How to construct a table of values for Dirichlet characters?

[Math100]

Homework Statement

Construct a table of values for all the nonprincipal Dirichlet characters modulo $16$.

Relevant Equations

A Dirichlet character modulo $k$ is an arithmetic function $\chi:\mathbb{N}\rightarrow \mathbb{C}$ satisfying
(1) $\chi(n+k)=\chi(n), \forall n\in\mathbb{N}$
(2) $\chi(mn)=\chi(m)\chi(n), \forall m, n\in\mathbb{N}$
(3) $\chi(n)\neq0\Leftrightarrow (n, k)=1$.

The principal character modulo $k$ is the unique Dirichlet character $\chi_{1}$ such that $\chi_{1}(n)=1\Leftrightarrow (n, k)=1$.

If $\chi$ is not the principal character $\chi_{0}$ modulo $k$,
then $\left | \sum_{n\leq x}\chi(n) \right |\leq \varphi(k)$, for $x\geq 1$.
If $\chi=\chi_{0}$, then $\left | \sum_{n\leq x}\chi(n)-\frac{\varphi(k)}{k}x \right |\leq 2\varphi(k)$, for $x\geq 1$.

Since $\varphi(16)=8$, it follows that there are $8$ Dirichlet characters modulo $16$.

 

[Office_Shredder]

I just learned about these now, but maybe you can start by writing down which values of $n$ can be non zero, and what possible values $\chi(n)$ can actually take. I don't think there are that many choices.

 

[Math100]

I just learned about these now, but maybe you can start by writing down which values of $n$ can be non zero, and what possible values $\chi(n)$ can actually take. I don't think there are that many choices.

The principal character $\chi_{1}(n)=1$ if $(n, k)=1$, and $0$ if $(n, k)>1$.
So I think $\chi_{1}(n)=1$ for $n=1, 3, 5, 7, 9, 11, 13, 15$ because if $n$ is even, then the function would equal to $0$. But how should I find other values of $\chi_{2}(n), \chi_{3}(n), ..., \chi_{8}(n)$ for $n=1, 3, 5, 7, 9, 11, 13, 15$?

 

[Office_Shredder]

Do you know any formula that has to be true for, say $\chi(1)$? What about $\chi(3)$?

 

[Math100]

Do you know any formula that has to be true for, say $\chi(1)$? What about $\chi(3)$?

No, what's the formula for them?

 

[Office_Shredder]

No, what's the formula for them?

$\chi(mn)=\chi(m)\chi(n)$. This means $\chi$ is a group homomorphism from the units of $\mathbb{Z}$ mod n to $\mathbb{C}^x$, the complex numbers minus 0 under multiplication.This has powerful implications. The simplest one is what $\chi(1)$ can be while still satisfying $\chi(mn)=\chi(m)\chi(n)$. Pick some choices of $m$ and $n$ that might be useful.

 

[Math100]

$\chi(n)^{\varphi(k)}=1$ whenever $(n, k)=1$.
So $\chi_{i}(3)\in\left \{ 1, -1, i, -i \right \}$. Same thing for $\chi_{i}(5), \chi_{i}(7), ..., \chi_{i}(15)$. But how to determine each value for $\chi_{i}(n)$?

 

[Office_Shredder]

I think you actually get $\chi(3)^8=1$ so it can be any of the 8 roots of unity. Let's say $\chi(3)=\zeta$
Then you have $\chi(9)=\chi(3*3)$. Try to figure out what it is in terms of $\zeta$. Then try adding another factor of 3 and see if you can pin down the next term.

 

[Math100]

Since the order doesn't matter, I got $\chi_{1}(3)=1, \chi_{2}(3)=-1, \chi_{3}(3)=i, \chi_{4}(3)=-i, \chi_{5}(3)=1, \chi_{6}(3)=-1, \chi_{7}(3)=i, \chi_{8}(3)=-i$. So this means $\chi_{1}(9)=1, \chi_{2}(9)=1, \chi_{3}(9)=-1, \chi_{4}(9)=-1, \chi_{5}(9)=1, \chi_{6}(9)=1, \chi_{7}(9)=-1, \chi_{8}(9)=-1$. Because $3^2=9$. Is this right? If so, then what about $\chi(5), \chi(7), \chi(11), \chi(13), \chi(15)$?

