aa r X i v : . [ m a t h . G T ] A p r LEFT ORDERABLE SURGERIES OF DOUBLE TWIST KNOTS
ANH T. TRAN
Abstract.
A rational number r is called a left orderable slope of a knot K ⊂ S ifthe 3-manifold obtained from S by r -surgery along K has left orderable fundamentalgroup. In this paper we consider the double twist knots C ( k, l ) in the Conway notation.For any positive integers m and n , we show that if K is a double twist knot of theform C (2 m, − n ), C (2 m + 1 , n ) or C (2 m + 1 , − n ) then there is an explicit unboundedinterval I such that any rational number r ∈ I is a left orderable slope of K . Introduction
The motivation of this paper is the L-space conjecture of Boyer, Gordon and Watson[BGW] which states that an irreducible rational homology 3-sphere is an L-space if andonly if its fundamental group is not left orderable. Here a rational homology 3-sphere Y isan L-space if its Heegaard Floer homology c HF( Y ) has rank equal to the order of H ( Y ; Z ),and a non-trivial group G is left orderable if it admits a total ordering < such that g < h implies f g < f h for all elements f, g, h in G . A knot K in S is called an L-space knotif it admits a positive Dehn surgery yielding an L-space. It is known that non-torusalternating knots are not L-space knots, see [OS]. In view of the L-space conjecture,this would imply that any non-trivial Dehn surgery along a non-torus alternating knotproduces a 3-manifold with left orderable fundamental group.A rational number r is called a left orderable slope of a knot K ⊂ S if the 3-manifoldobtained from S by r -surgery along K has left orderable fundamental group. As men-tioned above, one would expect that any rational number is a left orderable slope of anynon-torus alternating knot. It is known that any rational number r ∈ ( − ,
4) is a leftorderable slope of the figure eight knot, and any rational number r ∈ [0 ,
4] is a left order-able slope of the hyperbolic twist knot 5 , see [BGW] and [HTe2] respectively. Considerthe double twist knot C ( k, l ) in the Conway notation as in Figure 1, where k, l denote thenumbers of horizontal half-twists with sign in the boxes. Here the sign of is positive inthe box k and is negative in the box l . Then the following results were shown in [HTe1, Tr]by using continuous families of hyperbolic SL ( R )-representations of knot groups. If m, n are integers ≥
1, any rational number r ∈ ( − n, m ) is a left orderable slope of C (2 m, n ).If m, n are integers ≥ r ∈ [0 , max { m, n } ) is a left orderableslope of C (2 m, − n ) and any rational number r ∈ [0 ,
4] is a left orderable slope of both
Mathematics Subject Classification . Primary 57M27, Secondary 57M25.
Key words and phrases.
Dehn surgery, left orderable, L-space, Riley polynomial, double twist knot. C (2 m, −
2) and C (2 , − n ). Note that C (2 ,
2) is the figure eight knot and C (4 , −
2) is thetwist knot 5 . Moreover C (2 , −
2) is the trefoil knot, which is the (2 , Figure 1.
The double twist knot/link C ( k, l ) in the Conway notation.In this paper, by using continuous families of elliptic SL ( R )-representations of knotgroups we extend the range of left orderable slopes of C (2 m, − n ). Moreover, we alsogive left orderable slopes of C (2 m + 1 , ± n ). Theorem 1.
