## Group actions

We have already seen abstract groups as permutation groups, or, more generally, as groups of bijections, in a number of contexts:

• $G$ acting on itself by (left) multiplication, when we associated with each $g\in G$ a bijection $L_g:G\to G$ by $L_g(x)=gx$
• $G$ acting on itself by conjugation, when we associated with each $g\in G$ a bijection $C_g:G\to G$ by $C_g(x)=gxg^{-1}$
• $G$ acting on the set of (left) cosets $G/H=\{gH\mid g\in G\}$ of a subgroup $H\leq G,$ when we associated with each $g\in G$ a bijection $M_g:G/H\to G/H$ by $M_g(xH)=gxH$

We have also seen matrix groups at work as permutation groups (e.g. on the set of subspaces), permutation groups becoming matrix groups (e.g. $S_n$ represented by $n\times n$ permutation matrices), etc.
What unites all these examples is the concept of a group action.

Let $G$ be a group and $\Omega$ be a set, with a group of bijections $S(\Omega).$ An action of $G$ on $\Omega$ is a homomorphism $\phi: G\to S(\Omega).$

There are many different types of actions, e.g. permutation actions when $\Omega$ is a finite set and $S(\Omega)$ the symmetric group (of all permutations of $\Omega,$
linear actions, when $\Omega$ is a vector space, and $S(\Omega)$ a group of its linear transformations (i.e. matrix groups), etc.

By the homomorphism theorem, $Im(\phi)\cong G/Ker(\phi).$ An action satisfying $Ker(\phi)=\{id\}$ is called faithful (or effective).
For instance, the action of $G$ on itself by (left) multiplication is always faithful, but the action on cosets of an $H\leq G$ need not be faithful. E.g. it will not be faithful if $G$ is commutative and $H\neq\{id\}.$

Orbits and point stabilisers
An action $\phi$ of $G$ on $\Omega$ defines an equivalence relation $\sim$ on $\Omega,$ as follows: $a\sim b$ iff there exists $g\in G$ such that $\phi(g)a=b.$ The equivalence classes of $\sim$ are called orbits (more precisely, orbits of $Im(\phi)$ on $\Omega$).
Let $\omega\in\Omega$ and $G\geq H=\{h\in G\mid \phi(h)\omega=\omega\}.$ Then $\phi(H)$ is called the stabiliser of $\omega$ in $\phi(G),$ and denoted $\phi(G)_\omega.$

Let $\omega'=g\omega$ for some $g\in \phi(G).$ What can be said about $\phi(G)_{\omega'}?$

$\phi(G)_{\omega'}=g\phi(G)_\omega g^{-1}.$

Indeed, the coset $g\phi(G)_\omega$ consists of the elements of $\phi(G)$ that map $\omega$ to $\omega'.$ Thus any $x\in H:=g\phi(G)_\omega g^{-1}$ fixes $\omega',$ and any $y$ that fixes $\omega'$ belongs to $H,$ as $g^{-1}yg\in \phi(G)_\omega.$

When we consider the action $C: G\to G$ of $G$ on itself by conjugation, the orbits are called conjugacy classes.
For instance, the conjugacy classes of $S_n,$ the symmetric group on $n$ letters, can be shown to be in 1-to-1 correspondence with the unordered partitions of $n$ into parts $1\leq p_1\leq p_2\leq\dots\leq p_k,$ where $n=p_1+\dots +p_k,$ as follows:
Let a permutation $\sigma=(i_{1,1},\dots,i_{1,p_1})\dots (i_{k,1},\dots,i_{k,p_k})$ be written in the cyclic notation. Then for any $\tau\in S_n$ one has

$\displaystyle \tau\sigma\tau^{-1}=(\tau(i_{1,1}),\dots,\tau(i_{1,p_1}))\dots (\tau(i_{k,1}),\dots,\tau(i_{k,p_k})),$

and so we can transform $\sigma$ to any permutation with cycles of the same lengths $p_1,\dots,p_k.$

Equivalent actions
When do two actions $\phi$ and $\psi$ of a group $G$ on, respectively, sets $\Omega$ and $\Sigma$ can be viewed as “the same”? Obviously we need a bijection $f:\Omega\to\Sigma$ to exist, that also “respects” $G,$ in the sense that $f(\phi(g)\omega)=\psi(g)f(\omega)$ for all $\omega\in\Omega$ and $g\in G.$ Such an $f$ is called a $G$equivariant bijection, and is said to define an isomorphism of actions.

An action that has just one orbit is called transitive.

Let $\phi$ be a transitive action of $G$ on $\Omega.$ Then $\phi$ is isomorphic to the action on $G/G_\omega=\{gG_\omega\mid g\in G\},$ for any $\omega\in\Omega,$ where we denoted by $G_\omega$ the preimage of $\phi(G)_\omega$ in $G.$

This isomorphism of actions is given by $f:\Omega\to G/G_\omega$ such that $f(\mu)=gG_\omega,$ where $\phi(g)(\omega)=\mu.$ Indeed,

1. $f$ is obviously a bijection
2. $\phi(h)\omega=\mu$ iff $\phi(g^{-1}h)\omega=\omega$ iff $g^{-1}h\in G_\omega$ iff $h\in gG_\omega$
3. $f$ is $G$-equivariant. Indeed, let $\phi(h)(\omega)=\mu.$ Then for any $g\in G$ one has $f(\phi(g)\mu)=f(\phi(gh)\omega)=(gh)G_\omega=g(hG_\omega)=\phi(g)f(\mu),$ as claimed.

