A derivation of a Riemann zeta function identity

Yesterday, I saw the following Riemann zeta function identity:

\displaystyle\sum_{n=1}^{\infty} \frac{\sigma_a(n)\sigma_b(n)}{n^s} = \frac{\zeta(s)\zeta(s-a)\zeta(s-b)\zeta(s-a-b)}{\zeta(2s-a-b)}.

I took some time to try to derive it myself and to my great pleasure, I succeeded.

Eventually, I realized that it suffices to show that

\{(dd_a, dd_b, d^2 n) : d_a | n, d_b | n : d, d_a, d_b, n \in \mathbb{Z}\}

and

\{(dd_a, dd_b, n) : dd_a d_b | n : d, d_a, d_b, n \in \mathbb{Z}\}

are equal as multisets. As sets, they are both representations of the set of 3-tuples of positive integers such that the third is a multiple of the least common multiple of the first two. In the latter one, the frequency of (a,b,c) is the number of d that divides both a and b such that ab | cd. In the other one, if we write (a,b,c) as (d_1 d_2 a', d_1 d_2 b', c) where \mathrm{gcd}(a', b') = 1, the ab | cd condition equates to d_1^2 d_2 a'b' | c, which corresponds to the number of d_1 dividing a and b and such that d_1^2 | c and with that, d_2a', d_2b' both dividing d_1^2 / c, which is the frequency of (a,b,c) via the former representation.

The coefficients \{a_n\} of the Dirichlet series of the LHS of that identity can be decomposed as follows:

a_n = \displaystyle\sum_{d^2 | n, d_a | \frac{n}{d^2}, d_b | \frac{n}{d^2}} (dd_a)^a (dd_b)^b.

The coefficients \{b_n\} of the Dirichlet series of the RHS of that identity are

b_n = \displaystyle\sum_{dd_a d_b | n} (dd_a)^a (dd_b)^b.

Observe how both are equivalent in that via the multiset equivalence proved above, n determines the same multiset of (dd_a, dd_b) for both and across that, the values of the same function (dd_a)^a (dd_b)^b are summed. Hence the two series are equal.

湾区游

上周底,又跑湾区一趟,为了面试,也为了玩。这次受益匪浅,拿到了比较好的工作,同时也再次得到学习数学的启发。我的一位不善学,疯疯癫癫的,苏犹半朋友,却建议我跟一位正在MIT学数学的美国IMO金牌得主聊天。此人我已被介绍过,通过一位“知名的高才生中间人”,可与此人未说任意有含量的话,从而想这牛人太忙,与我这庸人无话可说。出乎意料,介绍者却说他与牛人沟通频繁,并且将我加入他们俩人的脸书群,后来又加了一位不相信智商,反徐道辉的,女生物博士生,犹裔也。面试之前的晚上,我住在旧金山的宾馆,附近有一家“现代犹太博物馆”,好奇去看了看,楼上关门,楼下没啥好看的。我们在群里所讨论的好多与种族,文化,智商,和学科相关,还记得我曾经对数学尖子开玩笑,美国IMO选手,非亚裔,必犹裔,而他似乎未感到我的玩笑口气,回说他那界有一两位非犹太白人。他还强调自己很美国人,不是那种在以色列待过的。高中时,他选了中文为他的外语,以童话为他爱听的一首中文歌。学术上,给的感觉是专注,单一的纯数学本科生,非常肯定他会走学术道路。他的具体数学兴趣及倾向为组合数学和理论计算机和解析数论。那天晚上,问他知道那几个二次互反律(law of quadratic reciprocity)的证明,他回答一个引用Zolotarev’s lemma的,并且发了个链接。当时,我只知道高斯和(Gauss sum)的那个,而细节已经记不清楚了。我花了一小时细读那个证明,领悟后感觉漂亮无比,引用的工具极其简单。从来没有想到可以将这数论皇后的定理视为,表达为置换的奇偶的积,毕竟Legendre symbol给的是1或-1,与sgn一样,好妙啊。回顾透明Legendre symbol给的基本是在给循环阿贝尔群(\mathbb{Z}/{p\mathbb{Z}})^{\times}的元素奇偶,二次剩余和偶置换在他们对应母群都是指数为2的子群。这又让我想到高斯对于正十七边形可作图的证明也是引用指数2的子群,其由Galois theory对应于度数为2的域扩张。过两天,我又学到了Gauss lemma,就是\left(\frac{a}{p}\right) = (-1)^n, n\{a, 2a, \ldots, \frac{p-1}{2}a\}大于p/2的元素的数量。证明思路很直接,将\{a, 2a, \ldots, \frac{p-1}{2}a\}的大于p/2的元素负掉,可和其他元素从新凑成\{1, 2, \ldots, \frac{p-1}{2}\}. Eisenstein对该定理的证明,我以前知道其存在但没看懂的,引用类似于Gauss lemma的引理,思路及证明策略抓住,这次却清晰了然。二次互反律的美妙我之前无法欣赏,记得对此定理有过一种稀奇古怪,难以思议之感,是没有抓住并且悟觉它的美妙的对称结构。想起在Eisenstein的证明中,画了一个pq的格子,将-1的次数示为格子的下左象限所有的点数,此为(p-1)(q-1)/4的来源。

数论是纯数学比较活的分支,与现在被我看为形式化繁琐枯燥的测度论相反。回想起,我学测度论习题做不出来开始大大怀疑自己数学的能力并且对数学稍失去了兴趣。而数论从某种角度而言有相反的作用。昨天,我从新过了Bertrand’s postulate的源于Erdos的证,对此同样有新的见解,至少感觉是这样。这次很明确它是以好紧限的有易与素数次数相连的\binom{2n}{n}在假设n2n范围无素数情况下,导致未成立的不等式。证明里的关键观察是2n/3n之间的素数不整除,此有大大减小上限的作用。

我有想过回到学校做数学,可是此愿望看来还是非强于我对自己的能力的怀疑加上挣钱的“心里”需要。在所谓的工业界有了一定的见识是大大开阔了眼界,与那些学术界单纯做学问的人恰恰相反。正面看,这给了我更多的实用性思维,眼观,和技能,而反面,这使得我失去,以不可逆转的方式,利于做研究的隔离所排出的干扰。我小时看到人说离开学校再回去很难,现在知道为何人这么说。所以我的回到学校的企图又失败了,下一个工作来了,有新的实用性的挑战和机会。不过,数学我很可能业余还会学学,盼望在此将出前所未有的新视野。

Proof of fundamental theorem of arithmetic

I read it a couple days ago and actually remembered it this time in a way that I will never forget it. It invokes Euclid’s lemma, which states that if p | ab for p prime, then p | a or p | b, which can be proved using Bezout’s lemma. For existence, it does induction on the number of factors, with 1 as the trivial base case. For the non base case, wherein our number is composite, apply the inductive hypothesis on the factors. For uniqueness, assume two distinct factorizations: p_1p_2\ldots p_n = q_1q_2\ldots q_n. By Euclid’s lemma, each of the p_is divides and is thus equal to one of the q_is. Keep invoking Euclid’s lemma, canceling out a prime factor on each iteration and eventually we must end with 1 = 1 in order for the two sides to be equal.