 

[Office_Shredder]

What is $\chi(27)$? Don't just write down the first answer you think of and come back, think about it a bit more.

Also, you're still missing the fact that there are 8 choices for $\chi(3)$.

 

[fresh_42]

That looks promising.

Since the order doesn't matter, I got $\chi_{1}(3)=1, \chi_{2}(3)=-1, \chi_{3}(3)=i, \chi_{4}(3)=-i, \chi_{5}(3)=1, \chi_{6}(3)=-1, \chi_{7}(3)=i, \chi_{8}(3)=-i$. So this means $\chi_{1}(9)=1, \chi_{2}(9)=1, \chi_{3}(9)=-1, \chi_{4}(9)=-1, \chi_{5}(9)=1, \chi_{6}(9)=1, \chi_{7}(9)=-1, \chi_{8}(9)=-1$. Because $3^2=9$. Is this right? If so, then what about $\chi(5), \chi(7), \chi(11), \chi(13), \chi(15)$?

Let's have a look at the group we are talking about, namely the group of units of $\mathbb{Z}_{16}.$ This group has $\varphi (16)=8$ elements which are $G=\{1,3,5,7,9,11,13,15\}.$ The elements $3,5,11,13$ are of order four, and $7,9,15$ of order two. Since a Dirichlet character is a group homomorphism into the multiplicative group of complex numbers (see post #6), we have
$$
\chi(k)^2=\chi(k)\cdot \chi(k)=\chi(k\cdot k)=\chi(k^2)=\chi(1)=1\text{ for all }k\in \{7,9,15\},
$$
i.e. only two possible targets $\pm 1.$ We also get
$$
\chi(k)^4=\chi(k^4)=\chi(1)=1\text{ for all }k\in \{3,5,11,13\},
$$
and of course always $\chi(1)=1.$ Complex numbers of order four are only $\pm i ,$ but they can also map onto $\pm 1$ under $\chi.$

Start with elements of order two. There are only $2^2$ possibilities. Since $7\cdot 9 = 15,$ we only need the values for $7$ and $9.$

Before you start writing down all other combinations, first write down the multiplications in $G.$ E.g. if you set $\chi(9)=1$ then $\chi(3)= i$ isn't possible anymore: $1=\chi(9)\neq\chi(3)\cdot \chi(3)=i \cdot i = -1,$ a contradiction.

 

[Math100]

That looks promising.

Let's have a look at the group we are talking about, namely the group of units of $\mathbb{Z}_{16}.$ This group has $\varphi (16)=8$ elements which are $G=\{1,3,5,7,9,11,13,15\}.$ The elements $3,5,11,13$ are of order four, and $7,9,15$ of order two. Since a Dirichlet character is a group homomorphism into the multiplicative group of complex numbers (see post #6), we have
$$
\chi(k)^2=\chi(k)\cdot \chi(k)=\chi(k\cdot k)=\chi(k^2)=\chi(1)=1\text{ for all }k\in \{7,9,15\},
$$
i.e. only two possible targets $\pm 1.$ We also get
$$
\chi(k)^4=\chi(k^4)=\chi(1)=1\text{ for all }k\in \{3,5,11,13\},
$$
and of course always $\chi(1)=1.$ Complex numbers of order four are only $\pm i ,$ but they can also map onto $\pm 1$ under $\chi.$

Start with elements of order two. There are only $2^2$ possibilities. Since $7\cdot 9 = 15,$ we only need the values for $7$ and $9.$

Before you start writing down all other combinations, first write down the multiplications in $G.$ E.g. if you set $\chi(9)=1$ then $\chi(3)= i$ isn't possible anymore: $1=\chi(9)\neq\chi(3)\cdot \chi(3)=i \cdot i = -1,$ a contradiction.

