Enforcing invariance under isometries on a manifold that approximates data

Question

I have some data, say a finite set $\mathcal{S}$ of grayscale images of given dimension $m\times n$, seen as vectors in $\mathbb{R}^{m \times n}$, with each coordinate expressing the intensity of one pixel. I have a machine learning model that selects a connected sub-manifold $\mathcal{M} \subset \mathbb{R}^{m \times n}$ that contains all elements of $\mathcal{S}$. I can thus measure the geodesic distance on $\mathcal{M}$ between my images, which is a more meaningful distance than the Euclidean distance, assuming that $\mathcal{M}$ has been chosen in a suitable way. However, I would also like to enforce invariance of my data under a finite group $G$ of isometries of $\mathbb{R}^{m \times n}$. For instance, one image and its mirror image should be considered the same. My dataset $\mathcal{S}$ is preserved under these transformations, but $\mathcal{M}$ is not.

If it were, I could take the quotient $\mathcal{M}/G$ and it would at least be a connected metric space (or even a Riemannian manifold, if $G$ acts freely); I could thus measure distances between classes (elements of $\mathcal{M}/G$) as $$ d_{\mathcal{M}/G}(A,B) := \mathrm{min}_{a \in A, b \in B} \ d_\mathcal{M}(a,b),$$ as explained here (note that $G$ acts by isometries also on $\mathcal{M}$, as long as $\mathcal{M}$ is preserved by $G$).

But, as I mentioned, the problem is that $\mathcal{M}$ is not preserved by $G$. Is there any smart way to enlarge $\mathcal{M}$ to a "minimal" (or at least not too high-dimensional) manifold $\mathcal{N}$ that is preserved by $G$? I cannot just take the union of the translates of $\mathcal{M}$ under $G$, as these might intersect quite badly; but maybe there is some kind of "enveloppe" of these translates that I could take?

Or, more generally: is there any meaningful way to measure distances between the points of $\mathcal{S}$ while taking into account the structure of $\mathcal{M}$ and while also enforcing $G$-invariance?

Answer 1

I am not sure this "answer" is helpful or not, but I will share, hopefully you may find some value in it.

If the manifold $\mathcal{M}$ is given by an immersion/embedding $f : \mathbb{R}^k \to \mathbb{R}^{m \times n}$ and the finite isometry group $G$ of $\mathbb{R}^{m \times n}$ is compatible with the vector space structure of $\mathbb{R}^{m \times n}$, maybe you might be able to average the immersion, i.e. for any $x \in \mathbb{R}^k$ define the new map:

$$f_G(x) = \frac{1}{|G|}\sum_{g \,\in \,G}\, g\cdot f(x)$$

Then for any $h \in G$, $$h\cdot f_G(x) = h \cdot \left( \frac{1}{|G|}\sum_{g \,\in \,G}\, g\cdot f(x) \,\right) = \frac{1}{|G|}\sum_{g \,\in \,G}\, hg\cdot f(x) = \frac{1}{|G|}\sum_{hg \,\in \,G}\, hg\cdot f(x) = f_G(x)$$

One may need to check whether the new invariant map $f_G(x)$ is still an immersion/embedding.

If the manifold $\mathcal{M}$ is given as a submersion, i.e. as a solution to a system of equation $f(y) = 0$, then define $$f_G(y) = \frac{1}{|G|} \,\sum_{g \, \in \, G}\, f(g\cdot y)$$ and a new invariant manifold as $$f_G(y) = 0$$

Then for any $h \in G$, $$f_G(h\cdot y) = \frac{1}{|G|}\sum_{g \,\in \,G}\, f(gh\cdot y) = \frac{1}{|G|}\sum_{gh \,\in \,G}\, f(gh\cdot y) = f_G(y)$$

Again, one may need to check whether the new invariant map $f_G(y)$ is still an submersion, i.e. has full rank for all solutions.