# Facts about Lie Groups and Algebras

In Spring 2016 I was taking 18.757 Representations of Lie Algebras. Since I knew next to nothing about either Lie groups or algebras, I was forced to quickly learn about their basic facts and properties. These are the notes that I wrote up accordingly. Proofs of most of these facts can be found in standard textbooks, for example Kirillov.

## 1. Lie groups

Let ${K = \mathbb R}$ or ${K = \mathbb C}$, depending on taste.

Definition 1

A Lie group is a group ${G}$ which is also a ${K}$-manifold; the multiplication maps ${G \times G \rightarrow G}$ (by ${(g_1, g_2) \mapsto g_1g_2}$) and the inversion map ${G \rightarrow G}$ (by ${g \mapsto g^{-1}}$) are required to be smooth.

A morphism of Lie groups is a map which is both a map of manifolds and a group homomorphism.

Throughout, we will let ${e \in G}$ denote the identity, or ${e_G}$ if we need further emphasis.

Note that in particular, every group ${G}$ can be made into a Lie group by endowing it with the discrete topology. This is silly, so we usually require only focus on connected groups:

Proposition 2 (Reduction to connected Lie groups)

Let ${G}$ be a Lie group and ${G^0}$ the connected component of ${G}$ which contains ${e}$. Then ${G^0}$ is a normal subgroup, itself a Lie group, and the quotient ${G/G^0}$ has the discrete topology.

In fact, we can also reduce this to the study of simply connected Lie groups as follows.

Proposition 3 (Reduction to simply connected Lie groups)

If ${G}$ is connected, let ${\pi : \widetilde G \rightarrow G}$ be its universal cover. Then ${\widetilde G}$ is a Lie group, ${\pi}$ is a morphism of Lie groups, and ${\ker \pi \cong \pi_1(G)}$.

Here are some examples of Lie groups.

Example 4 (Examples of Lie groups)

• ${\mathbb R}$ under addition is a real one-dimensional Lie group.
• ${\mathbb C}$ under addition is a complex one-dimensional Lie group (and a two-dimensional real Lie group)!
• The unit circle ${S^1 \subseteq \mathbb C}$ is a real Lie group under multiplication.
• ${\text{GL }(n, K) \subset K^{\oplus n^2}}$ is a Lie group of dimension ${n^2}$. This example becomes important for representation theory: a representation of a Lie group ${G}$ is a morphism of Lie groups ${G \rightarrow \text{GL }(n, K)}$.
• ${\text{SL }(n, K) \subset \text{GL }(n, K)}$ is a Lie group of dimension ${n^2-1}$.

As geometric objects, Lie groups ${G}$ enjoy a huge amount of symmetry. For example, any neighborhood ${U}$ of ${e}$ can be “copied over” to any other point ${g \in G}$ by the natural map ${gU}$. There is another theorem worth noting, which is that:

Proposition 5

If ${G}$ is a connected Lie group and ${U}$ is a neighborhood of the identity ${e \in G}$, then ${U}$ generates ${G}$ as a group.

## 2. Haar measure

Recall the following result and its proof from representation theory:

Claim 6

For any finite group ${G}$, ${\mathbb C[G]}$ is semisimple; all finite-dimensional representations decompose into irreducibles.

Proof: Take a representation ${V}$ and equip it with an arbitrary inner form ${\left< -,-\right>_0}$. Then we can average it to obtain a new inner form

$\displaystyle \left< v, w \right> = \frac{1}{|G|} \sum_{g \in G} \left< gv, gw \right>_0.$

which is ${G}$-invariant. Thus given a subrepresentation ${W \subseteq V}$ we can just take its orthogonal complement to decompose ${V}$. $\Box$
We would like to repeat this type of proof with Lie groups. In this case the notion ${\sum_{g \in G}}$ doesn’t make sense, so we want to replace it with an integral ${\int_{g \in G}}$ instead. In order to do this we use the following:

Theorem 7 (Haar measure)

Let ${G}$ be a Lie group. Then there exists a unique Radon measure ${\mu}$ (up to scaling) on ${G}$ which is left-invariant, meaning

$\displaystyle \mu(g \cdot S) = \mu(S)$

for any Borel subset ${S \subseteq G}$ and “translate” ${g \in G}$. This measure is called the (left) Haar measure.

Example 8 (Examples of Haar measures)

• The Haar measure on ${(\mathbb R, +)}$ is the standard Lebesgue measure which assigns ${1}$ to the closed interval ${[0,1]}$. Of course for any ${S}$, ${\mu(a+S) = \mu(S)}$ for ${a \in \mathbb R}$.
• The Haar measure on ${(\mathbb R \setminus \{0\}, \times)}$ is given by

$\displaystyle \mu(S) = \int_S \frac{1}{|t|} \; dt.$

In particular, ${\mu([a,b]) = \log(b/a)}$. One sees the invariance under multiplication of these intervals.

• Let ${G = \text{GL }(n, \mathbb R)}$. Then a Haar measure is given by

$\displaystyle \mu(S) = \int_S |\det(X)|^{-n} \; dX.$

• For the circle group ${S^1}$, consider ${S \subseteq S^1}$. We can define

$\displaystyle \mu(S) = \frac{1}{2\pi} \int_S d\varphi$

across complex arguments ${\varphi}$. The normalization factor of ${2\pi}$ ensures ${\mu(S^1) = 1}$.

Note that we have:

Corollary 9

If the Lie group ${G}$ is compact, there is a unique Haar measure with ${\mu(G) = 1}$.

This follows by just noting that if ${\mu}$ is Radon measure on ${X}$, then ${\mu(X) < \infty}$. This now lets us deduce that

Corollary 10 (Compact Lie groups are semisimple)

${\mathbb C[G]}$ is semisimple for any compact Lie group ${G}$.

Indeed, we can now consider

$\displaystyle \left< v,w\right> = \int_G \left< g \cdot v, g \cdot w\right>_0 \; dg$

as we described at the beginning.

## 3. The tangent space at the identity

In light of the previous comment about neighborhoods of ${e}$ generating ${G}$, we see that to get some information about the entire Lie group it actually suffices to just get “local” information of ${G}$ at the point ${e}$ (this is one formalization of the fact that Lie groups are super symmetric).

To do this one idea is to look at the tangent space. Let ${G}$ be an ${n}$-dimensional Lie group (over ${K}$) and consider ${\mathfrak g = T_eG}$ the tangent space to ${G}$ at the identity ${e \in G}$. Naturally, this is a ${K}$-vector space of dimension ${n}$. We call it the Lie algebra associated to ${G}$.

Example 11 (Lie algebras corresponding to Lie groups)

• ${(\mathbb R, +)}$ has a real Lie algebra isomorphic to ${\mathbb R}$.
• ${(\mathbb C, +)}$ has a complex Lie algebra isomorphic to ${\mathbb C}$.
• The unit circle ${S^1 \subseteq \mathbb C}$ has a real Lie algebra isomorphic to ${\mathbb R}$, which we think of as the “tangent line” at the point ${1 \in S^1}$.

