J. Munkres. Topology (2013) Chapter 2: Topological Spaces and Continuous Functions. Supplementary Exercises: Topological Groups

Definition

A topological group \(G\) consists of a group \(G\) and a \(T_1\) topology on \(G\) such that the multiplication of \(G\) and the function mapping \(x\) to \(x^{-1}\) are continuous.

Exercise 1

Let \(G\) be a group with a \(T_1\) topology. Then \(G\) is a topological group if and only if the function \(G^2 \rightarrow G\) that maps \((x,y) to xy^{-1}\) is continuous.

Exercise 2

The following are topological groups: \((\mathbb{Z}, +)\), \((\mathbb{R}, +)\), \((\mathbb{R}_+, \cdot)\), \((\{z \in \mathbb{C} \mid |z| = 1\}, \cdot)\), \(GL_n(\mathbb{R}) \subseteq \mathbb{R}^{n^2}\).

Exercise 3

Let \(G\) be a topological group. Let \(H\) be a subspace of \(G\) that is also a subgroup of \(G\). Then \(H\) and \(\overline{H}\) are topological groups.

Solution

  • \(H\) is a topological group.
    • \(H\) is \(T_1\)
      • Any subspace of a \(T_1\) space is \(T_1\).
    • The function mapping \((x,y)\) to \(xy^{-1}\) is continuous.
      • It is the restriction of the function in \(G\) to \(H^2\).
  • \(\overline{H}\) is a subgroup of \(G\).
    • For all \(x,y \in \overline{H}\) we have \(xy \in \overline{H}\).
      • Let \(x,y \in \overline{H}\).
      • Let \(U\) be a neighbourhood of \(xy\). Prove: \(U\) intersects \(H\).
      • \(m^{-1}(U)\) is a neighbourhood of \((x,y)\), where \(m : G^2 \rightarrow G\) is the multiplication of \(G\).
      • Pick neighbourhoods \(V\) of \(x\) and \(W\) of \(y\) such that \(V \times W \subseteq m^{-1}(U)\)
      • Pick \(a \in V \cap H\) and \(b \in W \cap H\)
      • \(ab \in U \cap H\)
    • For all \(x \in \overline{H}\) we have \(x^{-1} \in \overline{H}\).
      • Let \(x \in \overline{H}\).
      • Let \(U\) be a neighbourhood of \(x^{-1}\). Prove: \(U\) intersects \(H\).
      • \(i^{-1}(U)\) is a neighbourhood of \(x\), where \(i : G \rightarrow G\) is the function that maps \(x\) to \(x^{-1}\).
      • Pick \(a \in i^{-1}(U) \cap H\)
      • \(a^{-1} \in U \cap H\)
  • \(\Box\)

Exercise 4

Let \(\alpha \in G\). Show that the maps \(f_\alpha, g_\alpha : G \rightarrow G\) defined by

\[ f_\alpha(x) = \alpha x, \qquad g_\alpha(x) = x \alpha \]

are homeomorphisms of \(G\). Conclude that \(G\) is a homogeneous space. (This means that for every pair \(x\), \(y\) of points of \(G\), there exists a homeomorphism of \(G\) onto itself that carries \(x\) to \(y\).)

Solution

  • \(f_\alpha\) is injective.
    • If \(\alpha x = \alpha y\) then \(x = \alpha^{-1} \alpha x = \alpha^{-1} \alpha y = y\).
  • \(f_\alpha\) is surjective.
    • For any \(y \in G\) we have \(y = f_\alpha(\alpha^{-1} y)\).
  • \(f_\alpha\) is continuous.
    • It is the composite of multiplication, the constant function \(\alpha\) and the identity function.
  • \(f_\alpha^{-1}\) is continuous.
    • \(f_\alpha^{-1} = f_{\alpha^{-1}}\)
  • \(g_\alpha\) is a homeomorphism.
    • Similar.
  • \(G\) is homogeneous.
    • Given \(x,y \in G\), we have \(f_{yx^{-1}}\) is a homeomorphism that maps \(x\) to \(y\).
  • \(\Box\)

Exercise 5

Let \(H\) be a subgroup of \(G\). If \(x \in G\), define \(xH = \{xh \mid h \in H\}\); this set is called a left coset of \(H\) in \(G\). Let \(G/H\) be the set of left cosets of \(H\) in \(G\); it is a partition of \(G\). Given \(G/H\) the quotient topology.

