Mixing
Fourier transform of the Klein group
$begingroup$
Obviously we have the elements of $G = lbrace e, a, b, crbrace;$ where $;c=ab$
Define:
$$chi : G rightarrow mathbb{Z}$$
How would I construct the transform for each of the elements of $G$?
What I mean is, for example take another group $G$ as $G = mathbb{Z_5} = lbrace0,1,2,3,4rbrace$
As defined above, let $chi : G rightarrow mathbb{Z} $
Define:
$$chi(0)=1$$
$$chi(1)=a$$
$$chi(2)=a^2$$
$$chi(3)=a^3$$
$$chi(4)=a^4$$
Let:
begin{align*}
chi_0(x) &= 1 \[1ex]
chi_1(x) &=
begin{cases}
1, & text{if } x = 0, \
a, & text{if } x = 1, \
a^2, & text{if } x = 2, \
a^3, & text{if } x = 3, \
a^4, & text{if } x = 4,
end{cases}
&
chi_3(x) &=
begin{cases}
1, & text{if } x = 0, \
a^3, & text{if } x = 1, \
a, & text{if } x = 2, \
a^4, & text{if } x = 3, \
a^2, & text{if } x = 4,
end{cases}
\[1ex]
chi_2(x) &=
begin{cases}
1, & text{if } x = 0, \
a^2, & text{if } x = 1, \
a^4, & text{if } x = 2, \
a, & text{if } x = 3, \
a^3, & text{if } x = 4,
end{cases}
&
chi_4(x) &=
begin{cases}
1, & text{if } x = 0, \
a^4, & text{if } x = 1, \
a^3, & text{if } x = 2, \
a^2, & text{if } x = 3, \
a, & text{if } x = 4.
end{cases}
end{align*}
Also define:
$hat f(chi) = sum {f}(a)bar{chi}(a)$ - The Fourier transform
$f(a)=frac{1}{rule{0pt}{0.65em} |G|}sum hat f(chi)chi_i(a)$ - The inverse Fourier transform
Then we can compute the transforms for each element of $G$.
Now my question is how would I do the same with the Klein group? I mean, can I proceed by doing this:
$$chi(e)=1$$
$$chi(a)=alpha$$
$$chi(b)=alpha^2$$
$$chi(c)=alpha^3$$
Then proceed like the group $mathbb{Z}_5$?
EDIT: I just checked and this is not possible as the Cayley table doesn't not match
EDIT: I have solved this problem now.
group-theory fourier-transform
asked Jan 23 at 15:44
A.EA.E
2249
$endgroup$
|
show 2 more comments
$begingroup$
Obviously we have the elements of $G = lbrace e, a, b, crbrace;$ where $;c=ab$
Define:
$$chi : G rightarrow mathbb{Z}$$
How would I construct the transform for each of the elements of $G$?
What I mean is, for example take another group $G$ as $G = mathbb{Z_5} = lbrace0,1,2,3,4rbrace$
As defined above, let $chi : G rightarrow mathbb{Z} $
Define:
$$chi(0)=1$$
$$chi(1)=a$$
$$chi(2)=a^2$$
$$chi(3)=a^3$$
$$chi(4)=a^4$$
Let:
begin{align*}
chi_0(x) &= 1 \[1ex]
chi_1(x) &=
begin{cases}
1, & text{if } x = 0, \
a, & text{if } x = 1, \
a^2, & text{if } x = 2, \
a^3, & text{if } x = 3, \
a^4, & text{if } x = 4,
end{cases}
&
chi_3(x) &=
begin{cases}
1, & text{if } x = 0, \
a^3, & text{if } x = 1, \
a, & text{if } x = 2, \
a^4, & text{if } x = 3, \
a^2, & text{if } x = 4,
end{cases}
\[1ex]
chi_2(x) &=
begin{cases}
1, & text{if } x = 0, \
a^2, & text{if } x = 1, \
a^4, & text{if } x = 2, \
a, & text{if } x = 3, \
a^3, & text{if } x = 4,
end{cases}
&
chi_4(x) &=
begin{cases}
1, & text{if } x = 0, \
a^4, & text{if } x = 1, \
a^3, & text{if } x = 2, \
a^2, & text{if } x = 3, \
a, & text{if } x = 4.
end{cases}
end{align*}
Also define:
$hat f(chi) = sum {f}(a)bar{chi}(a)$ - The Fourier transform
$f(a)=frac{1}{rule{0pt}{0.65em} |G|}sum hat f(chi)chi_i(a)$ - The inverse Fourier transform
Then we can compute the transforms for each element of $G$.