Example 12 (${\mathfrak{gl}(n, K)}$)

Let’s consider ${\text{GL }(n, K) \subset K^{\oplus n^2}}$, an open subset of ${K^{\oplus n^2}}$. Its tangent space should just be an ${n^2}$-dimensional ${K}$-vector space. By identifying the components in the obvious way, we can think of this Lie algebra as just the set of all ${n \times n}$ matrices.

This Lie algebra goes by the notation ${\mathfrak{gl}(n, K)}$.

Example 13 (${\mathfrak{sl}(n, K)}$)

Recall ${\text{SL }(n, K) \subset \text{GL }(n, K)}$ is a Lie group of dimension ${n^2-1}$, hence its Lie algebra should have dimension ${n^2-1}$. To see what it is, let’s look at the special case ${n=2}$ first: then

$\displaystyle \text{SL }(2, K) = \left\{ \begin{pmatrix} a & b \\ c & d \end{pmatrix} \mid ad - bc = 1 \right\}.$

Viewing this as a polynomial surface ${f(a,b,c,d) = ad-bc}$ in ${K^{\oplus 4}}$, we compute

$\displaystyle \nabla f = \left< d, -c, -b, a \right>$

and in particular the tangent space to the identity matrix ${\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}}$ is given by the orthogonal complement of the gradient

$\displaystyle \nabla f (1,0,0,1) = \left< 1, 0, 0, 1 \right>.$

Hence the tangent plane can be identified with matrices satisfying ${a+d=0}$. In other words, we see

$\displaystyle \mathfrak{sl}(2, K) = \left\{ T \in \mathfrak{gl}(2, K) \mid \text{Tr } T = 0. \right\}.$

By repeating this example in greater generality, we discover

$\displaystyle \mathfrak{sl}(n, K) = \left\{ T \in \mathfrak{gl}(n, K) \mid \text{Tr } T = 0. \right\}.$

## 4. The exponential map

Right now, ${\mathfrak g}$ is just a vector space. However, by using the group structure we can get a map from ${\mathfrak g}$ back into ${G}$. The trick is “differential equations”:

Proposition 14 (Differential equations for Lie theorists)

Let ${G}$ be a Lie group over ${K}$ and ${\mathfrak g}$ its Lie algebra. Then for every ${x \in \mathfrak g}$ there is a unique homomorphism

$\displaystyle \gamma_x : K \rightarrow G$

which is a morphism of Lie groups, such that

$\displaystyle \gamma_x'(0) = x \in T_eG = \mathfrak g.$

We will write ${\gamma_x(t)}$ to emphasize the argument ${t \in K}$ being thought of as “time”. Thus this proposition should be intuitively clear: the theory of differential equations guarantees that ${\gamma_x}$ is defined and unique in a small neighborhood of ${0 \in K}$. Then, the group structure allows us to extend ${\gamma_x}$ uniquely to the rest of ${K}$, giving a trajectory across all of ${G}$. This is sometimes called a one-parameter subgroup of ${G}$, but we won’t use this terminology anywhere in what follows.

This lets us define:

Definition 15

Retain the setting of the previous proposition. Then the exponential map is defined by

$\displaystyle \exp : \mathfrak g \rightarrow G \qquad\text{by}\qquad x \mapsto \gamma_x(1).$

The exponential map gets its name from the fact that for all the examples I discussed before, it is actually just the map ${e^\bullet}$. Note that below, ${e^T = \sum_{k \ge 0} \frac{T^k}{k!}}$ for a matrix ${T}$; this is called the matrix exponential.

Example 16 (Exponential Maps of Lie algebras)

• If ${G = \mathbb R}$, then ${\mathfrak g = \mathbb R}$ too. Then ${\gamma_x(t) = tx \in \mathbb R}$ (where ${t \in \mathbb R}$) is a morphism of Lie groups ${\gamma_x : \mathbb R \rightarrow G}$. Hence

$\displaystyle \exp : \mathbb R \rightarrow \underbrace{\mathbb R}_{=G} \qquad \exp(x) = \gamma_x(1) = x \in \mathbb R = G.$

In other words, the exponential map is the identity.

• Ditto for ${\mathbb C}$.
• For ${S^1}$ and ${x \in \mathbb R}$, the map ${\gamma_x : \mathbb R \rightarrow S^1}$ given by ${t \mapsto e^{itx}}$ works. Hence

$\displaystyle \exp : \mathbb R \rightarrow S^1 \qquad \exp(x) = \gamma_x(1) = e^{it} \in S^1.$

• For ${\mathop{\mathrm{GL}}(n, K)}$, the map ${\gamma_X : K \rightarrow \mathop{\mathrm{GL}}(n, K)}$ given by ${t \mapsto e^{tX}}$ works nicely (now ${X}$ is a matrix). (Note that we have to check ${e^{tX}}$ is actually invertible for this map to be well-defined.) Hence the exponential map is given by

$\displaystyle \exp : \mathfrak{gl}(n,K) \rightarrow \mathop{\mathrm{GL}}(n,K) \qquad \exp(X) = \gamma_X(1) = e^X \in \mathop{\mathrm{GL}}(n, K).$

• Similarly,

$\displaystyle \exp : \mathfrak{sl}(n,K) \rightarrow \mathop{\mathrm{SL}}(n,K) \qquad \exp(X) = \gamma_X(1) = e^X \in \mathop{\mathrm{SL}}(n, K).$

Here we had to check that if ${X \in \mathfrak{sl}(n,K)}$, meaning ${\mathop{\mathrm{Tr}} X = 0}$, then ${\det(e^X) = 1}$. This can be seen by writing ${X}$ in an upper triangular basis.

Actually, taking the tangent space at the identity is a functor. Consider a map ${\varphi : G_1 \rightarrow G_2}$ of Lie groups, with lie algebras ${\mathfrak g_1}$ and ${\mathfrak g_2}$. Because ${\varphi}$ is a group homomorphism, ${G_1 \ni e_1 \mapsto e_2 \in G_2}$. Now, by manifold theory we know that maps ${f : M \rightarrow N}$ between manifolds gives a linear map between the corresponding tangent spaces, say ${Tf : T_pM \rightarrow T_{fp}N}$. For us we obtain a linear map

$\displaystyle \varphi_\ast = T \varphi : \mathfrak g_1 \rightarrow \mathfrak g_2.$

In fact, this ${\varphi_\ast}$ fits into a diagram

Here are a few more properties of ${\exp}$:

• ${\exp(0) = e \in G}$, which is immediate by looking at the constant trajectory ${\phi_0(t) \equiv e}$.
• ${\exp'(x) = x \in \mathfrak g}$, i.e. the total derivative ${D\exp : \mathfrak g \rightarrow \mathfrak g}$ is the identity. This is again by construction.
• In particular, by the inverse function theorem this implies that ${\exp}$ is a diffeomorphism in a neighborhood of ${0 \in \mathfrak g}$, onto a neighborhood of ${e \in G}$.
• ${\exp}$ commutes with the commutator. (By the above diagram.)

