Character varieties of odd classical pretzel knots
Haimiao Chen

TL;DR
This paper computes the SL(2,C)-character varieties for odd classical pretzel knots and introduces a method to calculate their A-polynomials, advancing understanding of knot invariants.
Contribution
It provides explicit character varieties for odd classical pretzel knots and a novel method for computing their A-polynomials.
Findings
Explicit character varieties for all odd classical pretzel knots.
A new method for calculating A-polynomials of these knots.
Enhanced tools for studying knot invariants and their geometric properties.
Abstract
We determine the -character variety for each odd classical pretzel knot , and present a method for computing its A-polynomial.
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Character varieties of odd classical pretzel knots
Haimiao Chen 111Email: [email protected]
Beijing Technology and Business University, Beijing, China
Abstract
We determine the -character variety for each odd classical pretzel knot , and present a method for computing its A-polynomial.
Keywords: -representation, character variety, odd classical pretzel knot, A-polynomial
MSC2010: 57M25, 57M27
1 Introduction
For a link , let , with a tubular neighborhood. The -representation variety of is the set of representations , and the character variety of is
[TABLE]
where the character sends each to .
It is well-known that (see [4]) both and can be defined by a finite set of polynomial equations, and this is why they are so named. Actually, is the geometric-invariant-theoretic quotient of under the action of via conjugation. To be more plain, denoting the subset of consisting of irreducible representations by , it is also well-known that up to conjugacy, each is determined by . Call
[TABLE]
the irreducible character variety of . Our main concern is to study ; the characters of reducible representations are relatively easy to understand.
Since the seminal paper [4], much attention has been attracted to representations of 3-manifold groups into . One reason is that, representation/character variety encodes much topological and geometric information on the underlying 3-manifolds.
Among the invariants extracted from character variety, the A-polynomial (the definition is recalled in Section 4) proposed in [3] is rather important, in that, it plays a key role not only in the problem of exceptional fillings, but also in the celebrated AJ Conjecture which involves the behavior of quantum invariants.
Till now, character varieties have been found for the following links: torus knots [11], double twist knots [12], double twist links [14], -pretzel links [15]. In this paper, we aim to determine when is an odd classical pretzel knot. This is the first time to deal with a 3-parameter family of knots. We are faced with the complications caused by three non-commuting variables. However, we can still find a route to bypass the difficulty, reducing equalities between matrices to those of traces.
As the result shows, consists of a finite set of isolated points, several conics and an algebraic curve of high genus. Based on this, we present a method for explicitly computing the A-polynomial.
Acknowledgement
The author is supported by NSFC (Grant No. 11401014). He is grateful to Prof. Xun Yu at Tianjin University for beneficial conversations.
2 Preliminary
Let denote the set of matrices with entries in ; it is a 4-dimensional vector space over . Let denote the identity matrix.
Given and , take with and put
[TABLE]
note that the right-hand-side is unchanged when is replaced by . It is easy to verify that for all ,
[TABLE]
If with , then by Cayley-Hamilton Theorem,
[TABLE]
and inductively we can obtain that for all ,
[TABLE]
Lemma 2.1**.**
For any with , and , one has
[TABLE]
Proof.
The first identity follows from , and the second one can be deduced from
[TABLE]
∎
Lemma 2.2**.**
Given , let
[TABLE]
(i)* There exist with , , , and if and only if*
[TABLE]
(ii)* If (9) holds and are required to have no common eigenvector, then the ordered triple is unique up to simultaneous conjugacy.*
Here (i) is a special case of the statement next to Lemma 2.3 in [9], and (ii) follows from Lemma 2.4 in [9] (referred to Lemma 1.5.2 in [4]), restating that up to conjugacy, an irreducible representation of the free group of rank 3 is determined by the tuple satisfying (9). Also see [7].
For with , put
[TABLE]
3 The Character variety
3.1 Set up
For the Wirtinger presentation of a link , refer to Theorem 3.4 in [1]. Suppose a part of is an integral tangle as shown in Figure 1. Let denote the elements of corresponding to the labeled directed arcs. Since , we have
[TABLE]
Hence for all ,
[TABLE]
In this paper we consider the odd classical pretzel knot
[TABLE]
A diagram for is shown in Figure 2 when .
Let be a representation, and let . The following can be deduced using (15):
[TABLE]
Conversely, any satisfying these relations give rise to a representation.
Remark 3.1**.**
Since the ’s are conjugate to each other, none of them can be , unless is trivial. Let us assume to be nontrivial.
Notation 3.2**.**
To simplify the writing, by we mean for and for ; by we mean for and for . Similarly for other situations.
