4 The 6 2 knot
4.5 Verifying the assumption of the saddle point method
In this section, we verify the assumption of the saddle point method in Lemma 4.9. In order to show this lemma, we show Lemmas 4.4–4.8 in advance. As we mentioned in Remark 2.7, we verify the assumption for ˆV(t, s, u) instead of V(t, s, u), since V(t, s, u) converges uniformly to ˆV(t, s, u) on ∆′ in the form mentioned in Remark 2.7.
−
− 0
0 5
5
−5 5
10
10
−10 10
Figure 3: Contour lines of Re ˆV(0.8 +X√
−1,0.18 +Y√
−1,0.1)−ςR
In the proof of Lemma 4.9, as mentioned at the beginning of Section 4.4, it is sufficient to decrease Re ˆV(t, s, u) by 0.06, by pushingt, s, u into the imaginary directions. In order to calculate this concretely, putting
f(X, Y, Z) = Re ˆV(t+X√
−1, s+Y√
−1, u+Z√
−1)−ςR
as in Section 3.5, we consider the behavior of f at each fiber of the projection C3 →R3. Then, unlike Section 3.5, there are some (t, s, u) ∈R3 such that, fixing X and Z, f has a maximal point as a function of Y; for example, see Figure 3. We also note that the
Hessian of f is not positive where the contour lines are not convex, though we like to show that the Hesse matrix of f is positive definite at any critical point of f.
Lemma 4.4. Fixing X and Y, we regard f as a function of Z.
(1) If s+u≥ 12, then f is monotonically decreasing.
(2) If s+u < 12, then f has a unique minimal point at Z =g3(t, s, u), where g3(t, s, u) = 1
2πlog sin 2πs sin 2π(1
2 −s−u). In particular, this minimal point goes to ∞ as s+u→ 12 −0.
Proof. As a function of Z, the differential of f is presented by
∂f
∂Z = Arg (1−z)−2π(
s+u−1 2
),
where z =e2π√−1 (u+Z√−1). Hence, we can show the lemma in a similar way as the proof of Lemma 3.5.
When we regard f as a function of Z fixing X and Y, by Lemma 4.4,f has a unique minimal point at Z =g3(t, s, u); we note that this minimal point does not depend onX and Y. In order to consider the behavior of f, putting
f(X, Yˇ ) = f(
X, Y, g3(t, s, u))
, (68)
we consider the behavior of ˇf in each fiber of the projectionC3 →R3at (t, s, u)∈∆′ ⊂R3. Lemma 4.5. For each (t, s, u) ∈ ∆′ satisfying that t−12 < s+u < 12, fˇhas a unique minimal point and no other critical points.
Proof. The differentials of ˇf are presented by
∂fˇ
∂X = ∂f
∂X = Im
(−2 log(1−x) + log( 1− y
x
))+ 2π( t− 1
2 ),
∂fˇ
∂Y = ∂f
∂Y = Im
(
2 log(1−y)−log( 1− y
x
))−2π(
2s+u−1) , where x=e2π√−1 (t+X√−1) and y=e2π√−1 (s+Y√−1). Hence, since dXdx =−2πx,
∂2fˇ
∂X2 = Im(( 2 1−x +
y x2
1− yx ) dx
dX )
= 2πIm
(− 2x 1−x −
y x
1− yx )
= 2πIm
(− 2
1−x − 1 1−xy
) ,
where we obtain the last equality by changing the content of the parentheses by a real number, which does not change its imaginary part. Similarly, we have that
∂2fˇ
∂X∂Y = 2πIm 1
1− yx , ∂2fˇ
∂Y2 = 2πIm ( 2
1−y − 1 1− yx
) .
Therefore, the Hesse matrix of ˇf is presented by 2π
(2a1+b −b
−b 2a2+b )
, (69)
where we put
a1 = −Im 1
1−x, a2 = Im 1
1−y, b = −Im 1 1− yx .
Since 0.6 ≤t ≤ 0.9, Im (1−x) >0, and hence, a1 >0. Similarly, since 0.14≤ s ≤ 0.4, we have thata2 >0.
When t−s ≤ 12, we have that Im( 1− yx)
≥0, and hence, b≥0. Then, we can verify that the trace and the determinant of the Hesse matrix (69) are positive. Hence, the Hesse matrix (69) is positive definite. Therefore, by Lemma 4.7, ˇf has a unique minimal point and no other critical points, as required.