## 5. The commutator

Right now ${\mathfrak g}$ is still just a vector space, the tangent space. But now that there is map ${\exp : \mathfrak g \rightarrow G}$, we can use it to put a new operation on ${\mathfrak g}$, the so-called commutator.

The idea is follows: we want to “multiply” two elements of ${\mathfrak g}$. But ${\mathfrak g}$ is just a vector space, so we can’t do that. However, ${G}$ itself has a group multiplication, so we should pass to ${G}$ using ${\exp}$, use the multiplication in ${G}$ and then come back.

Here are the details. As we just mentioned, ${\exp}$ is a diffeomorphism near ${e \in G}$. So for ${x}$, ${y}$ close to the origin of ${\mathfrak g}$, we can look at ${\exp(x)}$ and ${\exp(y)}$, which are two elements of ${G}$ close to ${e}$. Multiplying them gives an element still close to ${e}$, so its equal to ${\exp(z)}$ for some unique ${z}$, call it ${\mu(x,y)}$.

One can show in fact that ${\mu}$ can be written as a Taylor series in two variables as

$\displaystyle \mu(x,y) = x + y + \frac{1}{2} [x,y] + \text{third order terms} + \dots$

where ${[x,y]}$ is a skew-symmetric bilinear map, meaning ${[x,y] = -[y,x]}$. It will be more convenient to work with ${[x,y]}$ than ${\mu(x,y)}$ itself, so we give it a name:

Definition 17

This ${[x,y]}$ is called the commutator of ${G}$.

Now we know multiplication in ${G}$ is associative, so this should give us some nontrivial relation on the bracket ${[,]}$. Specifically, since

$\displaystyle \exp(x) \left( \exp(y) \exp(z) \right) = \left( \exp(x) \exp(y) \right) \exp(z).$

we should have that ${\mu(x, \mu(y,z)) = \mu(\mu(x,y), z)}$, and this should tell us something. In fact, the claim is:

Theorem 18

The bracket ${[,]}$ satisfies the Jacobi identity

$\displaystyle [x,[y,z]] + [y,[z,x]] + [z,[x,y]] = 0.$

Proof: Although I won’t prove it, the third-order terms (and all the rest) in our definition of ${[x,y]}$ can be written out explicitly as well: for example, for example, we actually have

$\displaystyle \mu(x,y) = x + y + \frac{1}{2} [x,y] + \frac{1}{12} \left( [x, [x,y]] + [y,[y,x]] \right) + \text{fourth order terms} + \dots.$

The general formula is called the Baker-Campbell-Hausdorff formula.

Then we can force ourselves to expand this using the first three terms of the BCS formula and then equate the degree three terms. The left-hand side expands initially as ${\mu\left( x, y + z + \frac{1}{2} [y,z] + \frac{1}{12} \left( [y,[y,z]] + [z,[z,y] \right) \right)}$, and the next step would be something ugly.

This computation is horrifying and painful, so I’ll pretend I did it and tell you the end result is as claimed. $\Box$
There is a more natural way to see why this identity is the “right one”; see Qiaochu. However, with this proof I want to make the point that this Jacobi identity is not our decision: instead, the Jacobi identity is forced upon us by associativity in ${G}$.

Example 19 (Examples of commutators attached to Lie groups)

• If ${G}$ is an abelian group, we have ${-[y,x] = [x,y]}$ by symmetry and ${[x,y] = [y,x]}$ from ${\mu(x,y) = \mu(y,x)}$. Thus ${[x,y] = 0}$ in ${\mathfrak g}$ for any abelian Lie group ${G}$.
• In particular, the brackets for ${G \in \{\mathbb R, \mathbb C, S^1\}}$ are trivial.
• Let ${G = \text{GL }(n, K)}$. Then one can show that

$\displaystyle [T,S] = TS - ST \qquad \forall S, T \in \mathfrak{gl}(n, K).$

• Ditto for ${\text{SL }(n, K)}$.

In any case, with the Jacobi identity we can define an general Lie algebra as an intrinsic object with a Jacobi-satisfying bracket:

Definition 20

A Lie algebra over ${k}$ is a ${k}$-vector space equipped with a skew-symmetric bilinear bracket ${[,]}$ satisfying the Jacobi identity.

A morphism of Lie algebras and preserves the bracket.

Note that a Lie algebra may even be infinite-dimensional (even though we are assuming ${G}$ is finite-dimensional, so that they will never come up as a tangent space).

Example 21 (Associative algebra ${\rightarrow}$ Lie algebra)

Any associative algebra ${A}$ over ${k}$ can be made into a Lie algebra by taking the same underlying vector space, and using the bracket ${[a,b] = ab - ba}$.

## 6. The fundamental theorems

We finish this list of facts by stating the three “fundamental theorems” of Lie theory. They are based upon the functor

$\displaystyle \mathscr{L} : G \mapsto T_e G$

we have described earlier, which is a functor

• from the category of Lie groups
• into the category of finite-dimensional Lie algebras.

The first theorem requires the following definition:

Definition 22

A Lie subgroup ${H}$ of a Lie group ${G}$ is a subgroup ${H}$ such that the inclusion map ${H \hookrightarrow G}$ is also an injective immersion.

A Lie subalgebra ${\mathfrak h}$ of a Lie algebra ${\mathfrak g}$ is a vector subspace preserved under the bracket (meaning that ${[\mathfrak h, \mathfrak h] \subseteq \mathfrak h]}$).

Theorem 23 (Lie I)

Let ${G}$ be a real or complex Lie group with Lie algebra ${\mathfrak g}$. Then given a Lie subgroup ${H \subseteq G}$, the map

$\displaystyle H \mapsto \mathscr{L}(H) \subseteq \mathfrak g$

is a bijection between Lie subgroups of ${G}$ and Lie subalgebras of ${\mathfrak g}$.

Theorem 24 (The Lie functor is an equivalence of categories)

Restrict ${\mathscr{L}}$ to a functor

• from the category of simply connected Lie groups over ${K}$
• to the category of finite-dimensional Lie algebras over ${K}$.

Then

1. (Lie II) ${\mathscr{L}}$ is fully faithful, and
2. (Lie III) ${\mathscr{L}}$ is essentially surjective on objects.

If we drop the “simply connected” condition, we obtain a functor which is faithful and exact, but not full: non-isomorphic Lie groups can have isomorphic Lie algebras (one example is ${\text{SO }(3)}$ and ${\text{SU }(2)}$).

# Things Fourier

For some reason several classes at MIT this year involve Fourier analysis. I was always confused about this as a high schooler, because no one ever gave me the “orthonormal basis” explanation, so here goes. As a bonus, I also prove a form of Arrow’s Impossibility Theorem using binary Fourier analysis, and then talk about the fancier generalizations using Pontryagin duality and the Peter-Weyl theorem.

