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[EQUATION] by Equation ( ). (4) Assume that [MATH] with [MATH] . Then [MATH] [MATH] , and [MATH] by Theorem 2.11 (2). Thus [EQUATION] |
(5) If [MATH] with [MATH] [MATH] , then [MATH] [MATH] , and [MATH] . Thus [EQUATION] (6) If [MATH] with [MATH] [MATH] , then [MATH] [MATH] , and |
[MATH] . Thus [EQUATION] (7) If [MATH] with [MATH] [MATH] , then [MATH] [MATH] , and [MATH] . Thus [EQUATION] since [MATH] (8) Now we suppose [MATH] is not equal to any of the above cases. By following the proof of Lemma 2.12 and using Theorem 2.11 |
we conclude that [MATH] divides [MATH] . Moreover, every irreducible factor of [MATH] is a factor of [MATH] Since [MATH] is a product of pairwise non-associated irreducible factors, we conclude that [MATH] contains every irreducible factor of [MATH] |
precisely once. Thus the proof is completed. Corollary 2.14 Let [MATH] be different from [MATH] for all [MATH] and [MATH] Then [MATH] and [MATH] |
are relatively prime if and only if [MATH] . In particular, [MATH] and [MATH] are relatively prime whenever [MATH] Proof. The claim follows from Lemma 2.9 and Proposition 2.13 |
3. The shuffle map over commutative rings In this section let [MATH] be a unital commutative ring. We calculate the determinant of the shuffle map over [MATH] |
Definition 3.1 Let [MATH] be a finitely generated free module over [MATH] and let [MATH] be an [MATH] -module endomorphism of [MATH] We say that [MATH] is a free prebraided module of diagonal type over |
[MATH] if there exist a basis [MATH] of [MATH] and [MATH] with [EQUATION] Let [MATH] and let [MATH] be a free prebraided module of diagonal type with basis [MATH] . Let [MATH] and |
[MATH] . Assume that [EQUATION] for all [MATH] . Let [MATH] denote the [MATH] -fold tensor product of [MATH] over [MATH] and let |
[MATH] Note that [MATH] is a free module over [MATH] for all [MATH] For any [MATH] let [MATH] denote the set of words over [MATH] |
of degree [MATH] Let [MATH] be the standard basis of [MATH] . Then [MATH] admits a [MATH] -grading given by [MATH] , for all [MATH] . Thus for any [MATH] , the degree of |
[MATH] is [MATH] and we write [MATH] for the degree of any homogeneous element [MATH] of [MATH] For any [MATH] let [MATH] denote the [MATH] -homogeneous component of [MATH] of degree [MATH] |
For any [MATH] , let [MATH] denote the monoid which is [MATH] and relations (1) [MATH] for [MATH] with [MATH] , and (2) [MATH] for [MATH] |
Let [MATH] , and for any [MATH] let [EQUATION] A variant of the following equation appeared already in , Lemma 6.12] Lemma 3.2 For any [MATH] the following equation holds in [MATH] |
[EQUATION] Proof. It is easy to check that [EQUATION] for all [MATH] . Thus [EQUATION] Hence, [EQUATION] This proves the lemma. For any [MATH] let [MATH] denote the monoid ring of [MATH] over [MATH] and let |
[MATH] be the ring homomorphism such that [MATH] is the prebraiding [MATH] applied to the [MATH] -th and [MATH] -th tensor factors of |
[MATH] Let [MATH] be the ring homomorphism such that [EQUATION] Lemma 3.3 For any [MATH] with [MATH] and [MATH] we have [EQUATION] |
Moreover, if [MATH] is a field, [MATH] , and [MATH] , then [EQUATION] Proof. Consider the action of [MATH] on [MATH] given by [EQUATION] |
In any [MATH] -orbit of [MATH] there is a unique element [MATH] where [MATH] is a Lyndon word and [MATH] with [MATH] for each [MATH] |
To any [MATH] -orbit [MATH] of [MATH] we attach the submodule [MATH] of [MATH] [MATH] with [MATH] . Then [EQUATION] Let [MATH] be a [MATH] -orbit and let [MATH] and [MATH] be the Lyndon word such that [MATH] . Then [MATH] . Let |