(a)

Show that if \(\alpha \in G\), the map \(f_\alpha\) of the preceding exercise induces a homeomorphism of \(G / H\) carring \(xH\) to \(\alpha x H\). Conclude that \(G/H\) is a homogeneous space.

Solution

  • Define \(f : G/H \rightarrow G/H\) by \(f(xH) = \alpha x H\).
    • If \(xH = yH\) then \(x^{-1}y \in H\) and so \(x^{-1} \alpha^{-1} \alpha y \in H\), hence \(\alpha x H = \alpha y H\).
  • \(f\) is a homeomorphism.
    • \(f\) is bijective.
      • The function that maps \(xH\) to \(\alpha^{-1}xH\) is its inverse.
    • \(f\) is continuous.
      • For \(V \subseteq G/H\) open, we have \(p^{-1}(f^{-1}(V)) = f_\alpha^{-1}(p^{-1}(V))\) is open, hence \(f^{-1}(V)\) is open.
    • \(f^{-1}\) is continuous.
      • Similar.
  • \(G/H\) is homogeneous.
    • Given \(xH,yH \in G/H\), the function that maps any \(zH\) to \(yx^{-1}zH\) is a homeomorphism that maps \(xH\) to \(yH\).
  • \(\Box\)

(b)

Show that if \(H\) is a closed set in the topology of \(G\), then \(G/H\) is \(T_1\).

Solution

If \(H\) is closed, then any left coset \(xH\) is closed in \(G\) being the image of \(H\) under the homeomorphism from (a), hence \(\{xH\}\) is closed in \(G/H\).

(c)

Show that the quotient map \(p : G \rightarrow G / H\) is an open map.

Solution

  • Let \(U \subseteq G\) be open.
  • \(p^{-1}(p(U)) = \bigcup_{z \in H} \beta_z(U)\)
    • \[ \begin{align} x \in p^{-1}(p(U)) & \Leftrightarrow \exists y \in U. xH = yH \\ & \Leftrightarrow \exists y \in U. y^{-1}x \in H \\ & \Leftrightarrow \exists z \in H. \exists y \in U. z = y^{-1} x \\ & \Leftrightarrow \exists z \in H. \exists y \in U. x = yz \\ & \Leftrightarrow x \in \bigcup_{z \in H} \beta_z(U) \end{align} \]
  • \(p^{-1}(p(U))\) is open in \(G\)
    • Each \(\beta_z(U)\) is open (because \(\beta_z\) is a homeomorphism) hence \(p^{-1}(p(U))\) is a union of open sets, hence open.
  • \(p(U)\) is open in \(G/H\)
  • \(\Box\)

(d)

If \(H\) is a closed normal subgroup then \(G/H\) is a topological group.

Solution

  • Assume \(H\) is a closed normal subgroup.
  • \(G/H\) is \(T_1\).
    • From part (b)
  • The function that maps \((xH,yH)\) to \(xy^{-1}H\) is continuous.
    • Let \(f : (G/H)^2 \rightarrow G/H\) be the function that maps \((xH,yH)\) to \(xy^{-1}H\).
    • Let \(g : G^2 \rightarrow G\) be the function that maps \((x,y)\) to \(xy^{-1}\).
    • Let \(U\) be open in \(G/H\).
    • \(p^2 : G^2 \rightarrow (G/H)^2\) is a quotient map.
      • It is a surjective continuous open map by (c).
    • \((p^2)^{-1}(f^{-1}(U)) = g^{-1}(p^{-1}(U))\)
    • \((p^2)^{-1}(f^{-1}(U))\) is open.
    • \(f^{-1}(U)\) is open.
  • \(\Box\)

Exercise 7

If \(A\) and \(B\) are subsets of \(G\), let \(AB\) denote the set of all points \(ab\) for \(a \in A\) and \(b \in B\). Let \(A^{-1}\) denote the set of all points \(a^{-1}\) for \(a \in A\).

(a)

A neighbourhood \(V\) of the identity element \(e\) is said to be symmetric iff \(V = V^{-1}\). If \(U\) is a neighbourhood of \(e\), show that there is a symmetric neighbourhood \(V\) of \(e\) such that \(VV \subseteq U\).

