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Q&A

Do linear and group invariant functions allowed to go inside(?) integral operators?

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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 $Ax= \frac{1}{\mu(\mathfrak G)}\int_{\mathfrak G} f(\mathfrak g.x)d\mu(\mathfrak g)$.

Two 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.)

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If $f$ is a linear map (between finite-dimensional spaces), then it can be represented by multiplying by some matrix.

The fact that $f$ is $\mathfrak G$-invariant presumably means that this matrix is constant with respect to the $\mathfrak g$ used as the variable of integration.

It's easy to see that multiplying by a constant matrix is an operation you can pull out of an integral, because sums can be pulled out of integrals

$$ \int \left(f(x) + g(x)\right)\,dx = \left(\int f(x)\,dx\right) + \left(\int g(x)\,dx\right) $$

and constant factors can be pulled out of integrals

$$ \int c\,f(x)\,dx = c\int f(x)\,dx $$

and matrix multiplication is just an application of those operations.

And yes, you can say that $f$ and the integral commute with each other as operators.

(If, contra my first sentence, you aren't working with finite-dimensional spaces, then you have to tread much more carefully than I'm doing here, starting with specifying precisely what definition of integration you're using. The usual suspects still commute with linear functionals, though.)

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