[MATH] Then [EQUATION] is a basis of [MATH] , where [MATH] . Moreover, for all [MATH] [EQUATION] where [MATH] and [MATH] We obtain that the matrix of [MATH] with respect to the basis ( 11 is [MATH] , where |
[EQUATION] Therefore, [EQUATION] Hence [EQUATION] because of the decomposition of [MATH] in ( 10 ). If [MATH] is a field, then the matrix [MATH] above has corank [MATH] or [MATH] Moreover, [MATH] has corank [MATH] if and only if |
[MATH] , that is, if and only if [MATH] where [MATH] . This implies the last claim. Lemma 3.4 For any [MATH] with [MATH] we have |
[EQUATION] Moreover, if [MATH] is a field, [MATH] , and [MATH] , then [EQUATION] Proof. Let us consider the [MATH] -action on [MATH] given by |
[EQUATION] Then [MATH] In any [MATH] -orbit of [MATH] there is a unique element [MATH] , where [MATH] [MATH] is a Lyndon word, and [MATH] . Moreover, then [MATH] and [MATH] for each [MATH] |
with [MATH] Again, to any [MATH] -orbit [MATH] we attach the submodule [MATH] of [MATH] [MATH] , where [MATH] Then [EQUATION] Let [MATH] be a Lyndon word, [MATH] , and [MATH] Assume that [MATH] . Then [MATH] Let [MATH] Then the monomials |
[EQUATION] form a basis of [MATH] for the [MATH] -orbit [MATH] of [MATH] where [MATH] For any [MATH] one obtains that [EQUATION] |
where [MATH] and [MATH] for all [MATH] Thus the matrix of [MATH] with respect to the basis ( 13 ) is [MATH] , where [EQUATION] Hence, |
[EQUATION] This implies the first claim. If [MATH] is a field, then the matrix [MATH] above has corank [MATH] or [MATH] Moreover, [MATH] has corank [MATH] if and only if |
[MATH] , that is, if and only if [MATH] where [MATH] . This implies the last claim. Lemma 3.5 Let [MATH] with [MATH] . Then [EQUATION] |
Proof. From Lemma 3.2 we conclude that [EQUATION] Because of Lemma 3.3 and 3.4 the above equality is equivalent to [EQUATION] This implies the lemma. |
Proposition 3.6 Let [MATH] with [MATH] (1) If [MATH] with [MATH] then [MATH] (2) If there exist [MATH] with [MATH] , then [EQUATION] |
Proof. Claim (1) follows directly from the definition of [MATH] In order to prove part (2) of the Proposition it suffices to consider [MATH] and [MATH] In this case the claim follows from Lemma 3.5 and Lemma 2.12 |
4. Nichols algebras which are free algebras In the remaining part of this paper let [MATH] be a field, let [MATH] and let [MATH] be an [MATH] -dimensional braided vector space of diagonal type with basis [MATH] |
and braiding matrix [MATH] Let [MATH] and [MATH] denote the tensor algebra and the Nichols algebra of [MATH] , respectively. For the basic theory of Nichols algebras we refer to |
In this section we determine when [MATH] is a free algebra, that is, [MATH] For all [MATH] there is a unique group homomorphism [MATH] |
with [MATH] for all [MATH] . We also write [MATH] for the induced algebra maps [MATH] For all [MATH] let [MATH] be the representation of [MATH] introduced in Section as [MATH] and let |
[EQUATION] Then, by [EQUATION] Lemma 4.1 Let [MATH] with [MATH] (1) If [MATH] , then [MATH] (2) If [MATH] and [MATH] then [MATH] |
Proof. If [MATH] for some [MATH] [MATH] then the claim holds because of Proposition 3.6 (1). Assume now that there exist [MATH] with [MATH] Then Proposition 2.13 implies that |
[MATH] if and only if [MATH] Hence the lemma follows from Proposition 3.6 (2). Proposition 4.2 Let [MATH] with [MATH] (1) If [MATH] , then there is a non-trivial relation in |
[MATH] of degree [MATH] (2) If [MATH] for all [MATH] with [MATH] then there is no non-trivial relation in [MATH] of degree [MATH] |
