MTH3003 Lecture 12

This lecture continues our study of $S_n$ by introducing the signature function, which measures whether a permutation is “even” or “odd”. We connect this to the cycle structure from last time and prepare the ground for defining the alternating group via the kernel of the signature homomorphism.

The Signature Function via a Special Polynomial

We work in the symmetric group symmetric group, $S_n$ acting on variables $x_1,x_2,\dots ,x_n$ by permuting indices: for $g\in S_n$ we define $g{x}_k=x_{gk}$.

The Vandermonde-type Polynomial

Define the polynomial

$$\Delta =\prod _{i<j}(x_i-x_j)=((x_1-x_2)\cdots (x_1-x_n))((x_2-x_3)\cdots (x_2-x_n))\cdots ((x_{n-1}-x_n)).$$

This is a product over all unordered pairs $\{i,j\}$ with $i<j$, and in small cases we can write it explicitly.

Example

In $S_3$ we have $\Delta =(x_1-x_2)(x_1-x_3)(x_2-x_3)$.

A permutation $g\in S_n$ acts on $\Delta$ by sending each $x_k$ to $x_{gk}$, hence $\Delta$ is carried to another polynomial with the same factors, but possibly in a different order and with some factors $(x_i-x_j)$ flipped to $(x_j-x_i)$.

Definition of the Signature

The key fact is that every permutation sends $\Delta$ either to itself or to its negative.

Formally:

Important

Definition (Signature function). Let $g\in S_n$. Consider the action of $g$ on $\Delta$:

• If $g$ maps $\Delta$ to $\Delta$, define $\sigma (g)=1$.
• If $g$ maps $\Delta$ to $-\Delta$, define $\sigma (g)=-1$. We call $\sigma :S_n\to \{1,-1\}$ the signature function on $S_n$, and say $g$ is even if $\sigma (g)=1$ and odd if $\sigma (g)=-1$.

So $\sigma$ partitions $S_n$ into the even and odd permutations, and we will later see that the even ones form a very important subgroup.

Example

In $S_3$:

• $e$ sends $\Delta$ to $\Delta$ so $\sigma (e)=1$.
• $(12)$ sends $(x_1-x_2)$ to $(x_2-x_1)$ and fixes the other factors, so overall $\Delta \mapsto -\Delta$, hence $\sigma ((12))=-1$.
• $(13)$ similarly sends $\Delta$ to $-\Delta$, so $\sigma ((13))=-1$.

At this stage, the definition is conceptually nice but computationally horrible: we do not want to recompute the image of $\Delta$ every time we meet a permutation.

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Signature as a Homomorphism

A crucial structural property is that the signature function respects the group operation.

Important

Proposition. The set $\{1,-1\}$ forms a group under multiplication, and the function $\sigma :S_n\to \{1,-1\}$ is a group homomorphism.

Sketch of the proof. Fix $g,h\in S_n$. Since $\sigma (g)\in \{1,-1\}$, we have

$$(gh)\Delta =g(h\Delta )=g(\sigma (h)\Delta )=\sigma (h)g(\Delta )=\sigma (h)\sigma (g)\Delta =\sigma (g)\sigma (h)\Delta .$$

By definition of $\sigma$, this implies $\sigma (gh)=\sigma (g)\sigma (h)$, so $\sigma$ is a homomorphism.

This is the philosophical “point”: the signature is not just a random $\pm 1$ label, it is a group homomorphism with a kernel that will turn out to be the alternating group.

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Computing Signatures Using Cycle Decompositions

So far the definition is in terms of how $g$ acts on $\Delta$, which is not convenient for actual calculations. We now use the cycle notation description of permutations to get a closed formula.

Let $g\in S_n$ and write it as a product of disjoint cycles

$$g=c_1{c}_2\cdots c_m,$$

where cycle $c_i$ has length $r_i$ for $i=1,\dots ,m$.

Important

Corollary (Very important). For a cycle $c_i$ of length $r_i$ we have $\sigma (c_i)=(-1{)}^{r_i-1}$, and hence $\sigma (g)=\sigma (c_1)\sigma (c_2)\cdots \sigma (c_m)=(-1{)}^{r_1-1}\cdots (-1{)}^{r_m-1}$.

So each $r_i$-cycle is odd if $r_i$ is even, and even if $r_i$ is odd, and the signature of $g$ is the product of these contributions.

