Xiaochuan Liu's Weblog

博客搬家

Posted 2011/03/31 by liuxiaochuan
Categories: life

我决定独立建站，给博客搬家。感谢佐拉的帮助。新地址:

http://www.liuxiaochuan.org

回顾这几年写博客的历程，我收获很大。我写过乱七八糟的东西当中，少数文章确实有不小影响。其中一篇译作甚至被一个山东大学办的杂志抄袭，这可能算是体现出有影响的“好现象”吧。我有过很多张狂的言论，也认识了很多人，打开了自己的视野。正如学数学需要对很熟悉的理论反复的思考和理解，不断的重新“温故知新”，我发现生活中处处有类似的感觉。我个人也享受重新做一件有价值的事情的过程。我打算将之前写过的东西扔在原来的wordpress上不动。希望大家们到我的新站继续关注。如今的互联网允许每一个人都是自己的出版商，这是表达自己的机会。每个人都应该试试。

有一些老生常谈，我近来有比较深刻的切身体会，我想重复一遍，给自己提个醒儿。

1.不应该试图改变其他人。

2.试图改变自己，要充分意识到困难，而且往往积年累月也只改变一点点，甚至这一点点也需要非凡的毅力。

3.与人相处直接简单，礼貌谦逊，还有就是不卑不亢。

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Problem For Week Nine

Posted 2011/01/20 by liuxiaochuan
Categories: One Problem A Week

Problem Nine (Jan 19th): Let $V=R^n$ with standard orthorgonal basis $\epsilon_1,\epsilon_2,\cdots,\epsilon_n$ , and define $\Phi$ to be the set of all $2n(n-1)$ vectors $\pm\epsilon_i\pm\epsilon_j, (1\leq i<j\leq n)$ . Let W be the group generated by the $2n(n-1)$ reflections in $R^4$ sending $\alpha = \pm\epsilon_i\pm\epsilon_j$ to its negative while fixing pointwise the hyperplane that is orthorgonal with $\alpha$ . Prove that W is the semidirect product of the symmetric group $S_n$ and $\Bbb Z_2^{n-1}$ .

Solutions to last problem: $p$ be a prime, and let $\Bbb Z_p=\Bbb Z / p\Bbb Z$ denote the congruence classes in $\Bbb Z$ modulo $p$ . $\Bbb Z_p^*=\Bbb Z_p-\{0\}$ also admits multiplication, making $\Bbb Z_p$ a finite field.

Now we fix a $n$ and define $G_n=\{x^n:x\in \Bbb Z_p^*\}$ . Then $G_n$ is a subgroup of $\Bbb Z_p^*$ , and we can split $\Bbb Z_p^*$ as coset space $\Bbb Z_p^*=a_1G_n\cup a_2G_n\cup\cdots\cup a_rG_n$ .

Since $z\to z^n$ is a surjective homomorphism from $\Bbb Z_p^*$ to $G_n$ , we have from the the fundamental theorem on homomorphisms that $r$ is the number of solutions of the function $x^n=1$ in $\Bbb Z_p^*$ .

When $x \in a_i G_n$ , color x with $i$ . This gives an $r$ coloring of $\Bbb Z_p^* = \{1, 2,\cdots, p - 1\}$ . If $p-1 \geq S(n) \geq S(r)$ , then by theorem 2 there is a monochromatic Schur triple, that is, there are integers $x, y, z$ , none of which is divisible by p, such that $a_ix^n + a_iy^n \equiv a_iz^n \mod p$ , since $a_i$ is not divisible by $p$ it follows that $x_n + y_n = z_n \mod p$ , completing the proof.

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Problem for week eight

Posted 2011/01/09 by liuxiaochuan
Categories: One Problem A Week

Problem Eight (Jan. 9th, 2011): Prove theorem 1 using theorem 2.

Theorem 1. Let $n > 1$ . Then there exists an integer $S(n)$ such that for all primes $p > S(n)$ the congruence $x^n + y^n \equiv z^n \pmod p$ has a solution in the integers, such that $p$ does not divide $xyz$ .

Theorem 2. (Schur’s theorem)Let $r> 1$ . Then there is a natural number $S(r)$ , such as if $N > S(r)$ and if the numbers $\{1, 2, . . . ,N\}$ are colored with $r$ colors, then there are three of them $x, y, z$ of the same color satisfying the equation: $x + y = z$ .

Solutions to last problem: Suppose a abstract simplicial $K$ have $m+1$ vertices $a^0,a^1,\cdots,a^m$ . We say $a^0,\cdots,a^n$ are geometric independent when the vectors $a^1-a^0,a^2-a^0,\cdots,a^n-a^0$ are linearly independent. We first show the fact that we can find in $R^{2n+1}$ $m+1$ points such that any $q\leq 2n+2$ of them are geometric independent.

We induct on the number $k$ and suppose that we have already found points $A^0,A^1,\cdots, A^k$ such that any $2n+2$ (or less) vertices are geometric independent. Since any $2n+1$ (or less) vertices are geometric independent, they have to lie in a space of at most $2n$ dimension. These $2n$ dimensional subspaces cannot make full of the whole $R^{2n+1}$ . We then choose a point $A^{k+1}$ which doesn’t lie in any of these $2n$ dimensional subspaces.Thus we complete our induction by choosing an additional $A^{k+1}$ making any $2n+2$ (or less) vertices choosing from $A^0,A^1,\cdots, A^k$ are geometric independentm, completing the induction step.

