Unfolding of a global integral for $GL_n \times GL_m$

Let \((\pi, V_\pi)\) and \((\pi^\prime, V_{\pi^\prime})\) be cuspidal, unitary, irreducible automorphic representations of \(GL_n(\mathbb{A})\) and \(GL_m(\mathbb{A})\) respectively. We assume \(m < n\). Let \(\varphi \in V_\pi\) and \(\varphi \in V_{\pi^\prime}\). To pair \(\varphi\) and \(\varphi^\prime\) suitably together, we first need to project \(\varphi\) correspondingly.

Let \(\psi\) be a additive continuous automorphic character of \(\mathbb{A}\). We can extend it to a character of \(N_n(\mathbb{A})\), the standard Borel subgroup of \(GL_n(\mathbb{A})\), in the standard way: \[\psi(u) = \psi\left(\sum_{i=1}^{n-1}u_{i,i+1}\right),\] for \(u = (u_{i,j}) \in N_n(\mathbb{A})\). Let \(Y=Y_{n, m}\) be the standard unipotent radical associated to the partition \((m+1, 1, \dots, 1)\) of \(GL_n\), i.e., \[Y=\left\{\left(\begin{array}{cc}I_{m+1}&*\\ 0&u\end{array}\right): u \in N_{n-m-1}\right\} \subset N_n.\] Let \(P_{m+1}\) be the mirabolic subgroup of \(GL_{m+1}\), then for \(p \in P_{m+1}(\mathbb{A})\) define \[\mathbb{P}^n_m \varphi(p) = |\det p|^{-\frac{n-m-1}{2}}\int_{Y(k)\backslash Y(\mathbb{A})} \varphi\left(y\left(\begin{array}{cc} p& \\ & I_{n-m-1} \end{array}\right)\right)\psi^{-1}(y) dy.\]

Now we can pair \(\varphi\) and \(\varphi^{\prime}\) in the following way: \[I(s; \varphi, \varphi^\prime) = \int_{GL_m(k)\backslash GL_m(\mathbb{A})} \mathbb{P}^n_m \varphi\left(\begin{array}{cc} h& \\ &1\end{array}\right)\varphi^\prime(h)|\det h|^{s-\frac{1}{2}}dh.\]

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Nontrivial bound for a sum involving characters

To find a non-trivial bound for an expression is an essential task in analytic number theory. For example, to derive a asymptotic formula for the number of primes less than or equal to \(x\), i.e. \(\pi(x)\), we investigate into the function \[\Psi(x) = \sum_{\substack{n \le x \\ n = p^k}} \log p.\] Along the way to the asymptotic formula for \(\Psi(x)\), it’s important to bound the error terms so that they won’t overweight the dominant term. Nowadays, many mathematicians are interested in finding non-trivial bounds for weighted \(\Psi(x)\), i.e., \[\Psi(x) = \sum_{\substack{n \le x \\ n = p^k}} a_n \log p,\] where \(a_n \in \C\). If \(a_n = \chi(n)\) for some character \(\chi \text{ mod }q\), this is actually \(\Psi(x, \chi)\). Trivially, we can have \[\left\lvert \Psi(x) \right\rvert \le \sum_{\substack{n \le x \\ n = p^k}} |a_n| \log p.\] Therefore, it seems that we can just work with \(a_n \in \R_{\ge 0}\). Truly, this will give us a bound for \(\Psi(x)\), but it ignores all the cancellation encoded in the signs of \(a_n\), thus it won’t provide any good bounds.

Another kind of expressions that many mathematicians work with are those sums involving characters. The simplest example is \(\displaystyle \sum_{n=m+1}^{m+N} \chi(n)\) for a character \(\chi \text{ mod }q\). The trivial bound for it is \(N\), since \(|\chi(n)| \le 1\), which is certainly a coarse bound. Actually, we have

Theorem 1 (Pólga-Vinogradov inequality). Let \(\chi \text{ mod }q\) be a primitive character and \(q \ge 3\). Then \[\left\lvert \sum_{n=m+1}^{m+N} \chi(n) \right\rvert < \sqrt{q}\log q.\]

This non-trivial bound doesn’t depend on the length \(N\) at all, but only depends on the modulus \(q\). If the primitive condition is dropped, then it is bounded by \(2 \sqrt{q}\log q\).

In this post, I will give a non-trivial bound for the sum \(\displaystyle \sum_{n=1}^{N} \chi(n)\chi(n+1)\), which was originally asked by my friend Yongxiao, as a practice for my analytic number theory course.

