*In which we introduce the theory of characters of finite abelian groups, which we will use to compute eigenvalues and eigenvectors of graphs such as the cycle and the hypercube*

In the past lectures we have established the Cheeger inequalities

and the fact that the SpectralPartitioning algorithm, when given an eigenvector of , finds a cut such that . In the next lecture we will show that all such results are tight, up to constants, by proving that

- The dimension- hypercube has and , giving an infinite family of graphs for which , showing that the first Cheeger inequality is exactly tight.
- The -cycle has , and , giving an infinite family of graphs for which , showing that the second Cheeger inequality is tight up to a constant.
- There is an eigenvector of the 2nd eigenvalue of the hypercube , such that the SpectralPartitioning algorithm, given such a vector, outputs a cut of expansion , showing that the analysis of the SpectralPartitioning algorithm is tight up to a constant.

In this lecture we will develop some theoretical machinery to find the eigenvalues and eigenvectors of *Cayley graphs of finite Abelian groups*, a class of graphs that includes the cycle and the hypercube, among several other interesting examples. This theory will also be useful later, as a starting point to talk about algebraic constructions of expanders.

For readers familiar with the Fourier analysis of Boolean functions, or the discrete Fourier analysis of functions , or the standard Fourier analysis of periodic real functions, this theory will give a more general, and hopefully interesting, way to look at what they already know.

**1. Characters **

We will use additive notation for groups, so, if is a group, its unit will be denoted by , its group operation by , and the inverse of element by . Unless, noted otherwise, however, the definitions and results apply to non-abelian groups as well.

Definition 1 (Character)Let be a group (we will also use to refer to the set of group elements). A function is acharacterof if

- is a group homomorphism of into the multiplicative group .
- for every ,

Though this definition might seem to not bear the slightest connection to our goals, the reader should hang on because we will see next time that finding the eigenvectors and eigenvalues of the cycle is immediate once we know the characters of the group , and finding the eigenvectors and eigenvalues of the hypercube is immediate once we know the characters of the group .

Remark 1 (About the Boundedness Condition)If is a finite group, and is any element, thenand so if is a group homomorphism then

and so is a root of unity and, in particular, . This means that, for finite groups, the second condition in the definition of character is redundant. In certain infinite groups, however, the second condition does not follow from the first, for example defined as is a group homomorphism of into but it is not a character.

Just by looking at the definition, it might look like a finite group might have an infinite number of characters; the above remark, however, shows that a character of a finite group must map into -th roots of unity, of which there are only , showing a finite upper bound to the number of characters. Indeed, a much stronger upper bound holds, as we will prove next, after some preliminaries.

Lemma 2If is finite and is a character that is not identically equal to 1, then

*Proof:* Let be such that . Note that

where we used the fact that the mapping is a permutation. (We emphasize that even though we are using additive notation, the argument applies to non-abelian groups.) So we have

and since we assumed , it must be .

If is finite, given two functions , define the inner product

Lemma 3If are two different characters of a finite group , then

We will prove Lemma 3 shortly, but before doing so we note that, for a finite group , the set of functions is a -dimensional vector space, and that Lemma 3 implies that characters are orthogonal with respect to an inner product, and so they are linearly independent. In particular, we have established the following fact:

Corollary 4If is a finite group, then it has at most characters.

It remains to prove Lemma 3, which follows from the next two statements, whose proof is immediate from the definitions.

Fact 5If are characters of a group , then the mapping is also a character.

Fact 6If is a character of a group , then the mapping is also a character, and, for every , we have .

To complete the proof of Lemma 3, observe that:

- the function is a character;
- the assumption of the lemma is that there is an such that , and so, for the same element ,
- thus is a character that is not identically equal to 1, and so

Notice that, along the way, we have also proved the following fact:

Fact 7If is a group, then the set of characters of is also a group, with respect to the group operation of pointwise multiplication. The unit of the group is the character mapping every element to 1, and the inverse of a character is the pointwise conjugate of the character.

The group of characters is called thePontryagin dualof , and it is denoted by .

We now come to the punchline of this discussion.

Theorem 8If is a finite abelian group, then it has exactly characters.

