Some problems - HackMD

# Some problems ## 2021.07.15 ### Problem 1 Consider the Laplace functor $\Delta$ as a map from $\text{Hom}_k(\mathbb R^n)$ to $\text{Hom}_{k-2}(\mathbb R^n)$. Show 1. the map is onto. 2. $(x_1^2+\cdots+x_n^2)\text{Hom}_{k-2}(\mathbb R^n)\cap\text{Harm}_k(\mathbb R^n)=\{0\}$ *proof* Say $f=\prod x_i^{c_i}$. Define $m(f)=\max \{c_1,\dots,c_n\}$. If $m=n,n-1$, we know $f\in \text{Im }\Delta$. Now suppose for all $m(f)>M$, we have $f\in \text{Im }\Delta$. Consider $g=\prod x_i^{d_i}$ with $m(g)=M$. W.L.O.G., we may assume $d_1=M$. And we'll get $$ \Delta (x_1^2\cdot g)=x^{m+2}\Delta(\prod_{i=2}^n x_i^{d_i} )+(m+2)(m+1)\cdot g $$ By hypothesis, we have $x^{m+2}\Delta(\prod_{i=2}^n x_i^{d_i})\in \text{Im }\Delta$, this implies $g\in \text{Im }\Delta$. By induction downward, we know the map is surjective. Rather than showing the second statement, we show $$ \text{Hom}_k(\mathbb R^n)=\bigoplus_{0\leq j\leq \lfloor \frac k2\rfloor} (x_1^2+\cdots+x_n^2)^j\cdot\text{Harm}_{k-2j}(\mathbb R^n) $$ To prove this fact, we first note that there's a inner product on $\mathbb R[x_1,\dots,x_n]$, given by $$ \langle f ,g \rangle=\frac 1{A(S^{n-1})}\int_{S^{n-1}} f(x)g(x)~\text d\sigma(x) $$ where $\text d\sigma$ is the surface measure. Then we have the following fact **Fact** $$ \langle f,g\rangle=0\\\forall f\in \text{Harm}_k(\mathbb R^n),g\in \text{Harm}_l(\mathbb R^n),k\neq l $$ This comes from Green's identity. (*ref*: Approximation theory and harmonic analysis on spheres and balls. Feng Dai and Yuan Xu) **Theorem** $$ \text{Hom}_k(\mathbb R^n)=\bigoplus_{0\leq j\leq \lfloor \frac k2\rfloor} (x_1^2+\cdots+x_n^2)^j\cdot\text{Harm}_{k-2j}(\mathbb R^n) $$ For $k=0,1$, we have $\text{Hom}_k(\mathbb R^n)=\text{Harm}_k(\mathbb R^n)$. Suppose the statement holds for $k=0,\dots,m-1$. Consider $(x_1^2+\cdots+x_n^2)\cdot \text{Hom}_{m-2}(\mathbb R^n)$ as the subspace of $\text{Hom}_m(\mathbb R^n)$. And by above fact, we have $$ \langle (x_1^2+\cdots+x_n^2)f,g\rangle=0,\\ \forall f\in \text{Hom}_{m-2}(\mathbb R^n),g\in \text{Harm}_m(\mathbb R^n) $$ Then by (1), we know $$ \begin{align} \dim \text{Harm}_{m}(\mathbb R^n)+\dim \text{Hom}_{m-2}(\mathbb R^n)&= \dim \text{Hom }_{m}(\mathbb R^n) \end{align}$$Then we have the desired conclusion. **Remark** This tells us $$ \begin{align} h_{k}:&=\dim \text{Harm}_k(\mathbb R^n)\\ &=\dim \text{Hom}_k(\mathbb R^n)-\dim \text{Hom}_{k-2}(\mathbb R^n)\\ &=\binom{n-1+k}{k}-\binom{n-3+k}{k-2} \end{align} $$ ### Problem 2 Show T.F.A.E., 1. $X$ is a spherical $t$-design 2. $$\sum_{x\in X}f(x)=0,\forall f\in\text{Harm}_i(\mathbb R^n), 1\leq i\leq t $$ *proof* (1)$\implies$(2) $X$ is a spherical $t$-design means $$ \frac{1}{|X|}\sum_{x\in X}f(x)=\frac1{|S^{n-1}|}\int_{S^{n-1}} f(x)~\text d\sigma(x) $$ for all polynomial of degree $t$. Then by the fact mentioned in problem 1, we immediately know that $$ \sum_{x\in X} f(x)=0,\forall f\in \text{Harm}_i(\mathbb R^n),1\leq i\leq t $$(Since constant function $1$ is harmonic of degree $0$.) For the converse, for