In what follows, we let ${\mathbb T = \mathbb R/\mathbb Z}$ denote the “circle group”, thought of as the additive group of “real numbers modulo ${1}$”. There is a canonical map ${e : \mathbb T \rightarrow \mathbb C}$ sending ${\mathbb T}$ to the complex unit circle, given by ${e(\theta) = \exp(2\pi i \theta)}$.

Disclaimer: I will deliberately be sloppy with convergence issues, in part because I don’t fully understand them myself, and in part because I don’t care.

## 1. Synopsis

Suppose we have a domain ${Z}$ and are interested in functions ${f : Z \rightarrow \mathbb C}$. Naturally, the set of such functions form a complex vector space. We like to equip the set of such functions with an positive definite inner product. The idea of Fourier analysis is to then select an orthonormal basis for this set of functions, say ${(e_\xi)_{\xi}}$, which we call the characters; the indexing ${\xi}$ are called frequencies. In that case, since we have a basis, every function ${f : Z \rightarrow \mathbb C}$ becomes a sum

$\displaystyle f(x) = \sum_{\xi} \widehat f(\xi) e_\xi$

where ${\widehat f(\xi)}$ are complex coefficients of the basis; appropriately we call ${\widehat f}$ the Fourier coefficients. The variable ${x \in Z}$ is referred to as the physical variable. This is generally good because the characters are deliberately chosen to be nice “symmetric” functions, like sine or cosine waves or other periodic functions. Thus ${we}$ decompose an arbitrarily complicated function into a sum on nice ones.

For convenience, we record a few facts about orthonormal bases.

Proposition 1 (Facts about orthonormal bases)

Let ${V}$ be a complex Hilbert space with inner form ${\left< -,-\right>}$ and suppose ${x = \sum_\xi a_\xi e_\xi}$ and ${y = \sum_\xi b_\xi e_\xi}$ where ${e_\xi}$ are an orthonormal basis. Then

\displaystyle \begin{aligned} \left< x,x \right> &= \sum_\xi |a_\xi|^2 \\ a_\xi &= \left< x, e_\xi \right> \\ \left< x,y \right> &= \sum_\xi a_\xi \overline{b_\xi}. \end{aligned}

## 2. Common Examples

### 2.1. Binary Fourier analysis on ${\{\pm1\}^n}$

Let ${Z = \{\pm 1\}^n}$ for some positive integer ${n}$, so we are considering functions ${f(x_1, \dots, x_n)}$ accepting binary values. Then the functions ${Z \rightarrow \mathbb C}$ form a ${2^n}$-dimensional vector space ${\mathbb C^Z}$, and we endow it with the inner form

$\displaystyle \left< f,g \right> = \frac{1}{2^n} \sum_{x \in Z} f(x) \overline{g(x)}.$

In particular,

$\displaystyle \left< f,f \right> = \frac{1}{2^n} \sum_{x \in Z} \left\lvert f(x) \right\rvert^2$

is the average of the squares; this establishes also that ${\left< -,-\right>}$ is positive definite.

In that case, the multilinear polynomials form a basis of ${\mathbb C^Z}$, that is the polynomials

$\displaystyle \chi_S(x_1, \dots, x_n) = \prod_{s \in S} x_s.$

Thus our frequency set is actually the subsets ${S \subseteq \{1, \dots, n\}}$. Thus, we have a decomposition

$\displaystyle f = \sum_{S \subseteq \{1, \dots, n\}} \widehat f(S) \chi_S.$

Example 2 (An example of binary Fourier analysis)

Let ${n = 2}$. Then binary functions ${\{ \pm 1\}^2 \rightarrow \mathbb C}$ have a basis given by the four polynomials

$\displaystyle 1, \quad x_1, \quad x_2, \quad x_1x_2.$

For example, consider the function ${f}$ which is ${1}$ at ${(1,1)}$ and ${0}$ elsewhere. Then we can put

$\displaystyle f(x_1, x_2) = \frac{x_1+1}{2} \cdot \frac{x_2+1}{2} = \frac14 \left( 1 + x_1 + x_2 + x_1x_2 \right).$

So the Fourier coefficients are ${\widehat f(S) = \frac 14}$ for each of the four ${S}$‘s.

This notion is useful in particular for binary functions ${f : \{\pm1\}^n \rightarrow \{\pm1\}}$; for these functions (and products thereof), we always have ${\left< f,f \right> = 1}$.

It is worth noting that the frequency ${\varnothing}$ plays a special role:

Exercise 3

Show that

$\displaystyle \widehat f(\varnothing) = \frac{1}{|Z|} \sum_{x \in Z} f(x).$

### 2.2. Fourier analysis on finite groups ${Z}$

This is the Fourier analysis used in this post and this post. Here, we have a finite abelian group ${Z}$, and consider functions ${Z \rightarrow \mathbb C}$; this is a ${|Z|}$-dimensional vector space. The inner product is the same as before:

$\displaystyle \left< f,g \right> = \frac{1}{|Z|} \sum_{x \in Z} f(x) \overline{g}(x).$

Now here is how we generate the characters. We equip ${Z}$ with a non-degenerate symmetric bilinear form

$\displaystyle Z \times Z \xrightarrow{\cdot} \mathbb T \qquad (\xi, x) \mapsto \xi \cdot x.$

Experts may already recognize this as a choice of isomorphism between ${Z}$ and its Pontryagin dual. This time the characters are given by

$\displaystyle \left( e_\xi \right)_{\xi \in Z} \qquad \text{where} \qquad e_\xi(x) = e(\xi \cdot x).$

In this way, the set of frequencies is also ${Z}$, but the ${\xi \in Z}$ play very different roles from the “physical” ${x \in Z}$. (It is not too hard to check these indeed form an orthonormal basis in the function space ${\mathbb C^{\left\lvert Z \right\rvert}}$, since we assumed that ${\cdot}$ is non-degenerate.)

Example 4 (Cube roots of unity filter)

Suppose ${Z = \mathbb Z/3\mathbb Z}$, with the inner form given by ${\xi \cdot x = (\xi x)/3}$. Let ${\omega = \exp(\frac 23 \pi i)}$ be a primitive cube root of unity. Note that

$\displaystyle e_\xi(x) = \begin{cases} 1 & \xi = 0 \\ \omega^x & \xi = 1 \\ \omega^{2x} & \xi = 2. \end{cases}$

Then given ${f : Z \rightarrow \mathbb C}$ with ${f(0) = a}$, ${f(1) = b}$, ${f(2) = c}$, we obtain

$\displaystyle f(x) = \frac{a+b+c}{3} \cdot 1 + \frac{a + \omega^2 b + \omega c}{3} \cdot \omega^x + \frac{a + \omega b + \omega^2 c}{3} \cdot \omega^{2x}.$

In this way we derive that the transforms are

\displaystyle \begin{aligned} \widehat f(0) &= \frac{a+b+c}{3} \\ \widehat f(1) &= \frac{a+\omega^2 b+ \omega c}{3} \\ \widehat f(2) &= \frac{a+\omega b+\omega^2c}{3}. \end{aligned}