Now my question is how would I do the same with the Klein group? I mean, can I proceed by doing this:
$$chi(e)=1$$
$$chi(a)=alpha$$
$$chi(b)=alpha^2$$
$$chi(c)=alpha^3$$
Then proceed like the group $mathbb{Z}_5$?
EDIT: I just checked and this is not possible as the Cayley table doesn't not match
EDIT: I have solved this problem now.
group-theory fourier-transform
asked Jan 23 at 15:44
A.EA.E
2249
$endgroup$
$begingroup$
please look at the edited question, hopefully it makes it clearer.
$endgroup$
– A.E
Jan 28 at 15:07
$begingroup$
It would be helpful if you gave some context - where is this idea/notion coming from? Is it from a lecture course you are taking, or just something you are thinking about by yourself? If the latter, then where does the "standard" stuff stop? Or is it all standard?
$endgroup$
– user1729
Feb 11 at 16:13
$begingroup$
it was a basic thing that my supervisor asked me to do during my research. I was being silly and the problem is quite simple.
$endgroup$
– A.E
Feb 11 at 16:14
$begingroup$
If you want then you could write up your answer and post it here? [Or you could delete the question, which would be less helpful/interesting :-) ]
$endgroup$
– user1729
Feb 11 at 16:17
$begingroup$
I can but its trivial :)
$endgroup$
– A.E
Feb 11 at 16:18
|
show 2 more comments
$begingroup$
Obviously we have the elements of $G = lbrace e, a, b, crbrace;$ where $;c=ab$
Define:
$$chi : G rightarrow mathbb{Z}$$
How would I construct the transform for each of the elements of $G$?
What I mean is, for example take another group $G$ as $G = mathbb{Z_5} = lbrace0,1,2,3,4rbrace$
As defined above, let $chi : G rightarrow mathbb{Z} $
Define:
$$chi(0)=1$$
$$chi(1)=a$$
$$chi(2)=a^2$$
$$chi(3)=a^3$$
$$chi(4)=a^4$$
Let:
begin{align*}
chi_0(x) &= 1 \[1ex]
chi_1(x) &=
begin{cases}
1, & text{if } x = 0, \
a, & text{if } x = 1, \
a^2, & text{if } x = 2, \
a^3, & text{if } x = 3, \
a^4, & text{if } x = 4,
end{cases}
&
chi_3(x) &=
begin{cases}
1, & text{if } x = 0, \
a^3, & text{if } x = 1, \
a, & text{if } x = 2, \
a^4, & text{if } x = 3, \
a^2, & text{if } x = 4,
end{cases}
\[1ex]
chi_2(x) &=
begin{cases}
1, & text{if } x = 0, \
a^2, & text{if } x = 1, \
a^4, & text{if } x = 2, \
a, & text{if } x = 3, \
a^3, & text{if } x = 4,
end{cases}
&
chi_4(x) &=
begin{cases}
1, & text{if } x = 0, \
a^4, & text{if } x = 1, \
a^3, & text{if } x = 2, \
a^2, & text{if } x = 3, \
a, & text{if } x = 4.
end{cases}
end{align*}
Also define:
$hat f(chi) = sum {f}(a)bar{chi}(a)$ - The Fourier transform
$f(a)=frac{1}{rule{0pt}{0.65em} |G|}sum hat f(chi)chi_i(a)$ - The inverse Fourier transform
Then we can compute the transforms for each element of $G$.
Now my question is how would I do the same with the Klein group? I mean, can I proceed by doing this:
$$chi(e)=1$$
$$chi(a)=alpha$$
$$chi(b)=alpha^2$$
$$chi(c)=alpha^3$$
Then proceed like the group $mathbb{Z}_5$?
EDIT: I just checked and this is not possible as the Cayley table doesn't not match
EDIT: I have solved this problem now.
group-theory fourier-transform
asked Jan 23 at 15:44
A.EA.E
2249
$endgroup$
Obviously we have the elements of $G = lbrace e, a, b, crbrace;$ where $;c=ab$
Define:
$$chi : G rightarrow mathbb{Z}$$
How would I construct the transform for each of the elements of $G$?
