Math Coffee

打开中国金融市场，就是21世纪美国鹦鹉外交最富成就的卓越胜利。

这一点从当时美国银行（原美洲银行）首席财务官普莱斯的讲话中反映的十分清楚，他告诉人们，2005年6月美国银行投入中国建设银行30亿美元，短短2年后的今天，美国银行在中国建设银行的直接和潜在获利已达到320亿美元，足以抵消该行在次贷危机中损失的近40亿美元。普莱斯的讲话人让所有美国人都激动不已，仅仅参股中国银行2年就有高达10余倍的惊人回报，如果直接控股中国银行，回报该是何等辉煌，恐怕将会达到百倍千倍！

美国本来通过埃文斯提出的要求是，把外资对中国银行的控股比例由25%提高到49%，结果却是中国干脆取消了全部限制，成为金融领域完全不设防的国家。

转移法孩子不听话，有时可以采用转移法以转移孩子的注意力和兴趣。

商业价值挂帅是影响香港学风的第三大负面因素。在《香港的困境》一文中，郎咸平强调香港是一个商人治理的社会。他毫不客气地批评香港的执政精英"一不懂高层次宏观管理，二不懂高新技术，三尤其是不重视研究发展和长期规划，因为他们以前的成功和这些因素无关。"其实不仅政界如此，商业气息弥漫在香港社会的各个领域。在大学餐厅的饭桌上，你会听到教授们热心于讨论各自申请研究经费的数额，而不是某学者文章的好坏。香港人普遍信奉功利主义，既不热衷于学术研究，也很难理解其长远价值。在商业文化浸润下成长起来的香港学生，很早就学会压抑个性，将自己嵌入整齐划一的白领模式。他们在校园里举办社团活动，不仅西装革履，而且样式颜色都别无二致，活脱在上就职预科班。在这崇商抑文的环境中，很难想像哪个年轻人会狂热地投入到实验室或书堆里。

香港的大学资源很多，但比起美国来相对落后，特别是教授评选的学术标准，非常跟不上趟。老一代学者，在旧制度下养尊处优惯了，抵抗变革。这是学术死气沉沉的重要原因。

各位许多是经过香港的大学来美国求学的。在美国大学的中国研究生里，已经有香港帮了。大陆人能去香港上学才几年呀！这本身说明香港大学的成就。各位可能觉得香港的大学比美国还落后。我完全同意。所以才主张进一步美国化。不过，香港的大学比起大陆的大学来，领先恐怕越来越大了（清华北大除了在生源上有优势外，哪方面能和香港的学校比？）。这是最近十年美国化的结果。这一过程，一定要走到底才对。不知道各位什么看法。

其中有一句说：世界上没有任何一所一流大学座落在弹丸之地，香港不太可能产生世界一流大学----除非香港已经和中国融为一体

我相信象各位这些在美国大学里的香港帮，如果不经过香港的教育，事业上大部分恐怕没有这么顺．我倒是觉得，北大人有一股傲气很要不得．就是觉得自己老子天下第一，目空一切．我和在纽约大学教书的张旭东有过一场辩论，他嘲笑我＂不过拿了人家几个奖学金＂就感恩戴德．我则确实拿了人家一点钱就非常珍惜．毕竟这是人家的钱．香港的大学，对各位学术事业逐益不小，我对各位对香港学生的蔑视有些不安．人家父母勤奋劳作纳税，创造了这么繁荣稳定的社会，并且给钱请各位去读书．难道人家没有一点好的地方？我们看不到人家身上可以学的地方，是人家一无是处，还是我们有这北大教育培养出来的＂北大眼光＂，看不到人家的任何优点？这是否是因为中了北大的毒而不自知？

4 July, 2008 at 6:09 am Emmanuel Kowalski A remark concerning Lior’s remark: the function h(u) in the current (v4) version of the paper is _not_ the same as the one that was defined when T. Tao pointed out a problem with it. This earlier one (still visible on arXiv, v1) was defined in different ways depending on whether the idele had at most one or more than one non-unit component, and was therefore not invariant under multiplication by . (It is another problem with looking at such a paper if corrections as drastic as that are made without any indication of when and why).