When t−s > 12, putting b′ =−b, we have that b′ >0. Since any critical point of ˇf is in the domain (72) by Lemma 4.6 below, it is sufficient (by Lemma 4.7) to show that the Hesse matrix (69) is positive definite in the domain (72). Hence, it is sufficient to show that
(the trace of (69))
= 4π(a1+a2−b′) > 0, (70)
(the determinant of (69))
= 4π2(
(2a1−b′)(2a2−b′)−b′2)
= 8π2a1a2b′(2 b′ − 1
a1 − 1 a2
) > 0. (71)
We show that (71) ⇒(70), as follows. Suppose that (71) holds. Then, b1′ > 12(1
a1 +a1
2
). Since a1, a2 and b′ are positive, b1′ > a1
1 or b1′ > a1
2. Hence, b′ < a1 or b′ < a2. Therefore, (70) holds.
We show (71), as follows. Since x=e2π√−1 (t+X√−1), a1 is presented by a1 = −Im 1
1−x = −Im 1−x
1−x2 = − e−2πXsin 2πt
(1−e2π√−1te−2πX)(1−e−2π√−1te−2πX)
= − sin 2πt
e2πX +e−2πX −2 cos 2πt = sin 2π(1−t)
e2πX +e−2πX −2 cos 2π(1−t). Hence,
1
a1 = e2πX +e−2πX −2 cos 2π(1−t) sin 2π(1−t) . Similarly, we have that
1
a2 = e2πY +e−2πY −2 cos 2πs
sin 2πs , 1
b′ = e2π(X−Y)+e2π(Y−X)−2 cos 2π(1−t+s) sin 2π(1−t+s) .
Therefore, the differential of b2′ − a11 − a12 with respect to X is given by 1
2π · ∂
∂X (2
b′ − 1 a1 − 1
a2
) = 2· e2π(X−Y)−e2π(Y−X)
sin 2π(1−t+s) − e2πX−e−2πX sin 2π(1−t) . Since X >0, Y <0, 0.33≤1−t+s <0.5 and 0.1≤1−t≤0.4,
1 2π · ∂
∂X (2
b′ − 1 a1 − 1
a2
) > (
e2πX −e−2πX)( 2
sin 2π(1−t+s)− 1 sin 2π(1−t)
)
> (
e2πX −e−2πX)( 2
sin 2π·0.33−max{ 1
sin 2π·0.1, 1 sin 2π·0.4
})
> (
e2πX −e−2πX)( 2
0.876307... −max{ 1
0.587785..., 1 0.587785...
})
> (
e2πX −e−2πX)
·0.581004... > 0.
Hence, it is sufficient to show the required inequality whenX = 0. Similarly, sinceX >0, Y <0, 0.33≤1−t+s <0.5 and 0.14≤s ≤0.4,
− 1 2π · ∂
∂Y (2
b′ − 1 a1 − 1
a2
) = 2·e2π(X−Y)−e2π(Y−X)
sin 2π(1−t+s) −e−2πY −e2πY sin 2πs
> (
e−2πY −e2πY)( 2
sin 2π(1−t+s) − 1 sin 2πs
)
> (
e−2πY −e2πY)( 2
sin 2π·0.33 −max{ 1
sin 2π·0.14, 1 sin 2π·0.4
})
> (
e−2πY −e2πY)( 2
0.876307...−max{ 1
0.770513..., 1 0.587785...
})
> (
e−2πY −e2πY)
·0.581004... > 0.
Hence, it is sufficient to show the required inequality when Y = 0. When X =Y = 0, 1
2 (2
b′ − 1 a1 − 1
a2
) = 2·1−cos 2π(1−t+s)
sin 2π(1−t+s) − 1−cos 2π(1−t)
sin 2π(1−t) − 1−cos 2πs sin 2πs . Further, since 1−sin 2παcos 2πα = 2 sin2 sinπα2cosπαπα = tanπα,
1 2
(2 b′ − 1
a1 − 1 a2
) = 2 tanπ(1−t+s)−tanπ(1−t)−tanπs > 0.
Hence, we obtain (71), as required.
The following two lemmas are used in the proof of Lemma 4.5.
Lemma 4.6. Let fˇbe as defined in (68). If s < t−12 < s+u < 12, then any critical point of fˇis in the following domain,
{(X, Y)∈R2 X >0, Y <0}
. (72)
Proof. Let (X, Y) be a critical point of ˇf.
We show that X >0, as follows. Since ∂X∂fˇ = 0,
−2 Arg (1−x) + 2π( t− 1
2
) = −Arg( 1− y
x ),
where x = e2π√−1 (t+X√−1) and y/x = e−2π√−1 (t−s)e2π(X−Y). Since 12 < t −s ≤ 0.67, the right-hand side of the above formula is positive. Further, since 0.6 ≤ t ≤ 0.9, the left-hand side is monotonically increasing with respect to X, and is equal to 0 atX = 0.
Therefore,X >0.