What I mean is, for example take another group $G$ as $G = mathbb{Z_5} = lbrace0,1,2,3,4rbrace$
As defined above, let $chi : G rightarrow mathbb{Z} $
Define:
$$chi(0)=1$$
$$chi(1)=a$$
$$chi(2)=a^2$$
$$chi(3)=a^3$$
$$chi(4)=a^4$$
Let:
begin{align*}
chi_0(x) &= 1 \[1ex]
chi_1(x) &=
begin{cases}
1, & text{if } x = 0, \
a, & text{if } x = 1, \
a^2, & text{if } x = 2, \
a^3, & text{if } x = 3, \
a^4, & text{if } x = 4,
end{cases}
&
chi_3(x) &=
begin{cases}
1, & text{if } x = 0, \
a^3, & text{if } x = 1, \
a, & text{if } x = 2, \
a^4, & text{if } x = 3, \
a^2, & text{if } x = 4,
end{cases}
\[1ex]
chi_2(x) &=
begin{cases}
1, & text{if } x = 0, \
a^2, & text{if } x = 1, \
a^4, & text{if } x = 2, \
a, & text{if } x = 3, \
a^3, & text{if } x = 4,
end{cases}
&
chi_4(x) &=
begin{cases}
1, & text{if } x = 0, \
a^4, & text{if } x = 1, \
a^3, & text{if } x = 2, \
a^2, & text{if } x = 3, \
a, & text{if } x = 4.
end{cases}
end{align*}
Also define:
$hat f(chi) = sum {f}(a)bar{chi}(a)$ - The Fourier transform
$f(a)=frac{1}{rule{0pt}{0.65em} |G|}sum hat f(chi)chi_i(a)$ - The inverse Fourier transform
Then we can compute the transforms for each element of $G$.
Now my question is how would I do the same with the Klein group? I mean, can I proceed by doing this:
$$chi(e)=1$$
$$chi(a)=alpha$$
$$chi(b)=alpha^2$$
$$chi(c)=alpha^3$$
Then proceed like the group $mathbb{Z}_5$?
EDIT: I just checked and this is not possible as the Cayley table doesn't not match
EDIT: I have solved this problem now.
group-theory fourier-transform
group-theory fourier-transform
asked Jan 23 at 15:44
A.EA.E
2249
asked Jan 23 at 15:44
A.EA.E
2249
edited Feb 11 at 16:11
A.E
asked Jan 23 at 15:44
A.EA.E
2249
asked Jan 23 at 15:44
A.EA.E
2249
asked Jan 23 at 15:44
A.EA.E
2249
2249
$begingroup$
please look at the edited question, hopefully it makes it clearer.
$endgroup$
– A.E
Jan 28 at 15:07
$begingroup$
It would be helpful if you gave some context - where is this idea/notion coming from? Is it from a lecture course you are taking, or just something you are thinking about by yourself? If the latter, then where does the "standard" stuff stop? Or is it all standard?
$endgroup$
– user1729
Feb 11 at 16:13
$begingroup$
it was a basic thing that my supervisor asked me to do during my research. I was being silly and the problem is quite simple.
$endgroup$
– A.E
Feb 11 at 16:14
$begingroup$
If you want then you could write up your answer and post it here? [Or you could delete the question, which would be less helpful/interesting :-) ]
$endgroup$
– user1729
Feb 11 at 16:17
$begingroup$
I can but its trivial :)
$endgroup$
– A.E
Feb 11 at 16:18
|
show 2 more comments
$begingroup$
please look at the edited question, hopefully it makes it clearer.
$endgroup$
– A.E
Jan 28 at 15:07
$begingroup$
It would be helpful if you gave some context - where is this idea/notion coming from? Is it from a lecture course you are taking, or just something you are thinking about by yourself? If the latter, then where does the "standard" stuff stop? Or is it all standard?
$endgroup$
– user1729
Feb 11 at 16:13
$begingroup$
it was a basic thing that my supervisor asked me to do during my research. I was being silly and the problem is quite simple.
$endgroup$
– A.E
Feb 11 at 16:14
$begingroup$
If you want then you could write up your answer and post it here? [Or you could delete the question, which would be less helpful/interesting :-) ]
$endgroup$
– user1729
Feb 11 at 16:17
$begingroup$
I can but its trivial :)
$endgroup$
– A.E
Feb 11 at 16:18
$begingroup$
please look at the edited question, hopefully it makes it clearer.
$endgroup$
– A.E
Jan 28 at 15:07
$begingroup$
please look at the edited question, hopefully it makes it clearer.