Solution

  • Let \(m : G^2 \rightarrow G\) be the multiplication of \(G\)
  • \((e,e) \in m^{-1}(U)\)
  • Pick neighbourhoods \(V_1\), \(V_2\) of \(e\) such that \(V_1 \times V_2 \subseteq m^{-1}(U)\).
  • Let \(V' = V_1 \cap V_2\).
  • \(V'\) is a neighbourhood of \(e\) and \(V' V' \subseteq U\)
  • Let \(f : G^2 \rightarrow G\) be the function that maps \((x,y)\) to \(xy^{-1}\).
  • \((e,e) \in f^{-1}(V')\)
  • Pick a neighbourhood \(W\) of \(e\) such that \(W^2 \subseteq f^{-1}(V')\).
  • \(W W^{-1} \subseteq V'\)
  • Let \(V = W W^{-1}\)
  • \(V\) is a symmetric neighbourhood of \(e\) and \(VV \subseteq U\)
  • \(\Box\)

(b)

Show that \(G\) is Hausdorff. In fact, show that if \(x \neq y\), there is a neighbourhood \(V\) of \(e\) such that \(Vx\) and \(Vy\) are disjoint.

Solution

  • Let \(x,y \in G\). Assume \(x \neq y\).
  • Let \(U = G - \{xy^{-1}\}\).
  • \(U\) is a neighbourhood of \(e\).
  • Pick a symmetric neighbourhood \(V\) of \(e\) such that \(VV \subseteq U\)
  • Assume for a contradiction \(z \in Vx \cap Vy\)
  • Pick \(a,b \in V\) such that \(z = ax = by\)
  • \(xy^{-1} = a^{-1}b \in U\)
  • This is a contradiction.
  • \(\Box\)

(c)

Show that \(G\) satisfies the following separation axiom, which is called the regularity axiom: Given a closed set \(A\) and a point \(x \notin A\), there exist disjoint open sets \(U\) and \(V\) such that \(A \subseteq U\) and \(x \in V\).

Solution

  • Let \(W = G - Ax^{-1}\).
  • \(W\) is a neighbourhood of \(e\)
  • Pick a symmetric neighbourhood \(W'\) of \(e\) such that \(W'W' \subseteq W\).
  • Let \(U = W'A\) and \(V = W'x\)
  • \(U\) is open.
    • It is \(\bigcup_{a \in A} W'a\).
  • Assume for a contradiction \(z \in U \cap V\).
  • Pick \(w_1, w_2 \in W'\) and \(a \in A\) such that \(z = w_1 a = w_2 x\)
  • \(ax^{-1} = w_1^{-1}w_2 \in W\)
  • This is a contradiction.
  • \(\Box\)

(d)

Let \(H\) be a subgroup of \(G\). Show that \(G/H\) satisfies the regularity axiom.

Solution

  • Let \(p : G \rightarrow G/H\) be the quotient map.
  • Let \(C\) be a closed set in \(G/H\) and \(xH \in G/H - C\).
  • Let \(A = p^{-1}(C)\).
  • \(A\) is a saturated closed set and \(x \notin A\).
  • \(A = AH\)
  • Pick a symmetric neighbourhood \(V\) of \(e\) such that \(VA \cap Vx = \emptyset\).
    • Let \(W = G - Ax^{-1}\).
    • \(W\) is a neighbourhood of \(e\)
    • Pick a symmetric neighbourhood \(V\) of \(e\) such that \(VV \subseteq W\).
    • Assume for a contradiction \(z \in VA \cap Vx\)
    • Pick \(v_1, v_2 \in V\) and \(a \in A\) such that \(z = v_1 a = v_2 x\)
    • \(ax^{-1} = v_1^{-1}v_2 \in W\)
    • This is a contradiction.
  • \(p(VA) \cap p(Vx) = \emptyset\)
    • Assume for a contradiction \(yH \in p(VA) \cap p(Vx)\)
    • Pick \(v_1, v_2 \in V\), \(a \in A\) and \(h_1, h_2 \in H\) such that \(y = v_1 a h_1 = v_2 x h_2\)
    • \(v_2 x = v_1 a h_1 h_2^{-1} \in VAH = VA\)
    • This is a contradiction.
  • \(p(VA)\) is open
    • \(p\) is an open map by Exercise 5(c).
  • \(p(Vx)\) is open
    • \(p\) is an open map by Exercise 5(c).
  • \(C \subseteq p(VA)\)
  • \(xH \in p(Vx)\)
  • \(\Box\)