Suppose K is a double twist knot of the form C (2 m, − n ) , C (2 m + 1 , n ) or C (2 m + 1 , − n ) in the Conway notation for some positive integers m and n . Let LO K = ( −∞ , if K = C (2 m, − n ) , ( −∞ , n − if K = C (2 m + 1 , n ) , (3 − n, ∞ ) if K = C (2 m + 1 , − n ) and n ≥ . Then any rational number r ∈ LO K is a left orderable slope of K . Combining this with results in [HTe1, Tr], we conclude that if m and n are integers ≥ r ∈ ( −∞ , max { m, n } ) is a left orderable slope of C (2 m, − n )and any rational number r ∈ ( −∞ ,
4] is a left orderable slope of both C (2 m, −
2) and C (2 , − n ). In the subsequent paper [KTT] we will use continuous families of hyperbolicSL ( R )-representations of knot groups to extend the range of left orderable slopes of C (2 m + 1 , − n ). More specifically, we will show that any rational number r ∈ ( − n, m )is a left orderable slope of C (2 m + 1 , − n ) detected by hyperbolic SL ( R )-representationsof the knot group.We remark that in the case of C (2 m +1 , ± n ), where m and n are positive integers, Gao[Ga] independently obtains similar results. She proves a weaker result that any rationalnumber r ∈ ( −∞ ,
1) is a left orderable slope of C (2 m + 1 , n ) and a stronger result thatany rational number r ∈ ( − n, ∞ ) is a left orderable slope of C (2 m + 1 , − n ).As in [BGW, HTe1, HTe2, Tr, CD] the proof of Theorem 1 is based on the existence ofcontinuous families of elliptic SL ( R )-representations of the knot groups of double twistknots C (2 m, − n ) and C (2 m + 1 , ± n ) into SL ( R ) and the fact that ^ SL ( R ), which isthe universal covering group of SL ( R ), is a left orderable group. EFT ORDERABLE SURGERIES OF DOUBLE TWIST KNOTS 3
This paper is organized as follows. In Section 1, we study certain real roots of the Rileypolynomial of double twist knots C ( k, − p ), whose zero locus describes all non-abelianrepresentations of the knot group into SL ( C ). In Section 2, we prove Theorem 1.2. Real roots of the Riley polynomial
For a knot K in S , let G ( K ) denote the knot group of K which is the fundamentalgroup of the complement of an open tubular neighborhood of K .Consider the double twist knot/link C ( k, l ) in the Conway notation as in Figure 1,where k, l are integers such that | kl | ≥
3. Note that C ( k, l ) is the rational knot/linkcorresponding to continued fraction k + 1 /l . It is easy to see that C ( k, l ) is the mirrorimage of C ( l, k ) = C ( − k, − l ). Moreover, C ( k, l ) is a knot if kl is even and is a two-component link if kl is odd. In this paper, we only consider knots and so we can assumethat k > l = − p is even.Note that C ( k, − p ) is the mirror image of the double twist knot J ( k, p ) in [HS].Then, by [HS], the knot group of C ( k, − p ) has a presentation G ( C ( k, − p )) = h a, b | aw p = w p b i where a, b are meridians and w = ( ( ab − ) m ( a − b ) m if k = 2 m, ( ab − ) m ab ( a − b ) m if k = 2 m + 1 . Moreover, the canonical longitude of C ( k, − p ) corresponding to the meridian µ = a is λ = ( w p ( w p ) ∗ a − ε ) − , where ε = 0 if k = 2 m and ε = 2 p if k = 2 m + 1. Here, for a word u in the letters a, b we let u ∗ be the word obtained by reading v backwards.Suppose ρ : G ( C ( k, − p )) → SL ( C ) is a nonabelian representation. Up to conjugation,we may assume that(2.1) ρ ( a ) = (cid:20) M M − (cid:21) and ρ ( b ) = (cid:20) M − y M − (cid:21) where ( M, y ) ∈ C satisfies the matrix equation ρ ( aw p ) = ρ ( w p b ). It is known that thismatrix equation is equivalent to a single polynomial equation R C ( k, − p ) ( x, y ) = 0, where x = (tr ρ ( a )) and R K ( x, y ) is the Riley polynomial of K , see [Ri]. This polynomial canbe described via the Chebychev polynomials as follows.Let { S j ( v ) } j ∈ Z be the Chebychev polynomials in the variable v defined by S ( v ) = 1, S ( v ) = v and S j ( v ) = vS j − ( v ) − S j − ( v ) for all integers j . Note that S j ( v ) = − S − j − ( v )and S j ( ±
2) = ( ± j ( j + 1). Moreover, we have S j ( v ) = ( s j +1 − s − ( j +1) ) / ( s − s − ) for v = s + s − = ±
2. Using this identity one can prove the following.
Lemma 2.1.
For any integer j and any positive integer n we have (1) S j ( v ) − vS j ( v ) S j − ( v ) + S j − ( v ) = 1 . (2) S n ( v ) − S n − ( v ) = Q nj =1 (cid:0) v − (2 j − π n +1 (cid:1) . ANH T. TRAN (3) S n ( v ) + S n − ( v ) = Q nj =1 (cid:0) v − jπ n +1 (cid:1) . (4) S n ( v ) = Q nj =1 (cid:0) v − jπn +1 (cid:1) . The Riley polynomial of C ( k, − p ), whose zero locus describes all non-abelian repre-sentations of the knot group of C ( k, − p ) into SL ( C ), is R C ( k, − p ) ( x, y ) = S p ( t ) − zS p − ( t )where t = tr ρ ( w ) = ( y + 2 − x )( y − S m − ( y ) if k = 2 m, − ( y + 2 − x )( S m ( y ) − S m − ( y )) if k = 2 m + 1 , and z = ( y + 2 − x ) S m − ( y )( S m ( y ) − S m − ( y )) if k = 2 m, − ( y + 2 − x ) S m ( y )( S m ( y ) − S m − ( y )) if k = 2 m + 1 . Moreover, for the representation ρ : G ( C ( k, − p )) → SL ( C ) of the form (2.1) the imageof the canonical longitude λ = ( w p ( w p ) ∗ a − ε ) − has the form ρ ( λ ) = (cid:20) L ∗ L − (cid:21) , where L = − M − ( S m ( y ) − S m − ( y )) − M ( S m − ( y ) − S m − ( y )) M ( S m ( y ) − S m − ( y )) − M − ( S m − ( y ) − S m − ( y )) if k = 2 m and L = − M p M − S m ( y ) − M S m − ( y ) M S m ( y ) − M − S m − ( y ) if k = 2 m + 1 . See e.g. [Tr, Pe].Lemmas (2.2)–(2.4) below describe continuous families of real roots of the Riley poly-nomials of the double twist knots C (2 m, − n ), C (2 m + 1 , n ) and C (2 m + 1 , − n ) re-spectively, where m and n are positive integers. Lemma 2.2.
