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Proposition 12.1.1. A quasi-ordering \( \leq \) on \( X \) is a well-quasi-ordering if and only if \( X \) contains neither an infinite antichain nor an infinite strictly decreasing sequence \( {x}_{0} > {x}_{1} > \ldots \)
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Proof. The forward implication is trivial. Conversely, let \( {x}_{0},{x}_{1},\ldots \) be any infinite sequence in \( X \) . Let \( K \) be the complete graph on \( \mathbb{N} = \) \( \{ 0,1,\ldots \} \) . Colour the edges \( {ij}\left( {i < j}\right) \) of \( K \) with three colours: green if \( {x}_{i} \leq {x}_{j} \), red if \( {x}_{i} > {x}_{j} \), and amber if \( {x}_{i},{x}_{j} \) are incomparable. By Ramsey’s theorem (9.1.2), \( K \) has an infinite induced subgraph whose edges all have the same colour. If there is neither an infinite antichain nor an infinite strictly decreasing sequence in \( X \), then this colour must be green, i.e. \( {x}_{0},{x}_{1},\ldots \) has an infinite subsequence in which every pair is good. In particular, the sequence \( {x}_{0},{x}_{1},\ldots \) is good.
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Yes
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Lemma 12.3.1. Let \( {t}_{1}{t}_{2} \) be any edge of \( T \) and let \( {T}_{1},{T}_{2} \) be the components of \( T - {t}_{1}{t}_{2} \), with \( {t}_{1} \in {T}_{1} \) and \( {t}_{2} \in {T}_{2} \) . Then \( {V}_{{t}_{1}} \cap {V}_{{t}_{2}} \) separates \( {U}_{1} \mathrel{\text{:=}} \mathop{\bigcup }\limits_{{t \in {T}_{1}}}{V}_{t} \) from \( {U}_{2} \mathrel{\text{:=}} \mathop{\bigcup }\limits_{{t \in {T}_{2}}}{V}_{t} \) in \( G \) (Fig. 12.3.2).
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Proof. Both \( {t}_{1} \) and \( {t}_{2} \) lie on every \( t - {t}^{\prime } \) path in \( T \) with \( t \in {T}_{1} \) and \( {t}^{\prime } \in {T}_{2} \) . Therefore \( {U}_{1} \cap {U}_{2} \subseteq {V}_{{t}_{1}} \cap {V}_{{t}_{2}} \) by (T3), so all we have to show is that \( G \) has no edge \( {u}_{1}{u}_{2} \) with \( {u}_{1} \in {U}_{1} \smallsetminus {U}_{2} \) and \( {u}_{2} \in {U}_{2} \smallsetminus {U}_{1} \) . If \( {u}_{1}{u}_{2} \) is such an edge, then by (T2) there is a \( t \in T \) with \( {u}_{1},{u}_{2} \in {V}_{t} \) . By the choice of \( {u}_{1} \) and \( {u}_{2} \) we have neither \( t \in {T}_{2} \) nor \( t \in {T}_{1} \), a contradiction.
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Yes
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Lemma 12.3.3. Suppose that \( G \) is an \( {MH} \) with branch sets \( {U}_{h} \) , \( h \in V\left( H\right) \) . Let \( f : V\left( G\right) \rightarrow V\left( H\right) \) be the map assigning to each vertex of \( G \) the index of the branch set containing it. For all \( t \in T \) let \( {W}_{t} \mathrel{\text{:=}} \left\{ {f\left( v\right) \mid v \in {V}_{t}}\right\} \), and put \( \mathcal{W} \mathrel{\text{:=}} {\left( {W}_{t}\right) }_{t \in T} \) . Then \( \left( {T,\mathcal{W}}\right) \) is a tree-decomposition of \( H \) .
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Proof. The assertions (T1) and (T2) for \( \left( {T,\mathcal{W}}\right) \) follow immediately from the corresponding assertions for \( \left( {T,\mathcal{V}}\right) \) . Now let \( {t}_{1},{t}_{2},{t}_{3} \in T \) be as in (T3), and consider a vertex \( h \in {W}_{{t}_{1}} \cap {W}_{{t}_{3}} \) of \( H \) ; we show that \( h \in {W}_{{t}_{2}} \) . By definition of \( {W}_{{t}_{1}} \) and \( {W}_{{t}_{3}} \), there are vertices \( {v}_{1} \in {V}_{{t}_{1}} \cap {U}_{h} \) and \( {v}_{3} \in {V}_{{t}_{3}} \cap {U}_{h} \) . Since \( {U}_{h} \) is connected in \( G \) and \( {V}_{{t}_{2}} \) separates \( {v}_{1} \) from \( {v}_{3} \) in \( G \) by Lemma 12.3.1, \( {V}_{{t}_{2}} \) has a vertex in \( {U}_{h} \) . By definition of \( {W}_{{t}_{2}} \) , this implies \( h \in {W}_{{t}_{2}} \) .
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Yes
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Lemma 12.3.4. Given a set \( W \subseteq V\left( G\right) \), there is either a \( t \in T \) such that \( W \subseteq {V}_{t} \), or there are vertices \( {w}_{1},{w}_{2} \in W \) and an edge \( {t}_{1}{t}_{2} \in T \) such that \( {w}_{1},{w}_{2} \) lie outside the set \( {V}_{{t}_{1}} \cap {V}_{{t}_{2}} \) and are separated by it in \( G \) .
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Proof. Let us orient the edges of \( T \) as follows. For each edge \( {t}_{1}{t}_{2} \in T \) , define \( {U}_{1},{U}_{2} \) as in Lemma 12.3.1; then \( {V}_{{t}_{1}} \cap {V}_{{t}_{2}} \) separates \( {U}_{1} \) from \( {U}_{2} \) . If \( {V}_{{t}_{1}} \cap {V}_{{t}_{2}} \) does not separate any two vertices of \( W \) that lie outside it, we can find an \( i \in \{ 1,2\} \) such that \( W \subseteq {U}_{i} \), and orient \( {t}_{1}{t}_{2} \) towards \( {t}_{i} \) .\n\nLet \( t \) be the last vertex of a maximal directed path in \( T \) ; we claim that \( W \subseteq {V}_{t} \) . Given \( w \in W \), let \( {t}^{\prime } \in T \) be such that \( w \in {V}_{{t}^{\prime }} \) . If \( {t}^{\prime } \neq t \) , then the edge \( e \) at \( t \) that separates \( {t}^{\prime } \) from \( t \) in \( T \) is directed towards \( t \) , so \( w \) also lies in \( {V}_{{t}^{\prime \prime }} \) for some \( {t}^{\prime \prime } \) in the component of \( T - e \) containing \( t \) . Therefore \( w \in {V}_{t} \) by (T3).
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Yes
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Lemma 12.3.8. Any set of vertices separating two covers of a bramble also covers that bramble.
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Proof. Since each set in the bramble is connected and meets both of the covers, it also meets any set separating these covers.
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No
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Every graph \( G \) has a lean tree-decomposition of width \( \operatorname{tw}\left( G\right) \) .
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There is now a short proof of Theorem 12.3.10; see the notes. The fact that this theorem gives us a useful property of minimum-width tree-decompositions 'for free' has made it a valuable tool wherever tree-decompositions are applied.
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No
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Proposition 12.3.11. \( G \) is chordal if and only if \( G \) has a tree-decomposition into complete parts.
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Proof. We apply induction on \( \left| G\right| \) . We first assume that \( G \) has a tree-decomposition \( \left( {T,\mathcal{V}}\right) \) such that \( G\left\lbrack {V}_{t}\right\rbrack \) is complete for every \( t \in T \) ; let us choose \( \left( {T,\mathcal{V}}\right) \) with \( \left| T\right| \) minimal. If \( \left| T\right| \leq 1 \), then \( G \) is complete and hence chordal. So let \( {t}_{1}{t}_{2} \in T \) be an edge, and for \( i = 1,2 \) define \( {T}_{i} \) and \( {G}_{i} \mathrel{\text{:=}} G\left\lbrack {U}_{i}\right\rbrack \) as in Lemma 12.3.1. Then \( G = {G}_{1} \cup {G}_{2} \) by (T1) and (T2), and \( V\left( {{G}_{1} \cap {G}_{2}}\right) = {V}_{{t}_{1}} \cap {V}_{{t}_{2}} \) by the lemma; thus, \( {G}_{1} \cap {G}_{2} \) is complete. Since \( \left( {{T}_{i},{\left( {V}_{t}\right) }_{t \in {T}_{i}}}\right) \) is a tree-decomposition of \( {G}_{i} \) into complete parts, both \( {G}_{i} \) are chordal by the induction hypothesis. (By the choice of \( \left( {T,\mathcal{V}}\right) \), neither \( {G}_{i} \) is a subgraph of \( G\left\lbrack {{V}_{{t}_{1}} \cap {V}_{{t}_{2}}}\right\rbrack = {G}_{1} \cap {G}_{2} \), so both \( {G}_{i} \) are indeed smaller than \( G \) .) Since \( {G}_{1} \cap {G}_{2} \) is complete, any induced cycle in \( G \) lies in \( {G}_{1} \) or in \( {G}_{2} \) and hence has a chord, so \( G \) too is chordal.\n\nConversely, assume that \( G \) is chordal. If \( G \) is complete, there is nothing to show. If not then, by Proposition 5.5.1, \( G \) is the union of smaller chordal graphs \( {G}_{1},{G}_{2} \) with \( {G}_{1} \cap {G}_{2} \) complete. By the induction hypothesis, \( {G}_{1} \) and \( {G}_{2} \) have tree-decompositions \( \left( {{T}_{1},{\mathcal{V}}_{1}}\right) \) and \( \left( {{T}_{2},{\mathcal{V}}_{2}}\right) \) into complete parts. By Lemma 12.3.5, \( {G}_{1} \cap {G}_{2} \) lies inside one of those parts in each case, say with indices \( {t}_{1} \in {T}_{1} \) and \( {t}_{2} \in {T}_{2} \) . As one easily checks, \( \left( {\left( {{T}_{1} \cup {T}_{2}}\right) + {t}_{1}{t}_{2},{\mathcal{V}}_{1} \cup {\mathcal{V}}_{2}}\right) \) is a tree-decomposition of \( G \) into complete parts.
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Yes
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Corollary 12.3.12. \( \operatorname{tw}\left( G\right) = \min \{ \omega \left( H\right) - 1 \mid G \subseteq H;H \) chordal \( \} \) .
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Proof. By Lemma 12.3.5 and Proposition 12.3.11, each of the graphs \( H \) considered for the minimum has a tree-decomposition of width \( \omega \left( H\right) - 1 \) . Every such tree-decomposition induced one of \( G \) by Lemma 12.3.2, so \( \operatorname{tw}\left( G\right) \leq \omega \left( H\right) - 1 \) for every \( H \) . Conversely, let us construct an \( H \) as above with \( \omega \left( H\right) - 1 \leq \operatorname{tw}\left( G\right) \) . Let \( \left( {T,\mathcal{V}}\right) \) be a tree-decomposition of \( G \) of width \( \operatorname{tw}\left( G\right) \) . For every \( t \in T \) let \( {K}_{t} \) denote the complete graph on \( {V}_{t} \), and put \( H \mathrel{\text{:=}} \mathop{\bigcup }\limits_{{t \in T}}{K}_{t} \) . Clearly, \( \left( {T,\mathcal{V}}\right) \) is also a tree-decomposition of \( H \) . By Proposition 12.3.11, \( H \) is chordal, and by Lemma 12.3.5, \( \omega \left( H\right) - 1 \) is at most the width of \( \left( {T,\mathcal{V}}\right) \) , i.e. at most \( \operatorname{tw}\left( G\right) \) .
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Yes
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Proposition 12.4.1. A graph property \( \mathcal{P} \) can be expressed by forbidden minors if and only if it is closed under taking minors.
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Proof. For the ’if’ part, note that \( \mathcal{P} = {\operatorname{Forb}}_{ \preccurlyeq }\left( \overline{\mathcal{P}}\right) \), where \( \overline{\mathcal{P}} \) is the \( \overline{\mathcal{P}} \) complement of \( \mathcal{P} \) .
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No
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Proposition 12.4.2. A graph has tree-width \( < 3 \) if and only if it has no \( {K}^{4} \) minor.
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Proof. By Lemma 12.3.5, we have \( \operatorname{tw}\left( {K}^{4}\right) \geq 3 \) . By Proposition 12.3.6, (12.3.5) therefore, a graph of tree-width \( < 3 \) cannot contain \( {K}^{4} \) as a minor. (12.3.6)\n\nConversely, let \( G \) be a graph without a \( {K}^{4} \) minor; we assume that (12.3.11) \( \left| G\right| \geq 3 \) . Add edges to \( G \) until the graph \( {G}^{\prime } \) obtained is edge-maximal without a \( {K}^{4} \) minor. By Proposition 7.3.1, \( {G}^{\prime } \) can be constructed recursively from triangles by pasting along \( {K}^{2}\mathrm{\;s} \) . By induction on the number of recursion steps and Lemma 12.3.5, every graph constructible in this way has a tree-decomposition into triangles (as in the proof of Proposition 12.3.11). Such a tree-decomposition of \( {G}^{\prime } \) has width 2, and by Lemma 12.3.2 it is also a tree-decomposition of \( G \) .
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Yes
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Given a graph \( H \), the graphs without an \( H \) minor have bounded tree-width if and only if \( H \) is planar.
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To prove Theorem 12.4.3 we have to show that forbidding any planar graph \( H \) as a minor bounds the tree-width of a graph. In fact, we only have to show this for the special cases when \( H \) is a grid, because every planar graph is a minor of some grid. (To see this, take a drawing of the graph, fatten its vertices and edges, and superimpose a sufficiently fine plane grid.) It thus suffices to show the following:\n\nTheorem 12.4.4. (Robertson & Seymour 1986)\n\nFor every integer \( r \) there is an integer \( k \) such that every graph of tree-width at least \( k \) has an \( r \times r \) grid minor.\n\nOur proof of Theorem 12.4.4 proceeds as follows. Let \( r \) be given, and let \( G \) be any graph of large enough tree-width (depending on \( r \) ). We first show that \( G \) contains a large family \( \mathcal{A} = \left\{ {{A}_{1},\ldots ,{A}_{m}}\right\} \) of disjoint connected vertex sets such that each pair \( {A}_{i},{A}_{j} \in \mathcal{A} \) can be linked in \( G \) by a family \( {\mathcal{P}}_{ij} \) of many disjoint \( {A}_{i} - {A}_{j} \) paths avoiding all the other sets in \( \mathcal{A} \) . We then consider all the pairs \( \left( {{\mathcal{P}}_{ij},{\mathcal{P}}_{{i}^{\prime }{j}^{\prime }}}\right) \) of these path families. If we can find a pair among these such that many of the paths in \( {\mathcal{P}}_{ij} \) meet many of the paths in \( {\mathcal{P}}_{{i}^{\prime }{j}^{\prime }} \), we shall think of the paths in \( {\mathcal{P}}_{ij} \) as horizontal and the paths in \( {\mathcal{P}}_{{i}^{\prime }{j}^{\prime }} \) as vertical and extract a subdivision of an \( r \times r \) grid from their union. (This will be the difficult part of the proof, because these paths will in general meet in a less orderly way than they do in a grid.) If not, then for every pair \( \left( {{\mathcal{P}}_{ij},{\mathcal{P}}_{{i}^{\prime }{j}^{\prime }}}\right) \) many of the paths in \( {\mathcal{P}}_{ij} \) avoid many of the paths in \( {\mathcal{P}}_{{i}^{\prime }{j}^{\prime }} \) . We can then select one path \( {P}_{ij} \in {\mathcal{P}}_{ij} \) from each family so that these selected paths are pairwise disjoint. Contracting each of the connected sets \( A \in \mathcal{A} \) will then give us a \( {K}^{m} \) minor in \( G \), which contains the desired \( r \times r \) grid if \( m \geq {r}^{2} \) .
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Yes
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Theorem 12.4.4. (Robertson & Seymour 1986)\n\nFor every integer \( r \) there is an integer \( k \) such that every graph of tree-width at least \( k \) has an \( r \times r \) grid minor.
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Our proof of Theorem 12.4.4 proceeds as follows. Let \( r \) be given, and let \( G \) be any graph of large enough tree-width (depending on \( r \) ). We first show that \( G \) contains a large family \( \mathcal{A} = \left\{ {{A}_{1},\ldots ,{A}_{m}}\right\} \) of disjoint connected vertex sets such that each pair \( {A}_{i},{A}_{j} \in \mathcal{A} \) can be linked in \( G \) by a family \( {\mathcal{P}}_{ij} \) of many disjoint \( {A}_{i} - {A}_{j} \) paths avoiding all the other sets in \( \mathcal{A} \) . We then consider all the pairs \( \left( {{\mathcal{P}}_{ij},{\mathcal{P}}_{{i}^{\prime }{j}^{\prime }}}\right) \) of these path families. If we can find a pair among these such that many of the paths in \( {\mathcal{P}}_{ij} \) meet many of the paths in \( {\mathcal{P}}_{{i}^{\prime }{j}^{\prime }} \), we shall think of the paths in \( {\mathcal{P}}_{ij} \) as horizontal and the paths in \( {\mathcal{P}}_{{i}^{\prime }{j}^{\prime }} \) as vertical and extract a subdivision of an \( r \times r \) grid from their union. (This will be the difficult part of the proof, because these paths will in general meet in a less orderly way than they do in a grid.) If not, then for every pair \( \left( {{\mathcal{P}}_{ij},{\mathcal{P}}_{{i}^{\prime }{j}^{\prime }}}\right) \) many of the paths in \( {\mathcal{P}}_{ij} \) avoid many of the paths in \( {\mathcal{P}}_{{i}^{\prime }{j}^{\prime }} \) . We can then select one path \( {P}_{ij} \in {\mathcal{P}}_{ij} \) from each family so that these selected paths are pairwise disjoint. Contracting each of the connected sets \( A \in \mathcal{A} \) will then give us a \( {K}^{m} \) minor in \( G \), which contains the desired \( r \times r \) grid if \( m \geq {r}^{2} \) .
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Yes
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Lemma 12.4.6. Let \( k \geq 2 \) be an integer. Let \( T \) be a tree of maximum degree \( \leq 3 \) and \( X \subseteq V\left( T\right) \) . Then \( T \) has a set \( F \) of edges such that every component of \( T - F \) has between \( k \) and \( {2k} - 1 \) vertices in \( X \), except that one such component may have fewer vertices in \( X \) .
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Proof. We apply induction on \( \left| X\right| \) . If \( \left| X\right| \leq {2k} - 1 \) we put \( F = \varnothing \) . So assume that \( \left| X\right| \geq {2k} \) . Let \( e \) be an edge of \( T \) such that some component \( {T}^{\prime } \) of \( T - e \) has at least \( k \) vertices in \( X \) and \( \left| {T}^{\prime }\right| \) is as small as possible. As \( \Delta \left( T\right) \leq 3 \), the end of \( e \) in \( {T}^{\prime } \) has degree at most two in \( {T}^{\prime } \), so the minimality of \( {T}^{\prime } \) implies that \( \left| {X \cap V\left( {T}^{\prime }\right) }\right| \leq {2k} - 1 \) . Applying the induction hypothesis to \( T - {T}^{\prime } \) we complete the proof.
