I suppose I should add a caveat before turning my comment into an answer. In your definition, you have the $\exists j,k$ qualifier, which I believe means that $\mathcal C_n$ is the union of $n \choose 2$ tori, i.e. one for every choice of $j \neq k$. If you allow $j=k$ then I suppose you would have $\mathcal C_n = C_n$, but I doubt you meant that. Finally, if you take the statement $\exists j,k$ out of your set definition, you would have a single torus. With that caveat, I'll expand my comment.
The observation follows from your equation $x_{i_j} = x_{i_k}$, this is basically stating your set is the graph of a function.
A slightly different perspective on a related idea. Any $n \times n$ matrix $A$ with integer coefficients and $\det(A) = \pm 1$ induces an affine-linear automorphism of $\mathcal C_n$ (thought of as a torus).
Provided you are okay with that, notice that the matrix
$$\pmatrix{1 & 1 \\ 0 & 1}$$
sends $\{0\} \times [0,L]$ to $C_2$.
i.e. we have turned the equation $x_1=x_2$ into the functional expression $(x_1, x_2) = (x_2, x_2)$, turning $\mathcal C_2$ into the image of the function $x_2 \longmapsto (x_2,x_2)$, or the graph of $x_1(x_2) = x_2$, i.e. $x_1$ as a function of $x_2$ is equal to $x_2$.
You can do the same with $C_3$, writing it as the union of the graphs of the functions
$$\pmatrix{x_1\\x_2\\0} \longmapsto \pmatrix{x_1 \\x_2\\x_2}$$
$$\pmatrix{x_1\\0\\x_3} \longmapsto \pmatrix{x_1 \\x_1\\x_3}$$
$$\pmatrix{0\\x_2\\x_3} \longmapsto \pmatrix{x_3 \\x_2\\x_3}$$
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