and at $γ = S$ , where $c = 1$ and $d = 0$ , the law reads $E_{2} (- 1 / τ) = τ^{2} E_{2} (τ) - \frac{6 i}{π} τ$ , which differs from $τ^{2} E_{2} (τ)$ for $τ \neq 0$ , so $E_{2}$ is not a modular form of weight $2$ .

Depends on

Used by

FALSE: the weight-two Eisenstein series E₂ is a modular form False statement
The Jacobi product formula for the discriminant Lemma

Dependency tree · two levels

106 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

D. Zagier, Elliptic Modular Forms and Their Applications, in The 1-2-3 of Modular Forms (Universitext, Springer, 2008) (standard reference, not scraped)
C. T. McMullen, Advanced Complex Analysis, Math 213a course notes (Harvard, 2010) (standard reference, not scraped)