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Yes
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Lemma 12.4.7. Let \( G \) be a bipartite graph with bipartition \( \{ A, B\} \) , \( \left| A\right| = a,\left| B\right| = b \), and let \( c \leq a \) and \( d \leq b \) be positive integers. Assume that \( G \) has at most \( \left( {a - c}\right) \left( {b - d}\right) /d \) edges. Then there exist \( C \subseteq A \) and \( D \subseteq B \) such that \( \left| C\right| = c \) and \( \left| D\right| = d \) and \( C \cup D \) is independent in \( G \) .
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Proof. As \( \parallel G\parallel \leq \left( {a - c}\right) \left( {b - d}\right) /d \), fewer than \( b - d \) vertices in \( B \) have more than \( \left( {a - c}\right) /d \) neighbours in \( A \) . Choose \( D \subseteq B \) so that \( \left| D\right| = d \) and each vertex in \( D \) has at most \( \left( {a - c}\right) /d \) neighbours in \( A \) . Then \( D \) sends a total of at most \( a - c \) edges to \( A \), so \( A \) has a subset \( C \) of \( c \) vertices without a neighbour in \( D \) .
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Yes
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Lemma 12.4.8. Every tree \( T \) of order at least \( r\left( {r - 1}\right) \) contains a good \( r \) -tuple of vertices.
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Proof. Pick a vertex \( x \in T \) . Then \( T \) is the union of its subpaths \( {xTy} \) , where \( y \) ranges over its leaves. Hence unless one of these paths has at least \( r \) vertices, \( T \) has at least \( \left| T\right| /\left( {r - 1}\right) \geq r \) leaves. Since any path of \( r \) vertices and any set of \( r \) leaves gives rise to a good \( r \) -tuple in \( T \), this proves the assertion.
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Yes
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For every \( n \in \mathbb{N} \) there exists a \( k \in \mathbb{N} \) such that every graph \( G \) not containing \( {K}^{n} \) as a minor has a tree-decomposition whose torsos are \( k \) -nearly embeddable in a surface in which \( {K}^{n} \) is not embeddable.
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Note that there are only finitely many surfaces in which \( {K}^{n} \) is not embeddable. The set of those surfaces in the statement of Theorem 12.4.11 could therefore be replaced by just two surfaces: the orientable and the non-orientable surface of maximum genus in this set. Note also that the separators \( {V}_{t} \cap {V}_{{t}^{\prime }} \) in the tree-decomposition of \( G \) (for edges \( t{t}^{\prime } \) of the decomposition tree) have bounded size, e.g. at most \( {2k} + n \) , because they induce complete subgraphs in the torsos and these are \( k \) - nearly embeddable in one of those two surfaces.\n\nWe remark that Theorem 12.4.11 has only a qualitative converse: graphs that admit a decomposition as described can clearly have a \( {K}^{n} \) minor, but there exists an integer \( r \) depending only on \( n \) such that none of them has a \( {K}^{r} \) minor.\n\nTheorem 12.4.11, as stated above, is true also for infinite graphs (Diestel & Thomas 1999). There are also structure theorems for excluding infinite minors, and we state two of these.
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Yes
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Theorem 12.5.1. (Robertson & Seymour 1986-2004)\n\nThe finite graphs are well-quasi-ordered by the minor relation \( \\preccurlyeq \) .
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We shall give a sketch of the proof of the graph minor theorem at the end of this section.
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No
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Corollary 12.5.3. For every surface \( S \) there exists a finite set of graphs \( {H}_{1},\ldots ,{H}_{n} \) such that a graph is embeddable in \( S \) if and only if it contains none of \( {H}_{1},\ldots ,{H}_{n} \) as a minor.
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The proof of Corollary 12.5.3 does not need the full strength of the minor theorem. We shall give a direct proof, which runs as follows. The main step is to prove that the graphs in \( {\mathcal{K}}_{\mathcal{P}\left( S\right) } \) do not contain arbitrarily large grids as minors (Lemma 12.5.4). Then their tree-width is bounded (Theorem 12.4.4), so \( {\mathcal{K}}_{\mathcal{P}\left( S\right) } \) is well-quasi-ordered (Theorem 12.3.7) and therefore finite.
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No
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For every surface \( S \) there exists an integer \( r \) such that no graph that is minimal with the property of not being embeddable in \( S \) contains \( {H}^{r} \) as a topological minor.
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Proof. Let \( G \) be a graph that cannot be embedded in \( S \) and is minimal with this property. Our proof will run roughly as follows. Since \( G \) is minimally not embeddable in \( S \), we can embed it in an only slightly larger surface \( {S}^{\prime } \). If \( G \) contains a very large \( {H}^{r} \) grid, then by Lemma B. 6 some large \( {H}^{m} \) subgrid will be flat in \( {S}^{\prime } \), that is, the union of its faces in \( {S}^{\prime } \) will be a disc \( {D}^{\prime } \). We then pick an edge \( e \) from the middle of this \( {H}^{m} \) grid and embed \( G - e \) in \( S \). Again by Lemma B. 6, one of the rings of our \( {H}^{m} \) will be flat in \( S \). In this ring we can embed the (planar) subgraph of \( G \) which our first embedding had placed in \( {D}^{\prime } \); note that this subgraph contains the edge \( e \). The rest of \( G \) can then be embedded in \( S \) outside this ring much as before, yielding an embedding of all of \( G \) in \( S \) (a contradiction).
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Yes
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Theorem 1.5. Let \( K \) be an oriented knot in (oriented) \( {S}^{3} \), and let \( X \) be its exterior. Then \( {H}_{1}\left( X\right) \) is canonically isomorphic to the integers \( \mathbb{Z} \) generated by the class of a simple closed curve \( \mu \) in \( \partial N \) that bounds a disc in \( N \) meeting \( K \) at one point. If \( C \) is an oriented simple closed curve in \( X \), then the homology class \( \left\lbrack C\right\rbrack \in {H}_{1}\left( X\right) \) is \( \operatorname{lk}\left( {C, K}\right) \) . Further, \( {H}_{3}\left( X\right) = {H}_{2}\left( X\right) = 0 \) .
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Proof. This result is true in any reasonable homology theory with integer coefficients; indeed, it follows at once from the relatively sophisticated theorem of Alexander duality. The following proof uses the Mayer-Vietoris theorem, which relates the homology of two spaces to that of their union and intersection. As it has been assumed that all links are piecewise linearly embedded, it is convenient to think of simplicial homology and to suppose that \( X \) and \( N \) are sub-complexes of some triangulation of \( {S}^{3} \) . Consider then the following Mayer-Vietoris exact sequence for \( X \) and the solid torus \( N \) that intersect in their common torus boundary:\n\n\[ {H}_{3}\left( X\right) \oplus {H}_{3}\left( N\right) \rightarrow {H}_{3}\left( {S}^{3}\right) \rightarrow \cdots \]\n\n\[ \cdots \rightarrow {H}_{2}\left( {X \cap N}\right) \rightarrow {H}_{2}\left( X\right) \oplus {H}_{2}\left( N\right) \rightarrow {H}_{2}\left( {S}^{3}\right) \rightarrow \cdots \]\n\n\[ \cdots \rightarrow {H}_{1}\left( {X \cap N}\right) \rightarrow {H}_{1}\left( X\right) \oplus {H}_{1}\left( N\right) \rightarrow {H}_{1}\left( {S}^{3}\right) \rightarrow \cdots . \]\n\nNow, \( {H}_{3}\left( X\right) \oplus {H}_{3}\left( N\right) = 0 \) . This is because any connected triangulated 3-manifold with non-empty boundary deformation retracts to some 2-dimensional subcomplex (just \
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Yes
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Theorem 1.7. Let \( L \) be an oriented link of \( n \) components in (oriented) \( {S}^{3} \) and let \( X \) be its exterior. Then \( {H}_{2}\left( X\right) = {\bigoplus }_{n - 1}\mathbb{Z} \) . Further, \( {H}_{1}\left( X\right) \) is canonically isomorphic to \( {\bigoplus }_{n}\mathbb{Z} \) generated by the homology classes of the meridians \( \left\{ {\mu }_{i}\right\} \) of the individual components of \( L \) .
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Proof. The proof of this is just an adaptation of that of the previous theorem. Here \( N \) is now a disjoint union of \( n \) solid tori. The map \( {H}_{3}\left( {S}^{3}\right) \rightarrow {H}_{2}\left( {X \cap N}\right) \) is the map \( \mathbb{Z} \rightarrow {\bigoplus }_{n}\mathbb{Z} \) that sends 1 to \( \left( {1,1,\ldots ,1}\right) \), implying that \( {H}_{2}\left( X\right) = {\bigoplus }_{n - 1}\mathbb{Z} \) . Now \( {H}_{1}\left( {N \cap X}\right) = {\bigoplus }_{2n}\mathbb{Z} \) and \( {H}_{1}\left( N\right) = {\bigoplus }_{n}\mathbb{Z} \), and the map \( {H}_{1}\left( {N \cap X}\right) \rightarrow \) \( {H}_{1}\left( N\right) \oplus {H}_{1}\left( X\right) \) is still an isomorphism, so \( {H}_{1}\left( X\right) = {\bigoplus }_{n}\mathbb{Z} \) . The argument about the generators is as before.
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Yes
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Theorem 2.4. For any two knots \( {K}_{1} \) and \( {K}_{2} \) ,\n\n\[ g\left( {{K}_{1} + {K}_{2}}\right) = g\left( {K}_{1}\right) + g\left( {K}_{2}\right) . \]
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Proof. Firstly, suppose that \( {K}_{1} \) and \( {K}_{2} \), together with minimal genus Seifert surfaces \( {F}_{1} \) and \( {F}_{2} \), are situated far apart in \( {S}^{3} \) . Each \( {F}_{i} \) is a connected surface with non-empty boundary, so elementary homology theory shows that \( {F}_{1} \cup {F}_{2} \) does not separate \( {S}^{3} \) . Thus one can choose an arc \( \alpha \) from a point in \( {K}_{1} \) to a point in \( {K}_{2} \) that meets \( {F}_{1} \cup {F}_{2} \) at no other point and that intersects once a 2 -sphere separating \( {K}_{1} \) from \( {K}_{2} \) . The union of \( {F}_{1} \cup {F}_{2} \) with a \
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No
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Corollary 2.8. A knot can be expressed as a finite sum of prime knots.
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Proof. If a knot is not prime, it can be expressed as the sum of two knots of smaller genus. Now use induction on the genus.
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No
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Theorem 2.10. Suppose that a knot \( K \) can be expressed as \( K = P + Q \), where \( P \) is a prime knot, and that \( K \) can also be expressed as \( K = {K}_{1} + {K}_{2} \) . Then either\n\n(a) \( {K}_{1} = P + {K}_{1}^{\prime } \) for some \( {K}_{1}^{\prime } \), and \( Q = {K}_{1}^{\prime } + {K}_{2} \), or\n\n(b) \( {K}_{2} = P + {K}_{2}^{\prime } \) for some \( {K}_{2}^{\prime } \), and \( Q = {K}_{1} + {K}_{2}^{\prime } \) .
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Proof. Let \( \sum \) be a 2-sphere in \( {S}^{3} \), meeting \( K \) transversely at two points, that demonstrates \( K \) as the sum \( {K}_{1} + {K}_{2} \) . The factorisation \( K = P + Q \) implies that there is a 3-ball \( B \) contained in \( {S}^{3} \) such that \( B \cap K \) is an arc \( \alpha \) (with \( K \) intersecting \( \partial B \) transversely at the two points \( \partial \alpha \) ) so that the ball-arc pair \( \left( {B,\alpha }\right) \) becomes, on gluing a trivial ball-arc pair to its boundary, the pair \( \left( {{S}^{3}, P}\right) \) . As in the proof of Theorem 2.4, it may be assumed, after small movements of \( \sum \), that \( \sum \) intersects \( \partial B \) transversely in a union of simple closed curves disjoint from \( K \) . The immediate aim will be to reduce \( \sum \cap \partial B \) . Note that if this intersection is empty, then \( B \) is contained in one of the two components of \( {S}^{3} - \sum \), and the result follows at once.\n\nAs \( \sum \cap K \) is two points, any oriented simple closed curve in \( \sum - K \) has linking number zero or \( \pm 1 \) with \( K \) . Amongst the components of \( \sum \cap \partial B \) that have zero linking number with \( K \) select a component that is innermost on \( \sum \) (with \( \sum \cap K \) considered \
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Yes
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Corollary 2.11. Suppose that \( P \) is a prime knot and that \( P + Q = {K}_{1} + {K}_{2} \) . Suppose also that \( P = {K}_{1} \) . Then \( Q = {K}_{2} \) .
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Proof. By Theorem 2.10, there are two possibilities. The first is that for some \( {K}_{1}^{\prime }, P + {K}_{1}^{\prime } = {K}_{1} = P \) and \( Q = {K}_{1}^{\prime } + {K}_{2} \) . But then the genus of \( {K}_{1}^{\prime } \) must be zero, so \( {K}_{1}^{\prime } \) is the unknot and so \( Q = {K}_{2} \) . The second possibility is that for some \( {K}_{2}^{\prime }, P + {K}_{2}^{\prime } = {K}_{2} \) and \( Q = {K}_{2}^{\prime } + {K}_{1} \) . But then \( Q = {K}_{2}^{\prime } + P = {K}_{2} \) .
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Yes
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Theorem 2.12. Up to ordering of summands, there is a unique expression for a knot \( K \) as a finite sum of prime knots.
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Proof. Suppose \( K = {P}_{1} + {P}_{2} + \cdots + {P}_{m} = {Q}_{1} + {Q}_{2} + \cdots + {Q}_{n} \), where the \( {P}_{i} \) and \( {Q}_{i} \) are all prime. By the theorem, \( {P}_{1} \) is a summand of \( {Q}_{1} \) or of \( {Q}_{2} + \) \( {Q}_{3} + \cdots + {Q}_{n} \), and if the latter, then it is a summand of one of the \( {Q}_{j} \) for \( j \geq 2 \) , by induction on \( n \) . Of course if \( {P}_{1} \) is a summand of \( {Q}_{j} \), then \( {P}_{1} = {Q}_{j} \) . By the corollary, \( {P}_{1} \) and \( {Q}_{j} \) may then be cancelled from both sides of the equation, and the result follows by induction on \( m \) . Note that this induction starts when \( m = 0 \) . Then \( n = 0 \) because the unknot cannot be expressed as a sum of non-trivial knots (again by consideration of genus).
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Yes
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Lemma 3.2. If a diagram is changed by a Type I Reidemeister move, its bracket polynomial changes in the following way:\n\n\[ \langle {\tau }_{0} - \rangle = - {A}^{3}\langle \frown \rangle ,\;\langle - \sigma \rangle = - {A}^{-3}\langle \frown \rangle . \]
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Proof.\n\n\[ \langle {\tau }^{ - }\rangle = A\langle \widehat{\sigma }\rangle + {A}^{-1}\langle \tau \rangle \]\n\n\[ = \left( {A\left( {-{A}^{-2} - {A}^{2}}\right) + {A}^{-1}}\right) \langle \frown \rangle \text{.} \]\n\nThat produces the first equation; the second follows in the same way.
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Yes
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Lemma 3.3. If a diagram \( D \) is changed by a Type II or Type III Reidemeister move, then \( \langle D\rangle \) does not change. That is,\n\n(i) \( \langle \) , \( > < \rangle = \langle > < \rangle \) , (ii) \( \langle z < < \rangle = \langle z < < \rangle \) .\n\nHence \( \langle D\rangle \) is invariant under regular isotopy of \( D \) .
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Proof. (i)\n\n\[ \langle > < > \rangle = A\langle > > < \rangle + {A}^{-1}\langle > < \rangle \]\n\n\[ = - {A}^{-2}\langle \rangle \langle \rangle + \langle > \langle \rangle + {A}^{-2}\langle \rangle \langle \rangle . \]\n\n(ii)\n\n\[ \langle x < y\rangle = A\langle x < y\rangle + {A}^{-1}\langle x > y\rangle \]\n\n\[ = A\langle > < \rangle + {A}^{-1}\langle > \subset \rangle \]\n\n\[ = \langle > \leq < \rangle \text{. } \]\n\nHere the second line follows from the first by using (i) twice.
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No
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Theorem 3.5. Let \( D \) be a diagram of an oriented link \( L \) . Then the expression\n\n\[ \n{\left( -A\right) }^{-{3w}\left( D\right) }\langle D\rangle \n\]\n\nis an invariant of the oriented link \( L \) .
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Proof. It follows from Lemma 3.3 that the given expression is unchanged by Reidemeister moves of Types II and III; Lemma 3.2 and the above remarks on \( w\left( D\right) \) show it is unchanged by a Type I move. As any two diagrams of two equivalent links are related by a sequence of such moves, the result follows at once.
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Yes
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Proposition 3.7. The Jones polynomial invariant is a function\n\n\[ V : \\left\\{ {\\text{ Oriented links in }{S}^{3}}\\right\\} \\rightarrow \\mathbb{Z}\\left\\lbrack {{t}^{-1/2},{t}^{1/2}}\\right\\rbrack \]\n\n such that\n\n(i) \( V \) (unknot) \( = 1 \) ,\n\n(ii) whenever three oriented links \( {L}_{ + },{L}_{ - } \) and \( {L}_{0} \) are the same, except in the neighbourhood of a point where they are as shown in Figure 3.2, then\n\n\[ {t}^{-1}V\\left( {L}_{ + }\\right) - {tV}\\left( {L}_{ - }\\right) + \\left( {{t}^{-1/2} - {t}^{1/2}}\\right) V\\left( {L}_{0}\\right) = 0. \]
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Proof.\n\n\[ \\langle X\\rangle = A\\langle X\\rangle + {A}^{-1}\\langle X\\rangle \]\n\n\[ \\langle X\\rangle = {A}^{-1}\\langle X\\rangle + A\\langle X\\rangle . \]\n\nMultiplying the first equation by \( A \), the second by \( {A}^{-1} \), and subtracting gives\n\n\[ A\\langle > < \\rangle - {A}^{-1}\\langle > < \\rangle = \\left( {{A}^{2} - {A}^{-2}}\\right) \\langle \\rangle (\\rangle . \]\n\nThus, for the oriented links with diagrams as shown, using the fact that in those diagrams \( w\\left( {L}_{ + }\\right) - 1 = w\\left( {L}_{0}\\right) = w\\left( {L}_{ - }\\right) + 1 \), it follows that\n\n\[ - {A}^{4}V\\left( {L}_{ + }\\right) + {A}^{-4}V\\left( {L}_{ - }\\right) = \\left( {{A}^{2} - {A}^{-2}}\\right) V\\left( {L}_{0}\\right) . \]\n\nThe substitution \( {t}^{1/2} = {A}^{-2} \) gives the required answer.
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Yes
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Lemma 4.5. Let \( D \) be a non-split diagram for \( L \) . Suppose that \( F \) is a 2-sphere with the property that it separates the components of \( L \) ; then \( F \) can be replaced by another 2-sphere with the same property that is in standard position.