Finally, for an action $\phi$ of $G$ on $\Omega$ we can characterise $Ker(\phi)$ as

$\displaystyle Ker(\phi)=\bigcap_{g\in G} gG_\omega g^{-1},$ for an $\omega\in\Omega.$

Actions of matrix groups (a.k.a. linear actions)
Subgroups of $GL_n(\mathbb{F}),$ the group, with respect to matrix multiplication, of invertible $n\times n$ matrices with entries in a field $\mathbb{F},$ have a number of natural actions, that all come one way or another from the vectorspace $V\cong \mathbb{F}^n.$ Indeed, each $g\in GL_n( \mathbb{F} )$ provides a bijection $V\to V$ by $v\mapsto gv$ for any $v\in V.$ Moreover, such a bijection is even an automorphism of the vectorspace $V.$ (Automorphisms of vectorspaces are defined just as these for groups and rings: these are bijections preserving the corresponding algebraic structure (i.e. operations, etc)).

Apart from the action $v\mapsto gv$ on the vectors of $V,$ (called the natural action (by left multiplication)) there is another action on the vectors, namely, by $v\mapsto vg^{-1}$ (which one can equivalently writen as $v\mapsto (g^{-1})^T v$). It is called contragradient action.
(Note that the map $f$ given by $v\mapsto vg$ is not always an action, as $f(gg')=f(g')f(g),$ and so $f$ is not a homomorphism, unless we have a commutative group.)

Natural and contragradient actions give a simple example of two non-equivalent faithful (i.e. with trivial kernel) actions of one group, say, $G=GL_n(\mathbb{F} )$, with $n\geq 2,$ on the same set $V.$ Indeed, if $f:V\to V$ were an equivalence of these actions, then $f(gv)=(g^{-1})^T (f(v))$ for all $v\in V$ and $g\in G.$
Let $n=2$ and $f(v)=u$ so that $u$ and $v$ are linearly independent. In the basis $(v,u)$, take $g=\begin{pmatrix} 1&1\\ 0& 1\end{pmatrix}.$ Then $(g^{-1})^T=\begin{pmatrix} 1&0\\ -1& 1\end{pmatrix}.$ We see that $f(gv)=f(v)=u\neq (g^{-1})^T(u),$ a contradiction.
If, on the other hand, $f(v)$ and $v$ are always linearly dependent,
we take the same $g$ in the standard basis $(v,w)$ and see that $f(gv)=f(v)=\alpha v=(g^{-1})^T (f(v))=\alpha (g^{-1})^T (v),$ for some $0\neq \alpha\in\mathbb{F},$ again a contradiction. The argument showing non-equivalence in the case $n\geq 3$ is similar. For $n=1$ these two actions are isomorphic, as $GL_1(\mathbb{F})$ is commutative.

Another important class of actions is the action on $G\leq GL_n(\mathbb{F} )$ on subspaces of $V.$ For simplicity, let us consider the action on the set $\Gamma_k(V)$ of subspaces of dimension $k.$ Given a $k$-dimensional subspace $U\subset V$ and $g\in G,$ we define $g(U)=\{gu\mid u\in U\}.$ Then $g(U)$ is a $k$-dimensional subspace, as can be seen by choosing a basis $u_1,\dots, u_k$ of $U$ and observing that $gu_1,\dots,gu_k$ is a basis of $g(U).$ Thus $g$ induces a bijection on $\Gamma_k(V).$
When $G$ fixes a $U\in\Gamma_k(V),$ we can choose a basis in such a way, that $G$ becomes a group of $2\times 2$ block matrices, discussed earlier here.In this case $G$ has a normal subgroup acting trivially on $U,$ that is the kernel of the action of $G$ on the quotient space $V/U.$

The centre $Z=Z(GL_n(\mathbb{F} ))$ of $GL_n(\mathbb{F} )$ acts trivially on $\Gamma_k(V),$ as it preserves every 1-dimensional subspace (indeed, any $g\in Z$ equals $\lambda I,$ with $0\neq\lambda\in\mathbb{F}$).

Projective line.
The simplest nontrivial $\Gamma_k(V)$ is $\Gamma_1(V)$ when $n=2.$ This set is called projective line and denoted $\mathbb{F}P^1,$ and can be viewed as the set of pairs $\{(1:\alpha)\mid \alpha\in \mathbb{F}\}\cup\{(0:1)\},$ as follows: every 1-dimensional subspace $(x:y)$ of $V$ can be described by equation $y=\alpha x$ or by equation $x=0.$
To understand the action of an element $g$ of $G=GL_2(V)$ on $\mathbb{F}P^1,$ consider $g=\begin{pmatrix} a& b\\ c& d\end{pmatrix}$ and $g((x:y))=(ax+by:cx+dy)$ just as if we multiply a vector by $g,$ and then normalize either to $\left(1:\frac{cx+dy}{ax+by}\right)$ or to $(0:1),$ depending upon whether or not $ax+by=0.$

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### 3 Responses to “Group actions”

1. zsharon Says:

In regard to the section on the action of a subset G of the general linear group over F^n on a subspace of F^n, if g is an nxn matrix and u is vector in F^k where k<n, then how is gu defined? The dimensions aren't right.

• Dima Says:

The dimensions ARE right. Indeed, a k-subspace $S$ of $V=F^n$ is just a subset of $V.$
In order to comput how the action works, take a basis of $S,$ and apply the transformation (the action of which we are after) to each element of the basis. The result will be again a basis of a k-subspace of $V.$ To see this as a linear action, read on on Plucker coordinates.

• zsharon Says:

I see now. I had the wrong thing in mind. I was thinking of a subspace of V as consisting of a set of vectors with fewer coordinates, which is not correct (though it is isomorphic to such a vector space).

Sorry for the pointless reply, but at least you know someone is reading. 🙂