$endgroup$
– A.E
Jan 28 at 15:07
$begingroup$
It would be helpful if you gave some context - where is this idea/notion coming from? Is it from a lecture course you are taking, or just something you are thinking about by yourself? If the latter, then where does the "standard" stuff stop? Or is it all standard?
$endgroup$
– user1729
Feb 11 at 16:13
$begingroup$
It would be helpful if you gave some context - where is this idea/notion coming from? Is it from a lecture course you are taking, or just something you are thinking about by yourself? If the latter, then where does the "standard" stuff stop? Or is it all standard?
$endgroup$
– user1729
Feb 11 at 16:13
$begingroup$
it was a basic thing that my supervisor asked me to do during my research. I was being silly and the problem is quite simple.
$endgroup$
– A.E
Feb 11 at 16:14
$begingroup$
it was a basic thing that my supervisor asked me to do during my research. I was being silly and the problem is quite simple.
$endgroup$
– A.E
Feb 11 at 16:14
$begingroup$
If you want then you could write up your answer and post it here? [Or you could delete the question, which would be less helpful/interesting :-) ]
$endgroup$
– user1729
Feb 11 at 16:17
$begingroup$
If you want then you could write up your answer and post it here? [Or you could delete the question, which would be less helpful/interesting :-) ]
$endgroup$
– user1729
Feb 11 at 16:17
$begingroup$
I can but its trivial :)
$endgroup$
– A.E
Feb 11 at 16:18
$begingroup$
I can but its trivial :)
$endgroup$
– A.E
Feb 11 at 16:18
|
show 2 more comments
1 Answer
1
active
oldest
votes
$begingroup$
Since $K_4$ is abelian of order $4$, there are exactly four irreducible representations, all of dimension one. Since every element has order two, all values of the characters are $chi(g)= pm 1$. This implies the character table will take the form:
begin{align*}
begin{array}{c | c c c c }
& e & a & b & c\
hline
chi_1 & 1 & 1 & 1 & 1\
chi_2 & 1 & pm 1 & pm 1 & pm 1\
chi_3 & 1 & pm 1 & pm 1 & pm 1\
chi_4 & 1 & pm 1 & pm 1 & pm 1\
end{array}
end{align*}
The only way to make all the rows of this table orthogonal, is by taking:
begin{align*}
begin{array}{c | c c c c }
& e & a & b & c\
hline
chi_1 & 1 & 1 & 1 & 1\
chi_2 & 1 & 1 & -1 & -1\
chi_3 & 1 & -1 & 1 & -1\
chi_4 & 1 & -1 & -1 & 1\
end{array}
end{align*}
Now we are in a position to construct the Fourier transform of the Klein group. Using the character table above and definition of Fourier transform, we have:
$$hat{f}(chi_1) = f(e)+f(a)+f(b)+f(c)$$
$$hat{f}(chi_2) = f(e)+f(a)-f(b)-f(c)$$
$$hat{f}(chi_3) = f(e)-f(a)+f(b)-f(c)$$
$$hat{f}(chi_4) = f(e)-f(a)-f(b)+f(c)$$
Computing the inverse Fourier transform using definition of inverse Fourier transform, we have:
$$f(e) = frac{1}{4}[f(e)+f(e)+f(e)+f(e)] = f(e)$$
$$f(a) = frac{1}{4}[f(a)+f(a)+f(a)+f(a)] = f(a)$$
$$f(b) = frac{1}{4}[f(b)+f(b)+f(b)+f(b)] = f(b)$$
$$f(c) = frac{1}{4}[f(c)+f(c)+f(c)+f(c)] = f(c)$$
as expected.
answered Feb 11 at 16:26
A.EA.E
2249
$endgroup$
$begingroup$
You can use the array environment instead of tabular.
$endgroup$
– user1729
Feb 11 at 16:28
$begingroup$
I don't know how to use that unfortunately..
$endgroup$
– A.E
Feb 11 at 16:29
$begingroup$
why don't you try editing the solution? if possible that is.
