Averaging Functions Over Compact Groups

Why Average Over a Group?

A function may vary substantially along the points that a group action moves into one another. Averaging its values over those moves removes that variation while retaining the information shared across the orbit. On a compact group, normalized Haar measure supplies a canonical way to take this average: the result is continuous and unchanged by the action.

Let $G$ be a compact Hausdorff group, let $X$ be a compact Hausdorff space, and let $(g,x)\mapsto g\cdot x$ be a continuous action of $G$ on $X$. Thus $e\cdot x=x$ and $g\cdot(h\cdot x)=(gh)\cdot x$. Write $m_G$ for normalized Haar measure on $G$: it is a Borel probability measure, so $m_G(G)=1$, invariant under both left and right translations. The existence of normalized Haar measure on a compact group, and its right invariance, are standard results from Haar measure theory.

Definition: For a real- or complex-valued continuous function $f\in C(X)$, its group average is the function $Pf$ defined by $$ (Pf)(x)=\int_G f(g^{-1}\cdot x)\,dm_G(g),\qquad x\in X. $$ A function $u\in C(X)$ is invariant if $u(g\cdot x)=u(x)$ for all $g\in G$ and $x\in X$. The integral defining $Pf$ is finite: its integrand is continuous and bounded, and $m_G$ is a finite measure.

The inverse in the formula makes the change of variables for the action particularly direct. Since $G$ is compact, its normalized Haar measure is invariant on both sides, so the average does not depend on a preferred point in the group. The operator $P$ will turn out to preserve precisely the part of $f$ that is already invariant.

The Average Is Continuous and Invariant

Theorem: Under the assumptions above, $P$ maps $C(X)$ into $C(X)$, and $Pf$ is invariant for every $f\in C(X)$. Here invariance holds at every point, not just almost everywhere.

Proof. Fix $x_0\in X$ and $\varepsilon>0$. Define $F(g,x)=f(g^{-1}\cdot x)$. The action, group inversion, and $f$ are continuous, so $F$ is continuous on $G\times X$. For each $g_0\in G$, continuity at $(g_0,x_0)$ gives neighborhoods $U_{g_0}$ of $g_0$ and $V_{g_0}$ of $x_0$ such that, whenever $g\in U_{g_0}$ and $x\in V_{g_0}$, both $|F(g,x)-F(g_0,x_0)|<\varepsilon/2$ and $|F(g,x_0)-F(g_0,x_0)|<\varepsilon/2$. The neighborhoods $U_{g_0}$ cover $G$. Compactness gives a finite subcover $U_{g_1},\ldots,U_{g_r}$. Set $V=\bigcap_{j=1}^rV_{g_j}$, a neighborhood of $x_0$. For every $x\in V$ and every $g\in G$, choose a $j$ with $g\in U_{g_j}$. The triangle inequality gives $|F(g,x)-F(g,x_0)|<\varepsilon$. Since $m_G(G)=1$, $$ |(Pf)(x)-(Pf)(x_0)| \le \int_G |F(g,x)-F(g,x_0)|\,dm_G(g) \le \varepsilon. $$ Thus $Pf$ is continuous at $x_0$; as $x_0$ was arbitrary, $Pf\in C(X)$.

To prove invariance, fix $h\in G$ and $x\in X$. In the integral for $(Pf)(h\cdot x)$, make the substitution $g=hk$. Left invariance of $m_G$ permits this change of variables, and $(hk)^{-1}h=k^{-1}$. Therefore $$ (Pf)(h\cdot x)=\int_G f(g^{-1}\cdot(h\cdot x))\,dm_G(g) =\int_G f(k^{-1}\cdot x)\,dm_G(k) =(Pf)(x). $$ This proves invariance for every $h$ and $x$. $\square$

Averaging Is an Idempotent Projection

Theorem: The operator $P:C(X)\to C(X)$ is linear, satisfies $\|Pf\|_\infty\le\|f\|_\infty$, and obeys $P^2=P$. Its range is exactly the space $C(X)^G$ of invariant continuous functions.

Proof. Linearity follows from linearity of the integral. Since $m_G$ is a probability measure, for each $x\in X$, $$ |(Pf)(x)|\le\int_G|f(g^{-1}\cdot x)|\,dm_G(g)\le\|f\|_\infty. $$ Taking the supremum over $x$ proves the norm bound. The previous theorem shows that $Pf$ is invariant. If $u$ is invariant, then $u(g^{-1}\cdot x)=u(x)$ for all $g,x$, and hence $$ (Pu)(x)=\int_G u(x)\,dm_G(g)=u(x)m_G(G)=u(x). $$ Thus $P$ fixes every invariant function. Applying this to $u=Pf$, which is invariant, gives $P(Pf)=Pf$. The range of $P$ is contained in $C(X)^G$ because every average is invariant; conversely, every $u\in C(X)^G$ is $Pu$, so it belongs to the range. These two inclusions prove the range statement. $\square$

An idempotent linear operator that fixes its range is a projection. This projection need not be an orthogonal projection: no inner product is assumed. Its defining feature here is that it maps each function to an invariant one and leaves invariant functions unchanged.