So the order two elements for $7, 9, 15$ are $\chi_{1}(7)=1, \chi_{2}(7)=-1, \chi_{2}(7)=1, \chi_{3}(7)=-1, \chi_{4}(7)=1, \chi_{5}(7)=-1, \chi_{6}(7)=1, \chi_{7}(7)=-1, \chi_{8}(7)=1$? But what about $9, 15$?

 

[fresh_42]

You have the choice between $4$ possibilities:
\begin{array}{|c|c|}
\hline \chi(7) & \chi(9) \\
\hline 1 & 1 \\
\hline -1 & 1 \\
\hline 1 & -1 \\
\hline -1 & -1 \\
\hline
\end{array}
Choose one of them, e.g. $\chi(7)=-1\, , \, \chi(9)=1.$ Then $\chi(15)=\chi(7)\cdot \chi(9)=-1.$

You first need the multiplication table of $G.$
\begin{array}{|c|c|c|c|c|c|c|c|}
\hline 1 &3 & 5 & 7 & 9 &11 & 13 & 15 \\
\hline 3 & 9 & 15 & 5 & \ldots & \ldots & \ldots &\ldots \\
\hline 5 & \ldots & \ldots & \ldots & \ldots & \ldots & \ldots &\ldots\\
\hline 7 & \ldots & \ldots & \ldots & 15 & \ldots & \ldots &\ldots \\
\hline 9 & \ldots & \ldots & \ldots & \ldots & \ldots & \ldots &\ldots\\
\hline 11 & \ldots & \ldots & \ldots & \ldots & \ldots & \ldots &\ldots\\
\hline 13 & \ldots & \ldots & \ldots & \ldots & \ldots & \ldots &\ldots\\
\hline 15 & \ldots & \ldots & \ldots & \ldots & \ldots & \ldots &\ldots\\
\hline
\end{array}

$\chi(3),\chi(5),\chi(11),\chi(13) \in \{-1,+1,- i , + i \}$ which are too many possibilities to check them all. Fortunately, it isn't necessary to check them all. E.g. $\chi(3)\cdot\chi(3)=\chi(9)$ and if we are in the case $\chi(9)=-1$ then $\chi(3)\in \{\pm i\}$ are the only possibilities left. Then make two cases: $\chi(3)=- i$ that fixes e.g. $\chi(3)\cdot \chi(9)=\chi(11)= i$ and so on, then consider $\chi(3)=+ i$ etc.

 

[Math100]

You have the choice between $4$ possibilities:
\begin{array}{|c|c|}
\hline \chi(7) & \chi(9) \\
\hline 1 & 1 \\
\hline -1 & 1 \\
\hline 1 & -1 \\
\hline -1 & -1 \\
\hline
\end{array}
Choose one of them, e.g. $\chi(7)=-1\, , \, \chi(9)=1.$ Then $\chi(15)=\chi(7)\cdot \chi(9)=-1.$

You first need the multiplication table of $G.$
\begin{array}{|c|c|c|c|c|c|c|c|}
\hline 1 &3 & 5 & 7 & 9 &11 & 13 & 15 \\
\hline 3 & 9 & 15 & 5 & \ldots & \ldots & \ldots &\ldots \\
\hline 5 & \ldots & \ldots & \ldots & \ldots & \ldots & \ldots &\ldots\\
\hline 7 & \ldots & \ldots & \ldots & 15 & \ldots & \ldots &\ldots \\
\hline 9 & \ldots & \ldots & \ldots & \ldots & \ldots & \ldots &\ldots\\
\hline 11 & \ldots & \ldots & \ldots & \ldots & \ldots & \ldots &\ldots\\
\hline 13 & \ldots & \ldots & \ldots & \ldots & \ldots & \ldots &\ldots\\
\hline 15 & \ldots & \ldots & \ldots & \ldots & \ldots & \ldots &\ldots\\
\hline
\end{array}

$\chi(3),\chi(5),\chi(11),\chi(13) \in \{-1,+1,- i , + i \}$ which are too many possibilities to check them all. Fortunately, it isn't necessary to check them all. E.g. $\chi(3)\cdot\chi(3)=\chi(9)$ and if we are in the case $\chi(9)=-1$ then $\chi(3)\in \{\pm i\}$ are the only possibilities left. Then make two cases: $\chi(3)=- i$ that fixes e.g. $\chi(3)\cdot \chi(9)=\chi(11)= i$ and so on, then consider $\chi(3)=+ i$ etc.

