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Section 2.5 Group actions

Subsection 2.5.1

Definition 2.5.1. Group action, orbit, stabilizer.

Let \(G\) be a group and let \(X\) be a set. An action of the group \(G\) on the set \(X\) is a group homomorphism
\begin{equation*} \phi \colon G\to \Perm(X). \end{equation*}
We say that the group \(G\) acts on the set \(X\text{,}\) and we call \(X\) a \(G\)-space. For \(g\in G\) and \(x\in X\text{,}\) we write \(gx\) to denote \((\phi(g))(x)\text{.}\) 1  We write \(\Orb(x)\) to denote the set
\begin{equation*} \Orb(x)=\{gx\colon g\in G\}, \end{equation*}
called the orbit of \(x\text{,}\) and we write \(\Stab(x)\) to denote the set
\begin{equation*} \Stab(x) = \{g\in G\colon gx=x\}, \end{equation*}
called the stabilizer or isotropy subgroup 2  of \(x\text{.}\) A group action is transitive if there is only one orbit. A group action is faithful if the map \(G\to \Perm(X)\) has a trivial kernel.
Find the indicated orbits and stabilizers for each of the following group actions.
  1. \(D_4\) acts on the square \(X=\{(x,y)\in \R^2\colon -1\leq x,y\leq 1\}\) by rotations and reflections. What is the orbit of \((1,1)\text{?}\) What is the orbit of \((1,0)\text{?}\) What is the stabilizer of \((1,1)\text{?}\) What is the stabilizer of \((1,0)\text{?}\)
  2. The circle group \(S^1\) (see Subsection 2.1.3) acts on the two-sphere \(S^2\) by rotation about the \(z\)-axis. Given an element \(e^{i\alpha}\) in \(S^1\) a point \((\theta,\phi)\) in \(S^2\) (in spherical coordinates), the action is given by
    \begin{equation*} e^{i\alpha}\cdot (\theta,\phi)=(\theta,\phi+\alpha). \end{equation*}
    What is the orbit of \((\pi/4,\pi/6)\text{?}\) What is the orbit of the north pole \((0,0)\text{?}\) What is the stabilizer of \((\pi/4,\pi/6)\text{?}\) What is the stabilizer of the north pole?
  3. Any group \(G\) acts on itself by conjugation, that is, by \((\phi(g))(x)=gxg^{-1}=C_g(x)\) (see Exercise Describe the orbit and stabilizer of a group element \(x\text{.}\)
  1. \(\Orb((1,1))=\{(1,1),(1,-1),(-1,1),(-1,-1)\}\text{,}\) \(\Orb((1,0))=\{(1,0),(-1,0),(0,1),(0,-1)\}\text{,}\) \(\Stab((1,1))= \{R_0,F_{D'}\}\text{,}\) \(\Stab((1,0))=\{R_0,F_H\}\)
  2. \(\Orb(\pi/4,\pi/6)\) is the horizontal circle on \(S^2\) with "latitude" \(\pi/4\text{,}\) \(\Orb(0,0)=\{(0,0)\}\text{,}\) \(\Stab(\pi/4,\pi/6)=\{1\}\text{,}\) \(\Stab{(0,0)}=S^1\)
  3. \(\Orb(x)=\{gxg^{-1}\colon g\in G\}\text{,}\) \(\Stab(x)=C(x)\) (the centralizer of \(x\))
Show that the stabilizer of an element \(x\) in a \(G\)-space \(X\) is a subgroup of \(G\text{.}\)

Instructor’s solution for Checkpoint 2.5.3.