For , put
[TABLE]
The relations (17)–(19) are equivalent to
[TABLE]
Remark 3.3**.**
If is reducible, then taking conjugation if necessary, we may assume that the elements of the image of are all upper-triangular, and furthermore, for some . Since are conjugate to each other via upper-triangular elements, we have for some ; let . Then so that , and
[TABLE]
The condition (22) is equivalent to
[TABLE]
All reducible representations can be found this way. Note that is abelian if and only if .
3.2 Irreducible representations
In this subsection, is assumed to be irreducible; equivalently, have no common eigenvector.
Suppose
[TABLE]
Clearly,
[TABLE]
Let
[TABLE]
Then (9) can be rewritten as
[TABLE]
Remark 3.4**.**
Note that
[TABLE]
so if and only if with .
For each , denote
[TABLE]
[TABLE]
so that
[TABLE]
where in the last line, (8) is used.
Lemma 3.5**.**
* for at least one .*
Proof.
Assume on the contrary that for all .
If , then taking conjugation if necessary, we may assume . It follows from that or for some . Similar situation occurs for . Since is irreducible, we have for some . But then , a contradiction.
If , then taking conjugation if necessary, we may assume . It would follow from that are both upper-triangular, contradicting the irreducibility of . ∎
Remark 3.6**.**
Referred to Remark 3.3, the condition that for at least one is also sufficient for to be irreducible.
Representations of with were determined by the author in [2]; recall the result of Section 3 there, stated in a different form:
Proposition 3.7**.**
Suppose and (30) is satisfied. Then (22) holds if and only if one of the following cases occurs:
- •
, ;
- •
, ;
- •
, and there exist , such that and .
Lemma 3.8**.**
For each , if and only if or .
Proof.
By (32), if and only if
[TABLE]
which, by (5), is equivalent to
[TABLE]
i.e., . Hence if and only if either or . ∎
Proposition 3.9**.**
If and for some , then (22) holds if and only if
[TABLE]
Proof.
By Lemma 3.8, .
If (22) holds, then for all . Since , by Lemma 3.8 we have , implying . Let . Then , as otherwise would have a common eigenvector, so . Thus , which is equivalent to ; in particular, , so that . Moreover,
[TABLE]
If , then or , either implying , which contradicts as just obtained.
If (34) is satisfied, then by Lemma 3.8, . Taking conjugation if necessary, we may assume . Assume
[TABLE]
From we obtain , implying , hence . From we obtain . It follows from that . Thus , and , establishing (22). ∎
Proposition 3.10**.**
Suppose , for all , and (30) is satisfied. Let . Then (22) holds if and only if
[TABLE]
Proof.
Using (33) and which is a special case of (4), the following can be computed directly:
[TABLE]
If (22) holds, then
[TABLE]
due to and , from (37) and (38) we see that (40) is equivalent to (35).
Now suppose (35) holds (so that (40) is satisfied). Then for ,
[TABLE]
hence
[TABLE]
where in the last line we use . By (30),
[TABLE]
If , then (41) and (42) imply , contradicting Lemma 3.5. Hence , so that and
[TABLE]
Consequently,
[TABLE]
which is independent of . Thus (39) is established.
The proof is completed once are shown to form a basis of ; then (39), (40) will imply that for all , forcing .
Assume on the contrary that are linearly dependent, so are . Without loss of generality, assume . Then
[TABLE]
The first equation implies
[TABLE]
The following can be computed using (8):
[TABLE]
Then (33) can be re-written as
[TABLE]
where the ’s stand for coefficients that are irrelevant.
By (39), (40), , . Consequently,
[TABLE]
Comparing and for , we respectively obtain
[TABLE]
which lead to
[TABLE]
These force , so that . With (45), one can deduce from (46)–(49) that , contradicting Lemma 3.5. ∎
Remark 3.11**.**
It is worth highlighting some points contained in the proof. With (35) satisfied, we have the following implications:
- (i)
If and (30) holds, then , and , , as seen from (44); also, for some , as seen from (41) and (42). 2. (ii)
[TABLE]
conversely, if and for some , then (30) can follow from (53), since by (43),
[TABLE]
3.3 The character variety
The results of Proposition 3.7, Lemma 3.8, Proposition 3.9 and Proposition 3.10 are summarized as:
Theorem 3.12**.**
The irreducible character variety of can be embedded in and is the disjoint union of four parts:
[TABLE]
where
- •
, where consists of with
[TABLE]
and consists of with
[TABLE]
- •
, where consists of with
[TABLE]
- •
* consists of with*
[TABLE]
so has components, each being a conic;
- •
* consists of with*
[TABLE]
The dimensions are: , .
Some supplements are in order:
- •
; actually, we could have defined by replacing (56) and (57) respectively with (55) and
[TABLE]
and adding (53) which, by Remark 3.11 (ii), is redundant for .
- •
Each of the points in is isolated.
- •
consists of with
[TABLE]
We investigate in more detail. Recall Remark 3.11 (i) that .