*Proof:* We give a constructive proof. We know that every finite abelian group is isomorphic to a product of cyclic groups

so it will be enough to prove that

- the cyclic group has characters;
- if and are finite abelian groups with and characters, respectively, then their product has characters.

For the first claim, consider, for every , the function

Each such function is clearly a character ( maps to 1, is the multiplicative inverse of , and, recalling that for every integer , we also have ), and the values of are different for different values of , so we get distinct characters. This shows that has at least characters, and we already established that it can have at most characters.

For the second claim, note that if is a character of and is a character of , then it is easy to verify that the mapping is a character of . Furthermore, if and are two distinct pairs of characters, then the mappings and are two distinct characters of , because we either have an such that , in which case , or we have a such that , in which case . This shows that has at least characters, and we have already established that it can have at most that many

This means that the characters of a finite abelian group form an orthogonal basis for the set of all functions , so that any such function can be written as a linear combination

For every character , , and so the characters are actually a scaled-up orthonormal basis, and the coefficients can be computed as

Example 1 (The Boolean Cube)Consider the case , that is the group elements are , and the operation is bitwise xor. Then there is a character for every bit-vector , which is the functionEvery boolean function can thus be written as

where

which is the boolean Fourier transform.

Example 2 (The Cyclic Group)To work out another example, consider the case . Then every function can be written aswhere

which is the discrete Fourier transform.

**2. A Look Beyond **

Why is the term ”Fourier transform” used in this context? We will sketch an answer to this question, although what we say from this point on is not needed for our goal of finding the eigenvalues and eigenvectors of the cycle and the hypercube.

The point is that it is possible to set up a definitional framework that unifies both what we did in the previous section with finite Abelian groups, and the Fourier series and Fourier transforms of real and complex functions.

In the discussion of the previous section, we started to restrict ourselves to finite groups when we defined an inner product among functions .

If is an infinite abelian group, we can still define an inner product among functions , but we will need to define a measure over and restrict ourselves in the choice of functions. A measure over (a sigma-algebra of subsets of) is a Haar measure if, for every measurable subset and element we have , where . For example, if is finite, is a Haar measure. If , then is also a Haar measure (it is ok for a measure to be infinite for some sets), and if then the Lebesgue measure is a Haar measure. When a Haar measure exists, it is more or less unique up to multiplicative scaling. All *locally compact topological* abelian groups have a Haar measure, a very large class of abelian groups, that include all finite ones, , , and so on.

Once we have a Haar measure over , and we have defined an integral for functions , we say that a function is an element of if

For example, if is finite, then all functions are in , and a function is in if the series converges.

If , we can define their inner product

and use Cauchy-Schwarz to see that .

Now we can repeat the proof of Lemma 3 that for two different characters, and the only step of the proof that we need to verify for infinite groups is an analog of Lemma 2, that is we need to prove that if is a character that is not always equal to 1, then

and the same proof as in Lemma 2 works, with the key step being that, for every group element ,

because of the property of being a Haar measure.

We don’t have an analogous result to Theorem 8 showing that and are isomorphic, however it is possible to show that itself has a Haar measure , that the dual of is isomorphic to , and that if is continuous, then it can be written as the “linear combination”

where

In the finite case, the examples that we developed before correspond to setting and .

Example 3 (Fourier Series)The set of characters of the group with the operation of addition modulo 1 is isomorphic to , because for every integer we can define the functionand it can be shown that there are no other characters. We thus have the Fourier series for continuous functions ,

where

## 5 comments

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January 28, 2011 at 2:00 am

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January 28, 2011 at 6:15 am

Tyson WilliamsAt the end of the proof of Lemma 2, you are missing “= 0″.

January 29, 2011 at 1:17 pm

MarcoThank you for the notes Prof. Trevisan! A couple of typo corrections:

- Corollary 2 “it as” -> “it has”

- In the proof of Lemma 3 the definition of X is missing an “(x)” at the end

- In property two 2 of the proof of Thm 8 there is a bar missing in |\Gamma_2|

January 29, 2011 at 1:38 pm

lucathanks for the corrections!

April 17, 2011 at 6:13 am

Answering my own question on the Fourier Series | cartesian product[...] CS359G Lecture 5: Characters of Abelian Groups (lucatrevisan.wordpress.com) [...]