We use the 2-step subgroup test. First, note that \(\Stab(x)\) is not empty, because \(\Stab(x)\) contains \(e_G\) (the action \(G\to \Perm(X)\) is a group homomorphism, so \(e_G\) must go to the identity transformation on \(X\) by Property 1 of Proposition 2.4.5). Now suppose \(g,h\in \Stab(x)\text{.}\) Then we have
\begin{equation*} (gh)(x)=g(hx)=gx=x \end{equation*}
so \(gh\in \Stab(x)\text{.}\) Also, we have
\begin{equation*} x=(g^{-1}g)x =g^{-1}(gx)=g^{-1}x\text{,} \end{equation*}
so \(g^{-1}\in \Stab(x)\text{.}\) This completes the proof.

Instructor’s solution for Checkpoint 2.5.5.

  1. There are 3 types of orbits for the action of \(D_4\) on the square.
    • the 1-point set \(\{(0,0)\}\) (a single orbit)
    • a 4-point set, the vertices of a square, of the form \((\pm a,\pm a)\) or \(\{(\pm a,0),(0\pm a)\}\) (infinitely many orbits, one for each \(a\) in the range \(0\lt a\leq 1)\)
    • an 8-point set of the form \(\{(\pm a,\pm b),(\pm b,\pm a)\}\) (infinitely many orbits, one for each \(0\leq a\lt b\))
  2. The orbits are all the latitude circles with latitude \(\theta\text{,}\) for \(0\lt \theta\lt \pi/2\text{,}\) and two 1-point orbits, the north and south poles.
  3. For the action of a group on itself by conjugation, the orbits are conjugacy classes. How many orbits and how they look depends on the group.

Subsection 2.5.2

Exercises 2.5.3 Exercises

1. Actions of a group on itself.

Let \(G\) be a group. Here are three actions \(G\to \Perm(G)\) of \(G\) on itself. Left multiplication is given by
\begin{equation*} g\to L_g \end{equation*}
where \(L_g\) is given by \(L_g(h)=gh\text{.}\) Right inverse multiplication is given by
\begin{equation*} g\to R_g \end{equation*}
where \(R_g\) is given by \(R_g(h)=hg^{-1}\text{.}\) Conjugation is given by
\begin{equation*} g\to C_g \end{equation*}
where \(C_g\) is given by \(C_g(h)=ghg^{-1}\text{.}\)
  1. Show that, for \(g\in G\text{,}\) the maps \(L_g,R_g,C_g\) are elements of \(\Perm(G)\text{.}\)
  2. Show that each of these maps \(L,R,C\) is indeed a group action.
  3. Show that the map \(L\) is injective, so that \(G\approx L(G)\text{.}\)
Consequence of this exercise: Every group is isomorphic to a subgroup of a permutation group.

Instructor’s solution for Exercise

  1. We will show that \(L_g\) is invertible. Similar arguments show that \(R_g,C_g\) are invertible. We have
    \begin{equation*} L_{g^{-1}}(L_g(x))=g^{-1}gx=x \end{equation*}
    for all \(x\in G\text{.}\) A similar argument shows that \(L_g(L_{g^{-1}}(x))=x\) for all \(x\text{.}\) We conclude that \(L_g\) is invertible, with inverse \((L_g)^{-1}=L_{g^{-1}}\text{.}\) Thus \(L_g\) is an element of \(\Perm(G)\text{,}\) as desired.
  2. You just have to check that \(L_{gh}=L_g\circ L_h\text{,}\) \(R_{gh}=R_g\circ R_h\) (here you see why the inverse is needed!), and \(C_{gh}=C_g\circ C_h\text{.}\) Here is the argument for \(L\text{.}\) Similar arguments work for \(R,C\text{.}\) Let \(g,h\in G\text{.}\) We have \(L_g(L_h(x))=g(hx) = (gh)x = L_{gh}(x)\) for all \(x\in G\text{,}\) so we have \(L_g\circ L_h = L_{gh}\text{.}\)
  3. For injectivity, if \(L_g=L_h\text{,}\) then \(g=L_g(e)=L_h(e)=h\text{.}\) By the First Isomorphism Theorem, we conclude that \(G\approx L(G)\text{,}\) which is subgroup of \(\Perm(G)\text{.}\)

