Timeline for Iwasawa main conjectures vs Bloch-Kato conjectures
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27 events
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| Dec 16, 2010 at 18:08 | comment | added | Olivier | Bien sûr. Indique moi juste une adresse mail valide. | |
| Dec 16, 2010 at 14:16 | comment | added | Joël | Well, as we already discussed, the order of the p-adic $L$-function should be dim $H^1_{L_n}(Q,\rho(1-n))$ plus or minus trivial terms (e.g. local cohomology groups, or global H^0 i.e. terms computable at glance) Moreover, the Euler characteristic of a complex is equal to the Euler characteristic of its cohomology. If you have an explicit complex doing what you say it does, then its Euler characteristic will doubtlessly have the form mentioned in the above paragraph. If this is known to be true, which I believe too (that's why I asked), can you give me the formula or a precise reference? | |
| Dec 16, 2010 at 9:25 | comment | added | Olivier | But this is essentially hopeless if you define $H^{1}_{\mathcal L_{n}}(\mathbb\Q,\rho(1-n))$ as a subgroup of $H^{1}$ satisfying certain conditions. How are you going to account for the order of the trivial zeroes? Again, the conjectural framework does not predict that there should be such a Selmer group: it predicts that there should the Euler charcteristic of a complex (which in the end amounts to the dimension of a vector space which admits $H^{1}(\mathbb\Q,\rho)$ as a quotient, not as a superobject). Besides, much of this is known to be true: look at the articles of D.Benois for instance. | |
| Dec 16, 2010 at 1:32 | comment | added | Joël | You have exactly understood what I think -- and made me realize how badly I have explained it in the first place. And that's what I meant by "proably in a weaker sense than in BK". I have finished my course today, and I will try to read Jay's paper to see if I can find the correct Selmer condition from it. | |
| Dec 15, 2010 at 19:59 | comment | added | Emerton | Ah ... is this what you mean by "probably in a weaker sense than in BK"? | |
| Dec 15, 2010 at 19:58 | comment | added | Emerton | Dear Joel, Suppose you can find the correct Selmer condition $\mathcal L_n$ of my previous comment. Do you think that one could deduce from the IMC that the order of vanishing of the $L$-function at $n$ is precisely equal to the dimension of $H^1_{\mathcal L_n}(\mathbb Q,\rho(1-n))$, or do you think you would just get the weaker statement that the $L$-function vanishes if and only if the Selmer group is non-zero? (I am worried about non-semi-simplicity again.) | |
| Dec 15, 2010 at 18:58 | comment | added | Emerton | Dear Joel, I read the discussion between you and Olivier below his answer, and it was very interesting, and illuminating. I think I understand your question better now. Am I right in thinking that you believe that for each $n$ there is a Selmer condition, call it $\mathcal L_n$, such that the order of vanishing of the $p$-adic $L$-function at $n$ is provably (given the IMC) equal to the dimension of $H^1_{\mathcal L_n}(\mathbb Q,\rho(1−n))$? And that the aim of your question is to get some kind of "formula" for $\mathcal L_n$? | |
| Dec 14, 2010 at 21:08 | comment | added | Olivier | @Rob H. Greenberg conjecture, as you described it, is equivalent to the conjecture that the complex giving the local condition at $p$ are semi-simple. So general believable conjectures imply that the difference in order of vanishing should be exactly what we believe it is. However, I have never heard any of the (big or small) semi-simplicity conjectures described as within reach, so I am not sure this is really good news. | |
| Dec 14, 2010 at 20:27 | comment | added | Rob Harron | In his paper in the p-adic mondromy and BSD proceedings, Greenberg conjectures that (if you're H^0=0, so that the order of vanishing of the complex L-function is just conjectured to be the dimension of the H^1_f) the difference in the order of vanishing of the complex L-function and the p-adic L-function is exactly his expected order of the trivial zero of the p-adic L-function. This order is conjecturally a local computation. But yeah I think Jay had some more high brow and general way he could see the contribution of the trivial zero. | |
| Dec 14, 2010 at 17:54 | comment | added | Emerton | Dear Joel, Thanks very much for this example; somehow I was aware of all or most of the fact you recalled, but had never put them together and tried to reconcile them. I now have to reflect on all this, and on Olivier's answer below. Best wishes, Matt | |
| Dec 14, 2010 at 17:07 | answer | added | Olivier | timeline score: 12 | |
| Dec 14, 2010 at 7:18 | comment | added | Joël | Yet on the $p$-adic size, things are not as clear. Take $k$ negative as above, and even. Then it is an open question whether $\zeta_p(1-k)$ is $0$ or not. Since the Main Conjecture is known in this case, then if the main conjecture implies that the non-vanishing of $\zeta_p(1-k)$ is equivalent to the vanishing of a suitable of $Q_p(k)$, this suitable Selmer group can't be the $H^1_f$ since this one is known for $Q_p(k)$. Obviously, the relevant Selmer group in this case is the full cohomology group $H^1(Q,Q_p(k))$ whose dimension is not known to be $0$ or $1$ (for $k$ negative even, I recall) | |
| Dec 14, 2010 at 7:10 | comment | added | Joël | Dear Matt, reading Jay Pottharst's paper and asking him for explanations when needed is also my hope to resolve my confusion (if no one here gives me the solution ready-for-use before). I am not sure that the exceptional zeros are the only issue, at least if by that you mean the exceptional zeros due to the interpolation factor at $p$ as in Mazur-Tate-Teitelbaum. The example I have in mind is $H^1_f(Q,Q_p(k))$ for $k$ negative. Those spaces are known to be zero (Soulé) and this is in perfect harmony with the Bloch-Kato conjecture, since obviously $\zeta(1-k) \neq 0$ (for $1-k > 1$). | |