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Proof. (a) Suppose that \( C \) is amongst the \( n \) components of \( F \cap {S}_{ + } \) that do not bound disc components of \( F \cap {B}_{ + } \) . Choose \( C \) to be innermost on \( {S}_{ + } \) amongst such components. Then \( C \) is the boundary of a disc \( \Delta \) in \( {S}_{ + } \), and any component of \( F \cap {S}_{ + } \) contained in the interior of \( \Delta \) does bound a disc of \( F \cap {B}_{ + } \) . Thus if \( {\Delta }^{\prime } \) denotes a copy of \( \Delta \) displaced into \( {B}_{ + },{\Delta }^{\prime } \) can be chosen so that \( {\Delta }^{\prime } \cap F = \partial {\Delta }^{\prime } \) , \( \partial {\Delta }^{\prime } \) being a copy of \( C \) displaced along \( F \) into \( {B}_{ + } \) . Now \( \partial {\Delta }^{\prime } \) separates the sphere \( F \) into two discs \( {E}_{1} \) and \( {E}_{2} \) . Then \( {\Delta }^{\prime } \cup {E}_{1} \) or \( {\Delta }^{\prime } \cup {E}_{2} \) separates the components of \( L \) (because \( F \) did so). Let this new sphere be \( {F}^{\prime } \) . Then \( {F}^{\prime } \cap {S}_{ + } \) has fewer than \( n \) components not bounding discs in \( {F}^{\prime } \cap {B}_{ + } \), for either \( C \) is no longer part of that intersection or, if \( C \) is still present, \( C \) now bounds a disc. Furthermore, \( \left( {{F}^{\prime } \cap {B}_{ - }}\right) \subset \left( {F \cap {B}_{ - }}\right) \) . Thus, by repeating this, it may be assumed that \( F \) satisfies condition (A).\n\n(b) Let \( H \) be the upper hemisphere of the boundary of a bubble. \( H \) is a disc in \( {S}_{ + } \) that meets \( L \) in one over-pass arc and meets \( F \) in disjoint arcs all parallel to the over-pass. Let \( \delta \) be a diameter of \( H \) that intersects each of these arcs transversely at one point. The components of \( F \cap {S}_{ + } \) are disjoint simple closed curves on the sphere \( {S}_{ + } \) . If \( \delta \) meets one of these components at more than one point, then \( \delta \) must meet some such component at two points of \( \delta \cap F \cap {S}_{ + } \) that are consecutive along \( \delta \) . (This follows by considering the \
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Yes
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Lemma 5.4. Let \( D \) be a link diagram with \( n \) crossings. Then\n\n(i) \( M\langle D\rangle \leq n + 2\left| {{s}_{ + }D}\right| - 2 \), with equality if \( D \) is plus-adequate, and\n\n(ii) \( m\langle D\rangle \geq - n - 2\left| {{s}_{ - }D}\right| + 2 \), with equality if \( D \) is minus-adequate.
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Proof. (This is due, essentially, to Kauffman.) For any state \( s \) for \( D \) let\n\n\[ \langle D \mid s\rangle = {A}^{\mathop{\sum }\limits_{{i = 1}}^{n}s\left( i\right) }{\left( -{A}^{-2} - {A}^{2}\right) }^{\left| {sD}\right| - 1}, \]\n\nso that \( \langle D\rangle = \mathop{\sum }\limits_{s}\langle D \mid s\rangle \) . As \( \mathop{\sum }\limits_{{i = 1}}^{n}{s}_{ + }\left( i\right) = n \), it follows that \( M\left\langle {D \mid {s}_{ + }}\right\rangle = \) \( n + 2\left| {{s}_{ + }D}\right| - 2 \) . Now any state \( s \) can be achieved by starting with \( {s}_{ + } \) and changing, one at a time, the value of \( {s}_{ + } \) on selected integers that label the crossings. In other words, there exist states \( {s}_{0},{s}_{1},{s}_{2},\ldots ,{s}_{k} \) with \( {s}_{0} = {s}_{ + },{s}_{k} = s \) and \( {s}_{r - 1}\left( i\right) = {s}_{r}\left( i\right) \) for all \( i \in \{ 1,2,\ldots n\} \) except for a single integer \( {i}_{r} \) for which \( {s}_{r - 1}\left( {i}_{r}\right) = 1 \) and \( {s}_{r}\left( {i}_{r}\right) = - 1 \) . Then \( \mathop{\sum }\limits_{{i = 1}}^{n}{s}_{r}\left( i\right) = n - {2r} \) and, because \( {s}_{r - 1}D \) and \( {s}_{r}D \) are the same diagram except near one crossing of \( D,\left| {{s}_{r}D}\right| = \left| {{s}_{r - 1}D}\right| \pm 1 \) . Hence \( M\left\langle {D \mid {s}_{r - 1}}\right\rangle - M\left\langle {D \mid {s}_{r}}\right\rangle \) is 0 or 4 . Thus \( M\left\langle {D \mid {s}_{r}}\right\rangle \leq M\left\langle {D \mid {s}_{r - 1}}\right\rangle \), and so, for all \( s \) , it follows that\n\n\[ M\langle D \mid s\rangle \leq n + 2\left| {{s}_{ + }D}\right| - 2. \]\n\nIf \( D \) is plus-adequate, it is immediate that \( \left| {{s}_{1}D}\right| = \left| {{s}_{ + }D}\right| - 1 \), so that \( M\left\langle {D \mid {s}_{r}}\right\rangle \) decreases at the first step, when \( r \) changes from 0 to 1, and never rises thereafter. Thus \( M\langle D \mid s\rangle < n + 2\left| {{s}_{ + }D}\right| - 2 \) when \( s \neq {s}_{ + } \) . Hence, in summing to achieve \( \langle D\rangle \), the maximal degree term of \( \left\langle {D \mid {s}_{ + }}\right\rangle \) is never cancelled by a term from \( \langle D \mid s\rangle \) for any \( s \) . The second statement of the lemma is really just the reflection of the first; its proof can be achieved by applying the above to \( \bar{D} \) .
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Yes
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Lemma 5.6. Let \( D \) be a connected link diagram with \( {n}_{ \cdot } \) crossings. Then\n\n\[ \left| {{s}_{ + }D}\right| + \left| {{s}_{ - }D}\right| \leq n + 2 \]
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Proof. Use induction on \( n \) . The result is clearly true when \( n = 0 \) ; suppose it to be true for diagrams with \( n - 1 \) crossings. Select a crossing of \( D \) . For at least one of the two ways of replacing the crossing with two segments that do not cross, the resulting diagram \( {D}^{\prime } \) is connected. Suppose, with no loss of generality, that this is achieved by the positive way. Then \( {s}_{ + }D = {s}_{ + }{D}^{\prime } \) and \( \left| {{s}_{ - }D}\right| = \left| {{s}_{ - }{D}^{\prime }}\right| \pm 1 \) . Thus, using the induction hypothesis,\n\n\[ \left| {{s}_{ + }D}\right| + \left| {{s}_{ - }D}\right| = \left| {{s}_{ + }{D}^{\prime }}\right| + \left| {{s}_{ - }{D}^{\prime }}\right| \pm 1 \leq \left( {n - 1}\right) + 2 \pm 1 \leq n + 2. \]
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Yes
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Lemma 5.7. Let \( D \) be a connected \( n \) -crossing diagram.\n\n(i) If \( D \) is alternating, then \( \left| {{s}_{ + }D}\right| + \left| {{s}_{ - }D}\right| = n + 2 \).
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Proof. When \( D \) is alternating, \( \left| {{s}_{ + }D}\right| + \left| {{s}_{ - }D}\right| \) is the number of planar regions in the complement of \( D \) (as \( \left| {{s}_{ + }D}\right| \) is the number of black regions, \( \left| {{s}_{ - }D}\right| \) the number of white regions in a chessboard colouring). However, \( D \) is a four-valent planar graph, so consideration of the Euler number of the sphere shows that the number of regions is \( n + 2 \) (for the number of edges is \( {2n} \) ). Hence \( \left| {{s}_{ + }D}\right| + \left| {{s}_{ - }D}\right| = n + 2 \).
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Yes
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Theorem 5.9. Let \( D \) be a connected, \( n \) -crossing diagram of an oriented link \( L \) with Jones polynomial \( V\left( L\right) \) . Then\n\n(i) \( B\left( {V\left( L\right) }\right) \leq n \) ;\n\n(ii) if \( D \) is alternating and reduced, then \( B\left( {V\left( L\right) }\right) = n \) ;\n\n(iii) if \( D \) is non-alternating and a prime diagram, then \( B\left( {V\left( L\right) }\right) < n \) .
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Proof. Recall that under the substitution \( t = {A}^{-4} \) the Jones polynomial is given by \( V\left( L\right) = {\left( -A\right) }^{-{3w}\left( D\right) }\langle D\rangle \), so that \( {4B}\left( {V\left( L\right) }\right) = B\langle D\rangle = M\langle D\rangle - m\langle D\rangle \) (where \( M\langle D\rangle \) and \( m\langle D\rangle \) refer to powers of \( A \) ). Hence, by Lemmas 5.4 and 5.6,\n\n\[ \n{4B}\left( {V\left( L\right) }\right) \leq {2n} + 2\left| {{s}_{ + }D}\right| + 2\left| {{s}_{ - }D}\right| - 4 \leq {4n}.\n\]\n\nBut if \( D \) is alternating and reduced, then it is adequate, and the inequalities of Lemma 5.4 are then equalities. Then the first part of Lemma 5.7 implies that \( {4B}\left( {V\left( L\right) }\right) = {4n} \) . When \( D \) is prime and non-alternating, any diagram summand that is a non-trivial diagram of the unknot makes no contribution to the Jones polynomial but does contribute to the number of crossings. Thus, without loss of generality, it may be assumed that \( D \) is strongly prime. Then the strict inequality of Lemma 5.7 produces the required result.
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Yes
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Corollary 5.10. If a link \( L \) has a connected, reduced, alternating diagram of \( n \) crossings, then it has no diagram of less than n crossings; any non-alternating prime diagram for \( L \) has more than \( n \) crossings.
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Proof. The existence of the reduced alternating diagram for \( L \) implies, using Theorem 5.9 (ii), that \( B\left( {V\left( L\right) }\right) = n \) . If \( L \) has another diagram of \( m \) crossings, then Theorem 5.9 (i) implies that \( n = B\left( {V\left( L\right) }\right) \leq m \) . If this second diagram is non-alternating, then, by Theorem 5.9 (iii), \( n = B\left( {V\left( L\right) }\right) < m \) .
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Yes
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Lemma 5.12. If \( D \) is plus-adequate, then \( {D}^{r} \) is plus-adequate; if \( D \) is minus-adequate, then \( {D}^{r} \) is minus-adequate.
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Proof. The result is immediate, because \( {s}_{ + }\left( {D}^{r}\right) = {\left( {s}_{ + }D\right) }^{r} \) ; see Figure 5.4. If \( D \) is plus-adequate, no component of \( {s}_{ + }\left( {D}^{r}\right) \) abuts itself at a former crossing, as it runs parallel to a component of \( {s}_{ + }D \) which, itself, has that property.
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Yes
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Theorem 5.13. Let \( D \) and \( E \) be diagrams, with \( {n}_{D} \) and \( {n}_{E} \) crossings respectively, for the same oriented link \( L \). Suppose that \( D \) is plus-adequate; then\n\n\[ \n{n}_{D} - w\left( D\right) \leq {n}_{E} - w\left( E\right) \n\]
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Proof. Let \( \left\{ {L}_{i}\right\} \) be the components of \( L \), and let \( {D}_{i} \) and \( {E}_{i} \) be the subdiagrams of \( D \) and \( E \) corresponding to \( {L}_{i} \). Choose non-negative integers \( {\mu }_{i} \) and \( {v}_{i} \) such that for each \( i, w\left( {D}_{i}\right) + {\mu }_{i} = w\left( {E}_{i}\right) + {v}_{i} \). Change \( D \) to \( {D}_{ * } \) by changing each \( {D}_{i} \) to \( {D}_{*i} \) by adding to \( {D}_{i} \) a total of \( {\mu }_{i} \) positive kinks. Similarly, change \( E \) to \( {E}_{ * } \) by adding \( {v}_{i} \) positive kinks to \( {E}_{i} \) for each \( i \). Note that \( {D}_{ * } \) is still plus-adequate, \( w\left( {D}_{*i}\right) = \) \( w\left( {E}_{*i}\right) \), and \( w\left( {D}_{ * }\right) = w\left( {E}_{ * }\right) \), because the sum of the signs of crossings of distinct components is determined by the linking numbers of components of \( L \). Now \( {D}_{ * }^{r} \) and \( {E}_{ * }^{r} \) are diagrams of the same link, namely \( L \) with each \( {L}_{i} \) replaced by \( r \) copies with mutual linking number \( w\left( {D}_{*i}\right) \). Thus they have the same Jones polynomial. But they have the same writhe (namely, \( {r}^{2}w\left( {D}_{ * }\right) \) ), and so \( \left\langle {D}_{ * }^{r}\right\rangle = \left\langle {E}_{ * }^{r}\right\rangle \). Now by\n\nLemma 5.4,\n\n\[ \nM\left\langle {E}_{ * }^{r}\right\rangle \leq \left( {{n}_{E} + \mathop{\sum }\limits_{i}{v}_{i}}\right) {r}^{2} + 2\left( {\left| {{s}_{ + }E}\right| + \mathop{\sum }\limits_{i}{v}_{i}}\right) r - 2, \n\]\n\n\[ \nM\left\langle {D}_{ * }^{r}\right\rangle = \left( {{n}_{D} + \mathop{\sum }\limits_{i}{\mu }_{i}}\right) {r}^{2} + 2\left( {\left| {{s}_{ + }D}\right| + \mathop{\sum }\limits_{i}{\mu }_{i}}\right) r - 2, \n\]\n\nthe equality occurring since \( {D}_{ * }^{r} \) is plus-adequate. This is true for all \( r \), so, comparing coefficients of \( {r}^{2} \), \n\n\[ \n{n}_{D} + \mathop{\sum }\limits_{i}{\mu }_{i} \leq {n}_{E} + \mathop{\sum }\limits_{i}{v}_{i} \n\]\n\nso that \( {n}_{D} - \mathop{\sum }\limits_{i}w\left( {D}_{i}\right) \leq {n}_{E} - \mathop{\sum }\limits_{i}w\left( {E}_{i}\right) \). Hence, once again using the fact that the sum of the signs of crossings of distinct components is determined by linking numbers of \( L,{n}_{D} - w\left( D\right) \leq {n}_{E} - w\left( E\right) \).
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Yes
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Corollary 5.14. Let \( D \) and \( E \) be as above.\n\n(i) The number of negative crossings of \( D \) is less than or equal to the number of negative crossings of \( E \) .\n\n(ii) The number of positive crossings in a minus-adequate diagram is minimal.\n\n(iii) An adequate diagram has the minimal number of crossings.\n\n(iv) Two adequate diagrams of the same link (e.g. reduced alternating diagrams) have the same writhe.
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The corollary is just restating the theorem in different ways. An example of the use of the corollary is the two famous diagrams (the Perko pair), originally labelled \( {10}_{161} \) and \( {10}_{162} \), shown in Figure 3.1. The diagrams \( {10}_{161} \) and \( \overline{{10}_{162}} \) represent the same knot. Observe that \( w\left( {10}_{161}\right) = - 8 \) and \( w\left( \overline{{10}_{162}}\right) = - {10} \) . Inspection of the diagrams shows that \( \overline{{10}_{162}} \) is minus-adequate, the minimal number possible of positive crossings being zero. However, \( {10}_{161} \) is plus-adequate, and so any diagram must have at least nine negative crossings. As \( {10}_{161} \) has no diagram of less than ten crossings (from the classification tables), it is impossible to display the minimal number of positive crossings and the minimal number of negative crossings on the same diagram, and the two minima are achieved by the two given diagrams.
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Yes
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Theorem 6.1. Any two presentation matrices \( A \) and \( {A}_{1} \) for \( M \) differ by a sequence of matrix moves of the following forms and their inverses:\n\n(i) Permutation of rows or columns;\n\n(ii) Replacement of the matrix \( A \) by \( \left( \begin{array}{ll} A & 0 \\ 0 & 1 \end{array}\right) \) ;\n\n(iii) Addition of an extra column of zeros to the matrix \( A \) ;\n\n(iv) Addition of a scalar multiple of a row (or column) to another row (or column).
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Proof. Suppose that the matrices \( A \) and \( {A}_{1} \) correspond, with respect to some bases, to the maps \( \alpha \) and \( {\alpha }_{1} \) in the following presentations:\n\n\[ F\overset{\alpha }{ \rightarrow }E\overset{\phi }{ \rightarrow }M \rightarrow 0 \]\n\n\[ \downarrow \gamma \; \downarrow \beta \; \updownarrow 1 \]\n\n\[ {F}_{1}\overset{{\alpha }_{1}}{ \rightarrow }{E}_{1}\overset{{\phi }_{1}}{ \rightarrow }M \rightarrow 0 \]\n\nThe free base of \( E \) and the surjectivity of \( {\phi }_{1} \) can be used to construct a linear map \( \beta : E \rightarrow {E}_{1} \) so that \( {\phi }_{1}\beta = \phi \) . Similarly, the freeness of \( F \) and exactness at \( E \) and \( {E}_{1} \) produce a map \( \gamma : F \rightarrow {F}_{1} \) such that \( {\beta \alpha } = {\alpha }_{1}\gamma \) . If then \( \beta \) and \( \gamma \) are represented by matrices \( B \) and \( C \) with respect to the given bases, then \( {BA} = {A}_{1}C \) . A completely symmetrical argument produces maps \( {\beta }_{1} \) and \( {\gamma }_{1} \) with matrices \( {B}_{1} \) and \( {C}_{1} \) such that \( {B}_{1}{A}_{1} = A{C}_{1} \) . Letting \
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Yes
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Proposition 6.3. Suppose that \( F \) is a connected, compact, orientable surface with non-empty boundary, piecewise linearly contained in \( {S}^{3} \) . Then the homology groups \( {H}_{1}\left( {{S}^{3} - F;\mathbb{Z}}\right) \) and \( {H}_{1}\left( {F;\mathbb{Z}}\right) \) are isomorphic, and there is a unique nonsingular bilinear form
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Proof. The surface \( F \) is now embedded in \( {S}^{3} \) . As before, \( {H}_{1}\left( {F;\mathbb{Z}}\right) = \) \( {\bigoplus }_{{2g} + n - 1}\mathbb{Z} \) generated by \( \left\{ \left\lbrack {f}_{i}\right\rbrack \right\} \) . Let \( V \) be a regular neighbourhood of \( F \) in \( {S}^{3} \), so that \( V \) is just a 3-ball with \( \left( {{2g} + n - 1}\right) 1 \) -handles attached. The inclusion of \( F \) in \( V \) is a homotopy equivalence, and \( {H}_{1}\left( {\partial V;\mathbb{Z}}\right) = \left( {{\bigoplus }_{{2g} + n - 1}\mathbb{Z}}\right) \oplus \left( {{\bigoplus }_{{2g} + n - 1}\mathbb{Z}}\right) \) . For this, generators \( \left\{ {\left\lbrack {f}_{i}^{\prime }\right\rbrack : 1 \leq i \leq {2g} + n - 1}\right\} \) and \( \left\{ {\left\lbrack {e}_{i}\right\rbrack : 1 \leq i \leq {2g} + n - 1}\right\} \) can be chosen so that each \( {e}_{i} \) is the boundary of a small disc in \( V \) that meets \( {f}_{i} \) at one point, and the inclusion \( \partial V \subset V \) induces on homology a map sending \( \left\lbrack {f}_{i}^{\prime }\right\rbrack \) to \( \left\lbrack {f}_{i}\right\rbrack \) and \( \left\lbrack {e}_{i}\right\rbrack \) to zero. Furthermore, the orientations of the \( \left\{ {e}_{i}\right\} \) can be chosen so that \( \operatorname{lk}\left( {{e}_{i},{f}_{j}}\right) = {\delta }_{ij} \) (the Krönecker delta). This all relates to the homology of the standard inclusion of \( F \) in a standard handlebody \( V \) ; it is \( {S}^{3} - F \) that is of interest. Now, if \( {V}^{\prime } \) is the closure of \( {S}^{3} - V \), then the inclusion of \( {V}^{\prime } \) in \( {S}^{3} - F \) is a homotopy equivalence. The Mayer-Vietoris theorem for \( {S}^{3} \) expressed as the union of \( V \) and \( {V}^{\prime } \) asserts that the following sequence is exact:
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Yes
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Theorem 6.5. Let \( F \) be a Seifert surface for an oriented link \( L \) in \( {S}^{3} \) and let \( A \) be a matrix, with respect to any basis of \( {H}_{1}\left( {F;\mathbb{Z}}\right) \), for the corresponding Seifert form. Then \( {tA} - {A}^{\tau } \) is a matrix that presents the \( \mathbb{Z}\left\lbrack {{t}^{-1}, t}\right\rbrack \) -module \( {H}_{1}\left( {{X}_{\infty };\mathbb{Z}}\right) \) .