We show that Y <0, as follows. Since ∂Y∂fˇ = 0, 2 Arg (1−y) = Arg(
1− y x
)+ 2π(
2s+u−1) , where y = e2π√−1 (s+Y√−1) and y/x = e−2π√−1 (t−s)e2π(X−Y). Since Arg(
1− yx)
< 0 as mentioned above,
2 Arg (1−y) < 2π(
2s+u−1) . Hence,
2 Arg (1−y) + 2π(1 2−s)
< 2π(
s+u−1 2
).
From the assumption of the lemma, the right-hand side of the above formula is negative.
Further, since 0.14≤s≤0.4, the left-hand side is monotonically increasing with respect toY, and is equal to 0 atY = 0. Therefore, Y <0.
Lemma 4.7. For each (t, s, u) ∈ ∆′ satisfying that t − 12 < s+u, fˇ(X, Y) → ∞ as X2+Y2 → ∞.
Proof. From the definition of ˆV, we have that Re ˆV(
t+X√
−1, s+Y√
−1, u+Z√
−1)
= Re 1
2π√
−1
(−2 Li2(e2π√−1 (t+X√−1))−Li2(e2π√−1 (s−t+Y√−1−X√−1))
+ 2 Li2(e2π√−1 (s+Y√−1)) + Li2(e2π√−1 (u+Z√−1)) ) + 2π(
t− 1 2
)X−2π( s−1
2
)Y −2π(
s+u−1 2
)(Y +Z).
Since ˇf is defined from the above formula by fixing Z, it is sufficient to show that
Re 1
2π√
−1
(−2 Li2(e2π√−1 (t+X√−1)) + 2 Li2(e2π√−1 (s+Y√−1))
−Li2(e2π√−1 (s−t+Y√−1−X√−1)) )
+ 2π( t− 1
2
)X+ 2π(
1−2s−u) Y
goes to ∞ asX2+Y2 → ∞. Hence, putting F(X, Y) =
( {(t− 12)
X if X ≥0
−( t− 12)
X if X <0 )
+ ( {
(1−2s−u)Y if Y ≥0
−uY if Y <0
)
+
( {(t−s+12)
(X−Y) ifX ≥Y
0 if X < Y
) ,
(73)
by Lemma 2.2, it is sufficient to show that F(X, Y)→ ∞as X2+Y2 → ∞.
We show that F(X, Y)→ ∞ as X2 +Y2 → ∞, as follows. As for the first summand of (73), since t >0.5, this summand goes to ∞ as|X| → ∞. As for the second summand of (73), since 1−2s−u > 0 and u > 0, this summand goes to ∞ as |Y| → ∞. As for the third summand of (73), since t−s+ 12 > 0, this summand is non-negative. Hence, F(X, Y)→ ∞ as X2+Y2 → ∞, as required.
Lemma 4.8. Suppose that t−12 ≥s+u, and consider the flow from(X, Y) = (0,0)along the vector field (
− ∂X∂fˇ,−∂Y∂fˇ)
. Then, Y → −∞.
Proof. The differential of ˇf with respect to X is presented by
∂fˇ
∂X = ∂f
∂X = −2 Arg (1−x) + Arg( 1− y
x
)+ 2π( t−1
2 ),
where x = e2π√−1 (t+X√−1) and yx = e−2π√−1 (t−s)e2π(X−Y). In particular, when X = 0, noting that Arg (1−x) = π(
t− 12) ,
∂fˇ
∂X
X=0
= Arg( 1− y
x
) < 0,
since 12 < t−s ≤ 0.7. Hence, the flow of the lemma goes in the domain {X > 0}. We suppose that X >0 in the following of this proof.
The differential of ˇf with respect toY is presented by
∂fˇ
∂Y = ∂f
∂Y = 2 Arg (1−y)−Arg( 1− y
x
)−2π(2s+u−1),
where y=e2π√−1 (s+Y√−1) and yx =e−2π√−1 (t−s)e2π(X−Y). Hence,
∂fˇ
∂Y
Y→−∞ = −2·2π(1 2−s)
+ 2π(
t−s−1 2
)−2π(2s+u−1) = 2π(
t−s−u−1 2
) ≥ 0.