Let . By (57) for ,
[TABLE]
Assertion 3.13**.**
On , if and only if and .
Proof.
The “if” part is trivial. Suppose . Then . If , then since , we have . If , then , which, due to , implies ; by (57) for , , hence also . Thus or . Then becomes . ∎
Decompose as the “singular” part and the “regular” part: let
[TABLE]
Obviously, consists of finitely many points, unless , in which case is a curve with two components, each being parameterized by .
Let , then for some . From (58), (59) we obtain
[TABLE]
With and substituted, (57) for can be re-written as one polynomial equation in , displaying as an affine algebraic curve. We may regard as a rational function on through (56); moreover, when , by Remark 3.11 (ii), the condition (56) is equivalent to
[TABLE]
4 The A-polynomial
Choose a meridian-longitude pair of , and denote the corresponding elements of by the same notations. Following [8] (see Page 303), let denote the subset of consisting of representations which send and to upper-triangular matrices. Define
[TABLE]
by setting (resp. ) to be the upper-left entry of (resp. ). Taking the Zariski closure of the image of , we obtain an affine algebraic curve , which is known to have the property that each component has dimension zero or one. The A-polynomial is defined to be the defining polynomial of the one-dimensional part of .
A-polynomial is notoriously difficult to compute. Till now, in the literary, formulas for A-polynomials have been obtained for few families of knots; see [5, 6, 10, 13] and the references therein.
Here we present a method for deriving a formula for the A-polynomial of an odd classical pretzel knot, leaving practical computations to possible future work.
For the pretzel knot , take . Let be a representation of as in Section 3, but this time we require .
For , let
[TABLE]
Observe that
[TABLE]
Let denote the image under of the longitude associated to (so ). Using (15), we find
[TABLE]
The 1-dimensional part of consists of and , the latter having components, so can be factorized as
[TABLE]
Note that are all known constants on each component of , thus the corresponding factor is easy to understand, as explained in Remark 4.2.
The main achievement of this section is concerned with the “hard” factor contributed by .
Suppose the upper-left entries of are , respectively, so that , .
Proposition 4.1**.**
If , then
[TABLE]
Proof.
It suffices to prove (70) when belongs to an open subset of . Let us assume and .
It is easy to see that
[TABLE]
Let . By (67), . By (68), , hence
[TABLE]
As a consequence,
[TABLE]
By (66),
[TABLE]
hence
[TABLE]
where
[TABLE]
By (66) again,
[TABLE]
hence
[TABLE]
Thus (73) implies
[TABLE]
Combining (72), (75) and (78), we obtain (70). ∎
Then can be obtained by computing the multi-variable resultant of the following (remembering ):
[TABLE]
Remark 4.2**.**
If lies in the -th component of , then (72), (74) and (80) are still valid (whenever ). Eliminating from these, we may obtain a polynomial in , which is the very .
Example 4.3**.**
As an illustration, consider the case . Then . From (79) for we obtain
[TABLE]
then use them to re-write (80), (81) as, respectively,
[TABLE]
Use (83), we can re-write (79) for as
[TABLE]
Then (for ) may be obtained from (82)–(84) by eliminating and .
The reference list from the paper itself. Each links out to its DOI / PubMed record.
- 1[1] G. Burde, H. Zieschang, Knots . de Gruyter Studies in Mathematics 5, second revised and extended edition, Walter de Gruyter, Berlin, New York, 2003.
- 2[2] H.-M. Chen, Trace-free SL ( 2 , ℂ ) SL 2 ℂ {\rm SL}(2,\mathbb{C}) -representations of Montesinos links . ar Xiv:1604.01326.
- 3[3] D. Cooper, M. Culler, H. Gillet, D. D. Long, P. B. Shalen, Plane curves associated to character varieties of 3-manifolds. Invent. Math. 118 (1994), 47–84.
- 4[4] M. Culler, P.B. Shalen, Varieties of group representations and splittings of 3-manifolds . Ann. Math. 117 (1983), no. 1, 109–146.
- 5[5] S. Garoufalidis, T.W. Mattman, The A-polynomial of the ( − 2 , 3 , 3 + 2 n ) 2 3 3 2 𝑛 (-2,3,3+2n) pretzel knots. New York J. Math. 17 (2011), 269–279.
- 6[6] J.-Y. Ham, J. Lee, An explicit formula for the A-polynomial of the knot with Conway’s notation C ( 2 n , 3 ) 𝐶 2 𝑛 3 C(2n,3) . J. Knot Theory Ramifications, 25 (2016), no. 10, 1650057 (9 pages).
- 7[7] R. Horowitz, Characters of free groups represented in the 2-dimensional special linear group . Comm. Pure Appl. Math. 25 (1972), 635–649.
- 8[8] D.D. Long, A.W. Reid, Integral points on character variety . Math. Ann. 325 (2003), 299–321.
