How do boundary conditions change during a spin transformation? I am currently reading the following review paper:
(1) Two Dimensional Model as a Soluble Problem for Many Fermions by Schultz et. al.
Equation (3.2), which is reproduced below, introduces the Jordan-Wigner transformations. $n$ corresponds to the site number. Suppose there are $N$ sites.
$$
c_n = \prod_{j = 1}^{n - 1} \exp(i\pi \sigma^+_j \sigma^-_j) \sigma^-_n \hspace{0.95in} c^\dagger_n = \sigma^+_n \prod_{j = 1}^{n - 1}\exp(-i\pi \sigma^+_j \sigma^-_j)
$$
Equations (3.12a) and (3.12b) provide some boundary conditions for these new operators $c$ and $c^\dagger$. These are produced below.
$$
 c_{N+1} = c_1 (\text{periodic}) \hspace{0.95in} c_{N+1} = -c_1 (\text{anti-periodic}) 
$$
I am trying to explore what these boundary conditions on $c$ imply about the boundary conditions for spins. To explore this, I used the following
\begin{align}
    \sigma^+_n \sigma^-_n &= \frac{1}{4}\big[ \sigma_n^x + i \sigma_n^y  \big]\big[ \sigma_n^x - i \sigma_n^y  \big] = \frac{1}{2}[1 + \sigma^z_n] \\
    \exp(i\pi\sigma^z_n/2) &= i\sigma^z_n
\end{align}
to conclude
\begin{align}
 c_n &= \prod_{j = 1}^{n - 1} \big[- \sigma_j^z\big]\sigma^-_n\\
    c^\dagger_n &=  \sigma^+_n \prod_{j = 1}^{n - 1}\big[ - \sigma^z_j\big]  
\end{align}
Thus we see that $c_{N+1} = (-1)^N\prod_{j = 1}^{N} \big[\sigma_j^z\big]\sigma^-_{N+1}$. Now it seems that $c_1 = \sigma_1^-$ First Question: Is my expression for $c_1$ correct?
Now suppose $N$ is even. Hence $c_{N+1} = \prod_{j = 1}^{N} \big[\sigma_j^z\big]\sigma^-_{N+1}$ Second Question: Can I say that $\prod_{j = 1}^{N} \big[\sigma_j^z\big] = 1$? If so, why? If not, why not?
In general: What Does Periodic Boundary Conditions for Fermions Say about Boundary Conditions for Spins?
 A: First, indeed $c_1=\sigma_1^-$ is correct.
Now to the general question.
Note that $\sigma_n^z=(-1)^{c_n^\dagger c_n}=1-2c_n^\dagger c_n$. By inverting the transformation, we can write
$$
\sigma_n^-=\prod_{j=1}^{n-1}(-1)^{c_n^\dagger c_n} c_n.
$$
In particular, take $n=N+1$, it becomes $\sigma_{N+1}^-=\prod_{j=1}^N (-1)^{c_j^\dagger c_j} c_{N+1}$.
Now observe that $(-1)^{\sum_{j=1}^N c_j^\dagger c_j}=(-1)^{N_f}$, where $N_f=\sum_{j=1}^N c_j^\dagger c_j$ is a conserved quantity, called the fermion parity, for any fermionic Hamiltonian with local interactions, so we can treat it as a c-number. In the spin representation, we have
$$
(-1)^{N_f}=\prod_{j=1}^N \sigma_j^z
$$
Again this must be a global symmetry of the spin model, otherwise the Jordan-Wigner transformation does not work.
Suppose we impose the following boundary condition on the fermion: $c_{N+1}=s c_1$ where $s=\pm 1$ for periodic/anti-periodic BCs. Then the boundary condition on the $\sigma_n^\pm$ operator becomes
$$
\sigma_{N+1}^\pm = (-1)^{N_f} s \sigma_1^\pm
$$
This relation defines the map between boundary conditions of the fermion and spin operators. If you impose PBC on $c$'s, the spins can still have P/AP BC depending on the total fermion parity/$Z_2$ charge.
A: This is not a complete answer.
I have a vague memory that this is discussed in papers by Eliot Montroll, Potts, and Ward back in the 1960s (papers that I read in the 1970's, and I've slept since then).  It is not straightforward.
In the case of Bosonizing free fermions at finite temperature using the Coleman-Mandlestam appraoch , the spatially periodic boson field corresponds to periodic boundary conditions for the fermions when the fermion number is odd and antiperiodic spatial fermion BC's when the fermion  number is even. Jordan Wigner is a bit different thought, because  the Ising Fermons are Majorana.
