# Thread: Characters: The usual suspects

1. ## Characters: The usual suspects

My group theory text (and other sources) say that the character of a n operator in a representation is the trace of the matrix corresponding to the operator. ie. Given a group operator $\Gamma$ in some representation we have $X = Tr( \Gamma )$.

But
My newest text says a function (in a space) ca be expanded into a sum/integral of characters $X = e^{i x \cdot \xi}$
$f(x) = \sum b_{\xi} e^{ix \cdot \xi}$
$f(x) = \int b( \xi ) e^{ix \cdot \xi}~d \xi$
(To put this into context the section is talking about inverse Fourier transforms and that the inverse FT diagonalizes the representation for f.)

Are these two character definitions related to each other in some way? Or are they representing (no pun intended) different concepts? My text, at least, uses the same symbol for both.

Thanks!
-Dan

2. Sorry, I do not understand what you said . But I can answer the question "are these charachter definitions the same". They do not have to do. I can give you two other references of the word charachter. In number theory a (Dirichlet) charachter is a homomorphism from the finite field to the complex numbers. In linear algebra it means: Let F be a field and G be a group, let f(G,F) be the set of all functions from G to F; a charachter in f(G,F) is a group homomorphism between G and F*. So it is very common in math to use the same word to represent a lot of different stuff.

3. Originally Posted by ThePerfectHacker
Sorry, I do not understand what you said . But I can answer the question "are these charachter definitions the same". They do not have to do. I can give you two other references of the word charachter. In number theory a (Dirichlet) charachter is a homomorphism from the finite field to the complex numbers. In linear algebra it means: Let F be a field and G be a group, let f(G,F) be the set of all functions from G to F; a charachter in f(G,F) is a group homomorphism between G and F*. So it is very common in math to use the same word to represent a lot of different stuff.
I was wondering if that might be the case. However since the book talks about both and uses the same symbol for them I suspect they are equivalent in some sense.

I'll be back later and post more of the text's inverse Fourier (actually Plancherel) transform. (As well as the source information in case that might help.)

-Dan

4. The connection here is that the irreducible unitary representations of the additive group of real numbers are all one-dimensional, and are given by $x\mapsto e^{ix\xi}\ (x\in\mathbb{R})$, where $\xi\in\mathbb{R}$. The trace of a one-dimensional representation is just its range. So the Fourier transform of a function on $\mathbb{R}$ is a weighted integral of the representation-theoretic characters of $\mathbb{R}$.

Edit. If it's a Fourier transform of a function of several variables then exactly the same thing holds, provided that $\mathbb{R}$ is replaced by $\mathbb{R}^n$ wherever it occurs, and the product $x\xi$ is replaced by a dot product $x.\xi$

5. Originally Posted by Opalg
The connection here is that the irreducible unitary representations of the additive group of real numbers are all one-dimensional, and are given by $x\mapsto e^{ix\xi}\ (x\in\mathbb{R})$, where $\xi\in\mathbb{R}$. The trace of a one-dimensional representation is just its range. So the Fourier transform of a function on $\mathbb{R}$ is a weighted integral of the representation-theoretic characters of $\mathbb{R}$.

Edit. If it's a Fourier transform of a function of several variables then exactly the same thing holds, provided that $\mathbb{R}$ is replaced by $\mathbb{R}^n$ wherever it occurs, and the product $x\xi$ is replaced by a dot product $x.\xi$
Okay, after a bit of effort I can make sense of this. However I still have a problem.

The text says:
Any function (in a suitable space) can be expanded into the direct sum/integral of characters $\chi = e^{ix \cdot \xi}$
$f(x) = \sum b_{\xi}e^{ix \cdot \xi}$ or $f(x) = \int b(\xi) e^{i x \cdot \xi} d \xi$
(Apologies, I mistyped this in my original post.)

Since the function f(x) does not need to take x into the space of complex numbers, how does this work? I should mention that I can understand the integral since the book specifies that the group under consideration is $\mathbb{R}^n, \mathbb{T}^n$. But no such specification was made for the discrete case. Is that what is meant by "in a suitable space?"

Thanks!
-Dan

6. Short of giving a whole lecture course on harmonic analysis, I can only scratch the surface of this topic.

The basic fact is that if G is any locally compact abelian group, the space of characters on G can also be made into a locally compact abelian group, called the dual group of G and usually denoted by $\hat{G}$. The dual of the circle group $\mathbb{T}$ is $\mathbb{Z}$, the group of integers. The dual of the real numbers is again the real numbers. The dual of $\mathbb{T}^n$ is $\mathbb{Z}^n$ and the dual of $\mathbb{R}^n$ is $\mathbb{R}^n$. In general, the dual of a compact group is a discrete group.