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Proof. Express \( {X}_{\infty } \) as the union of subspaces \( {Y}^{\prime } \) and \( {Y}^{\prime \prime } \), where \( {Y}^{\prime } = \mathop{\bigcup }\limits_{i}{Y}_{{2i} + 1} \) and \( {Y}^{\prime \prime } = \mathop{\bigcup }\limits_{i}{Y}_{2i} \) . Each of these subspaces is the disjoint union of countably many copies of \( Y \), and their intersection is the union of countably many copies of \( F \) . The homology of \( {X}_{\infty } \) will now be investigated, using the Mayer-Vietoris theorem, in terms of the homology of \( {Y}^{\prime } \) and \( {Y}^{\prime \prime } \) . The Mayer-Vietoris long exact sequence of homology groups comes from a short exact sequence of chain complexes in a standard way. In this case the exact sequence of chain complexes is the following (where \( {C}_{n} \) is the \( {n}^{th} \) chain group):\n\n\[ 0 \rightarrow {C}_{n}\left( {{Y}^{\prime } \cap {Y}^{\prime \prime }}\right) \overset{{\alpha }_{n}}{ \rightarrow }{C}_{n}\left( {Y}^{\prime }\right) \oplus {C}_{n}\left( {Y}^{\prime \prime }\right) \overset{{\beta }_{n}}{ \rightarrow }{C}_{n}\left( {X}_{\infty }\right) \rightarrow 0. \]\n\nNote that \( t \) interchanges \( {Y}^{\prime } \) and \( {Y}^{\prime \prime } \) so that the chain groups of these individual spaces are not modules over \( \mathbb{Z}\left\lbrack {{t}^{-1}, t}\right\rbrack \) ; however, each term in the above sequence is such a module. To achieve an exact sequence of homology modules, \( {\alpha }_{n} \) and \( {\beta }_{n} \) must be module maps with \( {\beta }_{n}{\alpha }_{n} = 0 \) . This is achieved if \( {\beta }_{n} \) is defined by \( {\beta }_{n}\left( {a, b}\right) = a + b \) and, for \( x \in {C}_{n}\left( {{Y}_{i - 1} \cap {Y}_{i}}\right) ,{\alpha }_{n} \) is defined by \( {\alpha }_{n}\left( x\right) = \left( {-x, x}\right) \in \) \( {C}_{n}\left( {Y}_{i - 1}\right) \oplus {C}_{n}\left( {Y}_{i}\right) \) . This short exact sequence of chain complexes of modules over \( \mathbb{Z}\left\lbrack {{t}^{-1}, t}\right\rbrack \) gives rise, in the usual way, to the following long exact sequence of homology modules:\n\n\[ \rightarrow {H}_{1}\left( {{Y}^{\prime } \cap {Y}^{\prime \prime };\mathbb{Z}}\right) \overset{{\alpha }_{ \star }}{ \rightarrow }{H}_{1}\left( {{Y}^{\prime };\mathbb{Z}}\right) \oplus {H}_{1}\left( {{Y}^{\prime \prime };\mathbb{Z}}\right) \overset{{\beta }_{ \star }}{ \rightarrow }{H}_{1}\left( {{X}_{\infty };\mathbb{Z}}\right) \rightarrow \]\n\n\[ \rightarrow {H}_{0}\left( {{Y}^{\prime } \cap {Y}^{\prime \prime };\mathbb{Z}}\right) \overset{{\alpha }_{ \star }}{ \rightarrow }{H}_{0}\left( {{Y}^{\prime };\mathbb{Z}}\right) \oplus {H}_{0}\left( {{Y}^{\prime \prime };\mathbb{Z}}\right) . \]\n\nNow \( F \) is, by definition of the term \
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Yes
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Theorem 6.10.\n\n(i) For any oriented link \( L,{\Delta }_{L}\left( t\right) \doteq {\Delta }_{L}\left( {t}^{-1}\right) \).\n\n(ii) For any (oriented) knot \( K,{\Delta }_{K}\left( 1\right) = \pm 1 \).
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Proof. (i) Suppose that \( A \) is an \( n \times n \) Seifert matrix for \( L \) . Then\n\n\[ \n{\Delta }_{L}\left( t\right) \doteq \det \left( {{tA} - {A}^{\tau }}\right) = \det \left( {t{A}^{\tau } - A}\right) = {\left( -t\right) }^{n}\det \left( {{t}^{-1}A - {A}^{\tau }}\right) \doteq {\Delta }_{L}\left( {t}^{-1}\right) .\n\]\n\n(ii) Let \( A \) be the Seifert matrix for \( K \) coming from a standard base of \( {2g} \) oriented curves \( \left\{ {f}_{i}\right\} \) on a genus \( g \) Seifert surface \( F \) as shown in Figure 6.1. Now, \( {\Delta }_{K}\left( 1\right) = \pm \det \left( {A - {A}^{\tau }}\right) \), but\n\n\[ \n{\left( A - {A}^{\tau }\right) }_{ij} = \operatorname{lk}\left( {{f}_{i}^{ - },{f}_{j}}\right) - \operatorname{lk}\left( {{f}_{i}^{ + },{f}_{j}}\right) ,\n\]\n\nand this is the algebraic number of intersections of \( {f}_{i} \) and \( {f}_{j} \) on the surface \( F \) . Hence \( \left( {A - {A}^{\tau }}\right) \) consists of \( g \) blocks of the form \( \left( \begin{matrix} 0 & 1 \\ - 1 & 0 \end{matrix}\right) \) down the diagonal and zeros elsewhere. The determinant of that is 1 .
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Yes
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Corollary 6.11. For any knot \( K \) ,\n\n\[{\Delta }_{K}\left( t\right) \doteq {a}_{0} + {a}_{1}\left( {{t}^{-1} + t}\right) + {a}_{2}\left( {{t}^{-2} + {t}^{2}}\right) + \cdots ,\]\n\nwhere the \( {a}_{i} \) are integers and \( {a}_{0} \) is odd.
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Proof. By Theorem 6.10(i), \( {\Delta }_{K}\left( t\right) \) can be written in the form \( {\Delta }_{K}\left( t\right) = {b}_{0} + \) \( {b}_{1}t + {b}_{2}{t}^{2} + \cdots + {b}_{N}{t}^{N} \), where \( {b}_{N - r} = \pm {b}_{r} \) with the same choice of sign for all \( r \) . If \( N \) were odd, \( {\Delta }_{K}\left( 1\right) \) would be even, which contradicts (ii) of the theorem. Hence \( N \) is even. If \( {b}_{N - r} = - {b}_{r} \) for all \( r \), then \( {b}_{N/2} = 0 \) and so \( {\Delta }_{K}\left( 1\right) = 0 \) , again a contradiction. Thus \( {b}_{N - r} = {b}_{r} \) for all \( r \) and \( {b}_{N/2} \) is odd, and so, within the indeterminacy of multiplication by units, \( {\Delta }_{K}\left( t\right) \) is of the required form.
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Yes
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Proposition 6.12. Let \( L \) be an oriented link. Then \( \bar{L} \) and \( \mathrm{r}L \), the reflection and the reverse of \( L \), have the same Alexander polynomial as \( L \) up to multiplication by units.
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Proof. If \( A \) is a Seifert matrix for \( L, - A \) is a Seifert matrix for \( \bar{L} \) and \( {A}^{\tau } \) is a Seifert matrix for \( \mathrm{r}L \) .
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Yes
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Proposition 6.13. If a knot \( K \) has genus \( g \), then \( {2g} \geq \) breadth \( {\Delta }_{K}\left( t\right) \) .
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Proof. Let \( F \) be a genus \( g \) Seifert surface for \( K \) . Then \( {tA} - {A}^{\tau } \) is a \( {2g} \times {2g} \) matrix, and so the degree in \( t \) of the polynomial \( \det \left( {{tA} - {A}^{\tau }}\right) \) is at most \( {2g} \) .
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Yes
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Proposition 6.14. Suppose an oriented link \( L \) bounds a disconnected oriented surface in \( {S}^{3} \) ; then \( {\Delta }_{L}\left( t\right) \) is the zero polynomial.
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Proof. Suppose \( \sum \) is a disconnected oriented surface with boundary \( L \) . Form a connected surface \( F \) by connecting the components of \( \sum \) together with thin \
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No
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In \( {S}^{3} \), let \( T \) be a standard, unknotted, solid torus that contains a knot \( K \). Let \( e : T \rightarrow {S}^{3} \) be an embedding of \( T \) onto a neighbourhood of a knot \( C \), so that e maps a longitude of \( T \) (coming from the inclusion of \( T \) in \( {S}^{3} \)) onto a longitude of \( C \). Then
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\[ {\Delta }_{eK}\left( t\right) \doteq {\Delta }_{K}\left( t\right) {\Delta }_{C}\left( {t}^{n}\right) \] where \( K \) represents \( n \) times a generator of \( {H}_{1}\left( T\right) \). Proof. Construct Seifert surfaces for the pattern knot \( K \) and the satellite \( {eK} \) in the following way: The unknotted solid torus \( T \) projects onto an annulus in the plane. Apply the Seifert method (Theorem 2.2) to the projection of \( K \), with some orientation, into this annulus. Seifert circuits in the annulus, connected by twisted strips at the crossings, are obtained. Cap off, with discs just above the annulus, any circuits that bound in the annulus; then use annuli to cap off adjacent pairs of curves that encircle the annulus in opposite directions. Add a vertical annulus to each remaining curve so that the result is an oriented surface \( F \) contained in \( T \), with \( \partial F \) being the union of \( K \) and \( n \) longitudes of \( T \) oriented in the same direction. A Seifert surface \( F \cup {nD} \) for \( K \) then consists of the union of \( F \) and \( n \) parallel copies of a spanning disc of \( T \). Similarly, a Seifert surface \( {eF} \cup {nG} \) for \( {eK} \) consists of the union of \( {eF} \) and \( n \) parallel copies of a genus \( g \) Seifert surface \( G \) of the companion knot \( C \) (this \( G \) being regarded as in the closure of \( {S}^{3} - {eT} \)). Note that if \( f \) is an oriented simple closed curve in \( T - K \), then \( \operatorname{lk}\left( {f, K}\right) = \) \( f \sqcap F \), where \
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Yes
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Theorem 6.17. Let \( K \) be a knot in \( {S}^{3} \) and let \( t : {X}_{\infty } \rightarrow {X}_{\infty } \) be the (covering) translation of \( {X}_{\infty } \) (the infinite cyclic cover of the exterior of \( K \) ). Then \( {H}_{1}\left( {{X}_{\infty };\mathbb{Q}}\right) \) is a finite-dimensional vector space over the field \( \mathbb{Q} \) . The characteristic polynomial of the linear map \( {t}_{ \star } : {H}_{1}\left( {{X}_{\infty };\mathbb{Q}}\right) \rightarrow {H}_{1}\left( {{X}_{\infty };\mathbb{Q}}\right) \) is, up to multiplication by a unit, equal to the Alexander polynomial of \( K \) .
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Proof. The ring \( \mathbb{Q}\left\lbrack {{t}^{-1}, t}\right\rbrack \) is a principal ideal domain. A proof of this, using the Euclidean algorithm, is much the same as the proof that shows the ring of ordinary polynomials over a field to be a principal ideal domain. Over \( \mathbb{Q}\left\lbrack {{t}^{-1}, t}\right\rbrack \) the module \( {H}_{1}\left( {{X}_{\infty };\mathbb{Q}}\right) \) is finitely presented by the matrix \( \left( {{tA} - {A}^{\tau }}\right) \) . However, over a principal ideal domain, any finitely presented module is just a direct sum of cyclic modules (see, for example, [38]). This is the same as saying that the module is presented by a square diagonal matrix. Thus \( {H}_{1}\left( {{X}_{\infty };\mathbb{Q}}\right) \) is presented by a matrix \( \operatorname{diag}\left( {{p}_{1},{p}_{2},\ldots ,{p}_{N}}\right) \), where \( {p}_{i} \in \mathbb{Q}\left\lbrack {{t}^{-1}, t}\right\rbrack \), and \( {H}_{1}\left( {{X}_{\infty };\mathbb{Q}}\right) \) is isomorphic as a module to \( {\bigoplus }_{i = 1}^{N}\left( {\mathbb{Q}\left\lbrack {{t}^{-1}, t}\right\rbrack /{p}_{i}}\right) \) . None of the \( {p}_{i} \) is zero, for then the Alexander polynomial, the determinant of the matrix, would be zero. However, for a knot \( K \) , \( {\Delta }_{K}\left( 1\right) = \pm 1 \) .\n\nConsider, then, a typical summand of the form \( \mathbb{Q}\left\lbrack {{t}^{-1}, t}\right\rbrack /p \) where, multiplying by a unit, it may be assumed that \( p = {a}_{0} + {a}_{1}t + {a}_{2}{t}^{2} + \cdots + {a}_{r}{t}^{r} \) with \( {a}_{r} = 1 \) . Over the field \( \mathbb{Q} \), the vector space \( \mathbb{Q}\left\lbrack {{t}^{-1}, t}\right\rbrack /p \) has a finite base \( \left\{ {1, t,{t}^{2},\ldots ,{t}^{r - 1}}\right\} \) , for the relation \
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Yes
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Lemma 7.4. A covering map \( p : E \rightarrow B \) has the path lifting property. That is, given a point \( {e}_{0} \in E \) and a continuous map \( f : \left\lbrack {0,1}\right\rbrack \rightarrow B \) such that \( f\left( 0\right) = p\left( {e}_{0}\right) \), there exists a unique continuous map \( \widehat{f} : \left\lbrack {0,1}\right\rbrack \rightarrow E \) such that \( \widehat{f}\left( 0\right) = {e}_{0} \) and \( p\widehat{f} = f \) .
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Proof. The space \( B \) is the union of open sets \( \{ V\} \), as in the definition of a covering. Thus, by the compactness of \( \left\lbrack {0,1}\right\rbrack \) there is a dissection \( 0 = {t}_{0} < {t}_{1} < \) \( {t}_{2} < \cdots < {t}_{n} = 1 \) so that \( f\left\lbrack {{t}_{i - 1},{t}_{i}}\right\rbrack \subset {V}_{i} \) for some such open set \( {V}_{i} \) . Assume that \( \widehat{f} \mid \left\lbrack {0,{t}_{i - 1}}\right\rbrack \) has been defined with \( \widehat{f}\left( {t}_{i - 1}\right) \in {W}_{i, j} \) where \( {W}_{i, j} \) is one of the open subsets of \( {p}^{-1}{V}_{i} \) for which \( p : {W}_{i, j} \rightarrow {V}_{i} \) is a homeomorphism. Define \( \widehat{f} \mid \left\lbrack {{t}_{i - 1},{t}_{i}}\right\rbrack \) to be equal to \( {\left( p \mid {W}_{i, j}\right) }^{-1}f \) . For the uniqueness, suppose \( \widehat{\phi } \) is a second lift of \( f \), with \( \widehat{\phi }\left( 0\right) = {e}_{0} \) . Let \( \tau = \sup \{ t : \widehat{\phi } \mid \left\lbrack {0, t}\right\rbrack = \widehat{f} \mid \left\lbrack {0, t}\right\rbrack \} \) ; by continuity, \( \widehat{\phi }\left( \tau \right) = \widehat{f}\left( \tau \right) \) . Then, if \( \tau < 1 \), the above argument shows that \( \widehat{\phi }\left( {\tau + \epsilon }\right) = \widehat{f}\left( {\tau + \epsilon }\right) \) for all sufficiently small \( \epsilon \), contradicting the definition of \( \tau \) .
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Yes
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Lemma 7.5. A covering map \( p : E \rightarrow B \) has homotopy-lifting property for paths. That is, given a continuous map \( \widehat{f} : \left\lbrack {0,1}\right\rbrack \times \{ 0\} \rightarrow E \) and a continuous map \( f : \left\lbrack {0,1}\right\rbrack \times \left\lbrack {0,1}\right\rbrack \rightarrow B \) such that \( f\left( {t,0}\right) = p\widehat{f}\left( {t,0}\right) \), there exists a unique continuous extension of \( \widehat{f} \) to \( \widehat{f} : \left\lbrack {0,1}\right\rbrack \times \left\lbrack {0,1}\right\rbrack \rightarrow E \) such that \( p\widehat{f} = f \) .
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Proof. The proof of this is entirely analogous to the proof of the previous lemma; here a dissection of the square \( \left\lbrack {0,1}\right\rbrack \times \left\lbrack {0,1}\right\rbrack \) into a mesh of small squares, each mapping into some \( {V}_{i} \), is used.
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No
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Proposition 7.7. Let \( p : E \rightarrow B \) be a covering map with base points \( {e}_{0} \in E \) and \( {b}_{0} \in B \), chosen so that \( p{e}_{0} = {b}_{0} \). Suppose \( X \) is a path-connected, locally path-connected, space with base point \( {x}_{0} \), and let \( f : \left( {X,{x}_{0}}\right) \rightarrow \left( {B,{b}_{0}}\right) \) be continuous. Then there exists a continuous map \( g : \left( {X,{x}_{0}}\right) \rightarrow \left( {E,{e}_{0}}\right) \) such that \( {pg} = f \) if and only if \[ {f}_{ \star }{\Pi }_{1}\left( {X,{x}_{0}}\right) \subset {p}_{ \star }{\Pi }_{1}\left( {E,{e}_{0}}\right) . \] When such a \( g \) exists, it is unique.
