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I encountered in my research on dynamical systems a problem, which considers for some $L>0$ on the $C_n=[0,L]^n$ the set $\mathcal{C}_n=\{(x_1,\ldots,x_n)\mid\exists j,k:\,x_{i_j}=x_{i_k}\}$. I am looking for an interpretation of $\mathcal{C}_n$ on the $n$-torus, when imposing periodic boundary conditions on $C_n$. For $n=2$ this is a simple closed curve which winds once around the torus. For $n=3$, I am already struggeling what $\mathcal{C}_3$ "looks like". In particular, I am wondering how curves of the form $\gamma:t\mapsto x+t\, e_j$, with $(e_j)_i=\mathrm{1}_{i=j}$, intersect with $\mathcal{C}_n$.

In general, literature recommendations on the $n$ torus are highly appreciated!

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    $\begingroup$ The $n=2$ case is very much like the generic case. i.e. $\mathcal C_n$ is always a torus of one-lower dimension. In fact, there is an affine-linear transformation of $C_n$ that turns $\mathcal C_n$ into $C_{n-1} \times \{0\}$. $\endgroup$ Commented Jan 8, 2023 at 9:56
  • $\begingroup$ Awesome! Could you give me a source for this result? I do not see a straight forward way to prove it but I am also not especially experienced in differential geometry... $\endgroup$ Commented Jan 8, 2023 at 12:50

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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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