So
\begin{array}{|c|c|c|c|c|c|c|c|}
\hline 1 & 3 & 5 & 7 & 9 & 11 & 13 & 15 \\
\hline 3 & 9 & 15 & 5 & 11 & 1 & 7 & 13 \\
\hline 5 & 15 & 9 & 3 & 13 & 7 & 1 & 11 \\
\hline 7 & 5 & 3 & 1 & 15 & 13 & 11 & 9 \\
\hline 9 & 11 & 13 & 15 & 1 & 3 & 5 & 7 \\
\hline 11 & 1 & 7 & 13 & 3 & 9 & 15 & 5 \\
\hline 13 & 7 & 1 & 11 & 5 & 15 & 9 & 3 \\
\hline 15 & 13 & 11 & 9 & 7 & 5 & 3 & 1 \\
\hline
\end{array}
?

 

[fresh_42]

So
\begin{array}{|c|c|c|c|c|c|c|c|}
\hline 1 & 3 & 5 & 7 & 9 & 11 & 13 & 15 \\
\hline 3 & 9 & 15 & 5 & 11 & 1 & 7 & 13 \\
\hline 5 & 15 & 9 & 3 & 13 & 7 & 1 & 11 \\
\hline 7 & 5 & 3 & 1 & 15 & 13 & 11 & 9 \\
\hline 9 & 11 & 13 & 15 & 1 & 3 & 5 & 7 \\
\hline 11 & 1 & 7 & 13 & 3 & 9 & 15 & 5 \\
\hline 13 & 7 & 1 & 11 & 5 & 15 & 9 & 3 \\
\hline 15 & 13 & 11 & 9 & 7 & 5 & 3 & 1 \\
\hline
\end{array}
?

Yes. You need it to compute $\chi(n)$ in case $n=a\cdot b.$ You have still many cases but I do not see a shortcut. Fix $\chi(7),\chi(9)$ and then all four cases for $\chi(3)$ which might have subcases for $\chi(11).$

 

[Math100]

At first, I got this.
\begin{array}{|c|c|c|c|c|c|c|c|c|}
\hline n & 1 & 3 & 5 & 7 & 9 & 11 & 13 & 15 \\
\hline \chi_{1}(n) & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \\
\hline \chi_{2}(n) & 1 & -1 & \Idots & -1 & 1 & -1 & \Idots & -1 \\
\hline \chi_{3}(n) & 1 & i & \Idots & 1 & -1 & -i & \Idots & -1 \\
\hline \chi_{4}(n) & 1 & -i & \Idots & -1 & -1 & i & \Idots & 1 \\
\hline \chi_{5}(n) & 1 & 1 & \Idots & 1 & 1 & 1 & \Idots & 1 \\
\hline \chi_{6}(n) & 1 & -1 & \Idots & -1 & 1 & -1 & \Idots & -1 \\
\hline \chi_{7}(n) & 1 & i & \Idots & 1 & -1 & -i & \Idots & -1 \\
\hline \chi_{8}(n) & 1 & -i & \Idots & -1 & -1 & i & \Idots & 1 \\
\hline
\end{array}

But this didn't seem to be right/correct, because I couldn't get the desired result at the end. So I made other attempts and finally got this.
\begin{array}{|c|c|c|c|c|c|c|c|c|}
\hline n & 1 & 3 & 5 & 7 & 9 & 11 & 13 & 15 \\
\hline \chi_{1}(n) & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \\
\hline \chi_{2}(n) & 1 & -1 & 1 & -1 & 1 & -1 & 1 & -1 \\
\hline \chi_{3}(n) & 1 & i & i & 1 & -1 & -i & -i & -1 \\
\hline \chi_{4}(n) & 1 & -i & i & -1 & -1 & i & -i & 1 \\
\hline \chi_{5}(n) & 1 & 1 & -1 & -1 & 1 & 1 & -1 & -1 \\
\hline \chi_{6}(n) & 1 & -1 & -1 & 1 & 1 & -1 & -1 & 1 \\
\hline \chi_{7}(n) & 1 & i & -i & -1 & -1 & -i & i & 1 \\
\hline \chi_{8}(n) & 1 & -i & -i & 1 & -1 & i & i & -1 \\
\hline
\end{array}