$endgroup$
– A.E
Feb 11 at 16:30
$begingroup$
Have done so :-)
$endgroup$
– user1729
Feb 11 at 16:33
$begingroup$
but the solution is trivial so I wouldn't bother unless you wish to present a neat answer for future viewers. Sweet :)
$endgroup$
– A.E
Feb 11 at 16:33
|
show 3 more comments
Your Answer
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1 Answer
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1 Answer
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active
oldest
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active
oldest
votes
$begingroup$
Since $K_4$ is abelian of order $4$, there are exactly four irreducible representations, all of dimension one. Since every element has order two, all values of the characters are $chi(g)= pm 1$. This implies the character table will take the form:
begin{align*}
begin{array}{c | c c c c }
& e & a & b & c\
hline
chi_1 & 1 & 1 & 1 & 1\
chi_2 & 1 & pm 1 & pm 1 & pm 1\
chi_3 & 1 & pm 1 & pm 1 & pm 1\
chi_4 & 1 & pm 1 & pm 1 & pm 1\
end{array}
end{align*}
The only way to make all the rows of this table orthogonal, is by taking:
begin{align*}
begin{array}{c | c c c c }
& e & a & b & c\
hline
chi_1 & 1 & 1 & 1 & 1\
chi_2 & 1 & 1 & -1 & -1\
chi_3 & 1 & -1 & 1 & -1\
chi_4 & 1 & -1 & -1 & 1\
end{array}
end{align*}
Now we are in a position to construct the Fourier transform of the Klein group. Using the character table above and definition of Fourier transform, we have:
$$hat{f}(chi_1) = f(e)+f(a)+f(b)+f(c)$$
$$hat{f}(chi_2) = f(e)+f(a)-f(b)-f(c)$$
$$hat{f}(chi_3) = f(e)-f(a)+f(b)-f(c)$$
$$hat{f}(chi_4) = f(e)-f(a)-f(b)+f(c)$$
Computing the inverse Fourier transform using definition of inverse Fourier transform, we have:
$$f(e) = frac{1}{4}[f(e)+f(e)+f(e)+f(e)] = f(e)$$
$$f(a) = frac{1}{4}[f(a)+f(a)+f(a)+f(a)] = f(a)$$
$$f(b) = frac{1}{4}[f(b)+f(b)+f(b)+f(b)] = f(b)$$
$$f(c) = frac{1}{4}[f(c)+f(c)+f(c)+f(c)] = f(c)$$
as expected.
answered Feb 11 at 16:26
A.EA.E
2249
$endgroup$
$begingroup$
You can use the array environment instead of tabular.
$endgroup$
– user1729
Feb 11 at 16:28
$begingroup$
I don't know how to use that unfortunately..
$endgroup$
– A.E
Feb 11 at 16:29
$begingroup$
why don't you try editing the solution? if possible that is.
$endgroup$
– A.E
Feb 11 at 16:30
$begingroup$
Have done so :-)
$endgroup$
– user1729
Feb 11 at 16:33
$begingroup$
but the solution is trivial so I wouldn't bother unless you wish to present a neat answer for future viewers. Sweet :)
$endgroup$
– A.E
Feb 11 at 16:33
|
show 3 more comments
$begingroup$
Since $K_4$ is abelian of order $4$, there are exactly four irreducible representations, all of dimension one. Since every element has order two, all values of the characters are $chi(g)= pm 1$. This implies the character table will take the form:
begin{align*}
begin{array}{c | c c c c }
& e & a & b & c\
hline
chi_1 & 1 & 1 & 1 & 1\
chi_2 & 1 & pm 1 & pm 1 & pm 1\
chi_3 & 1 & pm 1 & pm 1 & pm 1\
chi_4 & 1 & pm 1 & pm 1 & pm 1\
end{array}
end{align*}
The only way to make all the rows of this table orthogonal, is by taking:
begin{align*}
begin{array}{c | c c c c }
& e & a & b & c\
hline
chi_1 & 1 & 1 & 1 & 1\
chi_2 & 1 & 1 & -1 & -1\
chi_3 & 1 & -1 & 1 & -1\
chi_4 & 1 & -1 & -1 & 1\
end{array}
end{align*}
Now we are in a position to construct the Fourier transform of the Klein group. Using the character table above and definition of Fourier transform, we have:
$$hat{f}(chi_1) = f(e)+f(a)+f(b)+f(c)$$
$$hat{f}(chi_2) = f(e)+f(a)-f(b)-f(c)$$
$$hat{f}(chi_3) = f(e)-f(a)+f(b)-f(c)$$
$$hat{f}(chi_4) = f(e)-f(a)-f(b)+f(c)$$
Computing the inverse Fourier transform using definition of inverse Fourier transform, we have:
$$f(e) = frac{1}{4}[f(e)+f(e)+f(e)+f(e)] = f(e)$$
$$f(a) = frac{1}{4}[f(a)+f(a)+f(a)+f(a)] = f(a)$$
$$f(b) = frac{1}{4}[f(b)+f(b)+f(b)+f(b)] = f(b)$$
$$f(c) = frac{1}{4}[f(c)+f(c)+f(c)+f(c)] = f(c)$$
as expected.
answered Feb 11 at 16:26
A.EA.E
2249
$endgroup$
$begingroup$
You can use the array environment instead of tabular.