2. Cosets, revisited.

Let \(H\) be a subgroup of a group \(G\text{,}\) and consider the map
\begin{equation*} R\colon H\to \Perm(G) \end{equation*}
given by \(h\to R_h\text{,}\) where \(R_h(g)=gh^{-1}\) (this is the restriction of right inverse multiplication action in Exercise to \(H\)). Show that the orbits of this action of \(H\) on \(G\) are the same as the cosets of \(H\text{.}\) This shows that the two potentially different meanings of \(G/H\) (one is the set of cosets, the other is the set of orbits of the action of \(H\) on \(G\) via \(R\)), are in fact in agreement.

Instructor’s solution for Exercise

An orbit of the group action \(H\to \Perm(G)\) is a set of the form
\begin{equation*} \Orb(g)=\{gh^{-1}\colon h\in H\} \end{equation*}
for some \(g\in G\text{.}\) Consider the map \(x\to gx^{-1}g\text{.}\) For \(x=gh^{-1}\in \Orb(g)\text{,}\) we have \(i(gh^{-1})=g(gh^{-1})g=gh\text{,}\) so the image of \(i\) is contained in the coset \(gH\text{.}\) The map \(i\) is its own inverse:
\begin{equation*} i(i(x))= i(gxg^{-1})= g(gxg^{-1})g=x\text{,} \end{equation*}
so we conclude that \(i\colon \Orb(g)\to gH\) is bijective. Thus we have established that \(\Orb(g)=gH\text{,}\) as desired.

3. The natural action of a matrix group on a vector space.

Let \(G\) be a group whose elements are \(n\times n\) matrices with entries in a field \(\F\) and with the group operation of matrix multiplication. The natural action \(G\to \Perm(X)\) of \(G\) on the vector space \(X=\F^n\) is given by
\begin{equation*} g\to [v\to g\cdot v], \end{equation*}
where the "dot" in the expression \(g\cdot v\) is ordinary multiplication of a matrix times a column vector. Show that this is indeed a group action.

Instructor’s solution for Exercise

Let \(\phi\) denote the map \(\phi\colon G\to \Perm(X)\) given by
\begin{equation*} g\to [v\to g\cdot v], \end{equation*}
so that \((\phi(g))(v)=gv\) for \(g\in G\text{,}\) \(v\in X\text{.}\) To see that \(\phi\) is a homomorphism, we have,
\begin{equation*} (\phi(gh))(v)=(gh)v = g(hv) =(\phi(g)\circ\phi(h))(v) \end{equation*}
for all \(v\in X\text{.}\) [Note: the equality \((gh)v=g(hv)\) uses the fact that matrix multiplication is associative.] We conclude that \(\phi(gh)=\phi(g)\circ \phi(h)\text{,}\) as required.

Instructor’s solution for Exercise (proof of Proposition 2.5.4).

First, we check that \(\sim_G\) is an equivalence relation. For symmetry, we have \(x=e_Gx\) for all \(x\in X\text{.}\) For reflexivity, if \(y=gx\text{,}\) then \(x=g^{-1}y\text{.}\) For transitivity, if \(y=gx\) and \(z=hy\text{,}\) then \(z=hy=h(gx)=(hg)x\text{.}\) It follows that the set of equivalence classes \(X/\!\!\sim_G\) forms a partition of \(X\text{.}\) Since \(X/G=X/\!\!\sim_G\) by definition, it follows that orbits form a partition.