Every locally compact abelian group has an integral (called the Haar integral) associated with it, and every real- or complex-valued integrable function on the group has a "generalised Fourier transform." This is a continuous function on the dual group, defined by $\hat{f}(\xi) = \int_Gf(x)\xi(x)\,dx$, where $\xi(x)$ denotes the value of the character ξ at the element x of the group.

Since the dual group is also locally compact, it has its own dual, which turns out to be isomorphic to the original group (that's the Pontyagin duality theorem). There is a "Fourier inversion theorem" which says that, for a suitable class of functions, there is an inverse transform that reconstructs a function f on G from its Fourier transform $\hat{f}$ on the dual group.

7. Originally Posted by Opalg
Short of giving a whole lecture course on harmonic analysis, I can only scratch the surface of this topic.

The basic fact is that if G is any locally compact abelian group, the space of characters on G can also be made into a locally compact abelian group, called the dual group of G and usually denoted by $\hat{G}$. The dual of the circle group $\mathbb{T}$ is $\mathbb{Z}$, the group of integers. The dual of the real numbers is again the real numbers. The dual of $\mathbb{T}^n$ is $\mathbb{Z}^n$ and the dual of $\mathbb{R}^n$ is $\mathbb{R}^n$. In general, the dual of a compact group is a discrete group.

Every locally compact abelian group has an integral (called the Haar integral) associated with it, and every real- or complex-valued integrable function on the group has a "generalised Fourier transform." This is a continuous function on the dual group, defined by $\hat{f}(\xi) = \int_Gf(x)\xi(x)\,dx$, where $\xi(x)$ denotes the value of the character ξ at the element x of the group.

Since the dual group is also locally compact, it has its own dual, which turns out to be isomorphic to the original group (that's the Pontyagin duality theorem). There is a "Fourier inversion theorem" which says that, for a suitable class of functions, there is an inverse transform that reconstructs a function f on G from its Fourier transform $\hat{f}$ on the dual group.
(chuckles) Well, that's what I'm studying, so I guess I'll just have to keep at it until it gels a bit more.

Thanks again!

-Dan

8. Originally Posted by Opalg
(that's the Pontyagin duality theorem).
I been curious to look it up, but I got no results using that, it seems it should have been this. Interesting, did you know this guy was a blind mathematician? It is an excellent story, because he made (according to Wikipedia) contributions to geometry. This illustrates the power of math, one does not even have to see what he is doing, even in geometry (though seeing what one is doing only adds to the beauty).

9. Originally Posted by Opalg
Short of giving a whole lecture course on harmonic analysis, I can only scratch the surface of this topic.
Would you be explain to me, in elementary terms, what exactly is harmonic analysis?
I have heard answers from people which say it is a more generalized concept of Fourier series.
I can see that, for example, in a Sturm-Liouville problem the eigenfunctions are orthogonal.
And furthermore, we can expand a function into a series involving these eigenfunctions.
So that is what harmonic analysis is about? Dealing with a series of eigenfunctions?
Or is there more to it that I am missing?

10. Originally Posted by ThePerfectHacker
Would you be explain to me, in elementary terms, what exactly is harmonic analysis?
Harmonic analysis - Wikipedia, the free encyclopedia

The difficulty here is that harmonic analysis is not elementary, so it's hard to give an elementary explanation of it. I think of it as the extension of the ideas and techniques of Fourier theory from the groups R and Z to a wider class of groups such as Lie groups.

11. Originally Posted by Opalg
Harmonic analysis - Wikipedia, the free encyclopedia

The difficulty here is that harmonic analysis is not elementary, so it's hard to give an elementary explanation of it. I think of it as the extension of the ideas and techniques of Fourier theory from the groups R and Z to a wider class of groups such as Lie groups.
That's actually my problem. I have this new book about symmetries and Lagrangians and harmonic analysis and I thought "Hey! That'll be a good one to have in my bag of tricks." It's a much higher level book than I had thought and it's pulling in everything from group theory, algebra, topology, and tensor geometry. I've emptied half of my library (no joke) just trying to make sure I've got a good hold on the terms. It IS a good topic for me to learn but boy did I underestimate the task.... Hey, I need a good challenge from time to time.

-Dan