There exists a continuous real function y : [4 − / ( mn ) , → [2 , ∞ ) in thevariable x such that • y (4 − / ( mn )) = 2 and • R C (2 m, − n ) ( x, y ( x )) = 0 for all x ∈ [4 − / ( mn ) , .Proof. Let K = C (2 m, − n ). We have R K ( x, y ) = S n ( t ) − zS n − ( t ) where t = 2 + ( y + 2 − x )( y − S m − ( y ) ,z = 1 + ( y + 2 − x ) S m − ( y )( S m ( y ) − S m − ( y )) . Consider real numbers x ∈ [4 − / ( mn ) ,
4] and y ∈ [2 , ∞ ). Since y ≥ ≥ x −
2, we have t ≥ z ≥
1. This implies that zS n − ( t ) − S n − ( t ) ≥ S n − ( t ) − S n − ( t ) >
0, by Lemma2.1. The equation R K ( x, y ) = 0 is then equivalent to(2.2) (cid:0) S n ( t ) − zS n − ( t ) (cid:1)(cid:0) S n − ( t ) − zS n − ( t ) (cid:1) = 0 . EFT ORDERABLE SURGERIES OF DOUBLE TWIST KNOTS 5
Let P ( x, y ) denote the left hand side of equation (2.2). By Lemma 2.1, we have S n ( t ) − tS n ( t ) S n − ( t ) + S n − ( t ) = 1. This can be written as S n ( t ) S n − ( t ) = S n − ( t ) −
1. Fromthis and S n ( t ) + S n − ( t ) = tS n − ( t ) we get P ( x, y ) = ( z − tz + 1) S n − ( t ) − . By a direct calculation, using S m ( y ) + S m − ( y ) − yS m ( y ) S m − ( y ) = 1, we have z − tz + 1= ( z − − ( t − z = ( y + 2 − x ) S m − ( y )( S m ( y ) − S m − ( y )) − ( y + 2 − x )( y − S m − ( y ) (cid:2) y + 2 − x ) S m − ( y )( S m ( y ) − S m − ( y )) (cid:3) = ( y + 2 − x ) S m − ( y ) (cid:2) − x + ( y + 2 − x )( y − S m − ( y ) (cid:3) = ( y + 2 − x ) S m − ( y )( t + 2 − x ) . Hence P ( x, y ) = ( y + 2 − x ) S m − ( y )( t + 2 − x ) S n − ( t ) − l the Chebychev polynomial S l ( v ) = Q lj =1 ( v − jπl +1 ) is a strictly increasing function in v ∈ [2 , ∞ ). This implies that, for a fixed realnumber x ∈ [4 − / ( mn ) , t = 2 + ( y + 2 − x )( y − S m − ( y ) ≥ P ( x, y ) = ( y + 2 − x ) S m − ( y )( t + 2 − x ) S n − ( t ) − y ∈ [2 , ∞ ). Note that lim y →∞ P ( x, y ) = ∞ andlim y → + P ( x, y ) = P ( x,
2) = (4 − x ) m n − ≤ . Hence there exists a unique real number y ( x ) ∈ [2 , ∞ ) such that P ( x, y ( x )) = 0. Since P (4 − mn ,
2) = 0 we have y (4 − mn ) = 2. Finally, by the implicit function theorem y = y ( x ) is a continuous function in x ∈ [4 − / ( mn ) , (cid:3) Lemma 2.3.