Proof. (1) Assume that [MATH] Then [MATH] by Lemma 4.1 (1). Thus [MATH] by the definition of [MATH] and the claim follows from Equation ( 15 ). |
(2) Assume that the Nichols algebra [MATH] has a non-trivial relation in degree [MATH] Let [MATH] be such that [MATH] has a non-trivial relation in degree [MATH] and no non-trivial relation in any degree [MATH] . Let [MATH] . Then [MATH] |
[MATH] and [MATH] by ( 15 ) and by the definition of [MATH] Hence [MATH] by Lemma 4.1 (2). This proves (2). Based on the above proposition we obtain our first main Theorem as follows. |
Theorem 4.3 We have [MATH] if and only if [MATH] for all [MATH] with [MATH] Proof. The claim follows immediately from Proposition 4.2 |
Example 4.4 (Diophantine equation) Assume that the characteristic of [MATH] is neither [MATH] nor [MATH] and that [MATH] with [MATH] not a root of [MATH] |
and [MATH] For any [MATH] let [EQUATION] Then [MATH] in the following cases: (1) [MATH] [MATH] [MATH] (2) [MATH] [MATH] [MATH] [MATH] |
(3) [MATH] or [MATH] [MATH] [MATH] Moreover, for any other [MATH] with [MATH] [EQUATION] Hence, by Theorem 4.3 [MATH] if and only if there is no solution of the diophantine equation |
[MATH] with [MATH] [MATH] [MATH] We now provide concrete examples of Nichols algebras of diagonal type which are identified by this example as a free algebra. Let [MATH] and let [MATH] be positive integers with [MATH] Let [MATH] be not a root of [MATH] and let |
[MATH] with [MATH] and [MATH] with [MATH] . For any [MATH] with [MATH] we have [EQUATION] and hence [EQUATION] Assume that [MATH] and [MATH] . By symmetry of [MATH] without loss of generality we may assume that [MATH] Then |
[EQUATION] and [MATH] for [MATH] since [MATH] Hence [MATH] implies that [MATH] and hence [MATH] . However, [MATH] . Therefore [MATH] |
for all pairs [MATH] with [MATH] . Thus [MATH] 5. An upper bound on the dimension of the kernel of the shuffle map Let [MATH] [MATH] , and let [MATH] Let [MATH] be the free prebraided module of diagonal type over [MATH] |
with basis [MATH] and braiding matrix [MATH] such that [EQUATION] Assume that [MATH] Let [MATH] be an [MATH] -dimensional braided vector space over [MATH] |
with braiding matrix [MATH] and with basis [MATH] such that [EQUATION] There are unique ring homomorphisms [EQUATION] with [MATH] [MATH] |
for all [MATH] . We view them as evaluation at [MATH] [MATH] , and [MATH] , respectively. Correspondingly, we write [EQUATION] for any [MATH] and any [MATH] |
Let [MATH] be a basis of [MATH] For any [MATH] let [EQUATION] Lemma 5.1 Let [MATH] with [MATH] Let [MATH] be the ring homomorphism given by |
[EQUATION] Then there exists a ring homomorphism [MATH] such that [EQUATION] Proof. By Remark 2.4 and Lemma 2.7 there exists a ring automorphism [MATH] of [MATH] with |
[EQUATION] Let [MATH] be the ring homomorphism with [MATH] [MATH] for all [MATH] with [MATH] Then [EQUATION] is a ring homomorphism and |
[EQUATION] for all [MATH] with [MATH] Thus [MATH] Lemma 5.2 Let [MATH] with [MATH] . Assume that [MATH] Then [MATH] . Let [MATH] Then [MATH] |
is the unique irreducible factor [MATH] of [MATH] such that [MATH] Proof. The claim follows directly from Remark 2.4 and Lemma 2.8 |
Lemma 5.3 Let [MATH] with [MATH] Assume that [MATH] . Let [MATH] and [MATH] Then (1) [MATH] does not appear in the prime decomposition of the polynomial [MATH] , and |
(2) [MATH] appears [MATH] times in the prime decomposition of [MATH] where [EQUATION] and [EQUATION] Proof. Assume first that [MATH] for some [MATH] . Then [MATH] |
