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> Looking again at the description below, I think it's clear that the1> algebraic parts you calculate are the norms of the ones appearing in2> our paper (at least up to divisors of k!N\phi(N)). I assume this is3> the method used to calculate the "L(A,2)/ \Omega_A" referred to in4> Section 2.1 of something dated August 29th 2000 (William, the thing5> you gave me in Nottingham, with the table of examples).67I don't see whether or not this is true. The definition of8Omega_f^{-} in our paper is unclear to me, in the sense that I don't9see how one might actually compute it. Your Omega_f^{-} isn't10well-defined (it depends on a choice of delta), so I just don't know11whether what I call L(A,2)/Omega_A^{-} is the norm of what you would12call L(f,2)/Omega_f^{-}.1314In any case, for the purposes of "Constructing Visible Sha" I don't15think we need to know anything about L(A,2)/Omega_A^{-}. The only16reason we want to know about L(A,2)/Omega_A^{-} is to conclude that17the Bloch-Kato conjecture is making predictions that agree with what18we prove. But you amazingly deduce most of this in Section 5.19However...2021> are predicted by Bloch-Kato. Section 5 of our paper only proves that q22> divides it, not q^2, so I'll put something in Section 7 about this,2324Ahh, except for the square bit. So, if we understood the relationship25between L(A,2)/Omega_A^{-} and L(f,2)/Omega_f^{-}, then maybe we could26deduce that q^2 divides L(f,k/2)/vol_oo in our examples. Again, it's27not clear to me what the relationship is.2829> expanding on the comment at the end of the introduction. In each case,30> it is really a certain degree-1 prime divisor Q of q which we are31> interested in. If the other prime divisors of q have degree at least 332> then we lose nothing by passing to the norm, and know that our33> algebraic part is exactly divisible by Q^2. This is except in those34> cases where q^4 divides the norm, like 581k4, where we need the other35> factors to have degree 5 or more, but 3 or more would be enough to36> ensure that the congruences of modular forms are mod Q^2 rather than37> involving more than one prime divisor of q, or a degree 2 divisor. Do38> the other divisors of q always have large degree?3940I'm going to officially assume that L(A,k/2)/Omega_A^{+/-} equals41Norm(L(f,k/2)/Omega_f^{+/-}) for the rest of this discussion.4243In the first example, 127k4C, the factors of (43) in the degree-1744field have degrees 1, 3, 4, and 9.4546By section 5 of our paper, don't we know that q | L(A,k/2)/Omega_A^{+/-}?47If q^2 || L(A,k/2)/Omega_A^{+/-}, then Q.R || L(f,k/2)/Omega_f^{+/-}48where Norm(R) = q. So if Q is the only prime over (q) of degree 1,49then R must equal Q. Thus I think that one needs that the other primes50have degree at least 2 instead of degree at least 3. Again, I don't51know whether this holds in all of our examples, but I could check it.52Should I?5354> I think the only other thing that needs sorting out before our paper55> is ready is my footnote 2 on page 11. Do you have the Fourier56> coefficients for these forms so that this can be checked?5758I don't immediately know how to find the local L-factor at bad primes59for forms of higher weight (I know the definition, but not explicitly60enough to immediately compute it). Do you need anything more than the61Fourier coefficient of 567k4L at p=7? It satisfies the polynomial6263f7 = x^12 - 84*x^11 + 3234*x^10 - 75460*x^9 + 1188495*x^8 -6413311144*x^7 + 108707676*x^6 - 652246056*x^5 + 2853576495*x^4 -658877793540*x^3 + 18643366434*x^2 - 23727920916*x + 1384128720166= (x-7)^12 (mod 13).6768Thus a_7(f) = 7 (mod 13).6970I have an idea of what happens for weight 2 modular forms for71Gamma_0(N) in the level-raising context of Section 7.4. I think that72the order of the *geometric* component group at p (=7 in this example)73is always divisible by Q (|13 in this example). However, the Tamagawa74number at p is either a power of 2 or definitely divisible by Q,75depending on the sign of the Atkin-Lehner involution W_7 (since76Frobenious acts on the component group through -W_7). If g has sign77+1 (like in our example) and the sign of W_7 is +1, then Q doesn't78divide the Tamagawa number at p. If W_7 = -1, then I think it79definitely does. We have W_7 = -1 for 567k4L, so I'd *guess* by80analogy with the weight 2 situation that ord_q(c_7(2)) > 0. Are these81Bloch-Kato Tamagawa numbers orders of finite groups fixed under the82action of -AtkinLehner? (Maybe the groups we take the length of83on page 4?)8485The wording of the first sentence of the second paragraph of Section867.4 is confusing because it suggests that the relative levels of f and87g has something to do with the signs in the functional equation. What88do you think?8990William919293949596