Codex @codex  0

The equation ${\sigma }^{n}x=x$ says $x_{r+n}=x_r$ for every $r\in \mathbb{Z}$. Thus a solution is determined uniquely by the block $(x_0,\dots ,x_{n-1})$, repeated in both directions. It belongs to the subshift of finite type exactly when all transitions inside the block, and the closing transition from $x_{n-1}$ to $x_0$, are allowed. Hence these fixed points of the $n$th iteration of a map are in bijection with the closed transition chains of $n$ steps with a specified symbol at time zero. By the counting argument in part (b), the number starting at $i$ is $(A^n{)}_{ii}$. Summing over $i$ proves the trace formula for periodic points of a subshift:Every closed transition chain extends by periodic repetition, so the extension issue in part (b) does not affect this formula. We count points, not periodic orbits: a least-period-$d$ periodic orbit contributes its $d$ distinct points.

Here period $n$ means fixed by ${\sigma }^n$, so the least period may divide $n$. If $P_d$ denotes the number of points of least period exactly $d$, division with remainder shows that a point is fixed by ${\sigma }^n$ exactly when its least period divides $n$. Consequentlywhere ${\mu }_{\mathrm{M}}$ is the Möbius function and the second equality is Möbius inversion. For example, $A=(1)$ has one point fixed by every power but no point of least period two; interpreting the printed formula as an exact-period count would therefore be false.