any $f\in \text{Hom}_k(\mathbb R^n)$, by the decomposition in problem 1, we know we can write $f$ as $$ \sum_{0\leq i\leq \lfloor \frac k2\rfloor} (x_1^2+\cdots+x_n^2)^ig_i(x) $$where $g_i(x)$ is harmonic polynomial of degree $k-2i$. So we have $$ \sum_{x\in X}f(x)=\sum_{x\in X}\sum_ig_i(x) $$Since $||x||^2=1,\forall x\in X$. And for all harmonic function $g$ of degree $1\leq i\leq t$, we have $$ \frac 1{|X|}\sum_{x\in X} g(x)=0=\frac{1}{|S^{n-1}|}\int_{S^{n-1}} g(x)~\text d\sigma(x) $$ And for polynomials of degree $0$, i.e. constant, we have $$ \frac{1}{|X|}\sum_{x\in X}c=c=\frac{1}{|S^{n-1}|}\int_{S^{n-1}} c~\text d\sigma(x) $$This shows $X$ is a spherical $t$-design. ### Problem 3 For $n=2$, the $t+1$ vertices of a regular $(t+1)$-gon inscribed in $S^1(⊂ \mathbb R^2)$ form a $t$-design. *hint* For $n=2$, we have $$ \text{Harm}_k(\mathbb R^n)=\{\text{Re}((x+yi)^k),\text{Im}((x+yi)^k)\} $$For example, $$ \begin{align} \text{Harm}_1(\mathbb R^n)&=\{x,y\}\\ \text{Harm}_2(\mathbb R^n)&=\{x^2-y^2,2xy\}\\ \text{Harm}_3(\mathbb R^n)&=\{x^3-3xy,3x^2y-y^3\}\\ \end{align} $$ *proof* Consider in complex plane, the vertices of a regular $(t+1)$-gon is $V_{t+1}:=\{1,\zeta_{t+1},\dots,\zeta_{t+1}^{t}\}$. And observe the following fact $$ \sum_{z\in V} z^k=\begin{cases} 0&\text{if }1\leq k\leq t\\ t+1&\text{if }k=t+1 \end{cases} $$ Combine this with the fact $$ \int_{S^1} z^k =0 $$(Since $z^k$ is holomorphic), we get the conclusion. ### Problem 4 Show T.F.A.E 1. $X$ is a spherical $t$-design on $S^{n−1}$. 2. $H_l^\mathsf TH_0 = 0$ holds for $l = 1, 2, . . . , t$. 3. For any non-negative integers $k$ and $l$,with $0 ≤ k + l ≤ t$,$H_k^\mathsf TH_l = |X|\Delta_{k,l}$ holds. Here we define $\Delta_{k,k}$ to be the identity matrix and $\Delta_{k,l}$ is the zero matrix if $k\neq l$. ($\Delta_{k,l}$ is the matrix of size $h_k × h_l$.) 4. $H^\mathsf T_eH_e = |X|I$ and $H_e^\mathsf TH_r= 0$, where $e = \lfloor t/2\rfloor$ and $r=e−(−1)^t$. 5. $\displaystyle \sum_{(x,y)\in X\times X}Q_k(x\cdot y)=0,\text{ for }k=1,2,\dots,t$ Note that it's known that for any subser $X$ of $S^{n-1}$ and any non-negative integer, we have $$ \sum_{(x,y)\in X\times X}Q_{k}(x\cdot y)\geq 0 $$ *proof* (1)$\iff$(2) Observe that $H_0$ is a $|X|\times 1$ matrix, whose entries are all $1$. So the condition can be written as $$ \sum_{x\in X} \phi_{l,i}(x)\phi_{0}(x)=\sum_{x\in X}\phi_{l,i}(x)=0 $$which is just the statement of problem 2. (1)$\implies$(3) The $(i,j)$-entry of $H_k^\mathsf TH_l$ is $$ \begin{align} \sum_{x\in X} \phi_{l,i}(x)\phi_{k,j}(x)&=\frac{|X|}{|S^{n-1}|}\int_{S^{n-1}} \phi_{l,i}(x)\phi_{k,j}(x)~\text d\sigma(x)\\ &=|X|\cdot \langle \phi_{l,i},\phi_{k,j}\rangle \end{align} $$We know the integral of R.H.S is $0$ if $l\neq k$ by the lemma stated in problem 1. And if $l=k$, we know the term is nonzero if and only if $i=j$. This proves (3) (3)$\implies$(4) This implication is clear. (4)$\implies$(5) (5)$\iff$(2) By addition theorem on the sphere, we have $$ \begin{align} Q_k(x\cdot