# How to read permutation symbols like $(123)$?

I'd be grateful for some help reading permutation symbols such as $(123)$. Does it mean, when applied to a target sequence such as $(x y z w)$, "replace the element in the first slot of the target with the element in the second slotof the target, the element in the second slot of the target with the element in the third slot of the target, and the element in the third slot of the target with the element in the first slot of the target," resulting in $(y z x w)$? If so, applied twice, $(123)$ would produce $(z x y w)$, which would mean that $(123).(123)=(321)$?

If I'm getting it right, then I should imagine little leftward arrows inside the permutation symbol, as in $(\leftarrow 1 \leftarrow 2 \leftarrow 3)$, with the first arrow implicitly wrapping around to the last position of the permutation symbol.

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If $(123)$ is in "cycle notation", then this means that $1$ maps to $2$, $2$ maps to $3$, and $3$ (the last term in the cycle) maps to $1$. That is, those "little arrows" should point the other way.

But with composition, you have to be careful! It depends on whether you are composing right-to-left (like functions), or left-to-right. For instance, if you write $(12)(13)$, then composing left-to-right (apply $(12)$ first, then $(13)$) you get $(123)$, but if you compose right-to-left, you get $(132)$.

Which composition convention is being used depends on the author. But cycles are never, in my experience, read "right to left" themselves; that is, $(123)$ never represents $3\mapsto 2\mapsto 1\mapsto 3$.

If $(123)$ is in "one line notation", then this would mean that the permutation is applied to a 3 element set, with $1$ mapping to $1$, $2$ mapping to $2$, and $3$ mapping to $3$ (i.e., $(abc)$ means $1\mapsto a$, $2\mapsto b$, $3\mapsto c$). However, one-line notation is not common.

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thanks much. Very clear. I induced the "right-to-left" convention from reading the top of page 5 of Gilmore's "Lie Groups, Physics, and Geometry," where he writes "The symbol $(123)$ means that the first root, $z_1$, is replaced by $z_2$, $z_2$ is replaced by $z_3$, and $z_3$ is replaced by $z_1$." This made me very queasy, and it's why I asked the question here. I guess the larger point is that if there is more than one way to define or interpret some notation, someone will use the other way! – Reb.Cabin Aug 6 '11 at 18:23
ok, so now I understand Qiaochu's comment better. I think what's going on is that Gilmore intends the usual left-to-right cycle notation, just applied to the indices of the roots. Then $z1$ would be replaced by $z2$ because $1\rightarrow 2$, etc. – Reb.Cabin Aug 6 '11 at 18:48
@Reb.Cain: Yes. If your cycle is $(z_1z_2z_3\ldots z_n)$, then $z_1$ "goes to" (is replaced by) $z_2$; then $z_2$ is mapped to ("is replaced by") $z_3$, etc. – Arturo Magidin Aug 6 '11 at 21:21

The convention I'm familiar with is that $(123)$ means $1 \to 2 \to 3 \to 1$. Composition is just the usual composition of functions, and isn't described well by this example, so here's another example: $(12) \cdot (123) = (32)$.

Applying permutations to sequences is tricky. Are you permuting the entries of the sequence, or the indices of the entries? One is a left action, and the other is a right action, of the symmetric group, and in the latter case there are inverses you need to add in to get a left action.

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yes, I noticed the ambiguity of applying the permutation to the entries versus applying the permutation to the indices. This made me even more queasy than the left-right ambiguity, though I didn't mention it in my first question. It can be painful to have to decode the intention of an author who just slings out symbols without clearly resolving these ambiguities. Thanks for your clarity, here. – Reb.Cabin Aug 6 '11 at 18:27
Perhaps I can understand this best via matrices. If $(123)$ means "permute the entries," then a matrix rep could be $M(123)=[[0 0 1],[1 0 0],[0 1 0]]$. Multiplying on the left of the sequence-as-column-vector, $M(123)(z_1,z_2,z_3)^T=(z_3,z_1,z_2)^T$. The inverse or transpose, namely $[[0 1 0],[0 0 1],[1 0 0]]$, produces the permutation of indices: $M(123)^T(z_1,z_2,z_3)^T=(z_2,z_3,z_1)^T$. Perhaps, in general, the "permute-the-entries" matrix rep of any perm symbol $(\xi\eta\zeta)$ could be the matrix in which col $\xi$ has 1 in row $\eta$, col $\eta$ has 1 in col $\zeta$, etc.? – Reb.Cabin Aug 6 '11 at 23:04
oops, last line should be col $\eta$ has 1 in ROW $\zeta$, and completing the thought, col $\zeta$ has 1 in ROW $\xi$, and zeros everywhere else. – Reb.Cabin Aug 6 '11 at 23:12

Assuming $(1 2 3)$ refers to cycle notation then no, that's not the standard interpretation. It means "$1 to 2 to 3 to 1". Initially you shouldn't be thinking about the action of a permutation on some other set of symbols - think, more simply, about its action on the very symbols it comprises of (the numbers 1, 2, 3 in this case). A more general permutation could look like$(1 4 5)(2 6)$which means 1 goes to 4 who goes to 5 who goes back to 1, while 2 and 6 swap places and (implied) 3 stays put. - Generally$( 1 \ 2 \ 3)$means, the elements$1 \to 2$and$2 \to 3$and$3 \to 1$. That is the element$1$is sent to$2$and$2$is sent to$3$and so on. Suppose you to operate$(1 \ 2 \ 3)$with$(1 \ 3 \ 2)$then we do it this way: since$1$is sent to$2$in$(1 \ 2 \ 3)$we see where$2$is sent in$(1 \ 3 \ 2)$. Now$2$is sent to$1$is in$(1 \ 3 \ 2)$therefore$1$is sent to$1\$ in their product.

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