Construct nonnormal covering spaces of the Klein bottle by a Klein bottle and by a torus.
Solution
Goal: Construct explicit non-normal covering spaces of the Klein bottle \(K\) whose total spaces are a Klein bottle and a torus, by finding corresponding non-normal subgroups of \(\pi_1(K)\).
1. Presentation and algebraic structure of \(\pi_1(K)\):Proof:
1.1. The fundamental group of the Klein bottle has the presentation\[\pi_1(K) = \langle a, b \mid a b a^{-1} = b^{-1} \rangle.\]
1.2. Every element \(g \in \pi_1(K)\) can be written uniquely in normal form as \(g = a^m b^n\) for \(m, n \in \mathbb{Z}\).
1.3. Group multiplication follows the semidirect product rule \((a^m b^n)(a^p b^q) = a^{m+p} b^{(-1)^p n + q}\).
1.4. Conjugation of generators:\[b a b^{-1} = a (a^{-1} b a) b^{-1} = a b b^{-1} = a b^{-2} \quad \text{and} \quad a b a^{-1} = b^{-1}.\]
2. Non-normal covering of \(K\) by a Klein bottle:Proof:
2.1. Define the subgroup \(H_K = \langle a, b^3 \rangle \le \pi_1(K)\).
2.2. The elements of \(H_K\) are precisely \(\{a^m b^{3n} : m, n \in \mathbb{Z}\}\).
2.3. The generators satisfy \(a (b^3) a^{-1} = (a b a^{-1})^3 = (b^{-1})^3 = (b^3)^{-1}\), so\[H_K \cong \langle a, b^3 \mid a (b^3) a^{-1} = (b^3)^{-1} \rangle \cong \pi_1(K).\]
2.4. Since \(H_K \cong \pi_1(K)\), the corresponding connected covering space \(\widetilde{K}\) has \(\pi_1(\widetilde{K}) \cong \pi_1(K)\), so \(\widetilde{K}\) is a Klein bottle.
2.5. The index is \([\pi_1(K) : H_K] = 3\), so \(\widetilde{K} \to K\) is a 3-sheeted covering.
2.6. Non-normality check: Conjugate the generator \(a \in H_K\) by \(b \in \pi_1(K)\):\[b a b^{-1} = a b^{-2}.\]
2.7. The element \(a b^{-2}\) has \(b\)-exponent \(-2\). Since \(-2 \not\equiv 0 \pmod 3\), \(a b^{-2} \notin H_K\).
2.8. Thus \(b H_K b^{-1} \neq H_K\), so \(H_K\) is not a normal subgroup of \(\pi_1(K)\), and the 3-sheeted Klein bottle cover is non-normal.
3. Non-normal covering of \(K\) by a torus:Proof:
3.1. Define the subgroup \(H_T = \langle a^2 b, b^3 \rangle \le \pi_1(K)\).
3.2. The generators commute:\[(a^2 b) b^3 (a^2 b)^{-1} = a^2 b^4 b^{-1} a^{-2} = a^2 b^3 a^{-2} = (a^2 b a^{-2})^3 = (b^{(-1)^2})^3 = b^3.\]
3.3. Thus \(H_T \cong \mathbb{Z}^2\), so the covering space \(\widetilde{T}\) is a closed, connected, orientable surface with \(\pi_1(\widetilde{T}) \cong \mathbb{Z}^2\), which is the torus \(T^2\).
3.4. The elements of \(H_T\) are precisely \(\{(a^2 b)^m (b^3)^n = a^{2m} b^{m + 3n} : m, n \in \mathbb{Z}\}\).
3.5. The index is \([\pi_1(K) : H_T] = 6\), so \(\widetilde{T} \to K\) is a 6-sheeted covering.
3.6. Non-normality check: Conjugate the generator \(a^2 b \in H_T\) by \(a \in \pi_1(K)\):\[a (a^2 b) a^{-1} = a^2 (a b a^{-1}) = a^2 b^{-1}.\]
3.7. In the form \(a^{2m} b^{m + 3n}\), the element \(a^2 b^{-1}\) requires \(m = 1\) and \(1 + 3n = -1 \implies 3n = -2\), which has no integer solution \(n \in \mathbb{Z}\).
3.8. Thus \(a (a^2 b) a^{-1} = a^2 b^{-1} \notin H_T\), so \(a H_T a^{-1} \neq H_T\).
3.9. Therefore \(H_T\) is not a normal subgroup of \(\pi_1(K)\), and the 6-sheeted torus cover is non-normal.
4. Conclusion:Proof:The subgroup \(H_K = \langle a, b^3 \rangle\) gives a 3-sheeted non-normal Klein bottle covering space, and \(H_T = \langle a^2 b, b^3 \rangle\) gives a 6-sheeted non-normal torus covering space.