$endgroup$
– user1729
Feb 11 at 16:28
$begingroup$
I don't know how to use that unfortunately..
$endgroup$
– A.E
Feb 11 at 16:29
$begingroup$
why don't you try editing the solution? if possible that is.
$endgroup$
– A.E
Feb 11 at 16:30
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Have done so :-)
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– user1729
Feb 11 at 16:33
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but the solution is trivial so I wouldn't bother unless you wish to present a neat answer for future viewers. Sweet :)
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– A.E
Feb 11 at 16:33
|
show 3 more comments
$begingroup$
Since $K_4$ is abelian of order $4$, there are exactly four irreducible representations, all of dimension one. Since every element has order two, all values of the characters are $chi(g)= pm 1$. This implies the character table will take the form:
begin{align*}
begin{array}{c | c c c c }
& e & a & b & c\
hline
chi_1 & 1 & 1 & 1 & 1\
chi_2 & 1 & pm 1 & pm 1 & pm 1\
chi_3 & 1 & pm 1 & pm 1 & pm 1\
chi_4 & 1 & pm 1 & pm 1 & pm 1\
end{array}
end{align*}
The only way to make all the rows of this table orthogonal, is by taking:
begin{align*}
begin{array}{c | c c c c }
& e & a & b & c\
hline
chi_1 & 1 & 1 & 1 & 1\
chi_2 & 1 & 1 & -1 & -1\
chi_3 & 1 & -1 & 1 & -1\
chi_4 & 1 & -1 & -1 & 1\
end{array}
end{align*}
Now we are in a position to construct the Fourier transform of the Klein group. Using the character table above and definition of Fourier transform, we have:
$$hat{f}(chi_1) = f(e)+f(a)+f(b)+f(c)$$
$$hat{f}(chi_2) = f(e)+f(a)-f(b)-f(c)$$
$$hat{f}(chi_3) = f(e)-f(a)+f(b)-f(c)$$
$$hat{f}(chi_4) = f(e)-f(a)-f(b)+f(c)$$
Computing the inverse Fourier transform using definition of inverse Fourier transform, we have:
$$f(e) = frac{1}{4}[f(e)+f(e)+f(e)+f(e)] = f(e)$$
$$f(a) = frac{1}{4}[f(a)+f(a)+f(a)+f(a)] = f(a)$$
$$f(b) = frac{1}{4}[f(b)+f(b)+f(b)+f(b)] = f(b)$$
$$f(c) = frac{1}{4}[f(c)+f(c)+f(c)+f(c)] = f(c)$$
as expected.
answered Feb 11 at 16:26
A.EA.E
2249
$endgroup$
Since $K_4$ is abelian of order $4$, there are exactly four irreducible representations, all of dimension one. Since every element has order two, all values of the characters are $chi(g)= pm 1$. This implies the character table will take the form:
begin{align*}
begin{array}{c | c c c c }
& e & a & b & c\
hline
chi_1 & 1 & 1 & 1 & 1\
chi_2 & 1 & pm 1 & pm 1 & pm 1\
chi_3 & 1 & pm 1 & pm 1 & pm 1\
chi_4 & 1 & pm 1 & pm 1 & pm 1\
end{array}
end{align*}
The only way to make all the rows of this table orthogonal, is by taking:
begin{align*}
begin{array}{c | c c c c }
& e & a & b & c\
hline
chi_1 & 1 & 1 & 1 & 1\
chi_2 & 1 & 1 & -1 & -1\
chi_3 & 1 & -1 & 1 & -1\
chi_4 & 1 & -1 & -1 & 1\
end{array}
end{align*}
Now we are in a position to construct the Fourier transform of the Klein group. Using the character table above and definition of Fourier transform, we have:
$$hat{f}(chi_1) = f(e)+f(a)+f(b)+f(c)$$
$$hat{f}(chi_2) = f(e)+f(a)-f(b)-f(c)$$
$$hat{f}(chi_3) = f(e)-f(a)+f(b)-f(c)$$
$$hat{f}(chi_4) = f(e)-f(a)-f(b)+f(c)$$
Computing the inverse Fourier transform using definition of inverse Fourier transform, we have:
$$f(e) = frac{1}{4}[f(e)+f(e)+f(e)+f(e)] = f(e)$$
$$f(a) = frac{1}{4}[f(a)+f(a)+f(a)+f(a)] = f(a)$$
$$f(b) = frac{1}{4}[f(b)+f(b)+f(b)+f(b)] = f(b)$$
$$f(c) = frac{1}{4}[f(c)+f(c)+f(c)+f(c)] = f(c)$$
as expected.