Instructor’s solution for Exercise

Let \(x_0\in X\) and consider the function \(f_{x_0}\colon G\to X\) given by \(f_{x_0}(g)= gx_0\text{.}\) Applying Fact 1.4.5 to \(f_{x_0}\text{,}\) we have \(G/\!\!\sim_{f_{x_0}}=G/\Stab(x_0)\text{,}\) and \(f_{x_0}(G)=\Orb(x_0)\text{.}\)
Alternatively, one can check that the given map is well-defined, one-to-one, and onto. Let \(\pi\) denote the map \(\pi(g\Stab(x))=gx\text{.}\)
  • (\(\pi\) is well-defined) Suppose \(g\Stab(x)=h\Stab(x)\text{.}\) By Proposition 2.3.9, we have \(g^{-1}h\in \Stab(x)\text{.}\) Thus we have \(g^{-1}hx=x\text{,}\) so \(gx=hx\text{.}\) We conclude that \(\pi\) is onto.
  • (\(\pi\) is one-to-one) Suppose \(\pi(g\Stab(x))=gx=hx=\pi(h\Stab(x))\text{.}\) Then \(g^{-1}hx=x\text{,}\) so \(g^{-1}h\in \Stab(x)\text{,}\) so \(g\Stab(x)=h\Stab(x)\text{.}\) We conclude that \(\pi\) is one-to-one.
  • (\(\pi\) is onto) Given \(gx\in \Orb(x)\text{,}\) we have \(\pi(g\Stab(x))=gx\text{.}\) We conclude that \(\pi\) is onto.

6. The projective linear group action on projective space.

Let \(V\) be a vector space over a field \(\F\) (in this course, the base field \(\F\) is either the real numbers \(\R\) or the complex numbers \(\C\)). The group \(\F^\ast\) of nonzero elements in \(\F\) acts on the set \(V\setminus \!\{0\}\) of nonzero elements in \(V\) by scalar multiplication, that is, by the map \(\alpha \to [v\to \alpha v]\text{.}\) The set of orbits \((V\setminus\!\{0\})/\F^\ast\) is called the projectivization of \(V\text{,}\) or simply projective space, and is denoted \(\Proj(V).\)
  1. Let \(\sim_{\text{proj}}\) denote the equivalence relation that defines the orbits \((V\setminus \!\{0\})/\F^\ast\text{.}\) Verify that \(\sim_{\text{proj}}\) is given by \(x\sim_{\text{proj}} y\) if and only if \(x=\alpha y\) for some \(\alpha\in\F^\ast\text{.}\)
  2. Verify that the group \(GL(V)\) (the group of invertible linear transformations of \(V\)) acts on \(\Proj(V)\) by
    \begin{equation} g\cdot [{v}] = [g({v})]\tag{2.5.1} \end{equation}
    for \(g\in GL(V)\) and \({v}\in V\setminus\!\{0\}\text{.}\)
  3. Show that the kernel of the map \(GL(V)\to \Perm(\Proj(V))\) given by (2.5.1) is the subgroup \(K=\{\alpha\Id\colon \alpha\in \F^\ast\}\text{.}\)
  4. Conclude that the projective linear group \(PGL(V):=GL(V)/K\) acts on \(\Proj(V)\text{.}\)
  5. Show that \(\F^\ast\) acts on \(GL(V)\) by \(\alpha\cdot T=\alpha T\text{,}\) and that \(PGL(V)\approx GL(V)/\F^\ast\text{.}\)
  6. Let \(s\colon S^2 \to \extC\) denote the stereographic projection (see (1.3.6)). Show that the map \(\Proj(\C^2)\to S^2\) given by \([(\alpha,\beta)]\to s^{-1}(\alpha/\beta)\) if \(\beta\neq 0\) and given by \([(\alpha,\beta)]\to (0,0,1)\) if \(\beta=0\) is well-defined and is a bijection.