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2014年

我本来想着从Florida回到Columbus就为2014写点东西,可是一直拖延到了今天眼看着新的学期快开始了才急急忙忙着手这事。其实写年度总结之类的博文还是挺有趣的,能看看自己这一年以来做过什么趣事傻事、做出什么样的进步。

2014年我经历了哥村的零下20度,听了马友友的小提琴,暑假回国跟亲人好友见面,过了口语可以教书,选了导师成了Tango Club的President,第一次教书是教微积分1,现在在Florida的迪士尼等烟花。新年快乐。

这是我在2014年最后一天发的微博,发微博的时候我正在迪士尼外面的停车场等凌晨的烟花,心情实际上并不是兴奋而是有点因孤单而生出的伤感。迪士尼并不适合单身一人游玩,一个人行走在拥挤且快乐的人群里总是有点尴尬,以至于我在一个小岛的岸边静静地坐了一个下午,看着一艘轮船一圈又一圈地环绕着小岛。晚上八点半左右我就从迪士尼里出来了,回到停车场的巴士上了。让我想不到的是已经有人比我先回到了巴士上了,是一群男生,所以我也不是太惊讶了。这次出来Florida玩我是跟团的,只是团里的人我一个不认识。跟一群陌生的人出去玩是一件很有挑战性的事情:他们基本上都是三三两两认识的人参团的,我要做的第一件事情就是打进“敌人”内部,难度在于我要用我那蹩脚的英语进行这项任务。后来我是跟一个来自印度尼西亚的家庭混熟了,母亲是OSU的博士(可是看起来比我母亲大人还老啊),父亲和女儿跟了过来在美国这边生活。他们就坐在我对面,聊起来也实在方便,在旅途的车上,有人能对上话倒也不至于太无聊。这次出来玩了一个星期,除了在迪士尼那天之外,我还是玩得挺开心的。在旅途的最后一天,我还很幸运地见到了一位高中同学。说起这件事也是挺巧合的,最后一天我们再次来到了环球影城,我在冒险岛开心游荡着,很快就把我感兴趣的点玩遍了(single rider太贴心了)。在下午两点的时候,我就打算离开了,走到大门刷了一下朋友圈竟然发现一位高中同学刚发了一张环球影城里的黄油啤酒的图。于是我又跑到哈利波特那里坐了趟列车到环球影城那边与朋友见上一面。与朋友见面的地方是我前天去的时候没有去过的地方,我也在她的推荐下尝了那美味的黄油啤酒。这旅途的惊喜,让我感到十分高兴。

沙滩上的日落-2014.12.28@St Petersburg

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Induced representations and Mackey theory

Let \(G\) be a group. A linear representation of \(G\) is a pair \((\rho, V)\), where \(V\) is a finite dimensional vector space over \(\mathbb{C}\) and \(\rho: G \to GL(V)\) is a group homomorphism. Without ambiguity, we will call \(\rho\) a representation of \(G\). Let \((\rho_1, V_1)\) and \((\rho_2, V_2)\) be two representations of \(G\), an intertwining operator between them is a linear map \(T: V_1 \to V_2\) such that for any \(g \in G\), the following diagram commutes: \[\begin{equation}\label{eq:diagram} \require{AMScd}\begin{CD} V_1 @>T>> V_2 \\ @V{\rho_1(g)}VV @VV{\rho_2(g)}V \\ V_1 @>T>> V_2 \end{CD} \end{equation}\] Let \(\text{Hom}_G(\rho_1, \rho_2)\) be the space of all intertwining operators between \(\rho_1\) and \(\rho_2\).

Let \((\rho, V)\) be a representation of \(G\). Let \(H\) be a subgroup of \(G\), then we can restrict \(\rho\) to \(H\) to get a representation of \(H\). We will use \(\text{Res}^G_H \, \rho\) to denote the restricted representation of \(H\) from \(\rho\). Conversely, if \(\pi\) is a representation of \(H\), then we can construct a representation of \(G\) from \(\pi\), which is known as induced representation of \(G\) from \(\pi\), denoted as \(\text{Ind}_H^G \, \pi\). In this post, I will first talk about the precise description of induced representations and the relations between \(\text{Res}^G_H\) and \(\text{Ind}_H^G\). I will then discuss Mackey’s theorem, which dictates a further relation between restricted and induced representations.

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Tensor product of two linear maps

Let \(A \in End(V)\) and \(B \in End(W)\) be two linear maps. We can define naturally the tensor product \(A \otimes B\) of \(A\) and \(B\), from \(V \otimes W\) to \(V \otimes W\), sending \(v \otimes w\) to \(Av \otimes Bw\). In this post, I am going to realize \(A \otimes B\) as a matrix and relate the determinant and trace of \(A \otimes B\) to the ones of \(A\) and \(B\).

