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# System prepended metadata

title: Some problems

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# 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.

<br>
<br>

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
$$
\begin{align}
h_{k}:&=\dim \text{Harm}_k(\mathbb R^n)\\
&=\dim \text{Hom}_k(\mathbb  R^n)-\dim \text{Hom}_{k-2}(\mathbb R^n)\\
&=\binom{n-1+k}{k}-\binom{n-3+k}{k-2}
\end{align}
$$


<br>



### 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 $0$.)

<br>

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.

<br>

### 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.

<br>

### 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 
$$
\begin{align}
&\sum_{x\in X} \phi_{k,i}(x)=0,\forall k=1,\dots,t,1\leq i\leq h_k\\
\iff& \sum_{x\in X} \phi_{k,i}(x)^2=,\forall k=1,\dots,t,1\leq i\leq h_k\\
\iff&\sum_{i=1}^{h_k}\sum_{x\in X}\phi_{k,i}(x)^2=0\\
\iff& \sum_{i=1}^{h_j}(\sum_{x\in X} \phi_{k,i}(x) )(\sum_{y\in X}\phi_{k,i}(y))=0\\
\iff& \sum_{x\in X} Q_k(x\cdot y)=0
\end{align}
$$

<br>

## 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 $(x,y)\in R_k$.

<br>

### 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 $p_{ij}^k=p_{ji}^k$.

<br>
### 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,
$$
P(S_4)=\begin{pmatrix} 
1& 6& 8& 6&3\\
1&-6& 8&-6&3\\
1& 2& 0&-2&-1\\
1&-2& 0& 2&-1\\
1& 0&-4&0&3
\end{pmatrix}\\
P(A_5)=\begin{align}
\begin{pmatrix}
1&20&15&12&12\\
1&5&0&-3&-3\\
1&-4&3&0&0\\
1&0&-5&2(1+\sqrt 5)&2(1-\sqrt 5)\\
1&0&-5&2(1-\sqrt 5) &2(1+\sqrt 5)
\end{pmatrix}
\end{align}
$$