[MATH] [MATH] [MATH] , and [MATH] [MATH] Hence [MATH] [MATH] , and therefore [EQUATION] by Proposition 3.6 (1) and Remark 2.2 . Thus (2) holds in this case. Moreover, (1) is valid since [MATH] |
and [MATH] does not divide [MATH] Assume that there exists [MATH] with [MATH] Since [MATH] for all [MATH] and [MATH] , Propositions 3.6 (2) and |
2.13 imply that any irreducible factor of [MATH] is an irreducible factor of some [MATH] with [MATH] Thus, by Proposition 2.13 and Lemma 2.9 |
[MATH] and [MATH] are relatively prime, which proves (1). By Proposition 3.6 (2), for the proof of (2) it remains to determine the multiplicity of the irreducible factor [MATH] in [MATH] Let [MATH] with [MATH] . By Lemma 2.8 |
[MATH] has multiplicity [MATH] in [MATH] , except when [MATH] in which case it has multiplicity [MATH] By Definition 2.10 , there are |
[EQUATION] factors [MATH] in the numerator of [MATH] and [EQUATION] factors [MATH] in the denominator of [MATH] This proves the lemma. |
Lemma 5.4 Let [MATH] with [MATH] Suppose that [MATH] and [MATH] for all [MATH] Let [MATH] and [MATH] be ring homomorphisms as in Lemma 5.1 . Then the following hold. |
(1) [MATH] in [MATH] (2) the polynomials [MATH] and [MATH] are relatively prime in [MATH] , and (3) the factor [MATH] appears [MATH] times in the prime decomposition of |
[MATH] Proof. (1) By Lemma 5.1 , we have [EQUATION] As [MATH] for all [MATH] , we get [MATH] because of Proposition 4.2 (2) By definition, [MATH] . Thus (2) follows from (1). |
(3) Assume first that [MATH] for some [MATH] . Then [MATH] [MATH] by Proposition 3.6 (1), and hence [EQUATION] Since [MATH] in [MATH] and [MATH] , the irreducible factor [MATH] appears once in [MATH] . Moreover, [MATH] and |
[MATH] , see the first part of the proof of Lemma 5.3 Assume now that there exist [MATH] with [MATH] By (1), [MATH] does not appear in the prime decomposition of |
[MATH] Hence, by Proposition 3.6 (2), we have to determine the multiplicity [MATH] of [MATH] in the prime decomposition of [MATH] By the definition of [MATH] and by Lemma 2.12 |
[MATH] is a non-zero polynomial in [MATH] By Remark 2.4 and by Proposition 2.13 [MATH] is a product of polynomials of the form [MATH] with [MATH] . Hence |
[MATH] is a product of polynomials of the form [MATH] with [MATH] and the multiplicity of [MATH] with [MATH] in [MATH] is the same as the multiplicity of [MATH] in [MATH] Let [MATH] Then [MATH] |
divides [MATH] if and only if [MATH] Hence [MATH] is the multiplicity of [MATH] in [MATH] Therefore, by Proposition 3.6 (2) and by Lemma 5.3 |
[MATH] is the multiplicity of [MATH] in [MATH] , that is, [MATH] Proposition 5.5 Let [MATH] with [MATH] Suppose that [MATH] and that [MATH] for all [MATH] with [MATH] Then |
[EQUATION] Proof. Let [MATH] and [MATH] be ring homomorphisms as in Lemma 5.1 Let [MATH] be the Smith normal form of [MATH] which is a diagonal matrix. Then [MATH] by Lemma 5.4 (3), and hence there is no zero on the diagonal of [MATH] . Again by Lemma 5.4 (3), |
[MATH] appears [MATH] times in the prime decomposition of [MATH] . Hence [MATH] appears in at most [MATH] diagonal entries of [MATH] as an irreducible factor. Then |
[EQUATION] Hence the proposition holds. 6. The dimension of the kernel of shuffle map We use some notation and conventions from the previous section. So let us assume that [MATH] . Let [MATH] , let [MATH] , and let [MATH] be a braided vector space of diagonal type with basis [MATH] such that [MATH] for all [MATH] For e... |
be the integers defined in Lemma 5.3 (2). In this section we determine the dimension of the kernel of the shuffle map [MATH] for those |