There exists a continuous real function x : [2 , ∞ ) → (4 cos n − π n +2 , ∞ ) inthe variable y such that • x (2) < n − π n +2 , • lim y →∞ x ( y ) = ∞ and • R C (2 m +1 , n ) ( x ( y ) , y ) = 0 for all y ∈ [2 , ∞ ) .Proof. Let K = C (2 m + 1 , n ). We have R K ( x, y ) = S − n ( t ) − zS − n − ( t ) where t = 2 − ( y + 2 − x )( S m ( y ) − S m − ( y )) ,z = 1 − ( y + 2 − x ) S m ( y )( S m ( y ) − S m − ( y )) . Note that R K ( x, y ) = ( t − z ) S − n − ( t ) − S − n − ( t ) = S n ( t ) − ( t − z ) S n − ( t ).By Lemma 2.1 we have S n ( t ) − S n − ( t ) = n Y j =1 (cid:16) t − j − π n + 1 (cid:17) , ANH T. TRAN S n ( t ) + S n − ( t ) = n Y j =1 (cid:16) t − jπ n + 1 (cid:17) . Let t j = 2 cos jπ n +1 for j = 1 , · · · , n . By writing t j − = e iθ + e − iθ where θ = (2 j − π n +1 ,we have S n ( t j − ) = e i ( n +1) θ − e − i ( n +1) θ e iθ − e − iθ = sin (2 j − n +1) π n +1 sin (2 j − π n +1 = sin (cid:16) jπ − π + (2 j − π n +1) (cid:17) sin (2 j − π n +1 = ( − j − cos (2 j − π n +1) sin (2 j − π n +1 . This implies that ( − j − S n ( t j − ) >
0. Similarly, ( − j S n ( t j ) > y ≥
2. Let s j ( y ) = y + 2 − − t j ( S m ( y ) − S m − ( y )) for j = 1 , · · · , n . We alsolet s = y +2. Since − < t n < · · · < t < s n ( y ) < · · · < s ( y ) < y +2 = s ( y ).At x = s j − ( y ) we have t = t j − and so S n ( t ) = S n − ( t ). This implies that R K ( s j − ( y ) , y ) = (1 − ( t − z )) S n ( t j − )= − ( y + 2 − s j − ( y )) S m − ( y )( S m ( y ) − S m − ( y )) S n ( t j − ) . Since y ≥
2, by Lemma 2.1 we have S m ( y ) − S m − ( y ) ≥ S m (2) − S m − (2) = 1 and S m − ( y ) ≥ S m − (2) = m . Hence ( − j R K ( s j − ( y ) , y ) > ≤ j ≤ n we have R K ( s j ( y ) , y ) = (1 + t − z ) S n ( t j )= (cid:2) y + 2 − s j − ( y )) S m − ( y )( S m ( y ) − S m − ( y ) (cid:3) S n ( t j ) , which implies that ( − j R K ( s j ( y ) , y ) > ≤ j ≤ n −
1, since R K ( s j +1 ( y ) , y ) R K ( s j ( y ) , y ) < x j ( y ) ∈ ( s j +1 ( y ) , s j ( y )) such that R K ( x j ( y ) , y ) = 0 . Since R K ( s ( y ) , y ) = R K ( y + 2 , y ) = 1and R K ( s ( y ) , y ) < x ( y ) ∈ ( s ( y ) , s ( y )) such that R K ( x ( y ) , y ) = 0.Since R K ( x, y ) = zS n − ( t ) − S n − ( t ), we see that R K ( x, y ) is a polynomial of degree n in x for each fixed real number y ≥
2. This polyomial has exactly n simple real roots x ( y ) , · · · , x n − ( y ) satisfying x n − ( y ) < · · · < x ( y ) < y + 2, hence the implicit functiontheorem implies that each x j ( y ) is a continuous function in y ≥ x ( y ) = x n − ( y ) for y ≥
2, we have R K ( x ( y ) , y ) = 0. Moreover, since x ( y ) > s n − ( y ) = y + 2 − − (2 n − π n +1 ( S m ( y ) − S m − ( y )) we have lim y →∞ x ( y ) = ∞ and x ( y ) > − (cid:0) − (2 n − π n +1 (cid:1) = 4 cos n − π n +2 for y ≥ EFT ORDERABLE SURGERIES OF DOUBLE TWIST KNOTS 7
Finally, since x ( y ) < s n − ( y ) for all y ≥ x (2) < s n − (2) = 4 cos n − π n +2 . (cid:3) Lemma 2.4.