Exercise 5

Show that

$\displaystyle \widehat f(0) = \frac{1}{|Z|} \sum_{x \in Z} f(x).$

Olympiad contestants may recognize the previous example as a “roots of unity filter”, which is exactly the point. For concreteness, suppose one wants to compute

$\displaystyle \binom{1000}{0} + \binom{1000}{3} + \dots + \binom{1000}{999}.$

In that case, we can consider the function

$\displaystyle w : \mathbb Z/3 \rightarrow \mathbb C.$

such that ${w(0) = 1}$ but ${w(1) = w(2) = 0}$. By abuse of notation we will also think of ${w}$ as a function ${w : \mathbb Z \twoheadrightarrow \mathbb Z/3 \rightarrow \mathbb C}$. Then the sum in question is

\displaystyle \begin{aligned} \sum_n \binom{1000}{n} w(n) &= \sum_n \binom{1000}{n} \sum_{k=0,1,2} \widehat w(k) \omega^{kn} \\ &= \sum_{k=0,1,2} \widehat w(k) \sum_n \binom{1000}{n} \omega^{kn} \\ &= \sum_{k=0,1,2} \widehat w(k) (1+\omega^k)^n. \end{aligned}

In our situation, we have ${\widehat w(0) = \widehat w(1) = \widehat w(2) = \frac13}$, and we have evaluated the desired sum. More generally, we can take any periodic weight ${w}$ and use Fourier analysis in order to interchange the order of summation.

Example 6 (Binary Fourier analysis)

Suppose ${Z = \{\pm 1\}^n}$, viewed as an abelian group under pointwise multiplication hence isomorphic to ${(\mathbb Z/2\mathbb Z)^{\oplus n}}$. Assume we pick the dot product defined by

$\displaystyle \xi \cdot x = \frac{1}{2} \sum_i \xi_i x_i$

where ${\xi = (\xi_1, \dots, \xi_n)}$ and ${x = (x_1, \dots, x_n)}$.

We claim this coincides with the first example we gave. Indeed, let ${S \subseteq \{1, \dots, n\}}$ and let ${\xi \in \{\pm1\}^n}$ which is ${-1}$ at positions in ${S}$, and ${+1}$ at positions not in ${S}$. Then the character ${\chi_S}$ form the previous example coincides with the character ${e_\xi}$ in the new notation. In particular, ${\widehat f(S) = \widehat f(\xi)}$.

Thus Fourier analysis on a finite group ${Z}$ subsumes binary Fourier analysis.

### 2.3. Fourier series for functions ${L^2([-\pi, \pi])}$

Now we consider the space ${L^2([-\pi, \pi])}$ of square-integrable functions ${[-\pi, \pi] \rightarrow \mathbb C}$, with inner form

$\displaystyle \left< f,g \right> = \frac{1}{2\pi} \int_{[-\pi, \pi]} f(x) \overline{g(x)}.$

Sadly, this is not a finite-dimensional vector space, but fortunately it is a Hilbert space so we are still fine. In this case, an orthonormal basis must allow infinite linear combinations, as long as the sum of squares is finite.

Now, it turns out in this case that

$\displaystyle (e_n)_{n \in \mathbb Z} \qquad\text{where}\qquad e_n(x) = \exp(inx)$

is an orthonormal basis for ${L^2([-\pi, \pi])}$. Thus this time the frequency set ${\mathbb Z}$ is infinite. So every function ${f \in L^2([-\pi, \pi])}$ decomposes as

$\displaystyle f(x) = \sum_n \widehat f(n) \exp(inx)$

for ${\widehat f(n)}$.

This is a little worse than our finite examples: instead of a finite sum on the right-hand side, we actually have an infinite sum. This is because our set of frequencies is now ${\mathbb Z}$, which isn’t finite. In this case the ${\widehat f}$ need not be finitely supported, but do satisfy ${\sum_n |\widehat f(n)|^2 < \infty}$.

Since the frequency set is indexed by ${\mathbb Z}$, we call this a Fourier series to reflect the fact that the index is ${n \in \mathbb Z}$.

Exercise 7

Show once again

$\displaystyle \widehat f(0) = \frac{1}{2\pi} \int_{[-\pi, \pi]} f(x).$

Often we require that the function ${f}$ satisfies ${f(-\pi) = f(\pi)}$, so that ${f}$ becomes a periodic function, and we can think of it as ${f : \mathbb T \rightarrow \mathbb C}$.

### 2.4. Summary

We summarize our various flavors of Fourier analysis in the following table.

$\displaystyle \begin{array}{llll} \hline \text{Type} & \text{Physical var} & \text{Frequency var} & \text{Basis functions} \\ \hline \textbf{Binary} & \{\pm1\}^n & \text{Subsets } S \subseteq \left\{ 1, \dots, n \right\} & \prod_{s \in S} x_s \\ \textbf{Finite group} & Z & \xi \in Z, \text{ choice of } \cdot, & e(\xi \cdot x) \\ \textbf{Fourier series} & \mathbb T \text{ or } [-\pi, \pi] & n \in \mathbb Z & \exp(inx) \\ \end{array}$

In fact, we will soon see that all these examples are subsumed by Pontryagin duality for compact groups ${G}$.

## 3. Parseval and friends

The notion of an orthonormal basis makes several “big-name” results in Fourier analysis quite lucid. Basically, we can take every result from Proposition~1, translate it into the context of our Fourier analysis, and get a big-name result.

Corollary 8 (Parseval theorem)

Let ${f : Z \rightarrow \mathbb C}$, where ${Z}$ is a finite abelian group. Then

$\displaystyle \sum_\xi |\widehat f(\xi)|^2 = \frac{1}{|Z|} \sum_{x \in Z} |f(x)|^2.$

Similarly, if ${f : [-\pi, \pi] \rightarrow \mathbb C}$ is square-integrable then its Fourier series satisfies

$\displaystyle \sum_n |\widehat f(n)|^2 = \frac{1}{2\pi} \int_{[-\pi, \pi]} |f(x)|^2.$

Proof: Recall that ${\left< f,f\right>}$ is equal to the square sum of the coefficients. $\Box$

Corollary 9 (Formulas for ${\widehat f}$)

Let ${f : Z \rightarrow \mathbb C}$, where ${Z}$ is a finite abelian group. Then

$\displaystyle \widehat f(\xi) = \frac{1}{|Z|} \sum_{x \in Z} f(x) \overline{e_\xi(x)}.$

Similarly, if ${f : [-\pi, \pi] \rightarrow \mathbb C}$ is square-integrable then its Fourier series is given by

$\displaystyle \widehat f(n) = \frac{1}{2\pi} \int_{[-\pi, \pi]} f(x) \exp(-inx).$

Proof: Recall that in an orthonormal basis ${(e_\xi)_\xi}$, the coefficient of ${e_\xi}$ in ${f}$ is ${\left< f, e_\xi\right>}$. $\Box$
Note in particular what happens if we select ${\xi = 0}$ in the above!

