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$$\sigma = \begin{bmatrix} 1 &2 &3 &4 &5 &6 &7 &8 &9 \\ 1&5 &7 &4 &6 &9 &3 &2 &8 \end{bmatrix}$$

$$\sigma^{2} = \begin{bmatrix} 1 &2 &3 &4 &5 &6 &7 &8 &9 \\ 1& 6&3 &4 &9 &8 &7 &5 & 2 \end{bmatrix}$$

Given $$\sigma$$ I found $$\sigma^2.$$

I want to find $$\sigma^{-2}$$ but I'm being told that my permutation cycle for $\sigma^2$ is wrong. Very frustrating.

Please help... Thanks in advance..

JMP
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delta
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  • For the $\sigma$ you gave, your value for $\sigma^2$ looks correct to me. Who says it's wrong? – MJD Jun 11 '15 at 13:19
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    The column for 6 is missing in $\sigma^2$. That's probably just a typo. – Ethan Bolker Jun 11 '15 at 13:21
  • Well, I looked at a given solution for $$\sigma^{-2}$$ and since the inverse is the reverse of the elements in a cycle, the solution for the inverse of the inverse does not correspond to mine. I'm told that the solution for sigma-inverse is (29586) – delta Jun 11 '15 at 13:22
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    Aha, you are confusing the mapping notation for a permutation with the cycle notation. By the way, to get $\sigma^{-2}$ use $\sigma^{-2}$. – MJD Jun 11 '15 at 13:22
  • If my solution to $$\sigma^{-2}$$ is correct, then isn't the inverse just the reverse order of the elements? – delta Jun 11 '15 at 13:24
  • With your notation (and with the added column for $\sigma^2(6)$, the inverse is just the table flipped on its head (and the columns sorted). – Arthur Jun 11 '15 at 13:25
  • @Arthur I don't understand – delta Jun 11 '15 at 13:26
  • @delta I explained Arthur's comment below. – MJD Jun 11 '15 at 13:36
  • Your use of the word "cycle" makes me wonder whether there is a misunderstanding here. A cycle of a permutation is something never mentioned in your proposed value of $\sigma^2$. And also, there is a missing column -- the one headed by $6$ -- in your array representing $\sigma^2$. ${}\qquad{}$ – Michael Hardy Jun 11 '15 at 13:42

1 Answers1

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Permutations are represented in two ways. One is a verbose description of the mapping: $$\sigma = \begin{bmatrix} 1 &2 &3 &4 &5 &6 &7 &8 &9 \\ 1&5 &7 &4 &6 &9 &3 &2 &8 \end{bmatrix}$$ means that $1$ goes to $1$, $2$ goes to $5$, and so on.

We abbreviate this to cycle notation, which is more compact and also more revealing of important structure. This permutation abbreviates to $$(1)(25698)(37)(4)$$ which means the same thing as before, only more compactly. The $(25698)$ means that $2$ goes to $5$, $5$ goes to $6$, $6$ goes to $9$, $9$ goes to $8$, and $8$ goes back to $2$. Then we abbreviate further by dropping the $(1)$ and $(4)$, which can be inferred even if not written explicitly.

In the cycle notation, $\sigma^2$ is indeed written $(26859)$, as you should check. You are being asked to write $\sigma^{-2}$ in the compact cycle notation.

The inverse of a permutation $p$ is another permutation that un-does the effect of $p$. If $p$ takes $3$ to $7$, then $p^{-1}$ should take $7$ to $3$. Let's take a simple example:

$$p = \begin{bmatrix} 1 &2 &3 &4 &5 \\ 4&5 &2 &1 &3 \end{bmatrix}$$

The inverse of $p$, in mapping notation, is

$$p^{-1} = \begin{bmatrix} 4&5 &2 &1 &3 \\ 1 &2 &3 &4 &5 \\ \end{bmatrix}$$

because where $p$ took $2$ to $5$, $p^{-1}$ takes $5$ to $2$. As commented earlier, we have “flipped the notation upside down”. We usually arrange the columns in order:

$$p^{-1} = \begin{bmatrix} 1 &2 &3 &4 &5 \\ 4 & 3& 5 & 1 & 2 \\ \end{bmatrix}$$

In cycle notation, $p$ is written $(14)(253)$. Again, this says that $2$ goes to $5$. As you observed in the comments, you can find $p^{-1}$ by writing the cycle notation for $p$ backwards: $$p^{-1} = (352)(41)$$ which we would usually write as $$p^{-1} = (14)(235).$$ Again, notice that this says that $p^{-1}$ takes $5$ to $2$, as before.

MJD
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  • What I get and observed to be for $$\sigma^{2}$$ is (1)(26)(3)(4)(592)(7)(89) More compactly, we get (26)(592)(89) – delta Jun 11 '15 at 13:31
  • No, because that says that $\sigma^2$ takes $8$ to $9$. But as you correctly said in your original question, $\sigma^2$ actually takes $8$ to $5$. – MJD Jun 11 '15 at 13:36
  • It takes 8 to 5 and takes 5 to 9 and 9 to 2 and 2 to 6. Don't we 'linked' them? – delta Jun 11 '15 at 13:38
  • I used the same reasoning for $$\sigma$$ and for most other questions and it turned out to be correct. – delta Jun 11 '15 at 13:39
  • I don't know what you mean by "linked". Since $\sigma^2$ takes $8$ to $5$, $5$ to $9$, $9$ to $2$, $2$ to $6$, and $6$ to $8$, as you correctly said, you write $(85926)$, or $(26859)$, which means the same thing. – MJD Jun 11 '15 at 13:41
  • I did a careful check. I get (1)(26)(3)(4)(5926)(85) and removing 'redundancies', the form we have is (28596). But the order of the elements isn't an exact fit. I suppose order is of no significance? – delta Jun 11 '15 at 13:42
  • So order does not matter. That's probably the take home lesson... – delta Jun 11 '15 at 13:43
  • Order of the individual cycles matters, unless they are disjoint. (That is, unless they have no numbers in common.) So $(12)(34)$ is the same as $(34)(12)$, but $(12)(13)$ is different from $(13)(12)$. And order of the elements of a cycle matters, but the place in the cycle that you start does not matter. $(1234)$ is the same as $(2341)$ and $(3412)$, but not the same as $(1324)$. – MJD Jun 11 '15 at 13:44