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Proof. If \( g \) exists, then \( {p}_{ \star }{g}_{ \star } = {f}_{ \star } \), and the result is clear. Conversely, suppose \( {f}_{ \star }{\Pi }_{1}\left( {X,{x}_{0}}\right) \subset {p}_{ \star }{\Pi }_{1}\left( {E,{e}_{0}}\right) \). If \( x \in X \), choose a path \( \alpha : \left\lbrack {0,1}\right\rbrack \rightarrow X \) so that \( \alpha \left( 0\right) = {x}_{0} \) and \( \alpha \left( 1\right) = x \). By Lemma 7.4, the path \( {f\alpha } \) lifts to a path \( \widehat{f\alpha } : \left\lbrack {0,1}\right\rbrack \rightarrow \) \( E \) with \( \widehat{f\alpha }\left( 0\right) = {e}_{0} \). Note that if \( g \) exists as advertised, then \( g\left( x\right) = \widehat{f\alpha }\left( 1\right) \) by the uniqueness in Lemma 7.4, because \( {g\alpha } \) is a lift of \( {f\alpha } \). Thus if \( g \) exists, it is unique. Now define \( g \) by \( g\left( x\right) = \widehat{f\alpha }\left( 1\right) \). To check that is well defined, let \( \beta \) be another path in \( X \) from \( {x}_{0} \) to \( {x}_{1} \). Then \( {f}_{ \star }\left\lbrack {\alpha \cdot \bar{\beta }}\right\rbrack \in {f}_{ \star }{\Pi }_{1}\left( {X,{x}_{0}}\right) \subset {p}_{ \star }{\Pi }_{1}\left( {E,{e}_{0}}\right) \), so there exists a loop \( \gamma : \left\lbrack {0,1}\right\rbrack \rightarrow E \) with \( \gamma \left( 0\right) = {e}_{0} = \gamma \left( 1\right) \) so that \( {p\gamma } \) is homotopic, relative to \( \{ 0,1\} \), to \( f\left( {\alpha \cdot \bar{\beta }}\right) \). By Lemma 7.5 that homotopy can be lifted, relative to \( \{ 0,1\} \), so that (at the end of the homotopy) there is a loop \( \widetilde{\gamma } : \left\lbrack {0,1}\right\rbrack \rightarrow E \) with \( \widetilde{\gamma }\left( 0\right) = {e}_{0} = \widetilde{\gamma }\left( 1\right) \) such that \( p\widetilde{\gamma } = f\left( {\alpha \cdot \bar{\beta }}\right) \). Thus the lift of \( f\left( {\alpha \cdot \bar{\beta }}\right) \) starting at \( {e}_{0} \) is \( \widetilde{\gamma } \), a loop at \( {e}_{0} \). Hence \( p\widetilde{\gamma }\left( t\right) = {f\alpha }\left( {2t}\right) \) and \( p\widetilde{\gamma }\left( t\right) = {f\beta }\left( {2t}\right) \) for all \( 0 \leq t \leq 1/2 \). Thus \( \widehat{f\alpha }\left( 1\right) = \widetilde{\gamma }\left( {1/2}\right) = \widehat{f\beta }\left( 1\right) \), and so \( g \) is well defined. The continuity of \( g \) follows from the fact that \( X \) is locally path-connected, and so on sufficiently small open sets \( g \) is \( {p}^{-1}f \) .
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Yes
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Proposition 7.8. Suppose \( p : \left( {E,{e}_{0}}\right) \rightarrow \left( {B,{b}_{0}}\right) \) and \( {p}^{\prime } : \left( {{E}^{\prime },{e}_{0}^{\prime }}\right) \rightarrow \left( {B,{b}_{0}}\right) \) are two based coverings of \( B \) with the same group. Then these are equivalent in the sense that there exists a homeomorphism \( h : \left( {{E}^{\prime },{e}_{0}^{\prime }}\right) \rightarrow \left( {E,{e}_{0}}\right) \) such that \( {ph} = {p}^{\prime } \) .
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Proof. By Proposition 7.7, the map \( {p}^{\prime } \) lifts to a map \( h : \left( {{E}^{\prime },{e}_{0}^{\prime }}\right) \rightarrow \left( {E,{e}_{0}}\right) \) such that \( {ph} = {p}^{\prime } \) . Similarly, by Proposition 7.7 applied to the map \( p \) and covering \( {p}^{\prime } \), there is a map \( {h}^{\prime } : \left( {E,{e}_{0}}\right) \rightarrow \left( {{E}^{\prime },{e}_{0}^{\prime }}\right) \) such that \( {p}^{\prime }{h}^{\prime } = p \) . But then \( h{h}^{\prime } \) : \( \left( {E,{e}_{0}}\right) \rightarrow \left( {E,{e}_{0}}\right) \) is a lift of the map \( p \) with respect to the covering \( p \) . The identity map is another such lift. Hence, by the uniqueness of Proposition 7.7, \( h{h}^{\prime } \) is the identity. Similarly, \( {h}^{\prime }h \) is the identity, and so \( h \) and \( {h}^{\prime } \) are mutually inverse homeomorphisms.
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Yes
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Theorem 7.9. The covering space \( p : {X}_{\infty } \rightarrow X \) of the exterior \( X \) of an oriented link \( L \) does not depend on the choice of Seifert surface used in its construction. Further, the action of the infinite cyclic group on \( {X}_{\infty } \) is likewise independent of \( F \) .
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Proof. It is clear from the construction of \( {X}_{\infty } \) that a loop \( \alpha : \left\lbrack {0,1}\right\rbrack \rightarrow X \) lifts to a loop \( \widehat{\alpha } \) (that is, \( \widehat{\alpha }\left( 0\right) = \widehat{\alpha }\left( 1\right) \) ) in \( {X}_{\infty } \) provided \( \widehat{\alpha }\left( 0\right) \) and \( \widehat{\alpha }\left( 1\right) \) are in the same copy of \( Y \) . This is so if and only if \( \alpha \) intersects \( F \) zero times algebraically, for every time \( \alpha \) crosses \( F \), its lift moves from one copy of \( Y \) to an adjacent copy. Thus \( \alpha \) lifts to a loop if and only if the linking number of \( \alpha \) with \( L \) (that is, the sum of the linking numbers with the components of \( L \) ) is zero. Now, that statement is independent of the choice of Seifert surface for \( L \), so the group of the cover does not depend on \( F \) . Using the preceding proposition, the the first result follows at once. Consider the action by the infinite cyclic group \( \langle t\rangle \) on \( {X}_{\infty } \) . If \( \gamma : \left\lbrack {0,1}\right\rbrack \rightarrow {X}_{\infty } \) is any path from some point \( a \) to \( {ta} \), then, by the above reasoning, \( {p\gamma } \) is a loop in \( X \) having linking number 1 with \( L \) . Conversely the lift of any such loop in \( x \) is a path from some \( a \) to \( {ta} \) . Suppose \( {p}^{\prime } : {X}_{\infty }^{\prime } \rightarrow X \) is a second version of \( {X}_{\infty } \) constructed from Seifert surface \( {F}^{\prime } \) and \( {h}^{\prime } : {X}_{\infty } \rightarrow {X}_{\infty }^{\prime } \) is the homeomomorphism such that \( {p}^{\prime }{h}^{\prime } = p \) . Trivially \( {p}^{\prime }{h}^{\prime }\gamma = {p\gamma } \), so that \( {h}^{\prime }\gamma \), being a lift of the loop \( {p\gamma } \) with respect to the covering \( {p}^{\prime } \), is a path in \( {X}_{\infty }^{\prime } \) from a point to its \( t \) -translate. Hence \( t{h}^{\prime }\left( a\right) = {h}^{\prime }\left( {ta}\right) \), and the homeomorphism \( {h}^{\prime } \) preserves the \( t \) -action.
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Yes
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Theorem 7.10. If the rth elementary ideal of the Alexander module of a knot \( K \) is not the whole of \( \mathbb{Z}\left\lbrack {{t}^{-1}, t}\right\rbrack \), then \( K \) has unknotting number \( u\left( K\right) \geq r \) .
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As an example, consider the pretzel knot \( P\left( {3,3, - 3}\right) \) discussed in Example 6.9. There it was shown that the second elementary ideal of the Alexander module is not \( \mathbb{Z}\left\lbrack {{t}^{-1}, t}\right\rbrack \), and so \( u\left( {P\left( {3,3, - 3}\right) }\right) \geq 2 \) . It is easy to see that two crossing changes do undo the knot, and so \( u\left( {P\left( {3,3, - 3}\right) }\right) = 2 \) .
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Yes
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Theorem 7.13. Let \( B \) be a path-connected, locally path-connected, semi-locally simply connected space. Then there exists a simply connected space \( \widetilde{B} \) and covering map \( p : \widetilde{B} \rightarrow B \). Furthermore, the group \( {\Pi }_{1}\left( B\right) \) acts freely as a group of homeomorphisms on (the left of) \( \widetilde{B} \), the quotient map \( q : \widetilde{B} \rightarrow \widetilde{B}/{\Pi }_{1}\left( B\right) \) is a covering map and there is a homeomorphism \( h : \widetilde{B}/{\Pi }_{1}\left( B\right) \rightarrow B \) such that \( {hq} = p \).
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Proof. Let \( {b}_{0} \in B \) be a base point and let \( X \) be the set of all paths \( \alpha : \left\lbrack {0,1}\right\rbrack \rightarrow \) \( B \) such that \( \alpha \left( 0\right) = {b}_{0} \). Define an equivalence relation on \( \mathrm{X} \) by letting \( \alpha \sim \beta \) if and only if \( \alpha \left( 1\right) = \beta \left( 1\right) \) and \( \alpha \approx \beta \), where \
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No
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Theorem 7.14. Suppose that a group \( G \) acts as a group of homeomorphisms on a path-connected, locally path-connected, space \( Y \) . Suppose that each \( y \) belonging to \( Y \) has an open neighbourhood \( U \) such that \( U \cap {gU} = \varnothing \) for all \( g \in G - \{ 1\} \) . Then the quotient map \( q : Y \rightarrow Y/G \) is a covering map. If \( Y \) is simply connected, then \( {\Pi }_{1}\left( {Y/G}\right) \) is isomorphic to \( G \) .
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Proof. If \( y \in Y \), there is an open neighbourhood \( U \) of \( y \) such that \( U \cap {gU} = \varnothing \) for all \( g \in G - \{ 1\} \) . Now \( {q}^{-1}\left( {qU}\right) = \mathop{\bigcup }\limits_{{g \in G}}{gU} \) . This is open because each \( {gU} \) is open (because \( g \) is a homeomorphism). Hence \( {qU} \) is open in the quotient topology on \( Y/G \) . Similarly, if \( {U}^{\prime } \) is any open subset of \( U \), then \( q{U}^{\prime } \) is open. The map \( q : U \rightarrow {qU} \) is an injection because \( U \cap {gU} = \varnothing \) for all \( g \neq 1 \), and so it is a homeomophism. Of course, \( q{g}^{-1} = q \), so that \( q : {gU} \rightarrow {qU} \) is also a homeomorphism. Thus \( q \) is a covering map.\n\nSuppose now that \( Y \) is simply connected. Let \( {y}_{0} \) be a base point in \( Y \) and let \( g \) belong to \( G \) . Define a function \( \phi : G \rightarrow {\Pi }_{1}\left( {Y/G, q\left( {y}_{0}\right) }\right) \) as follows: Let \( \alpha \) be a path in \( Y \) from \( {y}_{0} \) to \( g{y}_{0} \) and let \( \phi \left( g\right) = \left\lbrack {q\alpha }\right\rbrack \) . If \( \beta \) is another such path, \( \alpha \approx \beta \) as \( Y \) is simply connected. So \( \left\lbrack {q\alpha }\right\rbrack = \left\lbrack {q\beta }\right\rbrack \), and \( \phi \) is well defined. Let \( {\alpha }_{1} \) be a path from \( {y}_{0} \) to \( {g}_{1}{y}_{0} \) and \( {\alpha }_{2} \) be a path from \( {y}_{0} \) to \( {g}_{2}{y}_{0} \) . Then \( {\alpha }_{1} \cdot {g}_{1}{\alpha }_{2} \) is a path from \( {y}_{0} \) to \( {g}_{1}{g}_{2}{y}_{0} \) . Thus \( \phi \left( {{g}_{1}{g}_{2}}\right) = \left\lbrack {q\left( {{\alpha }_{1} \cdot {g}_{1}{\alpha }_{2}}\right) }\right\rbrack = \left\lbrack {q\left( {\alpha }_{1}\right) \cdot q\left( {\alpha }_{2}\right) }\right\rbrack = \phi \left( {g}_{1}\right) \phi \left( {g}_{2}\right) \), and so \( \phi \) is a group homomorphism. The path lifting property of a covering (Lemma 7.4) implies at once that \( \phi \) is surjective, and the homotopy lifting property (Lemma 7.5) implies it is injective.
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Yes
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Theorem 8.4. Let \( A \) and \( B \) be Seifert matrices for an oriented link \( L \). Then \( A \) and B are S-equivalent.
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Proof. Suppose that \( A \) is an \( n \times n \) matrix corresponding to a Seifert surface \( F \), with respect to some base of \( {H}_{1}\left( {F;\mathbb{Z}}\right) \). Changing the base used for \( {H}_{1}\left( {F;\mathbb{Z}}\right) \) changes \( A \) to a matrix of the form \( {P}^{\tau }{AP} \), where \( P \) is the unimodular base-change matrix. Thus it suffices to check what happens when the Seifert surface is changed, and to do that it suffices, by Theorem 8.2, to check (with respect to any base) the effect of surgery along an arc. Suppose \( F \) is changed to \( {F}^{\prime } \) by surgery along an arc. A base for \( {H}_{1}\left( {{F}^{\prime };\mathbb{Z}}\right) \) can be chosen to be the homology classes of curves \( \left\{ {f}_{i}\right\} \) that constitute a base for \( {H}_{1}\left( {F;\mathbb{Z}}\right) \) together with the classes of a curve \( {f}_{n + 1} \) that goes once over the solid cylinder defining the surgery and of a curve \( {f}_{n + 2} \) around the middle of the cylinder (that is, \( {f}_{n + 2} = 1/2 \times \partial {D}^{2} \) in the notation of Definition 8.1). Then, because \( {f}_{n + 2} \) bounds a disc \( \left( {1/2 \times {D}^{2}}\right) \) that is disjoint from \( \bigcup \left\{ {{f}_{i} : i \leq n}\right\} \), \( \operatorname{lk}\left( {{f}_{n + 2}^{ \pm },{f}_{i}}\right) = 0 \) for all \( i \neq n + 1 \). Further, as \( {f}_{n + 1} \) meets this disc at one point in its boundary, choosing orientations carefully gives either \( \operatorname{lk}\left( {{f}_{n + 1}^{ + },{f}_{n + 2}}\right) = 0 \) and \( \operatorname{lk}\left( {{f}_{n + 1}^{ - },{f}_{n + 2}}\right) = 1 \), or \( \operatorname{lk}\left( {{f}_{n + 1}^{ + },{f}_{n + 2}}\right) = 1 \) and \( \operatorname{lk}\left( {{f}_{n + 1}^{ - },{f}_{n + 2}}\right) = 0 \). In the first case the new Seifert matrix is of the form\n\n\[ \left( \begin{matrix} A & \xi & 0 \\ ? & ? & 1 \\ 0 & 0 & 0 \end{matrix}\right) ,\;\text{ which is congruent to }\;\left( \begin{matrix} A & \xi & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{matrix}\right) .\n\nThe second case leads to a Seifert matrix of the form\n\n\[ \left( \begin{matrix} A & 0 & 0 \\ {\eta }^{\tau } & 0 & 0 \\ 0 & 1 & 0 \end{matrix}\right) \n\nIt follows from this theorem that any invariant well-defined on \( S \) -equivalence classes of square matrices of integers gives at once an invariant of oriented links.
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Yes
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Theorem 8.5. The Conway-normalised Alexander polynomial is a well-defined invariant of the oriented link \( L \) .
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Proof. It is only necessary to check the invariance of the Conway-normalised polynomial when \( A \) changes by \( S \) -equivalence. Firstly, note that\n\n\[ \det \left( {{t}^{1/2}{P}^{\tau }{AP} - {t}^{-1/2}{P}^{\tau }{A}^{\tau }P}\right) = {\left( \det P\right) }^{2}\det \left( {{t}^{1/2}A - {t}^{-1/2}{A}^{\tau }}\right) ,\]\n\nso that the normalised \( {\Delta }_{L}\left( t\right) \) is invariant under unimodular congruence. If now\n\n\[ B = \left( \begin{array}{lll} A & \xi & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{array}\right) \]\n\nthen\n\n\[ \left( {{t}^{1/2}B - {t}^{-1/2}{B}^{\tau }}\right) = \left( \begin{matrix} {t}^{1/2}A - {t}^{-1/2}{A}^{\tau } & {t}^{1/2}\xi & 0 \\ - {t}^{-1/2}{\xi }^{\tau } & 0 & {t}^{1/2} \\ 0 & - {t}^{-1/2} & 0 \end{matrix}\right) ,\]\n\nwhich has the same determinant as \( \left( {{t}^{1/2}A - {t}^{-1/2}{A}^{\tau }}\right) \) . Similarly, the other type of elementary enlargement of \( A \) has no effect on this determinant.
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Yes
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Theorem 8.6. For oriented links \( L \), the Conway-normalised Alexander polynomial \( {\Delta }_{L}\left( t\right) \in \mathbb{Z}\left\lbrack {{t}^{-\frac{1}{2}},{t}^{\frac{1}{2}}}\right\rbrack \) is characterised by\n\n(i) \( {\Delta }_{\text{unknot }}\left( t\right) = 1 \) ,\n\n(ii) whenever three oriented links \( {L}_{ + },{L}_{ - } \) and \( {L}_{0} \) are the same except in the neighbourhood of a point where they are as shown in Figure 3.2, then\n\n\[ \n{\Delta }_{{L}_{ + }} - {\Delta }_{{L}_{ - }} = \left( {{t}^{-1/2} - {t}^{1/2}}\right) {\Delta }_{{L}_{0}} \n\]
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Proof. Construct a Seifert surface \( {F}_{0} \) for \( {L}_{0} \) that meets the neighbourhood of the point in question as shown in Figure 8.1. The Seifert circuit method described in Chapter 2 will do this. Now form Seifert surfaces \( {F}_{ + } \) for \( {L}_{ + } \) and \( {F}_{ - } \) for \( {L}_{ - } \) by adding short twisted strips to \( {F}_{0} \) as also shown in Figure 8.1. Let \( {H}_{1}\left( {{F}_{0};\mathbb{Z}}\right) \) be\n\n\n\nFigure 8.1\n\ngenerated by the classes of oriented closed curves \( \left\{ {{f}_{2},{f}_{3},\ldots ,{f}_{n}}\right\} \), and for generators of \( {H}_{1}\left( {{F}_{ \pm };\mathbb{Z}}\right) \), take the classes of the same curves together with the class of an extra curve \( {f}_{1} \) that goes once along the twisted strip. If \( {A}_{0} \) is the resulting Seifert matrix for \( {L}_{0} \), the Seifert matrix for \( {L}_{ - } \) is of the form \( \left( \begin{matrix} N & {\xi }^{\tau } \\ \eta & {A}_{0} \end{matrix}\right) \) for some integer \( N \) and columns \( \xi \) and \( \eta \), whereas that for \( {L}_{ + } \) is \( \left( \begin{matrix} N - 1 & {\xi }^{\tau } \\ \eta & {A}_{0} \end{matrix}\right) \) . Consideration of \( \det \left( {{t}^{1/2}A - {t}^{-1/2}{A}^{\tau }}\right) \) when \( A \) is each of these three Seifert matrices immediately produces the required formula.
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Yes
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Proposition 8.7. For an oriented link \( L \) with \( \# L \) components, the Conway polynomial has the following properties.\n\n(i) If \( L \) is a split link, then \( {\nabla }_{L}\left( z\right) = 0 \) .
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Proof. (i) This follows from the stronger Proposition 6.14. However, it also follows at once by applying the skein formula to links \( {L}_{ + },{L}_{ - } \) and \( {L}_{0} \) shown in Figure 8.2. As \( {L}_{ + } \) and \( {L}_{ - } \) are here the same link, \( {\nabla }_{{L}_{0}}\left( z\right) = 0 \) .
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Yes
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Theorem 8.9. The \( \omega \) -signature \( {\sigma }_{\omega }\left( L\right) \) is well defined as an invariant of \( L \) .