Instructor’s solution for Exercise

  1. Apply Fact 1.4.4 (check that the action of \(g\) is constant on projective equivalence classes of vectors).
  2. Let \(\phi\colon GL(V)\to \Perm(\Proj(V))\) be the action defined by Equation (2.5.1). It is easy to check that \(K\subseteq \ker(\phi)\text{.}\) Containment in the other direction requires a short linear algebra argument. Suppose \(g\in \ker(\phi)\text{.}\) This means that for every \(v\in V\text{,}\) there is a nonzero scalar \(\lambda_v\) such that \(gv=\lambda_v v\) (that is, every vector in \(V\) is an eigenvector of the operator \(g\)). If \(\dim(V)=1\text{,}\) we are done. If \(\dim(V)\gt 1\text{,}\) choose independent vectors \(v,w\in V\text{.}\) We have \(\lambda_{v+w}(v+w)=g(v+w)=\lambda_v v+\lambda_w w\text{.}\) Because \(v,w\) are independent, it follows that \(\lambda_v=\lambda_w=\lambda_{v+w}\text{.}\) Thus it must be that there is a single \(\lambda\in \F^\ast\) such that \(gv=\lambda v\) for all \(v\in V\text{,}\) that is, \(g\) is an element of \(K\text{.}\)
  3. Apply the First Isomorphism Theorem (Corollary 2.4.7) to \(\phi\colon GL(V)\to \Perm(\Proj(V))\) to obtain
    \begin{equation*} GL(V)/K \approx \phi(G) \stackrel{\text{inclusion}}{\longrightarrow} \Perm(\Proj(V)). \end{equation*}
  4. Given \(\alpha,\beta\) in \(\F^\ast\) and \(T\) in \(GL(V)\text{,}\) the fact that \((\alpha \beta)T=\alpha(\beta T)\) verifies that the map \(\F^\ast \to \Perm(GL(V))\) given by \(\alpha\to [T\to \alpha T]\) is a homomorphism. The equivalence relation that corresponds to this action is given by \(T\sim_{\F^\ast} S\) if and only if \(S=\alpha T\) for some \(\alpha\in \F^\ast\text{.}\) Because \(S=\alpha T\) if and only if \(S=(\alpha \Id)T\text{,}\) the equivalence classes that define \(PGL(V)\) in the previous part of this problem are the same as the equivalence classes \(GL(V)/\F^\ast\text{.}\)
  5. Factor the given map as \(\Proj(\C^2)\to \extC\to S^2\) given by
    \begin{equation*} [(\alpha,\beta)] \to \alpha/\beta \to s^{-1}(\alpha/\beta). \end{equation*}
    Let \(\mu\) denote the map \([(\alpha,\beta)] \to \alpha/\beta \text{.}\) That \(\mu\) is well-defined follows from the observation that \((k\alpha)/(k\beta)=\alpha/\beta\) for all \(k\in \F^\ast\text{.}\) To see that \(\mu\) is bijective, we will exhibit an inverse map. Let \(\nu\colon \extC\to \Proj(\C^2)\) be given by \(\gamma \to [(\gamma,1)]\) for \(\gamma \in \C\) and \(\infty \to [(1,0)]\text{.}\) To check that \(\nu\circ\mu=\Id\text{,}\) for \(\beta\neq 0\text{,}\) we have
    \begin{equation*} (\nu\circ \mu)([(\alpha,\beta)])=\nu(\alpha/\beta) = [(\alpha/\beta,1)]=[(\alpha,\beta)] \end{equation*}
    and for \(\beta=0\text{,}\) we have
    \begin{equation*} (\nu\circ \mu)([(\alpha,0)])=\nu(\infty) = [(1,0)]=[(\alpha,0)]. \end{equation*}
    Reading the two displayed equations from right to left establishes that \(\mu\circ \nu=\Id\text{.}\) We conclude that \(\nu=\mu^{-1}\text{.}\) Using the fact that stereographic projection \(s\) is bijective, we have \(s^{-1}\circ \mu\) is bijective, as desired.
Other notations for \((\phi(g))(x)\) are \(g(x)\text{,}\) \(g\cdot x\text{,}\) and \(g.x\text{.}\)
It must be proved that \(\Stab(x)\) is indeed a subgroup of \(G\text{.}\) See Checkpoint 2.5.3 below.