Minimal coloring numbers on minimal diagrams of torus links Eri - - PowerPoint PPT Presentation
Introduction Results Proof Results2 Minimal coloring numbers on minimal diagrams of torus links Eri Matsudo Nihon University Institute of Natural Sciences Joint work with K. Ichihara (Nihon Univ.) and K.Ishikawa (RIMS, Kyoto Univ.) Nihon
Introduction Results Proof Results2
Nihon University Institute of Natural Sciences
Joint work with
Nihon University, Dec 19, 2019
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Z-coloring
Let L be a link, and D a diagram of L. Z-coloring A map γ : {arcs of D} → Z is called a Z-coloring on D if it satisfies the condition 2γ(a) = γ(b) + γ(c) at each crossing of D with the over arc a and the under arcs b and c.
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Z-colorable link
Z-colorable link L is Z-colorable if ∃ a diagram of L with a non-trivial Z-coloring, i.e., a Z-coloring which at least two colors. Remark L is Z-colorable if and only if det(L) = 0. Thus, any knot K is non-Z-colorable, for det(K) is odd. Minimal coloring number The minimal coloring number mincolZ(D) of a diagram D of L is defined as the minimum of the number of colors among non-trivial Z-colorings on D. The minimal coloring number mincolZ(L) of L is the minimum of mincolZ(D).
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Question (in 2016)
Let L be a Z-colorable link.
If L is non-splittable, then mincolZ(L) ≥ 4. Proposition. If the crossing number of L is at most 10, then mincolZ(L) = 4. Question: How many colors are enough to color? For any non-splittable Z-colorable link L, mincolZ(L) = 4?
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mincolZ(L)=4
For any non-splittable Z-colorable link L, mincolZ(L) = 4 holds.
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mincolZ(L)=4
For any non-splittable Z-colorable link L, mincolZ(L) = 4 holds. (Next) Problem. For a particular diagram D of a non-splittable Z-colorable link, how many colors are enough to color? i.e., mincolZ(D) =? Here we consider torus link & standard diagram.
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Torus link
Fact The torus link T(a, b) running a times meridionally and b times longitudinally is Z-colorable if a or b is even . Theorem [I.chihara-M., 2018] mincolZ(D) = 4 for the standard diagram D of T(pn, n) with n > 2, even and p = 0. The standard diagram
It is known to be a min- imal diagram if p ≥ q.
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Results
Theorem [Ichihara.-Ishikawa-M. (arXiv:1908.00857)] Let p, q, r be integers such that p and q are coprime with |p| ≥ q ≥ 1, r ≥ 2. Let D be the standard diagram of T(pr, qr). Suppose that T(pr, qr) is Z-colorable, i.e., pr or qr is even. Then, mincolZ(D) =
if r is even, 5 if r is odd.
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Proof of Theorem. [5 colors case: r is odd]
We assume that the colors are 0, 1, 2, 3 and derive a contradiction. Then, there are only crossings colored as;
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Proof of Theorem. [5 colors case: r is odd]
We assume that the colors are 0, 1, 2, 3 and derive a contradiction. Then, there are only crossings colored as; Thus the over arcs must be colored by 1 or 2. And the arcs of a component which has an arc colored by 0 or 2 (1
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Proof of Theorem. [5 colors case: r is odd, q = 1]
We may assume that the number of the over arcs colored by 1 is
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Proof of Theorem. [5 colors case: r is odd, q = 1]
We may assume that the number of the over arcs colored by 1 is
In the case of q = 1, since ∃ an over arc colored by 0 ⇒ a contradiction.
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Proof of Theorem. [5 colors case: r is odd, q ≥ 2]
In the case of q ≥ 2, by going through the r parallel arcs, the color 2 changes to 0, since there are odd over arcs colored by 1.
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Proof of Theorem. [5 colors case: r is odd, q ≥ 2]
The change of the colors are ex- pressed by using a linear function f. Then we see f(0) = 2 and f(2) = 0. That is, f(X) = −X + 2.
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Proof of Theorem. [5 colors case: r is odd, q ≥ 2]
From the Z-coloring is non-trivial, There exists an arc colored by 3. We see f(3) = −1. This gives a contradiction. ✷
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Results
We can have a complete classification of all the Z-colorings on the standard diagram of T(pr, qr). (Here the details are omitted.) A =
x1, . . . , xq defines a Z-coloring of D
We have A = {(a, . . . , a) | a ∈ Zr, ∆(a) = 0} if r is even, {(a, . . . , a) | a ∈ Zr} if r is odd, p is even, {(a, τ(a), a, . . . , τ(a)) | a ∈ Zr} if r is odd, q is even, where ∆(a) = a1 − a2 + · · · + (−1)rar ∈ Z and τ(a) = (−ai + 2∆(a))i ∈ Zr for a = (a1, . . . , ar) ∈ Zr.
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Example: r is even
T(4n, 8) (r = 4), a = (0 1 2 1)
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Example: r is odd, q is even
T(3n, 6) (r = 3, q = 2), a = (2 1 0), ∆(a) = 1
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Results
We can also have a complete classification of all the Z-colorings by
A(4) = (a1, . . . , aq) ∈ (Zr)q
x1, . . . , xq defines a Z-coloring of D with the four colors {0, 1, 2, 3}
Proposition 2.
We have A(4) =
01 ∪ A(4) 12 ∪ A(4) 23
where A(4)
01 = {(a1, . . . , ar) ∈ {0, 1}r | a1 = ar = 1, a2i = a2i+1 (i = 1, . . . , r/2 − 1)} ,
A(4)
12 = {(a1, . . . , ar) ∈ {1, 2}r | a2i−1 = a2i (i = 1, . . . , r/2)} ,
A(4)
23 = {(a1, . . . , ar) ∈ {2, 3}r | a1 = ar = 2, a2i = a2i+1 (i = 1, . . . , r/2 − 1)} . 16 / 19
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Example: A(4)
01
T(4n, 8) (r = 4), a = (1 0 0 1)
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Example: A(4)
12
T(4n, 8) (r = 4), a = (1 1 2 2)
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