Further, whenY ≤0, similarly as in the proof of Lemma 4.5, we have that 1
2π · ∂2fˇ
∂Y2 = Im ( 2
1−y − 1 1−xy
)
= 2 sin 2πs
e2πY +e−2πY −cos 2πs − sin 2π(1−t+s)
e2π(X−Y)+e2π(Y−X)−cos 2π(1−t+s)
> 2 sin 2πs−sin 2π(1−t+s) e2πY +e−2πY −cos 2πs ,
where we obtain the last inequality since 0.14≤s ≤1−t+s < 12. Further, since sin 2πs ≥ max{
sin 2π·0.14, sin 2π(1−t+s)}
= max{
0.770513... , sin 2π(1−t+s)}
> 1
2 sin 2π(1−t+s), we have that∂∂Y2fˇ2 >0. Hence, ∂Y∂fˇ is monotonically increasing. Therefore, since∂Y∂fˇ
Y→−∞ ≥ 0 as shown above, we have that ∂Y∂fˇ ≥0. Hence, by the flow of the lemma,Y → −∞. Lemma 4.9. When we apply Proposition 2.6 to (55), the assumption of Proposition 2.6 holds.
Proof. We show that there exists a homotopy ∆′δ (0≤ δ ≤ 2) between ∆′0 = ∆′ and ∆′2 such that
(t0, s0, u0)∈∆′2, (74)
∆′2− {(t0, s0, u0)} ⊂ {
(t, s, u)∈C3 Re ˆV(t, s, u)< ςR}
, (75)
∂∆′δ ⊂ {
(t, s, u)∈C3 Re ˆV(t, s, u)< ςR}
. (76)
For a sufficiently large R >0, we put ˆ
g3(t, s, u) = {
min{R, g3(t, s, u)} if s+u < 12,
R if s+u≥ 12,
whereg3(t, s, u) is given in Lemma 4.4. We note that, sinceg3(t, s, u)→ ∞ass+u→ 12−0, ˆ
g3(t, s, u) is continuous. We set the middle part ∆′1 of the homotopy by
∆′1 = {(
t, s, u+ ˆg3(t, s, u)√
−1)
∈C3 (t, s, u)∈∆′} .
Further, we define the internal part ∆′δ (0< δ < 1) of the homotopy by setting it along the flow from (t, s, u) determined by the vector field (
0,0,−∂Z∂f)
. By Lemma 4.4, such a flow goes to ∆′1. We note that, when s+u≥ 12, Re ˆV is sufficiently small on ∆′1. It is a problem how to move ∆′1 further to make Re ˆV smaller, when s+u < 12.
We consider the behavior of the flow from each point of ∆′1 determined by the vector field (
−∂X∂fˇ,−∂Y∂fˇ,0)
. Whent−12 < s+u, by Lemma 4.5, ˇf has a unique minimal point, and the flow goes to it; we put this minimal point to be (
g1(t, s, u), g2(t, s, u))
. When t− 12 ≥s+u, by Lemma 4.8, Y goes to −∞ by the flow. We put
ˆ
g2(t, s, u) = {
max{−R, g2(t, s, u)} if t− 12 < s+u,
−R if t− 12 ≥s+u.
We note that, sinceg2(t, s, u)→ −∞ast−s−u→ 12−0, ˆg2(t, s, u) is continuous. We put ˆ
g1(t, s, u) so that( ˆ
g1(t, s, u),gˆ2(t, s, u))
is on the flow. We set the ending of the homotopy by
∆′2 = {(
t+ ˆg1(t, s, u)√
−1, s+ ˆg2(t, s, u)√
−1, u+ ˆg3(t, s, u)√
−1)
∈C3 (t, s, u)∈∆′} . Further, we define the internal part ∆′δ (1< δ < 2) of the homotopy by setting it along the flow from (
t, s, u+ ˆg3(t, s, u)√
−1)
determined by the vector field (
− ∂X∂fˇ,−∂Y∂fˇ, 0) . We show (76), as follows. From the definition of ∆′,
∂∆′ ⊂ {
(t, s, u)∈C3 Re ˆV(t, s, u)< ςR} .
Further, by the construction of the homotopy, Re ˆV monotonically decreases by the ho-motopy. Hence, (76) holds.
We show (74) and (75), as follows. Consider the following functions F(t, s, u, X, Y, Z) = Re ˆV(
t+X√
−1, s+Y√
−1, u+Z√
−1) , h(t, s, u) = F(
t, s, u, ˆg1(t, s, u), gˆ2(t, s, u), ˆg3(t, s, u)) .
When t− 12 ≥ s+u or s+u ≥ 12, −h(t, s, u) is sufficiently large (because we let R be sufficiently large), and (75) holds in this case. When t− 12 < s+u < 12, similarly as in the proof of Lemma 3.9, we can show that, if (t, s, u) is a critical point of h(t, s, u), (t+g1(t, s, u)√
−1, s+g2(t, s, u)√
−1, u+g3(t, s, u)√
−1)
is a critical point of ˆV. Hence, by Lemma 4.2, h(t, s, u) has a unique maximal point at (t, s, u) = (Ret0,Res0,Reu0).
Therefore, (74) and (75) hold.