y)=\sum_{i=1}^{h_k} \phi_{k,i}(x)\phi_{k,i}(y) \end{align} $$ So $$ \begin{align} &\sum_{x\in X} \phi_{k,i}(x)=0,\forall k=1,\dots,t,1\leq i\leq h_k\\ \iff& \sum_{x\in X} \phi_{k,i}(x)^2=,\forall k=1,\dots,t,1\leq i\leq h_k\\ \iff&\sum_{i=1}^{h_k}\sum_{x\in X}\phi_{k,i}(x)^2=0\\ \iff& \sum_{i=1}^{h_j}(\sum_{x\in X} \phi_{k,i}(x) )(\sum_{y\in X}\phi_{k,i}(y))=0\\ \iff& \sum_{x\in X} Q_k(x\cdot y)=0 \end{align} $$ ## 2021.07.16 **Definition**(association scheme) $\mathfrak X=(X,\{R_i\}_{0\leq i\leq d})$ is an association scheme if 1. $\{R_i\}_{0\leq i\leq d}$ forms a partition of $X\times X$. 2. $\{R_0\}=I$. 3. $R_i^\mathsf T=R_{i'}$. 4. $p_{ij}^k:=|\{z\in X|(x,z)\in R_i,(z,y)\in R_j\}|$ is a constant for all $(x,y)\in R_k$. ### Problem 1 Let $G$ be a finite group and let $C_0=\{1\},C_1,\dots,C_d$ be the conjugacy classes of $G$. Set $X=G$ and define $(x,y)\in R_i\iff xy^{-1}\in C_i$. Then $(X,\{R_i\}_{0\leq i\leq d})$ forms a commutative associative scheme. *proof* The first two are obvious. The third condition holds since $$ \begin{align} &(x,y),(z,w)\in R_i\\ \iff& xy^{-1}=a\cdot zw^{-1}\cdot a^{-1}\\ \iff& yx^{-1}=a\cdot wz^{-1}\cdot a^{-1}\\ \iff& (y,x),(w,z)\in R_i \end{align} $$ Since $G$ is a finite group, the map $a\to ga$ is an automorphism of $G$. So for any $(z,w)\in R_i$, we can find $g\in G$ s.t. $(z,w)=(xg,yg)$. Now, for any $(x,y)\in R_k$, define $$ S(x,y)_{i,j}=\{z\in G| (x,z)\in R_i, (z,y)\in R_j\} $$Then observe the map $z\to zg$ is an bijection $S(x,y)_{i,j}$ to $S(x',y')_{i,j}$. $\forall z'\in S(x',y')_{i,j}$, $(x',z')=(xg,z'g^{-1}g)\in R_i\iff (x,z'g^{-1})\in R_i$. Similarly for $R_j$. So $z'\in S(x',y')_{i,j}\iff z'g^{-1}\in S(x,y)_{i,j}$. This verifies the fourth condition. Now, we want to show the associative scheme is commutative. i.e. $p_{ij}^k=p_{ji}^k$. Take $(x,y)\in R_k$, let $h=y^{-1}x$, $z'=zh$ we have $$ \begin{align} z\in S(x,y)_{i,j}&\iff xz^{-1}\in C_i,zy^{-1}\in C_j\\ &\iff yhh^{-1}z'^{-1}\in C_i,z'h^{-1}hx^{-1}\in C_j\\ &\iff z'\in S(x,y)_{j,i} \end{align} $$ So $z\to zh$ is a bijection between $S(x,y)_{i,j}$ and $S(x,y)_{ji}$, which shows $p_{ij}^k=p_{ji}^k$. ### Problem 2 Find the first eigenmatrices of the group association $S_3,S_3,A_5$. **Lemma** $$ P=\begin{pmatrix} \frac 1{f_1}&&&\\ &\frac1{f_2}&&\\ &&\ddots\\ &&&\frac1{f_d} \end{pmatrix} \mathbb T \begin{pmatrix} k_1&&&\\ &k_2&&\\ &&\ddots\\ &&&k_d \end{pmatrix} $$where $\mathbb T$ is the character table of $G$. Observe that $k_i=|C_i|$. So we have $$ \begin{align} P(S_3)&= \begin{pmatrix} 1&0&0\\ 0&1&0\\ 0&0&\frac 12 \end{pmatrix} \begin{pmatrix} 1&1&1\\ 1&-1&1\\ 2&0&-1 \end{pmatrix} \begin{pmatrix} 1&0&0\\ 0&3&0\\ 0&0&2 \end{pmatrix} \\&= \begin{pmatrix} 1&3&2\\ 1&-3&2\\ 1&0&-1 \end{pmatrix} \end{align} $$Similarly, $$ P(S_4)=\begin{pmatrix} 1& 6& 8& 6&3\\ 1&-6& 8&-6&3\\ 1& 2& 0&-2&-1\\ 1&-2& 0& 2&-1\\ 1& 0&-4&0&3 \end{pmatrix}\\ P(A_5)=\begin{align} \begin{pmatrix} 1&20&15&12&12\\ 1&5&0&-3&-3\\ 1&-4&3&0&0\\ 1&0&-5&2(1+\sqrt 5)&2(1-\sqrt 5)\\ 1&0&-5&2(1-\sqrt 5) &2(1+\sqrt 5) \end{pmatrix} \end{align} $$