Suppose n ≥ . Then there exists a continuous real function x : [2 , ∞ ) → (4 cos n − π n +2 , ∞ ) in the variable y such that • x (2) < n − π n +2 , • lim y →∞ x ( y ) = ∞ and • R C (2 m +1 , − n ) ( x ( y ) , y ) = 0 for all y ∈ [2 , ∞ ) .Proof. Let K = C (2 m + 1 , − n ). We have R K ( x, y ) = S n ( t ) − zS n − ( t ) where t = 2 − ( y + 2 − x )( S m ( y ) − S m − ( y )) ,z = 1 − ( y + 2 − x ) S m ( y )( S m ( y ) − S m − ( y )) . Fix a real number y ≥
2. Choose t j and s j ( y ) for 1 ≤ j ≤ n as in Lemma 2.3. Since R K ( s j − ( y ) , y ) = (1 − z ) S n ( t j − )= ( y + 2 − s j − ( y )) S m ( y )( S m ( y ) − S m − ( y )) S n ( t j − ) , we have ( − j − R K ( s j − ( y ) , y ) >
0. Hence, there exists x j ( y ) ∈ ( s j +1 ( y ) , s j − ( y )) suchthat R K ( x j ( y ) , y ) = 0 for each 1 ≤ j ≤ n − R K ( x, y ) = ( t − z ) S n − ( t ) − S n − ( t ) and noting that t − z = 1 + ( y + 2 − x )( S m ( y ) − S m − ( y )) S m − ( y ) , we see that R K ( x, y ) is a polynomial of degree n in x with negative highest coefficient foreach fixed real number y ≥
2. Since lim x →∞ R K ( x, y ) = −∞ and R K ( y + 2 , y ) = 1, thereexists x ( y ) ∈ ( y + 2 , ∞ ) such that R K ( x ( y ) , y ) = 0. For a fixed real number y ≥
2, thepolynomial R K ( x, y ) of degree n in x has exactly n simple real roots x ( y ) , · · · , x n − ( y )satisfying x n − ( y ) < · · · < x ( y ) < y + 2 < x ( y ), hence the implicit function theoremimplies that each x j ( y ) is a continuous function in y ≥ x ( y ) = x n − ( y ) for y ≥
2, we have R K ( x ( y ) , y ) = 0. Moreover, since x ( y ) > s n − ( y ) = y + 2 − − (2 n − π n +1 ( S m ( y ) − S m − ( y )) we have lim y →∞ x ( y ) = ∞ and x ( y ) > − (cid:0) − (2 n − π n +1 (cid:1) = 4 cos n − π n +2 for y ≥ x ( y ) < s n − ( y ) for all y ≥ x (2) < s n − (2) = 4 cos n − π n +2 . (cid:3) Proof of Theorem 1
Suppose K is a double twist knot of the form C (2 m, − n ), C (2 m + 1 , n ) or C (2 m +1 , − n ) in the Conway notation for some positive integers m and n . Let X be thecomplement of an open tubular neighborhood of K in S , and X r the 3-manifold obtained ANH T. TRAN from S by r -surgery along K . Recall thatLO K = ( −∞ ,
1) if K = C (2 m, − n ) , ( −∞ , n −
1) if K = C (2 m + 1 , n ) , (3 − n, ∞ ) if K = C (2 m + 1 , − n ) and n ≥ . An element of SL ( R ) is called elliptic if its trace is a real number in ( − , ρ : Z → SL ( R ) is called elliptic if the image group ρ ( Z ) contains anelliptic element of SL ( R ). In which case, since Z is an abelian group every non-trivialelement of ρ ( Z ) must also be elliptic.Using Lemmas 2.2–2.4 we first prove the following. Proposition 3.1.
