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| The '''signature of a knot''' is a [[topological invariant]] in [[knot theory]]. It may be computed from the [[Seifert surface]].
| | Macie Maskell is what my [http://thesaurus.com/browse/husband+loves husband loves] to call me but I never really liked that name. My friends say it's bad for me but the things i love doing is flower arranging but I'm thinking on starting something replacement. Some time ago he chose to reside Guam as well as will never move. I am currently a production and planning officer therefore don't think I'll transform anytime in the. My husband and I maintain a website. You'll probably decide to check it out here: http://www.pinterest.com/seodress/placerville-real-estate-your-realtor-for-everythin/ |
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| Given a [[knot (mathematics)|knot]] ''K'' in the [[3-sphere]], it has a [[Seifert surface]] ''S'' whose boundary is ''K''. The '''[[Seifert form]]''' of ''S'' is the pairing <math>\phi : H_1(S) \times H_1(S) \to \mathbb Z</math> given by taking the [[linking number]] <math>lk(a^+,b^-)</math> where <math>a, b \in H_1(S)</math> and <math>a^+, b^-</math> indicate the translates of ''a'' and ''b'' respectively in the positive and negative directions of the [[normal bundle]] to ''S''.
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| Given a basis <math>b_1,...,b_{2g}</math> for <math>H_1(S)</math> (where ''g'' is the genus of the surface) the Seifert form can be represented as a ''2g''-by-''2g'' [[Seifert matrix]] ''V'', <math>V_{ij}=\phi(b_i,b_j)</math>. The [[Symmetric bilinear form|signature]] of the matrix <math>V+V^\perp</math>, thought of as a symmetric bilinear form, is the signature of the knot ''K''.
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| [[Slice knot]]s are known to have zero signature.
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| ==The Alexander module formulation==
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| Knot signatures can also be defined in terms of the [[Alexander polynomial|Alexander module]] of the knot complement. Let <math>X</math> be the universal abelian cover of the knot complement. Consider the Alexander module to be the first homology group of the universal abelian cover of the knot complement: <math>H_1(X;\mathbb Q)</math>. Given a <math>\mathbb Q[\mathbb Z]</math>-module <math>V</math>, let <math>\overline{V}</math> denote the <math>\mathbb Q[\mathbb Z]</math>-module whose underlying <math>\mathbb Q</math>-module is <math>V</math> but where <math>\mathbb Z</math> acts by the inverse covering transformation. Blanchfield's formulation of [[Poincaré duality]] for <math>X</math> gives a canonical isomorphism <math>H_1(X;\mathbb Q) \simeq \overline{H^2(X;\mathbb Q)}</math> where <math>H^2(X;\mathbb Q)</math> denotes the 2nd cohomology group of <math>X</math> with compact supports and coefficients in <math>\mathbb Q</math>. The universal coefficient theorem for <math>H^2(X;\mathbb Q)</math> gives a canonical isomorphism with <math>Ext_{\mathbb Q[\mathbb Z]}(H_1(X;\mathbb Q),\mathbb Q[\mathbb Z])</math> (because the Alexander module is <math>\mathbb Q[\mathbb Z]</math>-torsion). Moreover, just like in the [[Poincaré duality|quadratic form formulation of Poincaré duality]], there is a canonical isomorphism of <math>\mathbb Q[\mathbb Z]</math>-modules <math> Ext_{\mathbb Q[\mathbb Z]}(H_1(X;\mathbb Q),\mathbb Q[\mathbb Z]) \simeq Hom_{\mathbb Q[\mathbb Z]}(H_1(X;\mathbb Q),[\mathbb Q[\mathbb Z]]/\mathbb Q[\mathbb Z] )</math>, where <math>[\mathbb Q[\mathbb Z]]</math> denotes the field of fractions of <math>\mathbb Q[\mathbb Z]</math>. This isomorphism can be thought of as a sesquilinear duality pairing <math>H_1(X;\mathbb Q) \times H_1(X;\mathbb Q) \to [\mathbb Q[\mathbb Z]]/\mathbb Q[\mathbb Z]</math> where <math>[\mathbb Q[\mathbb Z]]</math> denotes the field of fractions of <math>\mathbb Q[\mathbb Z]</math>. This form takes value in the rational polynomials whose denominators are the [[Alexander polynomial]] of the knot, which as a <math>\mathbb Q[\mathbb Z]</math>-module is isomorphic to <math>\mathbb Q[\mathbb Z]/\Delta K</math>. Let <math>tr : \mathbb Q[\mathbb Z]/\Delta K \to \mathbb Q</math> be any linear function which is invariant under the involution <math>t \longmapsto t^{-1}</math>, then composing it with the sesquilinear duality pairing gives a symmetric bilinear form on <math>H_1 (X;\mathbb Q)</math> whose signature is an invariant of the knot.
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| All such signatures are concordance invariants, so all signatures of [[slice knot]]s are zero. The sesquilinear duality pairing respects the prime-power decomposition of <math>H_1 (X;\mathbb Q)</math> -- ie: the prime power decomposition gives an orthogonal decomposition of <math>H_1 (X;\mathbb R)</math>. Cherry Kearton has shown how to compute the ''Milnor signature invariants'' from this pairing, which are equivalent to the ''Tristram-Levine invariant''.
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| ==References==
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| * C.Gordon, Some aspects of classical knot theory. Springer Lecture Notes in Mathematics 685. Proceedings Plans-sur-Bex Switzerland 1977.
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| * J.Hillman, Algebraic invariants of links. Series on Knots and everything. Vol 32. World Scientific.
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| * C.Kearton, Signatures of knots and the free differential calculus, Quart. J. Math. Oxford (2), 30 (1979).
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| * J.Levine, Knot cobordism groups in codimension two, Comment. Math. Helv. 44, 229-244 (1969)
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| * J.Milnor, Infinite cyclic coverings, J.G. Hocking, ed. Conf. on the Topology of Manifolds, Prindle, Weber and Schmidt, Boston, Mass, 1968 pp. 115–133.
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| * K.Murasugi, On a certain numerical invariant of link types, Trans. Amer. Math. Soc. 117, 387-482 (1965)
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| * A.Ranicki [http://www.maths.ed.ac.uk/~aar/slides/durham.pdf On signatures of knots] Slides of lecture given in Durham on 20 June 2010.
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| * H.Trotter, Homology of group systems with applications to knot theory, Ann. of Math. (2) 76, 464-498 (1962)
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| ==See also==
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| *[[Link concordance]]
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| {{Use dmy dates|date=September 2010}}
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| {{DEFAULTSORT:Signature Of A Knot}}
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| [[Category:Knot theory]]
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Macie Maskell is what my husband loves to call me but I never really liked that name. My friends say it's bad for me but the things i love doing is flower arranging but I'm thinking on starting something replacement. Some time ago he chose to reside Guam as well as will never move. I am currently a production and planning officer therefore don't think I'll transform anytime in the. My husband and I maintain a website. You'll probably decide to check it out here: http://www.pinterest.com/seodress/placerville-real-estate-your-realtor-for-everythin/