According to E(n) Equivariant Normalizing Flows,
Translation Invariance Recall that we want the distribution $p_V (V)$ to be translation invariant with respect to the overall location and orientation of positional coordinates $x$. For simplicity, let’s assume only a distribution $p_X(x)$ over positions and an invertible function $z = f(x)$. Translation invariance is defined as $p_X(x + t) = p_X(x)$ for all $t$: a constant function. However, this cannot be a distribution since it cannot integrate to one. Instead, we have to restrict $p_X$ to a subspace.
I take Normal distribution $N(0,1)$ as an example, and I know $N(0,1) \neq N(2,1)$ so it does not satisfy translation invariance. However, I could not understand why this paper also claims that However, this cannot be a distribution since it cannot integrate to one. Could someone give me an example?
Edited: Thanks to @geetha290krm, the above question has been solved in comments.
Further question: in the same paper
To construct a translation invariant $p_X$, we can restrict the data, flow $f_θ$ and prior $p_Z$ to a translation invariant linear subspace, for instance by centering the nodes so that their center of gravity is zero. Then the positions $x ∈ R^{M×n}$ lie on the $(M − 1) × n$-dimensional linear subspace defined by $\sum_{i=1}^{M} x_i = 0$. However, from a modelling perspective it is easier to represent node positions as M sets of coordinates that are n-dimensional in the ambient space. In short, we desire the distribution to be defined on the subspace, but with the representation of the nodes in the ambient space.
Does it mean that we could get a translation invariant distribution by centering the nodes so that their center of gravity is zero? Why above method could get a translation invariant distribution? Could someone give an example?