For each rational number r ∈ LO K \ { } there exists a representa-tion ρ : π ( X r ) → SL ( R ) such that ρ (cid:12)(cid:12) π ( ∂X ) : π ( ∂X ) ∼ = Z → SL ( R ) is an ellipticrepresentation.Proof. We first consider the case K = C (2 m, − n ). Let θ = arccos p − / (4 mn ). For θ ∈ (0 , θ ) ∪ ( π − θ , π ) we let x = 4 cos θ . Then x ∈ (4 − / ( mn ) , y : [4 − / ( mn ) , → [2 , ∞ )in Lemma 2.2. Let M = e iθ . Then x = 4 cos θ = ( M + M − ) . Since R K ( x, y ( x )) = 0there exists a non-abelian representation ρ : π ( X ) → SL ( C ) such that ρ ( a ) = (cid:20) M M − (cid:21) and ρ ( b ) = (cid:20) M − y ( x ) M − (cid:21) . Note that x is the square of the trace of a meridian. Moreover, the image of the canonicallongitude λ corresponding to the meridian µ = a has the form ρ ( λ ) = (cid:20) L ∗ L − (cid:21) , where L = − M − α − M βM α − M − β and α = S m ( y ( x )) − S m − ( y ( x )), β = S m − ( y ( x )) − S m − ( y ( x )). Note that α > β > y ( x ) > | L | = √ L ¯ L = 1, where ¯ L denotes the complex conjugate of L .Moreover, by a direct calculation, we haveRe( L ) = (cid:0) αβ − ( α + β ) cos 2 θ (cid:1) / | M α − M − β | , Im( L ) = ( α − β ) sin 2 θ/ | M α − M − β | . Note that Im( L ) > θ ∈ (0 , θ ) and Im( L ) < θ ∈ ( π − θ , π ). Let ϕ ( θ ) = ( arccos (cid:2)(cid:0) αβ − ( α + β ) cos 2 θ (cid:1) / | e iθ α − e − iθ β | (cid:3) if θ ∈ (0 , θ ) , − arccos (cid:2)(cid:0) αβ − ( α + β ) cos 2 θ (cid:1) / | e iθ α − e − iθ β | (cid:3) if θ ∈ ( π − θ , π ) . Then L = e iϕ ( θ ) . Note that ϕ ( θ ) ∈ (0 , π ) if θ ∈ (0 , θ ) and ϕ ( θ ) ∈ ( − π,
0) if θ ∈ ( π − θ , π ). EFT ORDERABLE SURGERIES OF DOUBLE TWIST KNOTS 9
The function f ( θ ) := − ϕ ( θ ) θ is a continuous function on each of the intervals (0 , θ ) and( π − θ , π ). As θ → + we have M → L = − M − α − MβMα − M − β → −
1, so ϕ ( θ ) → π . As θ → θ − we have x → − / ( mn ), y ( x ) → α, β →
1, so L = − M − α − MβMα − M − β → ϕ ( θ ) →
0. This implies thatlim θ → + − ϕ ( θ ) θ = −∞ and lim θ → θ − − ϕ ( θ ) θ = 0 . Hence the image of f ( θ ) on the interval (0 , θ ) contains the interval ( −∞ , θ → ( π − θ ) + − ϕ ( θ ) θ = 0 and lim θ → π − − ϕ ( θ ) θ = 1 , the image of f ( θ ) on the interval ( π − θ , π ) contains the interval (0 , r = pq is a rational number such that r ∈ ( −∞ , ∪ (0 , r = f ( θ ) = − ϕ ( θ ) θ for some θ ∈ (0 , θ ) ∪ ( π − θ , π ). Since M p L q = e i ( pθ + qϕ ( θ )) = 1, we have ρ ( µ p λ q ) = I . This means that the non-abelian representation ρ : π ( X ) → SL ( C ) extends to arepresentation ρ : π ( X r ) → SL ( C ). Finally, since 2 − y ( x ) <
0, a result in [Kh, page 786]implies that ρ can be conjugated to an SL ( R )-representation. Note that the restrictionof this representation to the peripheral subgroup π ( ∂X ) of the knot group is an ellipticrepresentation. This completes the proof of Proposition 3.1 for K = C (2 m, − n ).We now consider the case K = C (2 m + 1 , n ). Consider the continuous real function x : [2 , ∞ ) → (cid:16) (2 n − π n + 2 , ∞ (cid:17) in Lemma 2.3. Since x (2) < n − π n +2 and lim y →∞ x ( y ) = ∞ , there exists y ∗ > x ( y ∗ ) = 4 and 4 cos n − π n +2 < x ( y ) < y ∈ [2 , y ∗ ).For each y ∈ [2 , y ∗ ) we let θ ( y ) = arccos( p x ( y ) / θ (2) > (2 n − π n +2 , and for y ∈ [2 , y ∗ ) we have 0 < θ ( y ) < (2 n − π n +2 and x ( y ) = 4 cos θ ( y ). Since R K ( x ( y ) , y ) = 0 thereexists a non-abelian representation ρ : π ( X ) → SL ( C ) such that ρ ( a ) = (cid:20) M M − (cid:21) and ρ ( b ) = (cid:20) M − y M − (cid:21) , where M = e iθ ( y ) . Moreover, the image of the canonical longitude λ corresponding to themeridian µ = a has the form ρ ( λ ) = (cid:20) L ∗ L − (cid:21) , where L = − M − n M − γ − M δM γ − M − δ and γ = S m ( y ), δ = S m − ( y ). Note that γ > δ >