4 July, 2008 at 8:15 am Terence Tao Dear Lior, Emmanuel is correct. The old definition of h was in fact problematic for a large number of reasons (the author was routinely integrating h on the idele class group C, which is only well-defined if h was -invariant). Changing the definition does indeed fix the problem I pointed out (and a number of other issues too). But Connes has pointed out a much more serious issue, in the proof of the trace formula in Theorem 7.3 (which is the heart of the matter, and is what should be focused on in any future revision): the author is trying to use adelic integration to control a function (namely, h) supported on the ideles, which cannot work as the ideles have measure zero in the adeles. (The first concrete error here arises in the equation after (7.13): the author has made a change of variables on the idele class group C that only makes sense when u is an idele, but u is being integrated over the adeles instead. All subsequent manipulations involving the adelic Fourier transform Hh of h are also highly suspect, since h is zero almost everywhere on the adeles.)

More generally, there is a philosophical objection as to why a purely multiplicative adelic approach such as this one cannot work. The argument only uses the multiplicative structure of , but not the additive structure of k. (For instance, the fact that k is a cocompact discrete additive subgroup of A is not used.) Because of this, the arguments would still hold if we simply deleted a finite number of finite places v from the adeles (and from ). If the arguments worked, this would mean that the Weil-Bombieri positivity criterion (Theorem 3.2 in the paper) would continue to hold even after deleting an arbitrary number of places. But I am pretty sure one can cook up a function g which (assuming RH) fails this massively stronger positivity property (basically, one needs to take g to be a well chosen slowly varying function with broad support, so that the Mellin transforms at Riemann zeroes, as well as the pole at 1 and the place at infinity, are negligible but which gives a bad contribution to a single large prime (and many good contributions to other primes which we delete).)

Emmanuel Kowalski That’s an interesting point indeed, if one considers that the RH doesn’t work over function fields once we take out a point of a (smooth projective) curve — there arise zeros of the zeta function which are not on the critical line.

7 July, 2008 at 9:59 am javier Dear Terence, I am not sure I understand your “philosophical” complain on using only the multiplicative structure and not the additive one. This is essentially the philosophy while working over the (so over-hyped lately) field with one element, which apparently comes into the game in the description of the Connes-Bost system on the latest Connes-Consani-Marcolli paper (Fun with F_un). From an algebraic point of view, you can often recover the additive structure of a ring from the multiplicative one provided that you fix the zero. There is an explanation of this fact (using the language of monads) in the (also famous lately) work by Nikolai Durov “A new approach to Arakelov geometry (Section 4.8, on additivity on algebraic monads). By the way, I wanted to tell you that I think you are doing an impressive work with this blog and that I really enjoy learning from it, even if this is the very first time I’ve got something sensible to say :-)

6 July, 2008 at 7:44 pm Terence Tao Dear Chip, Actually, the product has a number of poles on the line , when s is a multiple of . Li’s approach to the RH was not to tackle it directly, but instead to establish the Weil-Bombieri positivity condition which is known to be equivalent to RH. However, the proof of that equivalence implicitly uses the functional equation for the zeta function (via the explicit formula). If one starts deleting places (i.e. primes) from the problem, the RH stays intact (at least on the half-plane ), but the positivity condition does not, because the functional equation has been distorted.

The functional equation, incidentally, is perhaps the one non-trivial way we do know how to exploit the additive structure of k inside the adeles, indeed I believe this equation can be obtained from the Poisson summation formula for the adeles relative to k. But it seems that the functional equation alone is not enough to yield the RH; some other way of exploiting additive structure is also needed, but I have no idea what it should be. [Revised, July 7:] Looking back at Li’s paper, I see now that Poisson summation was indeed used quite a few times, and in actually a rather essential way, so my previous philosophical objection does not actually apply here. My revised opinion is now that, beyond the issues with the trace formula that caused the paper to be withdrawn, there is another fundamental problem with the paper, which is that the author is in fact implicitly assuming the Riemann hypothesis in order to justify some facts about the operator E (which one can think of as a sort of Mellin transform multiplier with symbol equal to the zeta function, related to the operator on ). More precisely, on page 18, the author establishes that and asserts that this implies that , but this requires certain invertibility properties of E which fail if there is a zero off of the critical line. (A related problem is that the decomposition used immediately afterwards is not justified, because is merely dense in rather than equal to it.)

