description/proof of 2 possible meanings of permutation of sequence
Topics
About: set
The table of contents of this article
Starting Context
- The reader knows a definition of permutation of sequence.
Target Context
- The reader will have a description and a proof of the 2 possible meanings of permutation of sequence.
Orientation
There is a list of definitions discussed so far in this site.
There is a list of propositions discussed so far in this site.
Main Body
1: Structured Description
Here is the rules of Structured Description.
Entities:
\(\mathbb{N}\):
\(S\): \(\subseteq \mathbb{N}\)
\(f\): \(\in \{\text{ the sequences from } S\}\)
\(\sigma\): \(: S \to S\), \(\in \{\text{ the bijections }\}\)
\(\sigma (f)\): \(= f \circ \sigma\)
\(g\): \(= f \circ \sigma^{-1}\)
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Statements:
\(\sigma (f)\) is the permutation of \(f\) by \(\sigma\) by a definition of permutation of sequence, but some people may understand \(g\) by "permutation of \(f\) by \(\sigma\)"
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2: Proof
Whole Strategy: Step 1: see what \(\sigma (f)\) means; Step 2: see what \(g\) means.
Step 1:
Let us see what \(\sigma (f) := f \circ \sigma\) means.
For each \(s \in S\), \(\sigma (f) (s) = f (\sigma (s))\), which means that \(\sigma (f)_j = f (\sigma (s_j)) = f (\sigma_j)\) where \(s_j \in S\) is the \(j\)-th element of \(S\), which means that the sequence is \(f (\sigma_1), f (\sigma_2), ...\).
Note that when we regard the permutation as moving an item to a position, the permutation is moving the \(\sigma_j\) item to the \(j\)-th position.
For example, when \(S = \{1, 2, 3\}\), \(f = 4, 5, 6\), and \(\sigma: (1, 2, 3) \mapsto (3, 1, 2)\), \(\sigma (f) = f (\sigma (1)), f (\sigma (2)), f (\sigma (3)) = 6, 4, 5\), which is moving the \(\sigma (1) = 3\) item to the \(1\)-st position, the \(\sigma (2) = 1\) item to the \(2\)-nd position, and the \(\sigma (3) = 2\) item to the \(3\)-rd position.
Step 2:
The reason why we have described this proposition is that Step 1 may not be what some people understand as "permutation of \(f\) by \(\sigma\)".
They may understand it as moving the \(j\)-th item to the \(\sigma_j\) position.
Let us see that that is \(g\) not \(\sigma (f)\).
Let us denote the permutation that moves the \(j\)-th item to the \(\sigma_j\) position as \(h\).
\(f (s_j) = h (\sigma_j) = h (\sigma (s_j))\).
Taking \(t := \sigma (s_j)\), \(s_j = \sigma^{-1} (t)\), so, \(h (t) = f (s_j) = f (\sigma^{-1} (t)) = f \circ \sigma^{-1} (t)\).
So, \(h = f \circ \sigma^{-1} = g\) not \(= f \circ \sigma := \sigma (f)\).
For example, when \(S = \{1, 2, 3\}\), \(f = 4, 5, 6\), and \(\sigma: (1, 2, 3) \mapsto (3, 1, 2)\), which is the same with the above example, \(\sigma^{-1}: (1, 2, 3) \mapsto (2, 3, 1)\) and \(g = f (\sigma^{-1} (1)), f (\sigma^{-1} (2)), f (\sigma^{-1} (3)) = 5, 6, 4\), which is moving the \(1\)-st item to the \(3\)-rd position, the \(2\)-nd item to the \(1\)-st position, and the \(3\)-rd item to the \(2\)-nd position.
3: Note
This is not about which of the 2 is correct, but about the necessity of clarifying what is meant by 'permutation of \(f\) by \(\sigma\)' and being consistent.