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Proof. The signature of a Hermitian matrix is not changed by congruence (that fact is Sylvester's famous law of inertia), so it is only necessary to see whether the definition changes under an elementary enlargement of a Seifert matrix \( A \) .\n\nSuppose\n\n\[ B = \left( \begin{array}{lll} A & \xi & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{array}\right) \]\n\nthen\n\n\[ \left( {1 - \omega }\right) B + \left( {1 - \bar{\omega }}\right) {B}^{\tau } = \left( \begin{matrix} \left( {1 - \omega }\right) A + \left( {1 - \bar{\omega }}\right) {A}^{\tau } & \left( {1 - \omega }\right) \xi & 0 \\ \left( {1 - \bar{\omega }}\right) {\xi }^{\tau } & 0 & \left( {1 - \omega }\right) \\ 0 & \left( {1 - \bar{\omega }}\right) & 0 \end{matrix}\right) .\n\nAs \( \left( {1 - \omega }\right) \neq 0 \), the terms in \( \xi \) and \( {\xi }^{\tau } \) can be removed by congruence (subtracting multiples of the last row and column from predecessors), so that the signature of \( \left( {1 - \omega }\right) A + \left( {1 - \bar{\omega }}\right) {A}^{\tau } \) and the signature of \( \left( {1 - \omega }\right) B + \left( {1 - \bar{\omega }}\right) {B}^{\tau } \) differ by the signature of \( \left( \begin{matrix} 0 & \left( {1 - \omega }\right) \\ \left( {1 - \bar{\omega }}\right) & 0 \end{matrix}\right) \) . Of course, this last signature is zero, as the matrix clearly has one positive eigenvalue and one negative one. Consideration of the other type of elementary enlargement is exactly the same.\n\nNote that \( \left( {1 - \omega }\right) A + \left( {1 - \bar{\omega }}\right) {A}^{\tau } = - \left( {1 - \bar{\omega }}\right) \left( {{\omega A} - {A}^{\tau }}\right) \), so that the Hermitian matrix is non-singular except when \( \omega \) is a zero of the Alexander polynomial of \( L \) . In fact, it can be shown that for a fixed link \( L \), the invariant \( {\sigma }_{\omega }\left( L\right) \), when viewed as a function of \( \omega \), is continuous except at zeros of the Alexander polynomial. As signatures are integers, this means that \( {\sigma }_{\omega }\left( L\right) \) takes finitely many values as \( \omega \) varies on \( {S}^{1} \), with possible jumps at roots of \( {\Delta }_{L}\left( t\right) = 0 \) .
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Yes
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Theorem 8.10. If \( L \) is an oriented link in \( {S}^{3} \) and \( \bar{L} \) is its reflection, then for any unit complex number \( \omega \neq 1 \) , \[ {\sigma }_{\omega }\left( L\right) = - {\sigma }_{\omega }\left( \bar{L}\right) \]
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Proof. If \( A \) is a Seifert matrix for \( L \), then \( - A \) is a Seifert matrix for \( \bar{L} \)
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No
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Lemma 8.12. Suppose that for some knot \( K \) in \( {S}^{3} \), there is a flat surface \( F \) in \( {B}^{4} \) with \( F \cap {S}^{3} = \partial F \cap {S}^{3} = K \) . Then the inclusion map induces an isomorphism \( {H}_{1}\left( {{S}^{3} - K}\right) \rightarrow {H}_{1}\left( {{B}^{4} - F}\right) \cong \mathbb{Z}. \)
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Proof. Let \( N \), a copy of \( F \times {I}^{2} \), be a neighbourhood of \( F \) meeting \( {S}^{3} \) in \( \partial F \times {I}^{2} \) . The Mayer-Vietoris theorem gives an exact sequence\n\n\[ 0 = {H}_{2}\left( {B}^{4}\right) \rightarrow {H}_{1}\left( {F \times \partial {I}^{2}}\right) \rightarrow {H}_{1}\left( N\right) \oplus {H}_{1}\left( \overline{{B}^{4} - N}\right) \rightarrow {H}_{1}\left( {B}^{4}\right) = 0. \]\n\nExactness implies that the middle map of this must be an isomorphism. Of course,\n\n\[ {H}_{1}\left( {F \times \partial {I}^{2}}\right) = {H}_{1}\left( F\right) \oplus {H}_{1}\left( {\partial {I}^{2}}\right) ,\]\n\nand the \( {H}_{1}\left( F\right) \) component is mapped isomorphically to \( {H}_{1}\left( N\right) \) (and each is the direct sum of copies of \( \mathbb{Z} \) ); \( {H}_{1}\left( {\partial {I}^{2}}\right) \) is mapped to zero in \( {H}_{1}\left( N\right) \) . As \( {H}_{1}\left( {\partial {I}^{2}}\right) = \mathbb{Z} \) , it follows that \( {H}_{1}\left( \overline{{B}^{4} - N}\right) \) is also a copy of \( \mathbb{Z} \) . The map \( {H}_{1}\left( {\partial {I}^{2}}\right) \rightarrow {H}_{1}\left( \overline{{B}^{4} - N}\right) \) must send generator to generator, as otherwise a matrix representing the map in the above sequence will not have unit determinant. However, a generator of this copy of \( {H}_{1}\left( {\partial {I}^{2}}\right) \) is a meridian of the knot \( K \) . Thus the inclusion map from the knot exterior to \( \overline{{B}^{4} - N} \) induces an isomorphism on the first homology, and that is, up to adjustment by a small homotopy equivalence, the required statement.
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Yes
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Lemma 8.13. Suppose that \( {f}_{1} : {F}_{1} \rightarrow {B}^{4} \) and \( {f}_{2} : {F}_{2} \rightarrow {B}^{4} \) are maps, of orientable surfaces into the 4-ball, which have disjoint images. Suppose that on \( \partial {F}_{i} \) the map \( {f}_{i} \) is a homeomorphism onto a knot \( {K}_{i} \) in \( {S}^{3} = \partial {B}^{4} \) . Then \( \operatorname{lk}\left( {{K}_{1},{K}_{2}}\right) = 0 \) .
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Proof. After moving the maps into general position, it may be assumed that each \( {f}_{i} \) has only double points as singularities. That means that near the image of such a singularity in \( {B}^{4} \), the image of \( {F}_{i} \) looks like two standard planes in \( {\mathbb{R}}^{4} \) meeting in a point \( P \) . That is, near \( P \) it is the cone from \( P \) on a standard Hopf link (a non-trivial two-crossing link) in a copy of \( {S}^{3} \) . Replace the cone on that link with a Seifert surface of the link. This changes \( {F}_{i} \) by removing two discs and inserting an annulus, but there is no longer a point of self-intersection. There may also be points at which the image of \( {f}_{i} \) is locally knotted, points \( P \) near which the image is the cone on a knot in a copy of \( {S}^{3} \) ; replace that cone with a Seifert surface of the knot, changing \( {F}_{i} \) but gaining flatness. In this way it may be assumed that each \( {f}_{i} \) is an embedding onto a flat surface. Then the existence of \( {f}_{1}\left( {F}_{1}\right) \) asserts that \( {K}_{1} \) represents the zero homology class in \( {H}_{1}\left( {{B}^{4} - {F}_{2}}\right) \), and so, by the last lemma, \( {K}_{1} \) represents zero in \( {H}_{1}\left( {{S}^{3} - {K}_{2}}\right) \) .
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Yes
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Corollary 8.16. There is a base \( \left\lbrack {f}_{1}\right\rbrack ,\left\lbrack {f}_{2}\right\rbrack ,\ldots ,\left\lbrack {f}_{2g}\right\rbrack \) over \( \mathbb{Z} \) for \( {H}_{1}\left( {\partial M;\mathbb{Z}}\right) \) so that \( \left\lbrack {f}_{1}\right\rbrack ,\left\lbrack {f}_{2}\right\rbrack ,\ldots ,\left\lbrack {f}_{g}\right\rbrack \) map to zero in \( {H}_{1}\left( {M;\mathbb{Q}}\right) \) .
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Proof. One may consider \( {H}_{1}\left( {\partial M;\mathbb{Z}}\right) \) to be \( {\mathbb{Z}}^{2g} \subset {\mathbb{Q}}^{2g} = {H}_{1}\left( {\partial M;\mathbb{Q}}\right) \) . The \( g \) -dimensional subspace \( U \) of \( {\mathbb{Q}}^{2g} \), given by Lemma 8.15, has a base consisting of elements in \( {\mathbb{Z}}^{2g} \) . Let \( \widetilde{U} \) be the \( \mathbb{Z} \) -span of those elements. As a \( \mathbb{Z} \) -module \( {\mathbb{Z}}^{2g}/\widetilde{U} = \) \( A/\widetilde{U} \oplus B/\widetilde{U} \), where \( A \) and \( B \) are submodules of \( {\mathbb{Z}}^{2g}, A/\widetilde{U} \) is free and \( B/\widetilde{U} \) is a torsion module over \( \mathbb{Z} \) . Thus if \( b \in B \) then \( {nb} \in \widetilde{U} \) for some \( n \in \mathbb{Z} \) ; hence \( b \in U \) . Thus a \( \mathbb{Z} \) -base for \( B \) is a \( \mathbb{Q} \) -base for \( U \) and it extends, using a base of \( A/\widetilde{U} \), to a \( \mathbb{Z} \) -base of \( {\mathbb{Z}}^{2g} \) .
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Yes
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Proposition 8.17. Suppose that \( F \) is a genus \( g \) Seifert surface for a slice knot \( K \) in \( {S}^{3} \) . Then a base may be chosen for \( {H}_{1}\left( {F;\mathbb{Z}}\right) \) with respect to which the corresponding Seifert matrix has the form\n\n\[ \left( \begin{matrix} 0 & P \\ Q & R \end{matrix}\right) \]\n\nconsisting of a \( g \times g \) block of zeros together with \( g \times g \) blocks of integers \( P, Q \) and \( R \) .
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Proof. Let \( D \) be a slicing disc for \( K \) contained in \( {B}^{4} \) . By Lemma 8.14 there is contained in \( {B}^{4} \) a 3-manifold \( M \) having an \( M \times \left\lbrack {-1,1}\right\rbrack \) neighbourhood such that \( \partial M = D \cup F \) . Corollary 8.16 gives a certain base \( \left\lbrack {f}_{1}\right\rbrack ,\left\lbrack {f}_{2}\right\rbrack ,\ldots ,\left\lbrack {f}_{2g}\right\rbrack \) for \( {H}_{1}\left( {\partial M;\mathbb{Z}}\right) \) . It may be assumed that each \( \left\lbrack {f}_{i}\right\rbrack \) is represented by an oriented closed curve \( {f}_{i} \) in \( F \) . Consider the Seifert matrix \( A \) with respect to this basis. In the notation of Chapter \( 6,{A}_{ij} = \operatorname{lk}\left( {{f}_{i}^{ - },{f}_{j}}\right) \) . (If the \( {f}_{i} \) are not simple curves, they should here be changed by a very small amount in \( {S}^{3} \) to become simple so that \
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Yes
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Theorem 8.18. If \( K \) is a slice knot, then the Conway-normalised Alexander polynomial of \( K \) is of the form \( f\left( t\right) f\left( {t}^{-1}\right) \), where \( f \) is a polynomial with integer coefficients.
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Proof. Using the Seifert matrix of Proposition 8.17, the required Alexander polynomial is the determinant of\n\n\[ \left( \begin{matrix} 0 & {t}^{1/2}P - {t}^{-1/2}{Q}^{\tau } \\ {t}^{1/2}Q - {t}^{-1/2}{P}^{\tau } & {t}^{1/2}R - {t}^{-1/2}{R}^{\tau } \end{matrix}\right) ,\]\n\nwhich is \( \det \left( {{tP} - {Q}^{\tau }}\right) \det \left( {{t}^{-1}P - {Q}^{\tau }}\right) \).
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Yes
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Theorem 8.19. If \( K \) is a slice knot, then the signature of \( K \) is zero and, if the unit complex number \( \omega \) is not a zero of the Alexander polynomial, then \( {\sigma }_{\omega }\left( K\right) = 0 \) .
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Proof. This follows at once from the fact that the signature is zero for a quadratic form coming from a non-singular symmetric bilinear form that vanishes on a subspace of half the dimension of the space concerned. A similar result holds for Hermitian forms.
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Yes
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Theorem 9.1. Let \( {X}_{2} \) be the cyclic double cover of \( {S}^{3} \) branched over a link \( L \) and suppose that \( A \) is a Seifert matrix for \( L \) with respect to some orientation and some Seifert surface. Then \( {H}_{1}\left( {X}_{2}\right) \) is presented, as an abelian group, by the matrix \( \left( {A + {A}^{\tau }}\right) \) .
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Proof. In the above notation, \( {\widehat{X}}_{2} = {Y}_{0} \cup {Y}_{1} \), where \( {Y}_{0} \cap {Y}_{1} \) is two disjoint copies of \( F \) . A presentation of \( {H}_{1}\left( {\widehat{X}}_{2}\right) \) can be obtained from the following exact Mayer-Vietoris sequence:\n\n\[ \rightarrow {H}_{1}\left( {{Y}_{0} \cap {Y}_{1}}\right) \overset{{\alpha }_{ \star }}{ \rightarrow }{H}_{1}\left( {Y}_{0}\right) \oplus {H}_{1}\left( {Y}_{1}\right) \overset{{\beta }_{ \star }}{ \rightarrow }{H}_{1}\left( {\widehat{X}}_{2}\right) \rightarrow \]\n\n\[ \rightarrow {H}_{0}\left( {{Y}_{0} \cap {Y}_{1}}\right) \overset{{\alpha }_{ \star }}{ \rightarrow }{H}_{0}\left( {Y}_{0}\right) \oplus {H}_{0}\left( {Y}_{1}\right) . \]\n\nThe situation is here very similar to that of Theorem 6.5, and the same sign conventions will be used. There is now a homeomorphism \( t : {\widehat{X}}_{2} \rightarrow {\widehat{X}}_{2} \) with \( {t}^{2} = 1 \) which interchanges \( {Y}_{0} \) and \( {Y}_{1} \) . As in Theorem 6.5, one can take a base \( \left\{ \left\lbrack {f}_{i}\right\rbrack \right\} \) for \( {H}_{1}\left( F\right) \), with corresponding Seifert matrix \( A \) and dual base \( \left\{ \left\lbrack {e}_{i}\right\rbrack \right\} \) for \( {H}_{1}\left( Y\right) \) . Transferring to \( {\widehat{X}}_{2} \), this gives a base \( \left\{ \left\lbrack {f}_{i}\right\rbrack \right\} \cup \left\{ \left\lbrack {t{f}_{i}}\right\rbrack \right\} \) for \( {H}_{1}\left( {{Y}_{0} \cap {Y}_{1}}\right) \) (since \( {Y}_{0} \cap {Y}_{1} \) is two copies of \( F \) ), a base \( \left\{ \left\lbrack {e}_{i}\right\rbrack \right\} \) for \( {H}_{1}\left( {Y}_{0}\right) \) and a base \( \left\{ \left\lbrack {t{e}_{i}}\right\rbrack \right\} \) for \( {H}_{1}\left( {Y}_{1}\right) \) . Then, with respect to these bases, \( {\alpha }_{ \star } \) is represented by the matrix\n\n\[ \left( \begin{array}{rr} - A & {A}^{\tau } \\ {A}^{\tau } & - A \end{array}\right) \]\n\nSimilarly, using bases represented by single points, the map \( {H}_{0}\left( {{Y}_{0} \cap {Y}_{1}}\right) \rightarrow \) \( {H}_{0}\left( {Y}_{0}\right) \oplus {H}_{0}\left( {Y}_{1}\right) \) is represented by \( \left( \begin{array}{rr} - 1 & 1 \\ 1 & - 1 \end{array}\right) \) . Thus the kernel of this last map is a copy of \( \mathbb{Z} \), and (recalling the definition of the maps in the Mayer-Vietoris sequence) any loop in \( {\widehat{X}}_{2} \) that cuts each of the two components of \( {Y}_{0} \cap {Y}_{1} \) at one point maps to a generator of this copy of \( \mathbb{Z} \) .
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Yes
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Let \( {X}_{2} \) be the double cover of \( {S}^{3} \) branched over a link \( L \) . The order of the group \( {H}_{1}\left( {X}_{2}\right) \) is the modulus of the determinant of \( \left( {A + {A}^{\tau }}\right) \), that is\n\n\[ \left| {{H}_{1}\left( {X}_{2}\right) }\right| = \left| {\det \left( {A + {A}^{\tau }}\right) }\right| = \left| {{\Delta }_{L}\left( {-1}\right) }\right| . \]
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Proof. Any finitely generated abelian group can be expressed as a direct sum of cyclic groups. Thus it has as a presentation matrix a diagonal matrix, the entries on the diagonal being the orders of the summands, with the convention that an infinite group has order zero. By Theorem 6.1, the determinant of a square presentation matrix is unique up to multiplication by a unit (that is, by \( \pm 1 \) ), so the result follows at once. The statement about the Alexander polynomial then follows from Theorem 6.5.
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No
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Any Goeritz matrix for a link \( L \), associated with the white regions of a diagram of \( L \), represents, with respect to some base, the Gordon-Litherland form \[ {\mathcal{G}}_{F} : {H}_{1}\left( F\right) \times {H}_{1}\left( F\right) \rightarrow \mathbb{Z}, \] where \( F \) is the spanning surface for \( L \) given by the black regions of the diagram.
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Let the white regions, \( {R}_{0},{R}_{1},\ldots ,{R}_{n} \), of the diagram inherit an orientation from the sphere \( {S}^{2} \) in which they are assumed to lie; thus each \( \partial {R}_{i} \) has an orientation. Let \( {f}_{i} \) be the oriented simple closed curve in \( F \) that consists of \( \partial {R}_{i} \) pushed into the union of the black regions. Then \( \left\{ {\left\lbrack {f}_{i}\right\rbrack : 0 \leq i \leq n}\right\} \) forms a set of generators for \( {H}_{1}\left( F\right) \) ; any subset of \( n \) of the \( \left\{ \left\lbrack {f}_{i}\right\rbrack \right\} \) forms a base for \( {H}_{1}\left( F\right) \) . Suppose that the white regions \( {R}_{i} \) and \( {R}_{j} \) are both incident at a crossing \( c \) where \( \zeta \left( c\right) = + 1 \) . Then in the above notation, the curve or curves \( {p}^{-1}{f}_{j} \), namely the push-off of \( {f}_{j} \) from \( F \) locally to both sides of \( F \), meet \( {R}_{i} \) in a positive point of intersection and meet \( {R}_{j} \) in a negative point of intersection near to \( c \) . See Figure 9.4. The sign is positive if the orientation of the region is in the sense of a right-hand screw with respect to the orientation of \( {p}^{-1}{f}_{j} \) . The signs are reversed if \( \zeta \left( c\right) = - 1 \) . Thus for \( i \neq j,\operatorname{lk}\left( {{p}^{-1}{f}_{j},{f}_{i}}\right) = \sum \zeta \left( c\right) \), where the sum is over all crossings at which \( {R}_{i} \) and \( {R}_{j} \) come together, and \( \operatorname{lk}\left( {{p}^{-1}{f}_{j},{f}_{j}}\right) = - \sum \zeta \left( c\right) \) , the sum being over all \( c \) at which \( {R}_{j} \) is incident with other regions. Note the two points of \( {p}^{-1}{f}_{j} \cap {R}_{j} \) near a crossing at which \( {R}_{j} \) is incident with itself cancel each other. Hence the quadratic form \( {\mathcal{G}}_{F} \) is represented with respect to the base \( \left\lbrack {f}_{1}\right\rbrack ,\left\lbrack {f}_{2}\right\rbrack ,\ldots ,\left\lbrack {f}_{n}\right\rbrack \) by the Goeritz matrix of the diagram with the above labelling of the white regions.