0, since y > L = e iϕ ( y ) where ϕ ( y ) = (2 n − π − nθ ( y ) + arccos (cid:2)(cid:0) γδ − ( γ + δ ) cos 2 θ ( y ) (cid:1) / | e iθ ( y ) γ − e − iθ ( y ) δ | (cid:3) . Since (2 n − π n +2 < θ (2) < (2 n − π n +2 we have − π n +1 < ϕ (2) < π − π n +1 .As y → + , ρ approaches a reducible representation and so L → , ϕ ( y ) → ϕ (2) = k π for some integer k . Since − π n +1 < ϕ (2) < π − π n +1 , we must have ϕ (2) = 0. As y → ( y ∗ ) − , we have x ( y ) → M → , L = − M − n M − γ − MδMγ − M − δ → − θ ( y ) → + , ϕ ( y ) → (2 l − π for some integer l . Since(2 l − π = lim y → ( y ∗ ) − (2 n − π − nθ ( y )+ arccos (cid:2)(cid:0) γδ − ( γ + δ ) cos 2 θ ( y ) (cid:1) / | e iθ ( y ) γ − e − iθ ( y ) δ | (cid:3) = lim y → ( y ∗ ) − (2 n − π + arccos (cid:2)(cid:0) γδ − ( γ + δ ) cos 2 θ ( y ) (cid:1) / | e iθ ( y ) γ − e − iθ ( y ) δ | (cid:3) , we have (2 n − π ≤ (2 l − π ≤ (2 n − π . This implies that 2 l − n − ϕ ( y ) → (2 n − π as y → ( y ∗ ) − . Hence the image of g ( y ) := − ϕ ( y ) θ ( y ) on the interval (2 , y ∗ )contains the interval ( −∞ , θ ( y ) = π − θ ( y ) we have x ( y ) = 4 cos ( θ ( y )) and hence for each y ∈ [2 , y ∗ ) there exists a non-abelian representation ρ : π ( X ) → SL ( C ) such that ρ ( a ) = (cid:20) M M − (cid:21) and ρ ( b ) = (cid:20) M − y M − (cid:21) , where M = e iθ ( y ) . Moreover, the image of the canonical longitude λ corresponding to themeridian µ = a has the form ρ ( λ ) = (cid:20) L ∗ L − (cid:21) , where L = e iϕ ( y ) and ϕ ( y ) = − (2 n − π + 4 nπ − nθ ( y ) − arccos (cid:2)(cid:0) γδ − ( γ + δ ) cos 2 θ ( y ) (cid:1) / | e iθ ( y ) γ − e − iθ ( y ) δ | (cid:3) = − (2 n − π + 4 nθ ( y ) − arccos (cid:2)(cid:0) γδ − ( γ + δ ) cos 2 θ ( y )) (cid:14) | e iθ ( y ) γ − e − iθ ( y ) δ | (cid:3) . Since (2 n − π n +2 < θ (2) < (2 n − π n +2 we have − π + π n +1 < ϕ (2) < π n +1 .As y → + , ρ approaches a reducible representation and so L → , ϕ ( y ) → y → ( y ∗ ) − , we have x ( y ) → M → − , L = − M − n M − γ − MδMγ − M − δ → − θ ( y ) → π, ϕ ( y ) → − (2 n − π . This implies that the image of g ( y ) := − ϕ ( y ) θ ( y ) on theinterval (2 , y ∗ ) contains the interval (0 , n − C (2 m + 1 , n ) is similar to that for C (2 m, − n ).Lastly, we consider the case K = C (2 m + 1 , − n ) and n ≥
2. Consider the continuousreal function x : [2 , ∞ ) → (cid:16) (2 n − π n + 2 , ∞ (cid:17) EFT ORDERABLE SURGERIES OF DOUBLE TWIST KNOTS 11 in Lemma 2.4. Since x (2) < n − π n +2 and lim y →∞ x ( y ) = ∞ , there exists y ∗ > x ( y ∗ ) = 4 and 4 cos n − π n +2 < x ( y ) < y ∈ [2 , y ∗ ).For each y ∈ [2 , y ∗ ) we let θ ( y ) = arccos( p x ( y ) / θ (2) > (2 n − π n +2 , and for y ∈ [2 , y ∗ ) we have 0 < θ ( y ) < (2 n − π n +2 and x ( y ) = 4 cos θ ( y ). Since R K ( x ( y ) , y ) = 0 thereexists a non-abelian representation ρ : π ( X ) → SL ( C ) such that ρ ( a ) = (cid:20) M M − (cid:21) and ρ ( b ) = (cid:20) M − y M − (cid:21) , where M = e iθ ( y ) . Moreover, the image of the canonical longitude λ corresponding to themeridian µ = a has the form ρ ( λ ) = (cid:20) L ∗ L − (cid:21) , where L = − M n M − γ − M δM γ − M − δ and γ = S m ( y ), δ = S m − ( y ). Note that γ > δ >