Let \(V\) and \(W\) be two vector spaces over a field \(K\) with \(\dim V=n\) and \(\dim W=m\). Let \(e_1, \dots, e_n\) be a basis of \(V\) and let \(f_1, \dots, f_m\) a basis of \(W\). Under the basis, a linear map \(A: V \to V\) can be realized as an \(n \times n\) matrix \((a_{ij})_{1 \le i,j \le n}\), where \(a_{ij} \in K\). Similarly, a linear map \(B: W \to W\) can be realized as a \(m \times m\) matrix \((b_{kl})_{1 \le k, l \le m}\). Now, \(V \otimes W\), as a vector space, has basis \(\{e_i \otimes f_j: 1 \le i \le n, 1 \le j \le m\}\). And, \[\begin{align*} A \otimes B (e_i \otimes f_j) &= A(e_i) \otimes B(f_j) \\ &= \sum_{k}a_{ki}e_k \otimes \sum_{l}b_{lj}f_l \\ &= \sum_{k,l}a_{ki}b_{lj} e_k \otimes f_l. \end{align*}\]

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Locally closed subgroups are closed

In a topology group \(G\), an open subgroup \(H\) is also closed. The proof of this statement is not hard: \(G=\bigcup_{g_i} g_iH\) is a disjoint union of open left cosets, where \(\{g_i\}\) is a complete representatives set of \(G/H\). Then \(\bigcup_{g_i \ne 1}g_iH\) is open and \(H=G-\bigcup_{g_i \ne 1}g_iH\) is the complement of an open set, and therefore \(H\) is closed. In this post, I will prove a slightly more general theorem:

Theorem. Let \(G\) be a topological group. If \(H\) is a locally closed subgroup in \(G\), then \(H\) is closed.

We will see that an open subgroup \(H\) is locally closed, so it’s closed by the Theorem. Before the proof of the Theorem, let’s talk about locally closed sets first.

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Examples of split and non-split tori

Let \(T\) be an algebraic group. \(T\) is called an algebraic torus over \(k\), if \(T(E)\) is isomorphic to a finite direct product of copies of \(G_m(E)\) for some finite finite extension \(E\) of \(k\), where \(G_m\) is the multiplicative group. If \(E\) can be \(k\), then \(T\) is called a split torus over \(k\); otherwise, \(T\) is called a non split torus over \(k\). In this post, I am going to talk something about \(SO\) to give examples of non split and split tori.

Definition. Let $k$ be a field, and $V$ be a vector space over $k$. Let $q$ be a quadratic form on $V$. Define $$SO(V,q;k)=\{\gamma \in SL(V) : q(\gamma v)=q(v), \forall v \in V\}.$$

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The Mellin transform

Definition

Let \(\mathbb{R}^{+}\) be the set of positive real numbers. Given a function \(f\) on \(\mathbb{R}^{+}\), define the Mellin transform of \(f\), whenever it makes sense, as follow: \[\mathcal{M}(f)(s)=\int_{0}^{\infty} f(t)t^{s} \frac{dt}{t}. \; (1)\]

The very first example of the Mellin transform I have known is the gamma function, \[\Gamma(s) = \int_{0}^{\infty} e^{-t}t^{s} \frac{dt}{t},\] which is the Mellin transform of the function \(e^{-t}\).

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$k$ squares problem

Let’s fix a natural number \(k\). How many ways can we decompose \(n\) into \(k\) squares? It’s an interesting problem and it should help me understand automorphic forms better.

Let \(w_k(n)\) be the number of tuples \((n_1, \cdots, n_k)\in \mathbb{Z}^k\) such that \(n_1^2+\cdots+n_k^2=n\). Let \(g_k(q)\) be the generating function of \(w_k(n)\), i.e., \[g_k(q)=\sum_{n \in \mathbb{N}} w_k(n)q^n.\] Then, \[\begin{split} g_k(q) &= \sum_{n \in \mathbb{N}} w_k(n)q^n \\ &= \sum_{n \in \mathbb{N}} \sum_{\substack{n_1, \cdots, n_k \in \mathbb{Z} \\ n_1^2+\cdots+n_k^2=n}}q^n \\ &= \sum_{n_1, \cdots, n_k \in \mathbb{Z}} q^{n_1^2+\cdots+n_k^2} \\ &= \left(\sum_{n \in \mathbb{Z}} q^{n^2}\right)^k \\ &= (g_1(q))^k. \end{split}\]

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Self hypnosis and insomnia

I have been suffering from insomnia since I came back to Columbus. I thought it was just a matter of jet lag. But after a week or two, I began to realize that it is something else, something unknown yet, that matters. Insufficient sleep turns me into a zombie around 2 or 3 pm in the afternoon.

I had exactly the same problem last December. I tried many ways and finally figured out alcohol worked it out for me. However, alcohol has lost its power on me this time, soon after I finished my rum. Now I go to swimming and work out every other day to make me exhausted at the end of day. Theoretically, this should make me tired and then I could fall asleep easier. Reality always beats theory. It just worsen the consequence of insomnia.

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