Also, the second part of the question/problem asked to verify from my table that $\sum_{\chi \pmod {16}}\chi(3)=0$ and $\sum_{\chi \pmod {16}}\chi(11)=0$. So I think $\sum_{\chi \pmod {16}}\chi(3)=1+(-1)+i+(-i)+1+(-1)+i+(-i)=0$ and $\sum_{\chi \pmod {16}}\chi(11)=1+(-1)+(-i)+i+1+(-1)+(-i)+i=0$. Is this right/correct?

 

[fresh_42]

At first, I got this.
\begin{array}{|c|c|c|c|c|c|c|c|c|}
\hline n & 1 & 3 & 5 & 7 & 9 & 11 & 13 & 15 \\
\hline \chi_{1}(n) & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \\
\hline \chi_{2}(n) & 1 & -1 & \Idots & -1 & 1 & -1 & \Idots & -1 \\
\hline \chi_{3}(n) & 1 & i & \Idots & 1 & -1 & -i & \Idots & -1 \\
\hline \chi_{4}(n) & 1 & -i & \Idots & -1 & -1 & i & \Idots & 1 \\
\hline \chi_{5}(n) & 1 & 1 & \Idots & 1 & 1 & 1 & \Idots & 1 \\
\hline \chi_{6}(n) & 1 & -1 & \Idots & -1 & 1 & -1 & \Idots & -1 \\
\hline \chi_{7}(n) & 1 & i & \Idots & 1 & -1 & -i & \Idots & -1 \\
\hline \chi_{8}(n) & 1 & -i & \Idots & -1 & -1 & i & \Idots & 1 \\
\hline
\end{array}

But this didn't seem to be right/correct, because I couldn't get the desired result at the end. So I made other attempts and finally got this.
\begin{array}{|c|c|c|c|c|c|c|c|c|}
\hline n & 1 & 3 & 5 & 7 & 9 & 11 & 13 & 15 \\
\hline \chi_{1}(n) & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \\
\hline \chi_{2}(n) & 1 & -1 & 1 & -1 & 1 & -1 & 1 & -1 \\
\hline \chi_{3}(n) & 1 & i & i & 1 & -1 & -i & -i & -1 \\
\hline \chi_{4}(n) & 1 & -i & i & -1 & -1 & i & -i & 1 \\
\hline \chi_{5}(n) & 1 & 1 & -1 & -1 & 1 & 1 & -1 & -1 \\
\hline \chi_{6}(n) & 1 & -1 & -1 & 1 & 1 & -1 & -1 & 1 \\
\hline \chi_{7}(n) & 1 & i & -i & -1 & -1 & -i & i & 1 \\
\hline \chi_{8}(n) & 1 & -i & -i & 1 & -1 & i & i & -1 \\
\hline
\end{array}

Also, the second part of the question/problem asked to verify from my table that $\sum_{\chi \pmod {16}}\chi(3)=0$ and $\sum_{\chi \pmod {16}}\chi(11)=0$. So I think $\sum_{\chi \pmod {16}}\chi(3)=1+(-1)+i+(-i)+1+(-1)+i+(-i)=0$ and $\sum_{\chi \pmod {16}}\chi(11)=1+(-1)+(-i)+i+1+(-1)+(-i)+i=0$. Is this right/correct?

Looks good.

There are two theorems that you could check:
1.) Characters form a group.
2.) Characters are $\mathbb{C}-$linear independent.

 

[WWGD]

Are those Young Tableaux?

 

[fresh_42]

Are those Young Tableaux?

No.