0, since y > L = e iϕ ( y ) where ϕ ( y ) = − (2 n − π + 4 nθ ( y ) + arccos (cid:2)(cid:0) γδ − ( γ + δ ) cos 2 θ ( y ) (cid:1) / | e iθ ( y ) γ − e − iθ ( y ) δ | (cid:3) . Since (2 n − π n +2 < θ (2) < (2 n − π n +2 we have − π + π n +1 < ϕ (2) < π − (2 n − π n +1 .As y → + , ρ approaches a reducible representation and so L → , ϕ ( y ) → ϕ (2) = k π for some integer k . Since − π + π n +1 < ϕ (2) < π − (2 n − π n +1 , we must have ϕ (2) = 0.As y → ( y ∗ ) − , we have x ( y ) → M → , L = − M n M − γ − MδMγ − M − δ → − θ ( y ) → + , ϕ ( y ) → (2 l − π for some integer l . Since(2 l − π = lim y → ( y ∗ ) − − (2 n − π + 4 nθ ( y )+ arccos (cid:2)(cid:0) γδ − ( γ + δ ) cos 2 θ ( y ) (cid:1) / | e iθ ( y ) γ − e − iθ ( y ) δ | (cid:3) = lim y → ( y ∗ ) − − (2 n − π + arccos (cid:2)(cid:0) γδ − ( γ + δ ) cos 2 θ ( y ) (cid:1) / | e iθ ( y ) γ − e − iθ ( y ) δ | (cid:3) , we have − (2 n − π ≤ (2 l − π ≤ − (2 n − π . This implies that 2 l − − (2 n − ϕ ( y ) → − (2 n − π as y → ( y ∗ ) − . Hence the image of h ( y ) := − ϕ ( y ) θ ( y ) on the interval(2 , y ∗ ) contains the interval (0 , ∞ ).Similarly, with θ ( y ) = π − θ ( y ) we have x ( y ) = 4 cos ( θ ( y )) and hence for each y ∈ [2 , y ∗ ) there exists a non-abelian representation ρ : π ( X ) → SL ( C ) such that ρ ( a ) = (cid:20) M M − (cid:21) and ρ ( b ) = (cid:20) M − y M − (cid:21) , where M = e iθ ( y ) . Moreover, the image of the canonical longitude λ corresponding to themeridian µ = a has the form ρ ( λ ) = (cid:20) L ∗ L − (cid:21) , where L = e iϕ ( y ) and ϕ ( y ) = (2 n − π − nπ + 4 nθ ( y ) − arccos (cid:2)(cid:0) γδ − ( γ + δ ) cos 2 θ ( y ) (cid:1) / | e iθ ( y ) γ − e − iθ ( y ) δ | (cid:3) = (2 n − π − nθ ( y ) − arccos (cid:2)(cid:0) γδ − ( γ + δ ) cos 2 θ ( y ) (cid:1) / | e iθ ( y ) γ − e − iθ ( y ) δ | (cid:3) . Since (2 n − π n +2 < θ (2) < (2 n − π n +2 we have − π + (2 n − π n +1 < ϕ (2) < π − π n +1 .As y → + , ρ approaches a reducible representation and so L → , ϕ ( y ) → ϕ (2) = 0.As y → ( y ∗ ) − , we have x ( y ) → M → − , L = − M n M − γ − MδMγ − M − δ → − θ ( y ) → π, ϕ ( y ) → (2 n − π . This implies that the image of h ( y ) := − ϕ ( y ) θ ( y ) on theinterval (2 , y ∗ ) contains the interval ( − (2 n − , C (2 m + 1 , − n ) is similar to that for C (2 m, − n ). (cid:3) We now finish the proof of Theorem 1. Suppose r is a rational number such that r ∈ LO K . If r = 0, by Proposition 3.1, there exists a representation ρ : π ( X r ) → SL ( R )such that ρ (cid:12)(cid:12) π ( ∂X ) is an elliptic representation. This representation lifts to a representation˜ ρ : π ( X r ) → ^ SL ( R ), where ^ SL ( R ) is the universal covering group of SL ( R ). See e.g.[CD, Sec. 3.5] and [Va, Sec. 2.2]. Note that X r is an irreducible 3-manifold (by [HTh])and ^ SL ( R ) is a left orderable group (by [Be]). Hence, by [BRW], π ( X r ) is a left orderablegroup. Finally, 0-surgery along a knot always produces a prime manifold whose first Bettinumber is 1, and by [BRW] such manifold has left orderable fundamental group. Acknowledgements
The author has been partially supported by a grant from the Simons Foundation(
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