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 Gequivariant 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. 🙂

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