Operators and periodic boundary conditions Background:
In Ref. 1, a system of $N$ (identical) fermions is considered. The system is enclosed in a cubic box of volume $\Omega=L^3$ and periodic boundary conditions are employed, that is (I'll change and ease the notation a bit):
$$ \langle x_1\ldots x_j+L_i\ldots x_N|\Psi\rangle = \langle x_1\ldots x_j\ldots x_N|\Psi\rangle \tag{E.2}$$
for all $j=1,\ldots,N$ and $i=1,2,3$. It is then stated that:

Although these conditions seem natural they are not trivial in the
presence of many-body interactions. This is because the many-body interaction is invariant
under the simultaneous translation of all particles, i.e., $v(x − x^\prime) = v((x + L_i ) − (x^\prime +  L_i ))$
but not under the translation of a single particle. The Hamiltonian does not therefore have a symmetry compatible with the boundary conditions (E.2). To solve this problem we replace
the two-body interaction by $$v(x − x^\prime ) = \frac{1}{\Omega} \sum\limits_k\, \tilde v_k\,e^{ik(x-x^\prime)} \quad , \tag{E.3}$$ [...] With this replacement and the BvK boundary conditions $(\mathrm{E.2})$ the eigenvalue equation for the Hamiltonian $H$ becomes well-defined (of course the spatial
integrations in $(\mathrm{E.1})$ must be restricted to the box).

The relevant equation of the Hamiltonian is
$$H= \int\mathrm dx\,  \psi^\dagger(x) \, \left(-\nabla/2 -V(x)\right)\,\psi(x)\,\tag{E.1} + \frac{1}{2} \int \mathrm dx\,\mathrm dy \,v(x,y)\, \psi^\dagger(x)\,\psi^\dagger(y)\, \psi(y)\,\psi(x) \quad , $$
where, by the usual abuse of notation, $v(x,y)=v(x-y)$.

Question:
I wonder what exactly the authors mean by saying that The Hamiltonian does not therefore have a symmetry compatible with the boundary conditions $(\mathrm{E.2})$ and why the replacement is necessary.
My understanding is that the boundary conditions do not fix the domain of $H$, but instead mean that the single-particle Hilbert space is $\mathfrak h=L^2(\mathbb T^3)$ (neglecting spin) instead of $L^2(\Omega)$ and the corresponding $N$-particle space is $H_N:= \wedge^N \mathfrak h$, i.e. the $N$-fold antisymmetric tensor product. This is for example used (for a non-interacting system) in Ref. 2.
So why should the lack of a symmetry (I guess the authors mean that the Hamiltonian is not invariant under the application of the (discrete) translation operator) of the Hamiltonian be problematic?
Do the authors claim that $H$ in $(\mathrm{E.1})$ without the replacement of $v$ (but with integration restricted to the volume $\Omega$) is not an operator on $H_N$?
I guess/have the feeling that one can also rephrase the question for a non-interacting system (even of a single-particle) with a non-periodic external potential. See also this related MathSE question and answer.

References:
Ref. 1: Nonequilibrium many-body theory of quantum systems. Stefanucci and Leuuwen. Cambridge University Press. Appendix E, page 529.
Ref. 2: Mathematical Quantum Theory. Lecture notes 2019. Marcello Porta. Section 9.4.1, page 103. A PDF can be found here.
 A: Your intuition is right.  The salient detail can be understood by studying a single particle on a ring, with Hilbert space $L^2(\mathrm S^1)$.
The first thing to do is consider how we write down a function $\psi:\mathrm S^1\rightarrow \mathbb C$, for which there are at least two approaches.

*

*We could define two overlapping coordinate charts on $\mathrm S^1$ - say, $\theta : \mathrm S^1 \rightarrow (-\pi,\pi)$ and $\phi: \mathrm S^1\rightarrow (0,2\pi)$ - and then study $\psi_\theta := \psi \circ \theta^{-1}$ and $\psi_\phi := \psi \circ \phi^{-1}$.

*We could define the equivalence relation $\sim$ on the interval $[0,2\pi]$ such that $\forall x, x\sim x$ and additionally $0\sim 2\pi$.  From there, we could model $\mathrm S^1 = [0,2\pi]/\sim$.

Here we take the second approach.  Let $q:[0,2\pi] \rightarrow \big([0,2\pi]/\sim\big) \equiv \mathrm S^1$ be the canonical quotient map corresponding to our (nearly trivial) equivalence relation.

*

*Note that given any function $f:\mathrm S^1\rightarrow \mathbb C$ we may define a function $\tilde f:[0,2\pi] \rightarrow \mathbb C$ via $\tilde f := f\circ q$, and notice that $\tilde f(0)=\tilde f(2\pi)$.

*Similarly, given any $\tilde f:[0,2\pi]\rightarrow \mathbb C$ such
that $\tilde f(0)=\tilde f(2\pi)$, we may define a map $f:\mathrm S^1\rightarrow \mathbb C$ via $f = \tilde f \circ q^{-1}$, which is well-defined precisely because of the periodicity requirement.

So the takeaway is that the set of all functions from $\mathrm S^1\rightarrow \mathbb C$ is in one-to-one correspondence with the set of periodic functions from $[0,2\pi]\rightarrow \mathbb C$.

Having made this (possibly trivially obvious) point, we can proceed to construct $L^2(\mathrm S^1)$ as follows:
$$L_0^2 := \left\{ f :[0,2\pi] \rightarrow \mathbb C \ \bigg| \ f(0)=f(2\pi) \text{ and } \int_0^{2\pi} |f(x)|^2 \mathrm dx <\infty\right\}$$
$$\forall f,g \in L_0^2 : f \sim_{L^2}g \iff \int_0^{2\pi} |f(x)-g(x)|^2 = 0$$
$$L^2(\mathrm S^1) := L_0^2 / \sim_{L^2}$$
The key thing to note is that the periodicity requirement is baked into the definition of the Hilbert space from the start. Every function in $L_0^2$ is periodic, and so any representative of an equivalence class in $L^2(\mathrm S^1)$ is also periodic.  To put it differently, if you have a square-integrable function $f:[0,2\pi]\rightarrow \mathbb C$ such that $f(0)\neq f(2\pi)$, then it does not$^\ddagger$ correspond to an element of $L^2(\mathrm S^1)$.
As a result, given a multiplication operator
$$V:L^2(\mathrm S^1) \rightarrow L^2(\mathrm S^1)$$
$$\big(V\psi\big)(x) = v(x)\psi(x)$$
we must have that $v(0)=v(2\pi)$; otherwise the resulting vector would generically not belong to $L^2(\mathrm S^1)$, and $V$ would not be an operator on that space.
The extension to higher-dimensional spaces and to multiparticle systems is then straightforward.  When we study a particle on a compact space without boundary, we usually talk about functions defined on some compact region $R\in\mathbb R^n$ and then "glue" the edges of $R$ together via some equivalence relation(s). The result of this process - if we carry it through formally - is that periodicity is a fundamental prerequisite for the entire relevant Hilbert space, and that an object which maps a periodic function to a non-periodic function is not an acceptable operator.

$^\ddagger$This is in contrast to $L^2([0,2\pi])$, which does not have that requirement. In that case, we may impose periodicity on the domain of some operator, for example, but the full Hilbert space is not so restricted.