answered Feb 11 at 16:26
A.EA.E
2249
edited Feb 11 at 16:38
answered Feb 11 at 16:26
A.EA.E
2249
answered Feb 11 at 16:26
A.EA.E
2249
answered Feb 11 at 16:26
A.EA.E
2249
2249
$begingroup$
You can use the array environment instead of tabular.
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– user1729
Feb 11 at 16:28
$begingroup$
I don't know how to use that unfortunately..
$endgroup$
– A.E
Feb 11 at 16:29
$begingroup$
why don't you try editing the solution? if possible that is.
$endgroup$
– A.E
Feb 11 at 16:30
$begingroup$
Have done so :-)
$endgroup$
– user1729
Feb 11 at 16:33
$begingroup$
but the solution is trivial so I wouldn't bother unless you wish to present a neat answer for future viewers. Sweet :)
$endgroup$
– A.E
Feb 11 at 16:33
|
show 3 more comments
$begingroup$
You can use the array environment instead of tabular.
$endgroup$
– user1729
Feb 11 at 16:28
$begingroup$
I don't know how to use that unfortunately..
$endgroup$
– A.E
Feb 11 at 16:29
$begingroup$
why don't you try editing the solution? if possible that is.
$endgroup$
– A.E
Feb 11 at 16:30
$begingroup$
Have done so :-)
$endgroup$
– user1729
Feb 11 at 16:33
$begingroup$
but the solution is trivial so I wouldn't bother unless you wish to present a neat answer for future viewers. Sweet :)
$endgroup$
– A.E
Feb 11 at 16:33
$begingroup$
You can use the array environment instead of tabular.
$endgroup$
– user1729
Feb 11 at 16:28
$begingroup$
You can use the array environment instead of tabular.
$endgroup$
– user1729
Feb 11 at 16:28
$begingroup$
I don't know how to use that unfortunately..
$endgroup$
– A.E
Feb 11 at 16:29
$begingroup$
I don't know how to use that unfortunately..
$endgroup$
– A.E
Feb 11 at 16:29
$begingroup$
why don't you try editing the solution? if possible that is.
$endgroup$
– A.E
Feb 11 at 16:30
$begingroup$
why don't you try editing the solution? if possible that is.
$endgroup$
– A.E
Feb 11 at 16:30
$begingroup$
Have done so :-)
$endgroup$
– user1729
Feb 11 at 16:33
$begingroup$
Have done so :-)
$endgroup$
– user1729
Feb 11 at 16:33
$begingroup$
but the solution is trivial so I wouldn't bother unless you wish to present a neat answer for future viewers. Sweet :)
$endgroup$
– A.E
Feb 11 at 16:33
$begingroup$
but the solution is trivial so I wouldn't bother unless you wish to present a neat answer for future viewers. Sweet :)
$endgroup$
– A.E
Feb 11 at 16:33
|
show 3 more comments
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$begingroup$
please look at the edited question, hopefully it makes it clearer.
$endgroup$
– A.E
Jan 28 at 15:07
$begingroup$
It would be helpful if you gave some context - where is this idea/notion coming from? Is it from a lecture course you are taking, or just something you are thinking about by yourself? If the latter, then where does the "standard" stuff stop? Or is it all standard?
$endgroup$
– user1729
Feb 11 at 16:13
$begingroup$
it was a basic thing that my supervisor asked me to do during my research. I was being silly and the problem is quite simple.
$endgroup$
– A.E
Feb 11 at 16:14
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If you want then you could write up your answer and post it here? [Or you could delete the question, which would be less helpful/interesting :-) ]
$endgroup$
– user1729
Feb 11 at 16:17
$begingroup$
I can but its trivial :)
$endgroup$
– A.E
Feb 11 at 16:18