Then we can choose points $A^{0},A^{1},\cdots,A^{k+1}$ to corespond the abstract vertices $a^0,a^1,\cdots,a^{k+1}$ . We now verify they are the geomeric realization of the abstract simplicial $K$ . We only need to check the the intersection of any two simplices $\sigma_1, \sigma_2\in K$ is a face of both $\sigma_1$ and $\sigma_2$ . This is easily seen because there are totally no more than $2n+2$ points involved and thus can form a simplicial $\sigma'$ . This means $\sigma_1\cap \sigma_2$ is either empty set or a common face of $\sigma_1$ and $\sigma_2$ .

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Problem for week seven

Posted 2010/12/05 by liuxiaochuan
Categories: One Problem A Week

Problem Seven:(Dec 5th,2010) Suppose $K$ is an abstract simplicial complex of dimension $n$ . Show that $K$ has its gemotric realiztion in Euclidean space $R^{2n+1}$ .

Solutions to last problem: Denote the length of a permutation as $\ell(\pi)$ . On one hand, every transposition can change the number of invrsions by one. So one easily sees that the length of a permutation $\pi$ is no less than the number of “inversions” in $\pi$ . Read the rest of this post »

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Concave compositions

Posted 2010/11/14 by liuxiaochuan
Categories: combinatorics, Math.CO

George Andrews gives a new definition, namely, “concave compositins”. There are several different kinds of concave compositions. He then derived the generating functions of all these concave compositions. I wrote a post about this months ago.

Concave compositions of even length is the ordered pair of partitions with distinct parts with the property that they have the same length and that the last part are equal. We also note that zero part is allowed, too.

Recently, I figured out a combinatorial proof of the generating function of concave compositions of even length. I just submitted this work. I include the first draft of this paper here.

Comments: 2 Comments

Problem for week six

Posted 2010/11/14 by liuxiaochuan
Categories: combinatorics, One Problem A Week

(Due to my ~~business~~ laziness, this week is a little long, longer than a month.)

Problem 6: Denote the transpositions in the symmetric group $S_n$ as $(i,i+1)$ . Define the length of a permutation $\pi$ as the least number of transpositions that formulate the permutation: $\pi$ . Show that the length of a permutation $\pi$ is the number of “inversions”: the number of pairs $latx i<j$ for which $\pi(i)>\pi(j)$ .

Solution to last problem:

Define the rank of a partition $\lambda$ to be $r(\lambda)=a(\lambda)-\ell(\lambda)$ , which would be the difference of the length of the first part of $\lambda$ and the number of parts of $\lambda$ . Then the following finishes the proof.

$\sum_{n=1}^\infty q^{n(3n-1)/2}(1-q^n)$

$=\sum_{n=1}^\infty (\sum_{ \lambda\in \mathcal {D}_n \atop \lambda \text{ has odd number of parts}}(-1)^{r(\lambda)}$

$- \sum_{ \lambda\in \mathcal {D}_n\atop \lambda \text{ has even number of parts}}(-1)^{r(\lambda)} )q^n$

$=\sum_{n=1}^\infty (\sum_{ \lambda\in \mathcal {D}_n; \lambda \text{ has odd number of parts}\atop \text { the largest part is odd}}1$

$-\sum_{ \lambda\in \mathcal {D}_n; \lambda \text{ has odd number of parts}\atop \text{the largest part is odd}}1)q^n$

$-(\sum_{ \lambda\in \mathcal {D}_n; \lambda \text{ has even number of parts}\atop \text{the largest part is even}}1$

$- \sum_{ \lambda\in \mathcal {D}_n; \lambda \text{ has even number of parts}\atop \text{ the largest part is odd}}1)q^n$

$=\sum_{n=1}^\infty (\sum_{ \lambda\in \mathcal {D}_n; \lambda \text{ has odd number of parts}\atop\text { the largest part is odd}}1$

$+ \sum_{ \lambda\in \mathcal {D}_n; \lambda \text{ has even number of parts}\atop \text{ the largest part is odd}}1 )q^n$

$-(\sum_{ \lambda\in \mathcal {D}_n; \lambda \text{ has even number of parts}\atop \text{the largest part is even}}1$

$+\sum_{ \lambda\in \mathcal {D}_n; \lambda \text{ has odd number of parts}\atop \text{the largest part is odd}}1 )q^n$

$=\sum_{n=1}^\infty(\sum_{\lambda\in \mathcal {D}_n \atop \text{the largest part is odd}}1$

$-\sum_{\lambda\in \mathcal {D}_n\atop \text{the largest part is even}}1)q^n$

$=\sum_{n=1}^\infty(P_o(\mathcal {D}_n)-P_e(\mathcal {D}_n))q^n$

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Problem for week five

Posted 2010/10/07 by liuxiaochuan
Categories: combinatorics, One Problem A Week

Problem 5: (2010.10.07) Without applying Franklin’t involution, try to prove the following two lemmas from one to another.

Lemma 1: (Euler’s pentagonal number theorem)

$\prod_{n=1}^\infty(1-q^n)=1+\sum_{m=1}^\infty(-1)^mq^{\frac{1}{2}m(3m-1)}(1+q^m)$ Read the rest of this post »

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Xiaochuan Liu's Weblog

博客搬家

Problem For Week Nine

Problem for week eight

Problem for week seven

Concave compositions

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