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I am working through the book Geometric Deep Learning (https://arxiv.org/abs/2104.13478) and have hit the following formula (Chapter 3.5, page 27).

If $f$ is linear and $\mathfrak G$-invariant and $\mu(\mathfrak g)$ is Haar measure of the group $\mathfrak G$, then for all $x\in \mathcal{X}(\Omega)$, $$f(x) = \frac{1}{\mu(\mathfrak G)}\int_{\mathfrak G} f(\mathfrak g.x)d\mu(\mathfrak g) = f \left(\frac{1}{\mu(\mathfrak G)}\int_{\mathfrak G}(\mathfrak g.x)d\mu(\mathfrak g)\right)$$ which indicates that $F$ only depends on $x$ through the $\mathfrak G$-average $Ax= \frac{1}{\mu(\mathfrak G)}\int_{\mathfrak G} f(\mathfrak g.x)d\mu(\mathfrak g)$.

I have two quick questions.

  1. It doesn't seem obvious to me that the function $f$ being linear and $\mathfrak G$-invariant allows the integral $\int_{\mathfrak G}\cdot \, d\mu(\mathfrak g)$ to just go inside(?) the function $f$. The authors gave no explation as to how it is done. How is the formula derived?

  2. I don't know how this operation of integral $\int_{\mathfrak G}\cdot \, d\mu(\mathfrak g)$ 'going inside' the function $f$ is called. Would 'commute' be appropriate term? (For example, the function $f$ and integral $\int_{\mathfrak G}\cdot \, d\mu(\mathfrak g)$ commute with each other.)

Alp Uzman
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sepiabrown
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1 Answers1

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  1. This essentially follows from a standard approximation argument. The whole setup seems to be somewhat vague, and there are many subtleties involved in the general case the document seems to be indicating, but the main idea is this: say you have a probability space $(G,\mu)$, a Hilbert space $\mathcal{H}$ and a function $F:\mathcal{H}\to\mathbb{R}$. Let us put $\mathcal{H}=\mathbb{R}$. The claim is that if $F$ is linear, then for any $\phi\in L^2(G,\mu;\mathcal{H})$:

$$\int_G F(\phi(g))\, d\mu(g) = F\left(\int_G \phi(g)\,d\mu(g)\right).$$

The claim is true for $\phi=\mathbf{1}_A$ the indicator function of a measurable subset $A\subseteq G$ (by definition $\mathbf{1}_A(g)=1$ if $g\in A$ and $0$ otherwise); indeed we have

\begin{align*} \int_G F(\mathbf{1}_A(g))\, d\mu(g) =\int_A F(1) \,d\mu(g) = F(1) \mu(A) \stackrel{(\ast)}{=} F(\mu(A)) =F\left(\int_G \mathbf{1}_A(g)\,d\mu(g)\right), \end{align*}

where we used the linearity of $F$ for the equality marked with $(\ast)$. Next, again by the linearity of $F$ the claim is true for simple measurable functions (functions that are finite linear combinations of indicator functions, i.e. functions of the form $\sum_{i=1}^nc_i\mathbf{1}_{A_i}$). Finally one can approximate any $\phi\in L^2(G,\mu;\mathcal{H})$ by an increasing sequence of simple measurable functions, and the general result follows from the continuity of $F$.

In our case the Hilbert space $\mathcal{H}=\mathcal{X}(\Omega;\mathcal{C})$ is the space of square-integrable signals (according to p.11; where $\mu$ is not the same as the Haar measure on $G$), and $\phi:G\to \mathcal{H},$ $g\mapsto g. x$ for some anonymous $x\in\mathcal{X}(\Omega;\mathcal{C})$. Consequently there are some subtleties regarding the simple approximation and the continuity of $F$. The $G$-invariance of $F$ comes into play in the equality

$$F(x)=\int_G F(g.x) \,d\mu(g).$$

(Another subtlety is that it seems to be assumed that the group $G$ admits a finite Haar measure, or else there are again further subtleties involved with interpreting $\dfrac{1}{\mu(G)}$ properly.)

  1. Indeed, one can consider this as a commutation relation between $F$ and the integration along $G$-orbits, or as the book seems to prefer, averaging along $G$-orbits. If $\alpha_\bullet:G\curvearrowright\Omega$ is a $G$-action, then we have a $G$-action $K^\alpha_\bullet:G\curvearrowright\mathcal{X}(\Omega;\mathcal{C})$, $K^\alpha_g(x)=x\circ \alpha_{g}^{-1}$ on the signal space ("the Koopman action"; see e.g. Understanding the Definition of the Koopman Operator), and one has the averaging operator

$$A^\alpha:\mathcal{X}(\Omega;\mathcal{C})\to \mathcal{X}(\Omega;\mathcal{C}), x\mapsto \dfrac{1}{\mu(G)}\int_G K^\alpha_g(x)\,d\mu(g).$$

(Note that for any $x$, $A^\alpha(x)$ is $K^\alpha$-invariant, and if $x$ were $K^\alpha$-invariant to begin with then $A^\alpha(x)=x$; so that $A^\alpha$ is a projection operator onto the subspace of $K^\alpha$-invariant signals.)

(Along these lines, there seems to be a typo in the last paragraph on p.27: $A\circ U$ ought to take values in $\mathcal{X}(\Omega,\mathcal{C}'')$, instead of $\mathcal{C}''$.)

With this notation, the above claim becomes, for $F$ (continuous and) linear and $K^\alpha$-invariant:

$$A^\alpha( F(\phi) ) = F( A^\alpha(\phi) ),$$

that is,

$$A^\alpha\circ F=F\circ A^\alpha.$$

Alp Uzman
  • 10,742