Worked Example: Averaging a Function Over Circle Rotations

Let $G=X=S^1$, the unit circle, acting on itself by multiplication. Write a point of $S^1$ as $e^{it}$, and use normalized Haar measure $dm_G(e^{i\theta})=d\theta/(2\pi)$ for $0\le\theta<2\pi$. For $f(e^{it})=3+2\cos t-\sin(2t)$, compute $Pf$.

The action sends $e^{it}$ to $e^{i(t+\theta)}$, so $$ (Pf)(e^{it})=\frac1{2\pi}\int_0^{2\pi}\bigl(3+2\cos(t-\theta)-\sin(2t-2\theta)\bigr)\,d\theta. $$ As $\theta$ ranges from $0$ to $2\pi$, each trigonometric term completes an integer number of periods. In particular, $$ \int_0^{2\pi}\cos(t-\theta)\,d\theta =[-\sin(t-\theta)]_0^{2\pi}=0, $$ and $$ \int_0^{2\pi}\sin(2t-2\theta)\,d\theta =\left[\frac12\cos(2t-2\theta)\right]_0^{2\pi}=0. $$ The constant term integrates to $3(2\pi)$. Therefore $(Pf)(e^{it})=3$ for every $t$. Check: the result is constant, hence invariant under every rotation. Averaging it again gives $\frac1{2\pi}\int_0^{2\pi}3\,d\theta=3$, as required by $P^2=P$.

Finite Group Actions: Exact Averages

For a finite group $G$ with $N$ elements, normalized Haar measure assigns mass $1/N$ to each element. Thus the integral is an arithmetic mean: $$ (Pf)(x)=\frac1N\sum_{g\in G}f(g^{-1}\cdot x). $$ For finite groups, the average can be computed by listing the action on the point and summing the corresponding values.

Worked Example: Averaging Under Four Planar Rotations

Let the cyclic group of four rotations act on $\mathbb R^2$, with its usual topology, by rotations through $0,\pi/2,\pi,3\pi/2$. Give the group uniform probability measure and let $f(x,y)=x^2+2xy+3y^2$. Find $Pf$.

The orbit points are $(x,y)$, $(-y,x)$, $(-x,-y)$, and $(y,-x)$. Substitution into $f$ gives, respectively, $$ x^2+2xy+3y^2,\quad 3x^2-2xy+y^2,\quad x^2+2xy+3y^2,\quad 3x^2-2xy+y^2. $$ Their sum is $8x^2+8y^2$: the $xy$ terms cancel, and the coefficients of each square add to $8$. Dividing by the four group elements yields $$ (Pf)(x,y)=2x^2+2y^2. $$ Check: at $(1,0)$, the four function values are $1,3,1,3$, whose mean is $2$; the formula gives $2(1)^2+2(0)^2=2$. The result is unchanged by each of the four rotations because $x^2+y^2$ is unchanged.

Worked Example: Averaging Over Coordinate Permutations

Let the six permutations in the group $S_3$ act on $X=\mathbb R^3$ by permuting coordinates. For $f(x_1,x_2,x_3)=x_1^2+2x_2^2+3x_3^2$, compute the group average.

In the six terms of the average, each coordinate appears twice with coefficient $1$, twice with coefficient $2$, and twice with coefficient $3$. Consequently, the total coefficient of each $x_j^2$ before dividing by $6$ is $2(1)+2(2)+2(3)=12$. Dividing by $6$ gives coefficient $2$ for each square, so $$ (Pf)(x_1,x_2,x_3)=2x_1^2+2x_2^2+2x_3^2. $$ Check: at $(1,0,0)$, the six values are $1,1,2,2,3,3$, with mean $12/6=2$; the formula also gives $2$. The output is invariant because permuting the coordinates leaves the sum of their squares unchanged.

Normalization Matters

The projection conclusion uses $m_G(G)=1$, not merely translation invariance. If instead one uses a finite Haar measure $\lambda=c\,m_G$ with total mass $c>0$, the corresponding averaging operator $Q$ satisfies $Qu=c\,u$ for every invariant $u$. Since $Qf$ is invariant, it follows that $Q^2f=c\,Qf$, which is not the projection identity unless $c=1$ (or $Qf=0$).

Common mistake. Take $\lambda=2m_G$ and the constant function $f=1$. A mistaken calculation treats $\lambda$ as a probability measure and says $Q1=1$, hence $Q^21=1$. In fact, $$ Q1=\int_G1\,d\lambda=2,\qquad Q^21=Q(2)=\int_G2\,d\lambda=4. $$ Thus $Q^21=4\ne2=Q1$. The correction is to normalize Haar measure first: with $m_G(G)=1$, the average of $1$ is $1$, and the resulting operator is idempotent.

Group averaging is useful whenever the desired output must respect a symmetry. The theorem proves that the averaging operator enforces invariance while retaining any invariant component already present. In concrete problems, the integral over a compact group may reduce to a familiar trigonometric integral or, for a finite group, an exact finite sum.

Check Your Understanding

Use the definition of the group average and the projection theorem to answer the following.

Why does compactness of $G$ enter the proof that $Pf$ is continuous?
For a finite group with $N$ elements, write the averaging formula and explain why the factor $1/N$ is needed for idempotence.
Under the four rotations in the planar example, what is the average of $x^2-y^2$?
If $u$ is invariant, show directly from the integral formula that $Pu=u$.
For a Haar measure of total mass $c$, what identity does the corresponding averaging operator $Q$ satisfy in place of $Q^2=Q$?