By Shafarevich I.R. (ed.)

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**Example text**

Geometrically, the reason the funny subgroup Tl 2 arises is that 0(z, T) is 0 at the special point of order 2, namely, \ (T+l) € i / L / A - , and a b it is easy to check that ( J cF. i(T+l)to |(T'+l)mod 2 if and only if z |—> z/(c T+d) carries A r where f» = (a T+b)/(c T+d). (O, T) of one variable T. Note then that the functional equation of 0(0, T) reduces to: l *(0,(aT+b)/(cT+d)) = C(cT+d) 2 *(0,f) 8 where C 2 s 1, C as given in Theorem 7 . 1 . This will show that 0(0, T) is a modular form in T in the following sense: 37 Definition 9 .

Q f l _2_ . *nn(0f * (z) 00 11 One of the most important facts about the ^ -function is the differential equation that it satisfies. In fact, using the obvious facts (from the above equation) that (i)^(z) = $> (-z), (ii) the expansion of <£(z) at z = 0 begins with z -2 and (iii) the constant rigged so that this expansion has no constant term, it follows that: ^ (z) = J - + a z 2 z2 4 + bz +# #m ^ n e a r Therefore, 3 fr «(z) = - - 2 + 2az + 4bz + Z3 z = o# 27 and hence $'(z))2=4--^-16b+... z6 z* z6 z2 But 0 so that (^'(z) 2 -4^(z) 3 + 20 a ^(z) = -28b + .

Thus K • Mr Mod « © Mod, k k«ZZ+ is a graded ring, called the ring of modular forms of level N. 2. £ 2 (0, T), * 2 (0, T) and * (0, T) are modular forms of weight 1 & level 4. Proof. To start with, condition (a) for 0 2 (0, T) amounts to saying that C , QO the 8th root of 1, in the functional equation ( F ^ is 1 1 when (* j j ) « r ™ is immediate from the description of C (in fact, we only need c even and d = 1 (mod 4)). We can also verify immediately the bound (b)(ii) at co for S 40 2 £ (0, T).

### Algebraic geometry I-V by Shafarevich I.R. (ed.)

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