Corollary 10 (Plancherel theorem)

Let ${f : Z \rightarrow \mathbb C}$, where ${Z}$ is a finite abelian group. Then

$\displaystyle \left< f,g \right> = \sum_{\xi \in Z} \widehat f(\xi) \overline{\widehat g(\xi)}.$

Similarly, if ${f : [-\pi, \pi] \rightarrow \mathbb C}$ is square-integrable then

$\displaystyle \left< f,g \right> = \sum_n \widehat f(\xi) \overline{\widehat g(\xi)}.$

Proof: Guess! $\Box$

## 4. (Optional) Arrow’s Impossibility Theorem

As an application, we now prove a form of Arrow’s theorem. Consider ${n}$ voters voting among ${3}$ candidates ${A}$, ${B}$, ${C}$. Each voter specifies a tuple ${v_i = (x_i, y_i, z_i) \in \{\pm1\}^3}$ as follows:

• ${x_i = 1}$ if ${A}$ ranks ${A}$ ahead of ${B}$, and ${x_i = -1}$ otherwise.
• ${y_i = 1}$ if ${A}$ ranks ${B}$ ahead of ${C}$, and ${y_i = -1}$ otherwise.
• ${z_i = 1}$ if ${A}$ ranks ${C}$ ahead of ${A}$, and ${z_i = -1}$ otherwise.

Tacitly, we only consider ${3! = 6}$ possibilities for ${v_i}$: we forbid “paradoxical” votes of the form ${x_i = y_i = z_i}$ by assuming that people’s votes are consistent (meaning the preferences are transitive).

Then, we can consider a voting mechanism

\displaystyle \begin{aligned} f : \{\pm1\}^n &\rightarrow \{\pm1\} \\ g : \{\pm1\}^n &\rightarrow \{\pm1\} \\ h : \{\pm1\}^n &\rightarrow \{\pm1\} \end{aligned}

such that ${f(x_\bullet)}$ is the global preference of ${A}$ vs. ${B}$, ${g(y_\bullet)}$ is the global preference of ${B}$ vs. ${C}$, and ${h(z_\bullet)}$ is the global preference of ${C}$ vs. ${A}$. We’d like to avoid situations where the global preference ${(f(x_\bullet), g(y_\bullet), h(z_\bullet))}$ is itself paradoxical.

In fact, we will prove the following theorem:

Theorem 11 (Arrow Impossibility Theorem)

Assume that ${(f,g,h)}$ always avoids paradoxical outcomes, and assume ${\mathbf E f = \mathbf E g = \mathbf E h = 0}$. Then ${(f,g,h)}$ is either a dictatorship or anti-dictatorship: there exists a “dictator” ${k}$ such that

$\displaystyle f(x_\bullet) = \pm x_k, \qquad g(y_\bullet) = \pm y_k, \qquad h(z_\bullet) = \pm z_k$

where all three signs coincide.

The “irrelevance of independent alternatives” reflects that The assumption ${\mathbf E f = \mathbf E g = \mathbf E h = 0}$ provides symmetry (and e.g. excludes the possibility that ${f}$, ${g}$, ${h}$ are constant functions which ignore voter input). Unlike the usual Arrow theorem, we do not assume that ${f(+1, \dots, +1) = +1}$ (hence possibility of anti-dictatorship).

To this end, we actually prove the following result:

Lemma 12

Assume the ${n}$ voters vote independently at random among the ${3! = 6}$ possibilities. The probability of a paradoxical outcome is exactly

$\displaystyle \frac14 + \frac14 \sum_{S \subseteq \{1, \dots, n\}} \left( -\frac13 \right)^{\left\lvert S \right\rvert} \left( \widehat f(S) \widehat g(S) + \widehat g(S) \widehat h(S) + \widehat h(S) \widehat f(S) \right) .$

Proof: Define the Boolean function ${D : \{\pm 1\}^3 \rightarrow \mathbb R}$ by

$\displaystyle D(a,b,c) = ab + bc + ca = \begin{cases} 3 & a,b,c \text{ all equal} \\ -1 & a,b,c \text{ not all equal}. \end{cases}.$

Thus paradoxical outcomes arise when ${D(f(x_\bullet), g(y_\bullet), h(z_\bullet)) = 3}$. Now, we compute that for randomly selected ${x_\bullet}$, ${y_\bullet}$, ${z_\bullet}$ that

\displaystyle \begin{aligned} \mathbf E D(f(x_\bullet), g(y_\bullet), h(z_\bullet)) &= \mathbf E \sum_S \sum_T \left( \widehat f(S) \widehat g(T) + \widehat g(S) \widehat h(T) + \widehat h(S) \widehat f(T) \right) \left( \chi_S(x_\bullet)\chi_T(y_\bullet) \right) \\ &= \sum_S \sum_T \left( \widehat f(S) \widehat g(T) + \widehat g(S) \widehat h(T) + \widehat h(S) \widehat f(T) \right) \mathbf E\left( \chi_S(x_\bullet)\chi_T(y_\bullet) \right). \end{aligned}

Now we observe that:

• If ${S \neq T}$, then ${\mathbf E \chi_S(x_\bullet) \chi_T(y_\bullet) = 0}$, since if say ${s \in S}$, ${s \notin T}$ then ${x_s}$ affects the parity of the product with 50% either way, and is independent of any other variables in the product.
• On the other hand, suppose ${S = T}$. Then

$\displaystyle \chi_S(x_\bullet) \chi_T(y_\bullet) = \prod_{s \in S} x_sy_s.$

Note that ${x_sy_s}$ is equal to ${1}$ with probability ${\frac13}$ and ${-1}$ with probability ${\frac23}$ (since ${(x_s, y_s, z_s)}$ is uniform from ${3!=6}$ choices, which we can enumerate). From this an inductive calculation on ${|S|}$ gives that

$\displaystyle \prod_{s \in S} x_sy_s = \begin{cases} +1 & \text{ with probability } \frac{1}{2}(1+(-1/3)^{|S|}) \\ -1 & \text{ with probability } \frac{1}{2}(1-(-1/3)^{|S|}). \end{cases}$

Thus

$\displaystyle \mathbf E \left( \prod_{s \in S} x_sy_s \right) = \left( -\frac13 \right)^{|S|}.$

Piecing this altogether, we now have that

$\displaystyle \mathbf E D(f(x_\bullet), g(y_\bullet), h(z_\bullet)) = \left( \widehat f(S) \widehat g(T) + \widehat g(S) \widehat h(T) + \widehat h(S) \widehat f(T) \right) \left( -\frac13 \right)^{|S|}.$

Then, we obtain that

\displaystyle \begin{aligned} &\mathbf E \frac14 \left( 1 + D(f(x_\bullet), g(y_\bullet), h(z_\bullet)) \right) \\ =& \frac14 + \frac14\sum_S \left( \widehat f(S) \widehat g(T) + \widehat g(S) \widehat h(T) + \widehat h(S) \widehat f(T) \right) \widehat f(S)^2 \left( -\frac13 \right)^{|S|}. \end{aligned}