6 July, 2008 at 5:28 pm Chip Neville Terence, I have a question about your comment: “Because of this, the arguments would still hold if we simply deleted a finite number of finite places v from the adeles (and from k^*). … (basically, one needs to take g to be a well chosen slowly varying function with broad support, so that the Mellin transforms at Riemann zeroes, as well as the pole at 1 and the place at infinity, are negligible but which gives a bad contribution to a single large prime (and many good contributions to other primes which we delete).)” Does this mean that you would be considering the “reduced” (for lack of a better name) zeta function \prod 1/(1-1/p^{-s}), where the product is taken over the set of primes not in a finite subset S? If so, this “reduced” zeta function has the same zeroes as the standard Riemann zeta function, since the finite product \prod_S 1/(1-1/p^{-s}) is an entire function with no zeroes in the complex plane. Thus the classical situation in the complex plane seems to be very different in this regard from the situation with function fields over smooth projective curves alluded to by Emmanuel above. Does anyone have an example of an infinite set S and corresponding reduced zeta function with zeroes in the half plane Re z > 1/2? A set S of primes p so that \sum_S 1/p^{1/2} converges will not do, since \prod_S 1/(1-1/p^{-s}) is holomorphic in the half plane Re z > 1/2 with no zeroes there. Perhaps a set S of primes P thick enough so that \sum_S 1/p^{1/2} diverges, but thin enough so that \sum_S 1/p converges, might do. This seems to me to be a delicate and difficult matter. I hope these questions do not sound too foolish.

7 July, 2008 at 11:01 am Terence Tao Dear Javier, I must confess I do not understand the field with one element much at all (beyond the formal device of setting q to 1 in any formula derived using and seeing what one gets), and don’t have anything intelligent to say on that topic. Regarding my philosophical objection, the point was that if one deleted some places from the adele ring A and the multiplicative group (e.g. if k was the rationals, one could delete the place 2 by replacing with the group of non-zero rationals with odd numerator and denominator) then one would still get a perfectly good “adele” ring in place of A, and a perfectly good multiplicative group in place of (which would be the invertible elements in the ring of rationals with odd denominator), but somehow the arithmetic aspects of the adeles have been distorted in the process (in particular, Poisson summation and the functional equation get affected). The Riemann hypothesis doesn’t seem to extend to this general setting, so that suggests that if one wants to use adeles to prove RH, one has to somehow exploit the fact that one has all places present, and not just a subset of such places. Now, Poisson summation does exploit this very fact, and so technically this means that my objection does not apply to Li’s paper, but I feel that Poisson summation is not sufficient by itself for this task (just as the functional equation is insufficient to resolve RH), and some further exploitation of additive (or field-theoretic) structure of k should be needed. I don’t have a precise formalisation of this feeling, though.

7 July, 2008 at 1:22 pm Gergely Harcos Dear Terry, you are absolutely right that Poisson summation over k inside A is the (now) standard way to obtain the functional equation for Hecke L-functions. This proof is due to Tate (his thesis from 1950), you can also find it in Weil’s Basic Number Theory, Chapter 7, Section 5.

Babak Hi Terrance, A few months ago I stumbled upon an interesting differential equation while using probability heuristics to explore the distribution of primes. It’s probably nothing, but on the off-chance that it might mean something to a better trained mind, I decided to blog about it: http://babaksjournal.blogspot.com/2008/07/differential-equation-estimating.html -Babak

15 July, 2008 at 7:57 am michele I think that the paper of Prof. Xian-Jin Li will be very useful for a future and definitive proof of the Riemann hypothesis. Furthermore, many mathematics contents of this paper can be applied for further progress in varios sectors of theoretical physics (p-adic and adelic strings, zeta strings).