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Yes
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Corollary 9.5. The determinant of \( L,\left| {{\Delta }_{L}\left( {-1}\right) }\right| \), is equal to \( \left| {\det G}\right| \), where \( G \) is any Goertiz matrix for \( L \) .
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The proof of this is immediate from the last three theorems. It follows that \( \left| {\det G}\right| \) is an invariant of \( L \), and, as a Goeritz matrix is often easy to write down, it can be a useful invariant.
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No
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Theorem 9.6. Suppose that \( {L}_{ + },{L}_{ - },{L}_{0} \) and \( {L}_{\infty } \) are four links that have identical diagrams except near a point where they are as shown in Figure 9.5. Then\n\n\[ \n{\left( \det {L}_{ + }\right) }^{2} + {\left( \det {L}_{ - }\right) }^{2} = 2\left( {{\left( \det {L}_{0}\right) }^{2} + {\left( \det {L}_{\infty }\right) }^{2}}\right) .\n\]
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Proof. The diagram shows the four links together with connected shaded spanning surfaces \( {F}_{i} \) for \( i = + , - ,0,\infty \) . These can always be constructed by using Seifert’s method (see Chapter 2) for \( {F}_{0} \) and adding bands to get the other three surfaces. The four surfaces are taken to be identical outside the areas shown. Take closed curves in \( {F}_{0} \) representing a base of \( {H}_{1}\left( {F}_{0}\right) \) and, for bases of \( {H}_{1}\left( {F}_{i}\right) \) for \( i = + , - ,\infty \), take the classes of the extra curves shown in the diagrams (the ends of them are joined up outside the diagrams) together with the set of curves already chosen for \( {F}_{0} \) . Matrices \( {M}_{i} \) for the Gordon-Litherland forms \( {\mathcal{G}}_{{F}_{i}} \) with respect to these bases are of the following form:\n\n\[ \n{M}_{\infty } = \left( \begin{array}{rr} n & \rho \\ {\rho }^{\tau } & {M}_{0} \end{array}\right) ,\;{M}_{ \pm } = \left( \begin{matrix} n \mp 1 & \rho \\ {\rho }^{\tau } & {M}_{0} \end{matrix}\right) .\n\]\n\nThus\n\n\[ \n\det {M}_{ \pm } = \det {M}_{\infty } \mp \det {M}_{0} \n\]\n\nSquaring and adding give the required result.
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Yes
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Theorem 9.7. Let \( {X}_{r} \) be the cyclic \( r \) -fold cover of \( {S}^{3} \) branched over an \( n \) - component oriented link \( L \), and suppose that \( A \) is a Seifert matrix for \( L \) coming from a genus \( g \) Seifert surface. Then \( {H}_{1}\left( {X}_{r}\right) \) is presented, as an abelian group, by the \( r \times \left( {r + 1}\right) \) matrix of blocks\n\n\[ \left( \begin{matrix} - {A}^{\tau } & & & & A & B \\ A & - {A}^{\tau } & & & & B \\ & A & - {A}^{\tau } & & & B \\ & & \ddots & \ddots & & \vdots \\ & & & A & - {A}^{\tau } & B \end{matrix}\right) ,\]\n\nwhere \( B \) is the \( \left( {{2g} + n - 1}\right) \times \left( {n - 1}\right) \) matrix \( \left( \begin{array}{l} 0 \\ I \end{array}\right) \) .
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Assuming that a \
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No
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The order of the first homology group of \( {X}_{r} \), the cyclic \( r \) -fold cover of \( {S}^{3} \) branched over \( L \), is given by\n\n\[ \left| {{H}_{1}\left( {X}_{r}\right) }\right| = \left| {\mathop{\prod }\limits_{{v = 1}}^{{r - 1}}{\Delta }_{L}\left( {e}^{{2\pi }\imath \frac{v}{r}}\right) }\right| . \]
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Assuming that a \
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No
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Lemma 10.5. Suppose that \( L \) and \( {L}^{\prime } \) are oriented links having property \( \left( \star \right) \) which are the same except near one point, where they are as shown in Figure 10.1; then \( \mathcal{A}\left( L\right) = \mathcal{A}\left( {L}^{\prime }\right) \)
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Proof. The two segments shown on one of the two sides of Figure 10.1 must belong to the same component of the link. Suppose, without loss of generality, it is the two segments on the left side. Then using the Seifert circuit method of Theorem 2.2, a Seifert surface can be constructed for the left link that meets the neighbourhood of the point in question in the way indicated by the shading. Adding a band to that produces a Seifert surface for the right link as indicated. Now, as these two surfaces just differ by a band added to the boundary, the \( \mathbb{Z}/2\mathbb{Z} \) -homology of the second surface is just that of the first surface with an extra \( \mathbb{Z}/2\mathbb{Z} \) summand. However, that summand is in the image of the homology of the boundary of the surface; this image is disregarded (by means of the quotienting) in construction of the quadratic form that gives the Arf invariant.
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Yes
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Theorem 10.7. Let \( K \) be a knot. Then \( \mathcal{A}\left( K\right) \equiv {a}_{2}\left( K\right) \) modulo 2, where \( {a}_{2}\left( K\right) \) is the coefficient of \( {z}^{2} \) in the Conway polynomial \( {\nabla }_{K}\left( z\right) \) . The Arfinvariant of \( K \) is related to the Alexander polynomial by\n\n\[ \mathcal{A}\left( K\right) = \left\{ \begin{array}{ll} 0 & \text{ if }{\Delta }_{K}\left( {-1}\right) \equiv \pm 1\text{ modulo }8, \\ 1 & \text{ if }{\Delta }_{K}\left( {-1}\right) \equiv \pm 3\text{ modulo }8. \end{array}\right. \]\n\nIf \( K \) is a slice knot, then \( \mathcal{A}\left( K\right) = 0 \) .
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Proof. The formula \( \mathcal{A}\left( {L}_{ + }\right) - \mathcal{A}\left( {L}_{ - }\right) \equiv \operatorname{lk}\left( {L}_{0}\right) \) modulo 2, valid when \( {L}_{ + } \) has one component, allows calculation of \( \mathcal{A}\left( K\right) \) from \( \mathcal{A} \) (unknot) \( = 0 \) . However, this gives the same answer as the calculation, modulo 2, of \( {a}_{2}\left( K\right) \) using Proposition 8.7 (v). With the Conway normalisation, \( {\Delta }_{K}\left( {-1}\right) = {\nabla }_{K}\left( {-{2i}}\right) \) . However, \( {\nabla }_{K}\left( z\right) = \) \( 1 + {a}_{2}\left( K\right) {z}^{2} + {a}_{4}\left( K\right) {z}^{4} + \cdots \), and so \( {\nabla }_{K}\left( {-{2i}}\right) \equiv 1 - 4{a}_{2}\left( K\right) \) modulo 8 . Thus, modulo 8,\n\n\[ {\nabla }_{K}\left( {-{2i}}\right) \equiv \left\{ \begin{array}{ll} 1 & \text{ if }{a}_{2}\left( K\right) \equiv 0{\;\operatorname{modulo}\;2}, \\ - 3 & \text{ if }{a}_{2}\left( K\right) \equiv 1{\;\operatorname{modulo}\;2}. \end{array}\right. \]\n\nThis gives the required result. As remarked after Theorem 8.19, if \( K \) is a slice knot then \( {\Delta }_{K}\left( {-1}\right) \equiv \pm 1 \) modulo 8, and so, from the above discussion, \( \mathcal{A}\left( K\right) = 0 \) .
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Yes
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Theorem 11.2. Let \( X \) be the exterior of a knot \( K \) in \( {S}^{3} \) . If \( K \) is not the unknot, then the inclusion map induces an injection \( {\Pi }_{1}\left( {\partial X}\right) \rightarrow {\Pi }_{1}\left( X\right) \) .
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Proof. Suppose \( {\Pi }_{1}\left( {\partial X}\right) \rightarrow {\Pi }_{1}\left( X\right) \) is not injective. Then, by the loop theorem, there is an embedding \( e : {D}^{2} \rightarrow X \) sending \( \partial {D}^{2} \) into the torus \( \partial X \), to a simple closed curve not homotopically trivial in the torus. Now \( e\left( {\partial {D}^{2}}\right) \) is certainly the boundary of the disc \( e\left( {D}^{2}\right) \) and so represents a non-trivial element of the kernel of the map \( {H}_{1}\left( {\partial X}\right) \rightarrow {H}_{1}\left( X\right) \) ; the longitude of the knot \( K \) (with either orientation) is the only simple closed curve representing an element in this kernel (see Definition 1.6). The longitude is parallel to \( K \) in a small solid torus neighbourhood of \( K \), so expanding the disc \( e\left( {D}^{2}\right) \) by an annulus gives a disc embedded in \( {S}^{3} \) with \( K \) as its boundary. Thus \( e\left( {D}^{2}\right) \) when so expanded is a Seifert surface for \( K \) . This shows that \( K \) is unknotted.
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Yes
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Corollary 11.3. A knot \( K \) is the unknot if and only if \( {\Pi }_{1}\left( {{S}^{3} - K}\right) \) is infinite cyclic.
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Proof. If \( {\Pi }_{1}\left( {{S}^{3} - K}\right) \) is isomorphic to \( \mathbb{Z} \), there can be no injection \( {\Pi }_{1}\left( {\partial X}\right) \rightarrow \) \( {\Pi }_{1}\left( X\right) \) (as \( {\Pi }_{1}\left( {\partial X}\right) \) is isomorphic to \( \mathbb{Z} \oplus \mathbb{Z} \) ).
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No
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Corollary 11.4. Let \( {X}_{1} \) and \( {X}_{2} \) be the exteriors of two non-trivial knots and let M be a 3-manifold formed by identifying their boundaries together using any homeomorphism. Then the inclusion into \( M \) of the torus \( T \) that comes from the identified boundaries induces an injection \( {\Pi }_{1}\left( T\right) \rightarrow {\Pi }_{1}\left( M\right) \) .
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Proof. This follows at once from the above theorem and from the Van Kam-pen theorem, which describes how fundamental groups behave when a space is described as a union of subspaces.
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No
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Theorem 11.6. If \( K \) is a knot in \( {S}^{3} \) any map \( {S}^{2} \rightarrow {S}^{3} - K \) is homotopic to a constant map (that is, \( {\Pi }_{2}\left( {{S}^{3} - K}\right) = 0 \) ).
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Proof. If the statement is false then, by the sphere theorem, there exists a piecewise linear embedding \( e : {S}^{2} \rightarrow {S}^{3} - K \) that is not homotopic to a constant in \( \left( {{S}^{3} - K}\right) \) . Then, by the Schönflies theorem, \( e\left( {S}^{2}\right) \) separates \( {S}^{3} \) into two components, the closure of each of which is a ball with boundary \( e\left( {S}^{2}\right) \) . The knot \( K \), being connected and disjoint from \( e\left( {S}^{2}\right) \), lies in one of these balls, so \( e \) is homotopic to a constant using the other ball.
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Yes
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Theorem 11.7. If \( K \) is a knot in \( {S}^{3} \), any map \( {S}^{r} \rightarrow {S}^{3} - K \) is homotopic to a constant map (that is, \( {\Pi }_{r}\left( {{S}^{3} - K}\right) = 0 \) ) for all \( r \geq 2 \) .
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Proof. Let \( X \) be the exterior of \( K \) and let \( \widetilde{X} \) be the universal cover of \( X \) . Thus \( \widetilde{X} \) is the simply connected cover of \( X \), it is acted upon by \( {\Pi }_{1}\left( X\right) \), and the quotient of \( \widetilde{X} \) by this action is \( X \) . The operation of lifting maps and homotopies from \( X \) to \( \widetilde{X} \) shows that, for \( r \geq 2,{\Pi }_{r}\left( X\right) = 0 \) if and only if \( {\Pi }_{r}\left( \widetilde{X}\right) = 0 \) (or equivalently just use the homotopy long exact sequence of the covering). So certainly \( {\Pi }_{2}\left( \widetilde{X}\right) = 0 \) . Now the third homology of any non-compact connected 3-manifold is zero. A simplicial argument for this uses the fact that any 3-cycle would be a finite sum of oriented 3-simplexes, a neighbourhood of the union of those 3-simplexes is a compact 3-manifold \( N \) with non-empty boundary which can be taken to be connected; any such \( N \) deformation retracts to a 2-dimensional complex (by collapsing 3- simplexes from the boundary), and so \( {H}_{3}\left( N\right) = 0 \) . Of course, \( \widetilde{X} \) is non-compact because \( {\Pi }_{1}\left( X\right) \) is infinite (as \( {H}_{1}\left( X\right) \) is infinite), and so each simplex has infinitely many different lifts in \( \widetilde{X} \) . Thus \( {H}_{3}\left( \widetilde{X}\right) = 0 \) and \( {H}_{r}\left( \widetilde{X}\right) = 0 \) for \( r > 3 \), as then \( X \) has no \( r \) -simplex and so its \( {r}^{\text{th }} \) chain group is zero. Now, for a simply connected cell complex, the Hurewicz isomorphism theorem asserts that the first non-vanishing homology group and the first non-vanishing homotopy group occur in the same dimension and are isomorphic. Thus \( {\Pi }_{r}\left( \widetilde{X}\right) = 0 \) for all \( r \), and so \( \widetilde{X} \) is a contractible space. The above remark about lifting ensures that \( {\Pi }_{r}\left( X\right) = 0 \) for \( r \geq 2 \) .
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Yes
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Theorem 11.9. If \( {K}_{1} \) and \( {K}_{2} \) are prime knots in \( {S}^{3} \) and \( {\Pi }_{1}\left( {{S}^{3} - {K}_{1}}\right) \) and \( {\Pi }_{1}\left( {{S}^{3} - }\right. \) \( \left. {K}_{2}\right) \) are isomorphic groups, then \( \left( {{S}^{3} - {K}_{1}}\right) \) and \( \left( {{S}^{3} - {K}_{2}}\right) \) are homeomorphic spaces.
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Thus, for prime knots, the knot group determines the complement of the knot. It is by no means obvious that this means that the knots are the same. Perhaps the homeomorphism might send a meridian to a non-meridian. That this is not so is the substance of one of the most impressive results in knot theory of the 1980's. It is due to Gordon and J. Luecke [37] and the proof is lengthy and intricate:
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No
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Lemma 12.2. Suppose that \( U \) and \( V \) are 3-manifolds with homeomorphic boundaries, and that \( {h}_{0} : \partial U \rightarrow \partial V \) and \( {h}_{1} : \partial U \rightarrow \partial V \) are isotopic homeomorphisms. Then \( U{ \cup }_{{h}_{0}}V \) and \( U{ \cup }_{{h}_{1}}V \) are homeomorphic.
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Proof. Choose ([113],[47]) a collar neighbourhood \( C \) of \( \partial U \) in \( U \) ; \( C \) is a neighbourhood of \( \partial U \) homeomorphic to \( \partial U \times \left\lbrack {0,1}\right\rbrack \), with \( \partial U \) identified with \( \partial U \times 0 \) . A homeomorphism \( f : U{ \cup }_{{h}_{0}}V \rightarrow U{ \cup }_{{h}_{1}}V \) can be constructed by defining \( f \) to be the identity on \( \left( {U - C}\right) \cup V \) and on \( C \) defining \( f\left( {x, t}\right) = \) \( \left( {{h}_{1}^{-1}{h}_{t}x, t}\right) \) .
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Yes
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Lemma 12.5. Suppose oriented simple closed curves \( p \) and \( q \), contained in the interior of the surface \( F \), intersect transversely at precisely one point. Then \( p{ \sim }_{\tau }q \) .
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Proof. The first diagram of Figures 12.3 shows the intersection point of \( p \) and \( q \) and also a simple closed curve \( {C}_{1} \) that runs parallel to, and is slightly displaced from, \( q \) . Similarly, \( {C}_{2} \) is a slightly displaced copy of \( p \) . The second diagram shows \( {\tau }_{1}p \), where \( {\tau }_{1} \) is a twist about \( {C}_{1} \) . The third diagram shows \( {\tau }_{2}{\tau }_{1}p \), where \( {\tau }_{2} \) is a twist about \( {C}_{2} \) . In this diagram \( {\tau }_{2}{\tau }_{1}p \) has a doubled-back portion that can easily be moved by a homeomorphism isotopic to the identity (that is, a slide in \( F \) ) to change \( {\tau }_{2}{\tau }_{1}p \) to \( q \) .
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Yes
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Lemma 12.6. Suppose that oriented simple closed curves \( p \) and \( q \) contained in the interior of the surface \( F \) are disjoint and that neither separates \( F \) (that is, \( \left\lbrack p\right\rbrack \neq 0 \neq \left\lbrack q\right\rbrack \) in \( \left. {{H}_{1}\left( {F,\partial F}\right) }\right) \) . Then \( p{ \sim }_{\tau }q \) .
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Proof. Consideration of the surface obtained by cutting \( F \) along \( p \cup q \) shows at once that there is a simple closed curve \( r \) in \( F \) that intersects each of \( p \) and \( q \) transversely at one point. Then, by Lemma 12.5, \( p{ \sim }_{\tau }r{ \sim }_{\tau }q \) .
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Yes
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Proposition 12.7. Suppose that oriented simple closed curves \( p \) and \( q \) are contained in the interior of the surface \( F \) and that neither separates \( F \) . Then \( p{ \sim }_{\tau }q \) .
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Proof. Changing \( q \) by means of a homeomorphism of \( F \) that is (close to and) isotopic to the identity, it can be assumed that \( p \) and \( q \) intersect transversely at \( n \) points. The proof is by induction on \( n \) ; Lemmas 12.5 and 12.6 start the induction, so assume that \( n \geq 2 \) and that the result is true for less that \( n \) points of intersection.\n\nLet \( A \) and \( B \) be consecutive points along \( p \) of \( p \cap q \) . Suppose firstly that \( p \) leaves \( A \) on one side of \( q \) and returns to \( B \) from the other side of \( q \) . Let \( r \) be a simple closed curve in \( F \) that starts near \( A \), follows close to \( p \) until near \( B \) and then returns to its start in a neighbourhood of \( q \) . As shown in the first diagram of Figure 12.4, \( r \) can be chosen so that \( p \cap r \) contains less than \( n \) points and \( q \cap r \) is one point. Hence \( p{ \sim }_{\tau }r \) by the induction hypothesis, and \( r{ \sim }_{\tau }q \) by Lemma 12.5.\n\nSuppose now that \( p \) leaves \( A \) on one side of \( q \) and returns to \( B \) from the same side of \( q \) . Let \( {r}_{1} \) and \( {r}_{2} \) be the two simple closed curves shown in the second\n\n \n\nFigure 12.4\n\ndiagram of Figure 12.4. Each starts near \( A \), proceeds near \( p \) until close to \( B \) and then back to its start following near to \( q \) . However, \( {r}_{1} \) starts on the right of \( p \) and \( {r}_{2} \) starts on the left. Now in \( {H}_{1}\left( {F,\partial F}\right) ,\left\lbrack {r}_{1}\right\rbrack - \left\lbrack {r}_{2}\right\rbrack = \left\lbrack q\right\rbrack \), and hence at least one of \( {r}_{1} \) and \( {r}_{2} \) does not separate (as \( \left\lbrack q\right\rbrack \neq 0 \) ). Let that curve be defined to be \( r \) . Then \( r \) is disjoint from \( q \), so \( r{ \sim }_{\tau }q \) by Lemma 12.6 and, as \( r \cap p \) has at most \( n - 2 \) points, \( p{ \sim }_{\tau }r \) by the induction hypothesis.