| Dec 14, 2010 at 2:25 | comment | added | Emerton | Dear Joel, I have wondered the same question (what is the relationship between the Bloch--Kato Selmer groups, defined for pst $\rho$, and the more Selmer groups appearing in (say) ordinary Iwasawa theory. My (perhaps naive) picture was that they were more or less the same, perhaps taking into account issues with exceptional zeroes (whose manifestation on the algebraic side always confuses me). I also imagined that reading Nekovar, and perhaps Jay Pottharst, would resolve my confusion. What do they have to say on this issue? | |
| Dec 13, 2010 at 20:20 | comment | added | sibilant | @Joel: You would need even more general twists than this -- you would at least need to twist by all characters of $Z_p^\times$. (The zeroes of the $p$-adic $L$-function are just some random characters in weight space.) With that said, the data of all Selmer groups of $\rho$ twisted by any of these characters is nearly the same as the this $Q_p/Z_p$-Selmer group over the cyclotomic $Z_p$-extension. The only difference that remains is a $\mu$-invariant. Indeed, writing down a $Q_p/Z_p$-Selmer group requires a choice of lattice in $\rho$ and this can affect the $\mu$-invariant. | |
| Dec 13, 2010 at 19:41 | comment | added | Joël | @Emerton: Yes, I knew this issue that the MC will only implies a lower bound of $1$ on "the Selmer group" even if the $p$-adic $L$-functions vanishes as an high order, at least until some conjecture on the semi-simplicty of the Selmer group is solved. But actually, what I don't know is even more basic: what Selmer group are we talking about in this context? | |
| Dec 13, 2010 at 19:28 | comment | added | Joël | @Rob: You may be right, but I don't understand why it is necessary to work with the $\Q_p/\Z_p$-style Selmer group to see this phenomenon. You can as well twist your representation over $\mathbb{Q}_p$ by a non-inetgral power of the cyclotomic character (say a power $x \in Z_p$ congruent to $0 \pmod{p-1}$ to be sure). In your context, wouldn't we see the nonvanishing of the Selmer group of the twisted representation over $Q_p$ as well? For example, in your situation, wouldn't we have that $H^1(G,\rho^\ast \otimes \omega_p^{x}) \neq 0$, for $\omega_p$ the cyclotomic character, and | |
| Dec 13, 2010 at 18:38 | comment | added | Emerton | I don't think that the Iwasawa MC for ordinary ell. curves implies $p$-adic BSD. Indeed, as Joel surely knows, there is the problem that the $p$-adic $L$-function could have a double root at $s = 1$, but the Selmer group might still be of rank one (non-semiplicity of the Selmer group of $Q_{\infty}$). It seems (based on work of Joel with Gaetan Chenevier, and also on annoucned work of Skinner and Urban) that constructing Selmer groups for some fixed $\rho$ of weight one of dimension $> 1$ can be harder than proving the Iwasawa MC. | |
| Dec 13, 2010 at 18:04 | comment | added | sibilant | $p$-adic $L$-function might have its zeroes at some random characters which are not integral powers of the cyclotomic character. | |
| Dec 13, 2010 at 18:03 | comment | added | sibilant | @Joel: Thinking about elliptic curves, one reason for looking at Selmer groups of $Q_p/Z_p$-type representations is that these keep track of both Mordell-Weil and Tate-Shafarevich. I think that such a Selmer group over the cyclotomic $Z_p$-extension of say $Q$ has more information than the Selmer groups of $\rho(n)$ as $n$ varies over $Z$. Indeed, imagine a situation where the complex $L$-series does not vanish at any integer. Then the Selmer group of $\rho(n)$ should vanish for all $n$ by Bloch-Kato. However, there is no need for the $Q_p/Z_p$-Selmer group to vanish. Indeed, the... | |
| Dec 13, 2010 at 17:30 | comment | added | Joël | Thanks Chris for your comments. Your views are compatible with mine :-). Essentially looking at the Selmer group of $\rho$ over the cyclotomic $p$-extensions tower of $K$ is the same thing as looking at the Selmer group of the family of twists of $\rho$ by powers of the cyclotomic character. That's why I formulated Bloch-Kato with integral Tate twists. You Iwasawa-guys twist by more general $p$-adic power of the cyclotomic characters, but in particular specializing at integral twist you shoudl get something looking like Bloch-Kato's conjecture, with a $p$-adic, not complex, $L$-function. | |
| Dec 13, 2010 at 17:18 | history | edited | Alex B. | CC BY-SA 2.5 |
Added some tags and fixed a typo
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| Dec 13, 2010 at 16:46 | comment | added | Chris Wuthrich | My views, presented here as a few comments rather than an answer, is that Iwasawa theory will always talk about Bloch-Kato type conjectures in a full tower of fields, not just over $K$. And it will be the p-adic rather than the complex L-function. So with some additional conjectures, Iwasawa main conjectures will say something about Bloch-Kato conjectures, i.e. they are in some way compatible. | |
| Dec 13, 2010 at 16:42 | comment | added | Chris Wuthrich | The Iwasawa main conjecture for ordinary elliptic curves + the non-degeneracy of the p-adic height + the finiteness of Sha imply the p-adic BSD. To get BSD one would also need that the order of vanishing the complex and the p-adic L-function agree. | |
| Dec 13, 2010 at 16:40 | comment | added | Chris Wuthrich | Perrin-Riou's book on p-adic L-functions is written for general p-adic representations and she has a view on Bloch-Kato. | |
| Dec 13, 2010 at 16:40 | comment | added | Chris Wuthrich | Who is a comathematician here ??? :-) | |
| Dec 13, 2010 at 15:49 | history | asked | Joël | CC BY-SA 2.5 |