Comments

Post a Comment

Popular posts from this blog

J. Munkres. Topology (2013) Chapter 4: Countability and Separation Axioms. 32: Normal Spaces

Theorem 32.1 Every second countable regular space is normal. Proof Let \(X\) be a second countable regular space. Pick a countable basis \(\mathcal{B}\). Let \(A\) and \(B\) be disjoint closed sets in \(X\). Pick a countable covering \(\{U_1, U_2, \ldots\}\) of \(A\) by open sets whose closures are disjoint from \(B\). Let \(\{U_1, U_2, \ldots\} = \{U \in \mathcal{B} \mid \overline{U} \cap B = \emptyset\}\). Prove: this covers \(A\). Let \(x \in A\). Prove: there exists \(U \in \mathcal{B}\) such that \(x \in U\) and \(\overline{U} \cap B = \emptyset\). Pick disjoint open sets \(U\) and \(V\) such that \(x \in U\) and \(B \subseteq V\). By regularity. Pick \(U' \in \mathcal{B}\) such that \(x \in U' \subseteq U\). \(x \in U'\) and \(\overline{U'} \cap B \subseteq \overline{U} \cap B \subseteq (X - V) \cap B = \emptyset\) Pick a countable covering \(\{V_1, V_2, \ldots\}\) of \(B\) by open sets whose ...
Read more

J. Munkres. Topology (2013) Chapter 3: Connectedness and Compactness. 27: Compact Subspaces of the Real Line

Theorem 27.1 Let \(X\) be a linearly ordered set with the least upper bound property in the order topology. Every closed interval in \(X\) is compact. Proof Let \(a,b \in X\) with \(a Let \(\mathcal{A}\) be a covering of \([a,b]\) by sets open in \(X\). \( \langle 1 \rangle 1 \) For any \(x \in [a,b)\), there exists \(y \in (x,b]\) such that \([x,y]\) can be covered by at most two elements of \(\mathcal{A}\). Let \(x \in [a,b)\). Pick \(U \in \mathcal{A}\) such that \(x \in U\). Pick \(y > x\) such that \([x,y) \subseteq U\). Assume w.l.o.g. \(y \leq b\). Pick \(V \in \mathcal{A}\) such that \(y \in V\). \([x,y]\) is covered by \(U\) and \(V\). Let \(C = \{y \in (a,b] \mid [a,y] \text{ can be covered by finitely many elements of } \mathcal{A} \}\). Let \(c = \sup C\). \(C\) is nonempty. Pick \(y \in (a,b]\) such that \([a,y]\) can be covered by at most two elements of \(\mathcal{A}\). \(\...
Read more

J. Munkres. Topology (2013) Chapter 4: Countability and Separation Axioms. 31: The Separation Axioms

Definition A topological space \(X\) is regular iff it is \(T_1\) and, for any point \(x \in X\) and closed set \(B\) such that \(x \notin B\), there exist disjoint open sets \(U\) and \(V\) with \(x \in U\) and \(B \subseteq V\). A topological space \(X\) is normal iff it is \(T_1\) and, for any disjoint closed sets \(A\) and \(B\), there exist disjoint sets \(U\) and \(V\) with \(A \subseteq U\) and \(B \subseteq V\). Lemma 31.1 Let \(X\) be a \(T_1\) space. \(X\) is regular if and only if, for every point \(x \in X\) and neighbourhood \(U\) of \(x\), there exists a neighbourhood \(V\) of \(x\) such that \(\overline{V} \subseteq U\). \(X\) is normal if and only if, for every closed set \(A\) and open set \(U\) including \(A\), there exists an open set \(V\) including \(A\) such that \(\overline{V} \subseteq U\). Proof If \(X\) is regular then, for every point \(x \in X\) and neighbourhood \(U\) of \(x\), there exists a neighbourhood \(V\) of \(x\) such th...
Read more