[MATH] with [MATH] [MATH] and [MATH] for all [MATH] with [MATH] Proposition 6.1 Let [MATH] with [MATH] Suppose that [MATH] and [MATH] for all [MATH] |
with [MATH] . Then [EQUATION] Proof. Since [MATH] for all [MATH] with [MATH] it follows from Proposition 4.2 that [MATH] is injective. Hence |
[EQUATION] because of Lemma 3.2 Therefore [EQUATION] that is, [EQUATION] because of Lemmas 3.3 and 3.4 This proves the Proposition. |
Theorem 6.2 Let [MATH] with [MATH] . Assume that [MATH] and [MATH] for all [MATH] with [MATH] . Then [EQUATION] Proof. The theorem follows immediately from Propositions 5.5 and Proposition 6.1 |
Remark 6.3 Let [MATH] with [MATH] . Assume [MATH] for all [MATH] with [MATH] Then [MATH] for all [MATH] with [MATH] by the definition of [MATH] . Moreover, |
[MATH] for all [MATH] with [MATH] by Proposition 4.2 (1). Assume now that [MATH] Then [MATH] by Proposition 4.2 (2). Thus [EQUATION] |
by Theorem 6.2 and by the bijectivity of the maps [MATH] for [MATH] with [MATH] Example 6.4 Here we give an example of a Nichols algebra of diagonal type where in some degree one has two defining relations. |
Let [MATH] be the two-dimensional braided vector space of diagonal type with basis [MATH] and braiding matrix [MATH] , such that |
[EQUATION] and [MATH] where [MATH] is a primitive fifth root of unity. Let [MATH] . Then [MATH] and [EQUATION] Thus [MATH] and [MATH] |
Let us check that [MATH] for all [MATH] with [MATH] [EQUATION] Moreover, [MATH] for [MATH] since [MATH] . Hence [MATH] for all [MATH] with [MATH] We now calculate [MATH] and [MATH] . By definition, [EQUATION] since [MATH] [MATH] [MATH] [EQUATION] Hence [MATH] |
# Source: arxiv 1806.05977 # Title: Understanding Complex Systems: From Networks to Optimal Higher-Order Models # Sections: all # Downloaded: 2026-03-03T05:16:44.194613+00:00 |
Understanding Complex Systems: From Networks to Optimal Higher-Order Models Abstract To better understand the structure and function of complex systems, researchers often represent direct interactions between components in complex systems with networks, assuming that indirect influence between distant components can be... |
A long-standing goal of statistical physics is to understand emergent phenomena in complex systems that consist of a large number of interacting components. Such systems not only occur in condensed matter physics, but they are also widespread in other disciplines, and physicists have been able to contribute to a better... |
Such systems can be conveniently represented as graphs or networks , where nodes [MATH] represent the components of a system and links |
[MATH] capture the topology of direct pairwise interactions. The indirect influence between two components [MATH] and [MATH] can be studied based on sequences of direct interactions [MATH] that mediate the influence between [MATH] and [MATH] via a path |
[MATH] Building on this abstraction, network science has developed methods that help us to better understand the structure and function of complex systems. The success and popularity of these network science methods across disciplines rest on their broad applicability to relational data that capture pairwise interactio... |
also makes an important assumption, namely that the paths by which a system’s components indirectly influence each other can be understood based on the transitive closure of direct, pairwise interactions. That is, most network science methods rest on the assumption that the existence of direct interactions [MATH] and [... |
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