Comparing this with the definition of ${D}$ gives the desired result. $\Box$

Now for the proof of the main theorem. We see that

$\displaystyle 1 = \sum_{S \subseteq \{1, \dots, n\}} -\left( -\frac13 \right)^{\left\lvert S \right\rvert} \left( \widehat f(S) \widehat g(S) + \widehat g(S) \widehat h(S) + \widehat h(S) \widehat f(S) \right).$

But now we can just use weak inequalities. We have ${\widehat f(\varnothing) = \mathbf E f = 0}$ and similarly for ${\widehat g}$ and ${\widehat h}$, so we restrict attention to ${|S| \ge 1}$. We then combine the famous inequality ${|ab+bc+ca| \le a^2+b^2+c^2}$ (which is true across all real numbers) to deduce that

\displaystyle \begin{aligned} 1 &= \sum_{S \subseteq \{1, \dots, n\}} -\left( -\frac13 \right)^{\left\lvert S \right\rvert} \left( \widehat f(S) \widehat g(S) + \widehat g(S) \widehat h(S) + \widehat h(S) \widehat f(S) \right) \\ &\le \sum_{S \subseteq \{1, \dots, n\}} \left( \frac13 \right)^{\left\lvert S \right\rvert} \left( \widehat f(S)^2 + \widehat g(S)^2 + \widehat h(S)^2 \right) \\ &\le \sum_{S \subseteq \{1, \dots, n\}} \left( \frac13 \right)^1 \left( \widehat f(S)^2 + \widehat g(S)^2 + \widehat h(S)^2 \right) \\ &= \frac13 (1+1+1) = 1. \end{aligned}

with the last step by Parseval. So all inequalities must be sharp, and in particular ${\widehat f}$, ${\widehat g}$, ${\widehat h}$ are supported on one-element sets, i.e. they are linear in inputs. As ${f}$, ${g}$, ${h}$ are ${\pm 1}$ valued, each ${f}$, ${g}$, ${h}$ is itself either a dictator or anti-dictator function. Since ${(f,g,h)}$ is always consistent, this implies the final result.

## 5. Pontryagin duality

In fact all the examples we have covered can be subsumed as special cases of Pontryagin duality, where we replace the domain with a general group ${G}$. In what follows, we assume ${G}$ is a locally compact abelian (LCA) group, which just means that:

• ${G}$ is a abelian topological group,
• the topology on ${G}$ is Hausdorff, and
• the topology on ${G}$ is locally compact: every point of ${G}$ has a compact neighborhood.

Notice that our previous examples fall into this category:

Example 13 (Examples of locally compact abelian groups)

• Any finite group ${Z}$ with the discrete topology is LCA.
• The circle group ${\mathbb T}$ is LCA and also in fact compact.
• The real numbers ${\mathbb R}$ are an example of an LCA group which is not compact.

### 5.1. The Pontryagin dual

The key definition is:

Definition 14

Let ${G}$ be an LCA group. Then its Pontryagin dual is the abelian group

$\displaystyle \widehat G \overset{\mathrm{def}}{=} \left\{ \text{continuous group homomorphisms } \xi : G \rightarrow \mathbb T \right\}.$

The maps ${\xi}$ are called characters. By equipping it with the compact-open topology, we make ${\widehat G}$ into an LCA group as well.

Example 15 (Examples of Pontryagin duals)

• ${\widehat{\mathbb Z} \cong \mathbb T}$.
• ${\widehat{\mathbb T} \cong \mathbb Z}$. The characters are given by ${\theta \mapsto n\theta}$ for ${n \in \mathbb Z}$.
• ${\widehat{\mathbb R} \cong \mathbb R}$. This is because a nonzero continuous homomorphism ${\mathbb R \rightarrow S^1}$ is determined by the fiber above ${1 \in S^1}$. (Covering projections, anyone?)
• ${\widehat{\mathbb Z/n\mathbb Z} \cong \mathbb Z/n\mathbb Z}$, characters ${\xi}$ being determined by the image ${\xi(1) \in \mathbb T}$.
• ${\widehat{G \times H} \cong \widehat G \times \widehat H}$.
• If ${Z}$ is a finite abelian group, then previous two examples (and structure theorem for abelian groups) imply that ${\widehat{Z} \cong Z}$, though not canonically. You may now recognize that the bilinear form ${\cdot : Z \times Z \rightarrow Z}$ is exactly a choice of isomorphism ${Z \rightarrow \widehat Z}$.
• For any group ${G}$, the dual of ${\widehat G}$ is canonically isomorphic to ${G}$, id est there is a natural isomorphism

$\displaystyle G \cong \widehat{\widehat G} \qquad \text{by} \qquad x \mapsto \left( \xi \mapsto \xi(x) \right).$

This is the Pontryagin duality theorem. (It is an analogy to the isomorphism ${(V^\vee)^\vee \cong V}$ for vector spaces ${V}$.)

### 5.2. The orthonormal basis in the compact case

Now assume ${G}$ is LCA but also compact, and thus has a unique Haar measure ${\mu}$ such that ${\mu(G) = 1}$; this lets us integrate over ${G}$. Let ${L^2(G)}$ be the space of square-integrable functions to ${\mathbb C}$, i.e.

$\displaystyle L^2(G) = \left\{ f : G \rightarrow \mathbb C \quad\text{such that}\quad \int_G |f|^2 \; d\mu < \infty \right\}.$

Thus we can equip it with the inner form

$\displaystyle \left< f,g \right> = \int_G f\overline{g} \; d\mu.$

In that case, we get all the results we wanted before:

Theorem 16 (Characters of ${\widehat G}$ forms an orthonormal basis)

Assume ${G}$ is LCA and compact. Then ${\widehat G}$ is discrete, and the characters

$\displaystyle (e_\xi)_{\xi \in \widehat G} \qquad\text{by}\qquad e_\xi(x) = e(\xi(x)) = \exp(2\pi i \xi(x))$

form an orthonormal basis of ${L^2(G)}$. Thus for each ${f \in L^2(G)}$ we have

$\displaystyle f = \sum_{\xi \in \widehat G} \widehat f(\xi) e_\xi$

where

$\displaystyle \widehat f(\xi) = \left< f, e_\xi \right> = \int_G f(x) \exp(-2\pi i \xi(x)) \; d\mu.$

The sum ${\sum_{\xi \in \widehat G}}$ makes sense since ${\widehat G}$ is discrete. In particular,

• Letting ${G = Z}$ gives “Fourier transform on finite groups”.
• The special case ${G = \mathbb Z/n\mathbb Z}$ has its own Wikipedia page.
• Letting ${G = \mathbb T}$ gives the “Fourier series” earlier.

### 5.3. The Fourier transform of the non-compact case

If ${G}$ is LCA but not compact, then Theorem~16 becomes false. On the other hand, it is still possible to define a transform, but one needs to be a little more careful. The generic example to keep in mind in what follows is ${G = \mathbb R}$.

In what follows, we fix a Haar measure ${\mu}$ for ${G}$. (This ${\mu}$ is no longer unique up to scaling, since ${\mu(G) = \infty}$.)

