Why is the complete state of a turing machine representable by a finite word? At the moment I'm just reading through this article on the word problem for groups (https://projecteuclid.org/euclid.bams/1183548590), and I'm wondering about a certain snippet.
On page 40 the article describes a word that can encode the state of a turing machine, namely $S_{k_{u}}...S_{k_{1}}q_{i}S_{j_{1}}...S_{j_{v}}$ where $S_{k_{u}}...S_{k_{1}}S_{j_{1}}...S_{j_{v}}$ is the tape expression, $q_{i}$ is the current internal state of the turing machine, and $S_{j_{1}}$ is the currently scanned symbol.
I'm just wondering why this word is necessarily finite. Why can't the input of the turing machine have an infinite amount of non-zero characters?
I'm just wondering why this word is necessarily finite. Why can't the input of the turing machine have an infinite amount of non-zero characters?


No, it can not.

When the Turing machine starts, the tape contains the input (a finite-length word) followed by an infinite amount of blank symbols (often written as #). It is common to simply omit the trailing infinitely many blanks, and just represent a finite prefix instead, adding blanks as needed when the head moves right past the end of the tape.

When the machine runs, in each step it can at most (over-)write over one tape symbol, and move right by at most one position. This is an essential feature of any Turing machine.

Hence, after finitely many steps, we still have a finite amount of non-blank symbols. The tape again contains infinitely many trailing blanks (usually not represented), so it can be represented by a suitable finite prefix.