Example

Let $g=(15742)(129)$. Then the cycle lengths are $5$ and $3$, so $\sigma (g)=(-1{)}^{5-1}(-1{)}^{3-1}=1$, meaning $g$ is even.

This formulation makes it very easy to compute signatures once we know the cycle shape of a permutation.

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Behaviour under Inverses and Dependence on Cycle Shape

We next understand how the signature behaves when we invert a permutation and what data it depends on.

Important

Proposition. If $g\in S_n$ is a permutation, then $\sigma (g^{-1})=\sigma (g)$.

The proof is given on the handout and runs first for cycles, then extends to permutations written as products of disjoint cycles.

• A cycle $c$ of length $r$ has signature $\sigma (c)=(-1{)}^{r-1}$.
• Its inverse $c^{-1}$ is just the same cycle written backwards, so still has length $r$, hence $\sigma (c^{-1})=(-1{)}^{r-1}=\sigma (c)$.
• For $g=c_1{c}_2\cdots c_m$, we have $g^{-1}=c_1^{-1}{c}_2^{-1}\cdots c_m^{-1}$, and the homomorphism property then yields $\sigma (g^{-1})=\sigma (g)$.

Note

The signature of a permutation depends only on its cycle type, i.e. the multiset of cycle lengths. Two permutations with the same cycle shape have the same signature.

This is immediate from the formula $\sigma (g)={\prod }_i(-1{)}^{r_i-1}$, which only “sees” the lengths $r_i$.

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Worked Examples with Signatures

The handout provides a list of explicit examples to build intuition for even and odd permutations.

Example

Let $g=(17342)(956)(811)\in S_{12}$. The cycle lengths are $5,3,2$, so $\sigma (g)=(-1{)}^{5-1}(-1{)}^{3-1}(-1{)}^{2-1}=-1$, hence $g$ is odd.

Example

Let $h=(123)\in S_{12}$. This is a single $3$-cycle, so $\sigma (h)=(-1{)}^{3-1}=1$, hence $h$ is even.

When computing the signature of a product $gh$, we can either explicitly write $gh$ as a product of disjoint cycles and then apply the formula, or (preferably) use the homomorphism property.

Example

Consider $g,h$ as above and compute $gh$.

1. First write $gh$ as a product of disjoint cycles: $gh=(17342)(956)(811)(123)=(24)(37)(569)(811)$, so $\sigma (gh)=(-1{)}^{2-1}(-1{)}^{2-1}(-1{)}^{3-1}(-1{)}^{2-1}=-1$, and $gh$ is odd.
2. Alternatively, using the homomorphism property: $\sigma (gh)=\sigma (g)\sigma (h)=(-1)(1)=-1$, again showing $gh$ is odd.

Two further quick consequences are highlighted:

• Every $3$-cycle is even, since $\sigma ((abc))=(-1{)}^{3-1}=1$.
• Any product of even permutations is even, because the product of $1$‘s in $\{1,-1\}$ is still $1$.

These become very useful when recognising elements of the alternating group later.

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Pre-Lecture Notes from University Notes

• Action of $S_n$ on variables $x_1,\dots ,x_n$ by $g{x}_k=x_{gk}$, setting up the polynomial viewpoint.
• Definition of the Vandermonde-like polynomial $\Delta ={\prod }_{i<j}(x_i-x_j)$ and observation that any $g\in S_n$ sends $\Delta$ to either $\Delta$ or $-\Delta$.
• Definition of the signature function $\sigma :S_n\to \{1,-1\}$, classifying permutations as even or odd depending on the image of $\Delta$.
• Proof that $\sigma$ is a group homomorphism, via the calculation $(gh)\Delta =\sigma (g)\sigma (h)\Delta$.
• Formula for the signature of a cycle: a cycle of length $r$ has signature $\sigma (c)=(-1{)}^{r-1}$, giving $\sigma (g)$ as a product over the cycle lengths in the disjoint cycle decomposition.
• Consequences: signature depends only on the cycle shape; inverse permutations have the same signature; every $3$-cycle is even; products of even permutations remain even.
• Multiple worked examples in $S_{12}$ illustrating how to compute signatures directly from cycle decompositions and using the homomorphism property for products.
• Next time we will use the signature homomorphism to define and study the alternating group $A_n$ as the kernel of $\sigma$, and explore its basic properties and importance.