One considers this time the space ${L^1(G)}$ of absolutely integrable functions. Then one directly defines the Fourier transform of ${f \in L^1(G)}$ to be

$\displaystyle \widehat f(\xi) = \int_G f \overline{e_\xi} \; d\mu$

imitating the previous definitions in the absence of an inner product. This ${\widehat f}$ may not be ${L^1}$, but it is at least bounded. Then we manage to at least salvage:

Theorem 17 (Fourier inversion on ${L^1(G)}$)

Take an LCA group ${G}$ and fix a Haar measure ${\mu}$ on it. One can select a unique dual measure ${\widehat \mu}$ on ${\widehat G}$ such that if ${f \in L^1(G)}$, ${\widehat f \in L^1(\widehat G)}$, the “Fourier inversion formula”

$\displaystyle f(x) = \int_{\widehat G} \widehat f(\xi) e_\xi(x) \; d\widehat\mu.$

holds almost everywhere. It holds everywhere if ${f}$ is continuous.

Notice the extra nuance of having to select measures, because it is no longer the case that ${G}$ has a single distinguished measure.

Despite the fact that the ${e_\xi}$ no longer form an orthonormal basis, the transformed function ${\widehat f : \widehat G \rightarrow \mathbb C}$ is still often useful. In particular, they have special names for a few special ${G}$:

• If ${G = \mathbb R}$, then ${\widehat G = \mathbb R}$, and this construction gives the poorly named “(continuous) Fourier transform”.
• If ${G = \mathbb Z}$, then ${\widehat G = \mathbb T}$, and this construction gives the poorly named “DTFT..

### 5.4. Summary

In summary,

• Given any LCA group ${G}$, we can transform sufficiently nice functions on ${G}$ into functions on ${\widehat G}$.
• If ${G}$ is compact, then we have the nicest situation possible: ${L^2(G)}$ is an inner product space with ${\left< f,g \right> = \int_G f \overline{g} \; d\mu}$, and ${e_\xi}$ form an orthonormal basis across ${\widehat \xi \in \widehat G}$.
• If ${G}$ is not compact, then we no longer get an orthonormal basis or even an inner product space, but it is still possible to define the transform

$\displaystyle \widehat f : \widehat G \rightarrow \mathbb C$

for ${f \in L^1(G)}$. If ${\widehat f}$ is also in ${L^1(G)}$ we still get a “Fourier inversion formula” expressing ${f}$ in terms of ${\widehat f}$.

We summarize our various flavors of Fourier analysis for various ${G}$ in the following. In the first half ${G}$ is compact, in the second half ${G}$ is not.

$\displaystyle \begin{array}{llll} \hline \text{Name} & \text{Domain }G & \text{Dual }\widehat G & \text{Characters} \\ \hline \textbf{Binary Fourier analysis} & \{\pm1\}^n & S \subseteq \left\{ 1, \dots, n \right\} & \prod_{s \in S} x_s \\ \textbf{Fourier transform on finite groups} & Z & \xi \in \widehat Z \cong Z & e( i \xi \cdot x) \\ \textbf{Discrete Fourier transform} & \mathbb Z/n\mathbb Z & \xi \in \mathbb Z/n\mathbb Z & e(\xi x / n) \\ \textbf{Fourier series} & \mathbb T \cong [-\pi, \pi] & n \in \mathbb Z & \exp(inx) \\ \hline \textbf{Continuous Fourier transform} & \mathbb R & \xi \in \mathbb R & e(\xi x) \\ \textbf{Discrete time Fourier transform} & \mathbb Z & \xi \in \mathbb T \cong [-\pi, \pi] & \exp(i \xi n) \\ \end{array}$

You might notice that the various names are awful. This is part of the reason I got confused as a high school student: every type of Fourier series above has its own Wikipedia article. If it were up to me, we would just use the term “${G}$-Fourier transform”, and that would make everyone’s lives a lot easier.

## 6. Peter-Weyl

In fact, if ${G}$ is a Lie group, even if ${G}$ is not abelian we can still give an orthonormal basis of ${L^2(G)}$ (the square-integrable functions on ${G}$). It turns out in this case the characters are attached to complex irreducible representations of ${G}$ (and in what follows all representations are complex).

The result is given by the Peter-Weyl theorem. First, we need the following result:

Lemma 18 (Compact Lie groups have unitary reps)

Any finite-dimensional (complex) representation ${V}$ of a compact Lie group ${G}$ is unitary, meaning it can be equipped with a ${G}$-invariant inner form. Consequently, ${V}$ is completely reducible: it splits into the direct sum of irreducible representations of ${G}$.

Proof: Suppose ${B : V \times V \rightarrow \mathbb C}$ is any inner product. Equip ${G}$ with a right-invariant Haar measure ${dg}$. Then we can equip it with an “averaged” inner form

$\displaystyle \widetilde B(v,w) = \int_G B(gv, gw) \; dg.$

Then ${\widetilde B}$ is the desired ${G}$-invariant inner form. Now, the fact that ${V}$ is completely reducible follows from the fact that given a subrepresentation of ${V}$, its orthogonal complement is also a subrepresentation. $\Box$

The Peter-Weyl theorem then asserts that the finite-dimensional irreducible unitary representations essentially give an orthonormal basis for ${L^2(G)}$, in the following sense. Let ${V = (V, \rho)}$ be such a representation of ${G}$, and fix an orthonormal basis of ${e_1}$, \dots, ${e_d}$ for ${V}$ (where ${d = \dim V}$). The ${(i,j)}$th matrix coefficient for ${V}$ is then given by

$\displaystyle G \xrightarrow{\rho} \mathop{\mathrm{GL}}(V) \xrightarrow{\pi_{ij}} \mathbb C$

where ${\pi_{ij}}$ is the projection onto the ${(i,j)}$th entry of the matrix. We abbreviate ${\pi_{ij} \circ \rho}$ to ${\rho_{ij}}$. Then the theorem is:

Theorem 19 (Peter-Weyl)

Let ${G}$ be a compact Lie group. Let ${\Sigma}$ denote the (pairwise non-isomorphic) irreducible finite-dimensional unitary representations of ${G}$. Then

$\displaystyle \left\{ \sqrt{\dim V} \rho_{ij} \; \Big\vert \; (V, \rho) \in \Sigma, \text{ and } 1 \le i,j \le \dim V \right\}$

is an orthonormal basis of ${L^2(G)}$.

Strictly, I should say ${\Sigma}$ is a set of representatives of the isomorphism classes of irreducible unitary representations, one for each isomorphism class.

In the special case ${G}$ is abelian, all irreducible representations are one-dimensional. A one-dimensional representation of ${G}$ is a map ${G \hookrightarrow \mathop{\mathrm{GL}}(\mathbb C) \cong \mathbb C^\times}$, but the unitary condition implies it is actually a map ${G \hookrightarrow S^1 \cong \mathbb T}$, i.e. it is an element of ${\widehat G}$.