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Yes
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Corollary 12.8. Let \( {p}_{1},{p}_{2},\ldots ,{p}_{n} \) be disjoint simple closed curves in the interior of \( F \) the union of which does not separate \( F \) . Let \( {q}_{1},{q}_{2},\ldots ,{q}_{n} \) be another set of curves with the same properties. Then there is a homeomorphism \( h \) of \( F \) that is in the group generated by twists, so that \( h{p}_{i} = {q}_{i} \) for each \( i = 1,2,\ldots, n \) .
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Proof. Suppose inductively that such an \( h \) can be found so that \( h{p}_{i} = {q}_{i} \) for each \( i = 1,2,\ldots, n - 1 \) . Apply Proposition 12.7 to \( h{p}_{n} \) and \( {q}_{n} \) in \( F \) cut along \( {q}_{1} \cup {q}_{2} \cup \ldots \cup {q}_{n - 1} \)
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Yes
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Lemma 12.12. Any closed connected orientable 3-manifold has a Heegaard splitting.
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Proof. This is similar to the first part of the proof of Theorem 8.2. Take a triangulation of \( M \) as a simplicial complex \( K \) . The vertices of the first derived subdivision \( {K}^{\left( 1\right) } \) of \( K \) are the barycentres \( \widehat{A} \) of the simplexes \( A \) of \( K \) . The second derived subdivision \( {K}^{\left( 2\right) } \) of \( K \) is, of course, just \( {\left( {K}^{\left( 1\right) }\right) }^{\left( 1\right) } \) . The 1-skeleton of \( K \) (that is, the sub-complex consisting of the 0 -simplexes and 1 -simplexes of \( K \) ), being a graph, has, as intimated above, for its simplicial neighbourhood in \( {K}^{\left( 2\right) } \), a handlebody. The closure of the complement of this is the simplicial neighbourhood in \( {K}^{\left( 2\right) } \) of another graph. That graph, called the dual 1-skeleton of \( K \), is the sub-complex \( \mathop{\bigcup }\limits_{A}{C}_{A} \) of \( {K}^{\left( 1\right) } \), where the union is over all 3-simplexes \( A \), and \( {C}_{A} \) is the cone with vertex \( \widehat{A} \) on the barycentres of the 2-dimensional faces of \( A \) . Thus \( {K}^{\left( 2\right) } \) is expressed as the union of two handlebodies that intersect in their common boundary, and this is the required Heegaard splitting.
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Yes
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Theorem 12.13. Let \( M \) be a closed connected orientable 3-manifold. There exists finite sets of disjoint solid tori \( {T}_{1}^{\prime },{T}_{2}^{\prime },\ldots ,{T}_{N}^{\prime } \) in \( M \) and \( {T}_{1},{T}_{2},\ldots ,{T}_{N} \) in \( {S}^{3} \) such that \( M - { \cup }_{1}^{N}\operatorname{Int}\left( {T}_{i}^{\prime }\right) \) and \( {S}^{3} - { \cup }_{1}^{N}\operatorname{Int}\left( {T}_{i}\right) \) are homeomorphic.
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Proof. By Lemma 12.12, \( M \) has a Heegaard splitting, so for handlebodies \( U \) and \( V \) of some genus \( g \), and some homeomorphism \( h : \partial U \rightarrow \partial V, M = \) \( U{ \cup }_{h}V \) . Let \( {p}_{1}^{\prime },{p}_{2}^{\prime },\ldots ,{p}_{g}^{\prime } \) be disjoint simple closed curves in \( \partial U \), that bound disjoint discs in \( U \) and let \( {q}_{1},{q}_{2},\ldots {q}_{g} \) be disjoint simple closed curves in \( \partial V \) (one around each \
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No
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Theorem 12.15. Two framed links in \( {S}^{3} \) give, by surgery, the same oriented 3- manifold if and only if they are related by a sequence of moves of two types. In a move of type 1, an extra unknotted component, unlinked from all other components, with framing 1 or -1 is added to or removed from the link. In a move of type 2, any two components that are, together with their framing curves, contained in a doubly punctured disc (itself possibly knotted up and linked with other components) in \( {S}^{3} \) , as on the left of Figure 12.6, can be changed to the two curves on the right, the new framing curves again being on the punctured disc.
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For the proof, which uses 4-dimensional Cerf theory, refer to [65]. If one considers the surgery information as a recipe for adding 2-handles on to a 4-ball to create a 4-manifold with the 3-manifold as its boundary, a move of type 2 corresponds to sliding one 2-handle over another. A type 1 move changes the 4-manifold by taking the connected sum with a complex projective plane (oriented in either way), or by removing such a summand. Neither manoeuvre changes the boundary of the 4-manifold. The two moves of Theorem 12.15 can be, and indeed have been, explored at length to give many examples of different framed links representing the same manifold [65].
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No
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Lemma 13.2. Suppose that \( {A}^{4} \) is not a \( {k}^{\text{th }} \) root of unity for \( k \leq n \) . Then there is a unique element \( {f}^{\left( n\right) } \in T{L}_{n} \) such that\n\n(i) \( {f}^{\left( n\right) }{e}_{i} = 0 = {e}_{i}{f}^{\left( n\right) } \) for \( 1 \leq i \leq n - 1 \) ,\n\n(ii) \( \left( {{f}^{\left( n\right) } - 1}\right) \) belongs to the algebra generated by \( \left\{ {{e}_{1},{e}_{2},\ldots ,{e}_{n - 1}}\right\} \) ,\n\n(iii) \( {f}^{\left( n\right) }{f}^{\left( n\right) } = {f}^{\left( n\right) } \) and\n\n(iv) \( {\Delta }_{n} = {\left( -1\right) }^{n}\left( {{A}^{2\left( {n + 1}\right) } - {A}^{-2\left( {n + 1}\right) }}\right) /\left( {{A}^{2} - {A}^{-2}}\right) \) .
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Proof. Note that if \( {f}^{\left( n\right) } \) exists, \( \mathbf{1} - {f}^{\left( n\right) } \) is the identity of the algebra generated by \( \left\{ {{e}_{1},{e}_{2},\ldots ,{e}_{n - 1}}\right\} \), and so \( {f}^{\left( n\right) } \) is then certainly unique. Let \( {f}^{\left( 0\right) } \) be the empty diagram (so that \( {\Delta }_{0} = 1 \) ), let \( {f}^{\left( 1\right) } = \mathbf{1} \), and inductively assume that \( {f}^{\left( 2\right) },{f}^{\left( 3\right) },\ldots ,{f}^{\left( n\right) } \) have been defined with the above properties (i),(ii),(iii) and (iv). Observe that (i) and (ii) immediately imply (iii) and that this generalises to the identity shown in Figure 13.4 provided that \( \left( {i + j}\right) \leq n \) .\n\n\n\nFigure 13.4\n\nNow consider the element \( x \), say, of \( T{L}_{n - 1} \) shown at the start of Figure 13.5. The identity of Figure 13.4 implies that \( {f}^{\left( n - 1\right) }x = x \) . But \( {f}^{\left( n - 1\right) }x \) is, by (i), just some scalar multiple \( \lambda \) of \( {f}^{\left( n - 1\right) } \) (because \( x \) is a linear sum of 1 ’s and products of \( {e}_{i} \) ’s); the trick of placing squares in the plane and joining points on the left to points on the right, in the standard way, implies that the scalar \( \lambda \) is \( {\Delta }_{n}/{\Delta }_{n - 1} \) .\n\n\n\nFigure 13.5\n\nSuppose now that \( {A}^{4k} \neq 1 \) for \( k \leq n + 1 \), so that \( {\Delta }_{k} \neq 0 \) for \( k \leq n \) . Define \( {f}^{\left( n + 1\right) } \in T{L}_{n + 1} \) inductively by the equation of Figure 13.6.\n\n\n\nFigure 13.6\n\nProperties (i) and (ii) (and hence (iii)) for \( {f}^{\left( n + 1\right) } \) follow immediately, except perhaps for the fact that \( {f}^{\left( n + 1\right) }{e}_{n} = 0 \) . However, Figure 13.7 shows, using the identities of Figure 13.5 and Figure 13.4, why that also is true.\n\n\n\nFigure 13.7
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Yes
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Lemma 13.4. In \( \mathcal{S}\left( {{S}^{1} \times I,2\text{points}}\right) ,{a\omega } - {b\omega } \) is a linear sum of two elements, each of which contains a copy of \( {f}^{\left( r - 1\right) } \) . (That is, each of the two elements is the image of \( {f}^{\left( r - 1\right) } \) under some map \( T{L}_{r - 1} \rightarrow \mathcal{S}\left( {{S}^{1} \times I,2\text{points}}\right) \) formed by including a square into an annulus and joining up boundary points in some way.)
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Proof. Consider the inclusion, shown in Figure 13.10, of the \( T{L}_{n + 1} \) recurrence relation of Figure 13.6 into the annulus.\n\n\n\nFigure 13.10\n\nThe top boundary points on either side of the square are joined to the two points on the annulus boundary, and the other \( n \) points on the left of the square are joined to the \( n \) on the right by parallel arcs encircling the annulus. As in the proof of Lemma 13.2, the two small squares in the final diagram of Figure 13.10 can be slid together (using \( {f}^{\left( n\right) }{f}^{\left( n\right) } = {f}^{\left( n\right) } \) ) to become one square, and the equality can then be rearranged to become that of Figure 13.11. Sum these equalities from \( n = 0 \) to \( n = r - 2 \) (here \( {\Delta }_{-1} = 0 \) ). The right-hand side is \( {a\omega } \) . Rotate each annulus of Figure 13.11 through \( \pi \) and sum again. The right-hand side is now \( {b\omega } \) . The left-hand sides of the formulae so obtained are almost the same; recalling that \( {\Delta }_{-1} = 0 \), the difference of these left-hand sides is the difference of the first term of Figure 13.11, when \( n = r - 2 \), and its rotation; in each is a copy of \( {f}^{\left( r - 1\right) } \) .\n\n\n\nFigure 13.11
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Yes
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Lemma 13.5. Suppose that \( A \) is chosen so that \( {A}^{4} \) is a primitive \( {r}^{\text{th }} \) root of unity, \( r \geq 3 \) . Suppose that \( D \) is a planar diagram of a link of \( n \) (ordered) components. Suppose that \( {D}^{\prime } \) is another such diagram, obtained from \( D \) by a Kirby type 2 move, in which a parallel of the first component of \( D \) is joined by some band to another component (or, equivalently, a segment of the second component is moved up to and over the first). Then\n\n\[< \omega ,\ldots ,\;{ > }_{D} = < \omega ,\ldots ,\;{ > }_{{D}^{\prime }}.\]
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Proof. It must be checked that the elements of \( \mathcal{S}\left( {\mathbb{R}}^{2}\right) \), produced as described above from \( D \) and from \( {D}^{\prime } \), with \( \omega \) as the \
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No
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Theorem 13.8. Suppose that a closed oriented 3-manifold \( M \) is obtained by surgery on a framed link that is represented by a planar diagram D. Let \( {b}_{ + } \) be the number of positive eigenvalues and \( {b}_{ - } \) be the number negative eigenvalues of the linking matrix of this link. Suppose \( r \geq 3 \) and that \( A \) is a primitive \( 4{r}^{\text{th }} \) root of unity. Then\n\n\[ < \omega ,\omega ,\ldots ,\omega { > }_{D} < \omega { > }_{{U}_{ + }}^{-{b}_{ + }} < \omega { > }_{{U}_{ - }}^{-{b}_{ - }} \]\n\nis a well-defined invariant of \( M \) .
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Proof. Note that \( A \) is a primitive \( 4{r}^{th} \) root of unity, and so, by Lemma 13.7, \( < \omega { > }_{{U}_{ + }} \) and \( < \omega { > }_{{U}_{ - }} \) are non-zero. It follows from the Corollary 13.6 and the preceding remarks about the linking matrix that the given expression is invariant under Kirby type 2 moves. The last two factors make it invariant under Kirby type 1 moves, and regular isotopy of \( D \) just induces regular isotopies of all the diagrams used in defining the expression.
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Yes
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Theorem 13.8. Suppose that a closed oriented 3-manifold \( M \) is obtained by surgery on a framed link that is represented by a planar diagram D. Let \( {b}_{ + } \) be the number of positive eigenvalues and \( {b}_{ - } \) be the number negative eigenvalues of the linking matrix of this link. Suppose \( r \geq 3 \) and that \( A \) is a primitive \( 4{r}^{\text{th }} \) root of unity. Then\n\n\[< \omega ,\omega ,\ldots ,\omega { > }_{D} < \omega { > }_{{U}_{ + }}^{-{b}_{ + }} < \omega { > }_{{U}_{ - }}^{-{b}_{ - }}\]\n\nis a well-defined invariant of \( M \) .
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Proof. Note that \( A \) is a primitive \( 4{r}^{th} \) root of unity, and so, by Lemma 13.7, \( < \omega { > }_{{U}_{ + }} \) and \( < \omega { > }_{{U}_{ - }} \) are non-zero. It follows from the Corollary 13.6 and the preceding remarks about the linking matrix that the given expression is invariant under Kirby type 2 moves. The last two factors make it invariant under Kirby type 1 moves, and regular isotopy of \( D \) just induces regular isotopies of all the diagrams used in defining the expression.
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Yes
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Lemma 13.9. Suppose \( r \geq 3 \) and \( A \) is a primitive \( 4{r}^{\text{th }} \) root of unity. The element of \( T{L}_{n} \) shown in Figure 13.13 is the zero map of outsides if \( 1 \leq n \leq r - 2 \) . When \( n = 0 \), the element acts as multiplication by \( \langle \omega {\rangle }_{U} \) .
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Proof. Consider first the element of \( T{L}_{n} \) that consists of \( {f}^{\left( n\right) } \) encircled by one simple closed curve. This is shown in Figure 13.14 for \( n = 4 \) . Figure 13.14 shows a calculation for that element. Firstly one crossing is removed in the two standard ways, the results being multiplied by \( A \) and \( {A}^{-1} \) and added. The two elements obtained are then simplified by removing kinks and multiplying by \( - {A}^{\pm 3} \) . Now, in the two resulting elements, removal of any of the crossings depicted in one of the standard ways gives zero (as \( {f}^{\left( n\right) }{e}_{i} = 0 \) ), so only the other standard way need be considered. It follows that \( {f}^{\left( n\right) } \) encircled by one simple closed curve is equal, in \( T{L}_{n} \), to \( \left( {-{A}^{2\left( {n + 1}\right) } - {A}^{-2\left( {n + 1}\right) }}\right) {f}^{\left( n\right) } \) . Now the element required in this Lemma is \( {f}^{\left( n\right) } \) encircled by an \( \omega \), regarded as a map of outsides. Let this be denoted \( x \) . A small single unknotted simple closed curve inserted into this changes \( x \), in the usual way, to \( \left( {-{A}^{-2} - {A}^{2}}\right) x \) . However, that small curve can be slid right over the \( \omega \) without (by Lemma 13.5) changing the map of outsides, and then removed altogether (by the preceding paragraph) at the cost of multiplying by \( \left( {-{A}^{2\left( {n + 1}\right) } - {A}^{-2\left( {n + 1}\right) }}\right) \) . Thus \( \left( {-{A}^{-2} - {A}^{2}}\right) x = \) \( \left( {-{A}^{2\left( {n + 1}\right) } - {A}^{-2\left( {n + 1}\right) }}\right) x \) . Hence either \( x = 0 \) or \( {A}^{2\left( {n + 1}\right) } = {A}^{2} \) or \( {A}^{2\left( {n + 1}\right) } = {A}^{-2} \) . The two latter possibilities do not occur for \( 1 \leq n \leq r - 2 \), as \( A \) is a primitive \( 4{r}^{th} \) root of unity, so then \( x = 0 \) . When \( n = 0 \), it is trivial that \( x \) acts as multiplication by \( \langle \omega {\rangle }_{U} \) because there is nothing but the curve labelled \( \omega \) to consider.
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Yes
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Lemma 14.2. The element of \( T{L}_{n} \) shown in Figure 14.2, which consists of the idempotent \( {f}^{\left( n\right) } \) with all its strands encircled by a parallel strands that join up the ends of an idempotent \( {f}^{\left( a\right) } \), is\n\n\[ \n{\left( -1\right) }^{a}\frac{{A}^{2\left( {n + 1}\right) \left( {a + 1}\right) } - {A}^{-2\left( {n + 1}\right) \left( {a + 1}\right) }}{{A}^{2\left( {n + 1}\right) } - {A}^{-2\left( {n + 1}\right) }}{f}^{\left( n\right) }.\n\]
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Proof. The \( a \) parallel strands and the idempotent \( {f}^{\left( a\right) } \) can, as explained in Chapter 13, be thought of as \( {S}_{a}\left( \alpha \right) \) contained in an annulus encircling the strands of \( {f}^{\left( n\right) } \), where \( {S}_{a} \) is the \( {a}^{\text{th }} \) Chebyshev polynomial. Now, as in the proof of Lemma 13.9, \( {f}^{\left( n\right) } \) with a single strand encircling it (to be thought of as \( \alpha \) in the annulus) is \( \left( {-{A}^{2\left( {n + 1}\right) } - {A}^{-2\left( {n + 1}\right) }}\right) {f}^{\left( n\right) } \) . Hence the element required here is \( {S}_{a}\left( {-{A}^{2\left( {n + 1}\right) } - {A}^{-2\left( {n + 1}\right) }}\right) {f}^{\left( n\right) } \) . This immediately gives the result using the remarks about Chebyshev polynomials after Lemma 13.2.
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Yes
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Lemma 14.3. Suppose \( A \) is a primitive \( 4{r}^{\text{th }} \) root of unity. Then\n\n\[< \omega { > }_{{U}_{ + }} = \frac{G}{2{A}^{\left( 3 + {r}^{2}\right) }\left( {{A}^{2} - {A}^{-2}}\right) },\]
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Proof. Recall that \( {U}_{ + } \) is the diagram of the unknot with one positive crossing,\n\n\[ \omega = \mathop{\sum }\limits_{{n = 0}}^{{r - 2}}{\Delta }_{n}{S}_{n}\left( \alpha \right) \text{ and }{\Delta }_{n} = \frac{{\left( -1\right) }^{n}\left( {{A}^{2\left( {n + 1}\right) } - {A}^{-2\left( {n + 1}\right) }}\right) }{{A}^{2} - {A}^{-2}}.\]\n\nSo use of Lemma 14.1 to remove the kink in each \( {S}_{n}\left( \alpha \right) \) shows that \( \langle \omega {\rangle }_{{U}_{ + }} \) is\n\n\[ \mathop{\sum }\limits_{{n = 0}}^{{r - 2}}{\Delta }_{n}^{2}{\left( -1\right) }^{n}{A}^{{n}^{2} + {2n}} = {\left( {A}^{2} - {A}^{-2}\right) }^{-2}\mathop{\sum }\limits_{{n = 0}}^{{r - 2}}{\left( -1\right) }^{n}{A}^{{n}^{2} + {2n}}{\left( {A}^{2\left( {n + 1}\right) } - {A}^{-2\left( {n + 1}\right) }\right) }^{2}.\]\n\nNow elementary manoeuvres of algebraic number theory (see [74] for example) show that the summation in the last term is \( \frac{1}{2}{A}^{-\left( {3 + {r}^{2}}\right) }\left( {{A}^{2} - {A}^{-2}}\right) \mathop{\sum }\limits_{{n = 1}}^{{4r}}{A}^{{n}^{2}} \) .
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Yes
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