Untying knots in 4D and Wedderburn's Theorem

TL;DR

This paper demonstrates that Wedderburn's Theorem on finite division rings implies all knots and links in smooth 4D manifolds are trivial, revealing a deep connection between algebra and topology.

Contribution

It establishes a novel link between algebraic structures and 4-dimensional topology by deriving topological triviality from algebraic properties.

Findings

All knots and links in smooth 4D manifolds are trivial.

Wedderburn's Theorem implies topological triviality in 4D.

Connects algebraic theorems with topological properties.

Abstract

It is proved that the Wedderburn Theorem on finite division rings implies that all knots and links in the smooth 4-dimensional manifolds are trivial.

Equations

$$\mathbf{I}\subseteq {\mathbf{I}}_{\max }\subset O_{\mathbb{K}}.$$

$$h:O_{\mathbb{K}}\to R$$

$$R\cong M_n(\mathcal{D}),$$

$$R\cong \mathcal{D}.$$

Full text

1. Introduction

Our brief note contains an algebraic proof of the otherwise known topological fact, that all knots and links in the smooth 4-dimensional manifolds can be untied, i.e. are trivial. The novelty is a surprising rôle of the Wedderburn Theorem [Maclagan-Wedderburn 1905] [2] in the 4-dimensional topology [4].

Recall that arithmetic topology studies a functor, $F$, between the 3-dimensional manifolds and the fields of algebraic numbers [Morishita 2012] [3]. Such a functor maps $3$-dimensional manifolds $\mathscr{M}^{3}$ to the algebraic number fields $K$, so that the knots (links, resp.) in $\mathscr{M}^{3}$ correspond to the prime ideals (ideals, resp.) in the ring of integers $O_{K}$.

The map $F$ extends to the smooth 4-dimensional manifolds $\mathscr{M}^{4}$ and the fields of hyper-algebraic numbers $\mathbb{K}$, i.e. fields with a non-commutative multiplication [4]. To formulate our result, denote by $O_{\mathbb{K}}$ the ring of integers of the field $\mathbb{K}$. A ring $R$ is called a domain, if $R$ has no zero divisors. The $R$ is called simple, if it has only trivial two-sided ideals. Our main result is the following theorem.

Theorem 1.1**.**

$O_{\mathbb{K}}$* is a simple domain.*

Remark 1.2*.*

Theorem 1.1 is false for the algebraic integers, since the domain $O_{K}$ is never simple.

Corollary 1.3**.**

Any knot or link in $\mathscr{M}^{4}$ is trivial.

Proof.

If $\mathscr{K}\subset\mathscr{M}^{4}$ ($\mathscr{L}\subset\mathscr{M}^{4}$, resp.) is a non-trivial knot (link, resp.), then $F(\mathscr{K})$ ($F(\mathscr{L})$, resp.) is a non-trivial two-sided prime ideal (two-sided ideal, resp.) in $O_{\mathbb{K}}$. The latter contradicts 1.1, since $O_{\mathbb{K}}$ is a simple ring. ∎

The paper is organized as follows. Section 2 contains a brief review of the preliminary results. Theorem 1.1 is proved in Section 3.

2. Preliminaries

2.1. Arithmetic topology

The arithmetic topology studies an interplay between 3-dimensional manifolds and number fields [Morishita 2012] [3]. Let $\mathfrak{M}^{3}$ be a category of closed 3-dimensional manifolds, such that the arrows of $\mathfrak{M}^{3}$ are homeomorphisms between the manifolds. Likewise, let $\mathbf{K}$ be a category of the algebraic number fields, where the arrows of $\mathbf{K}$ are isomorphisms between such fields. Let $\mathscr{M}^{3}\in\mathfrak{M}^{3}$ be a 3-manifold, let $S^{3}\in\mathfrak{M}^{3}$ be the 3-sphere and let $O_{K}$ be the ring of integers of $K\in\mathbf{K}$. An exact relation between 3-manifolds and number fields can be described as follows.

Theorem 2.1**.**

The exists a covariant functor $F:\mathfrak{M}^{3}\to\mathbf{K}$, such that:

(i) $F(S^{3})=\mathbf{Z}$;

(ii) each ideal $I\subseteq O_{K}=F(\mathscr{M}^{3})$ corresponds to a link $\mathscr{L}\subset\mathscr{M}^{3}$;

(iii) each prime ideal $I\subseteq O_{K}=F(\mathscr{M}^{3})$ corresponds to a knot $\mathscr{K}\subset\mathscr{M}^{3}$.

Denote by $\mathfrak{M}^{4}$ a category of all smooth 4-dimensional manifolds $\mathscr{M}^{4}$, such that the arrows of $\mathfrak{M}^{4}$ are homeomorphisms between the manifolds. Denote by $\mathfrak{K}$ a category of the hyper-algebraic number fields $\mathbb{K}$, such that the arrows of $\mathfrak{K}$ are isomorphisms between the fields. Theorem 2.1 extends to 4-manifolds as follows.

Theorem 2.2**.**

([4, Theorem 1.1])* The exists a covariant functor $F:\mathfrak{M}^{4}\to\mathfrak{K}$, such that the 4-manifolds $\mathscr{M}^{4}_{1},\mathscr{M}^{4}_{2}\in\mathfrak{M}^{4}$ are homeomorphic if and only if the hyper-algebraic number fields $F(\mathscr{M}^{4}_{1}),F(\mathscr{M}^{4}_{2})\in\mathfrak{K}$ are isomorphic.*

2.2. Wedderburn Theorem

Roughly speaking, Wedderburn’s Theorem says that finite non-commutative fields cannot exist [Maclagan-Wedderburn 1905] [2]. Namely, denote by $\mathscr{D}$ a division ring. Let $\mathbb{F}_{q}$ be a finite field for some $q=p^{r}$, where $p$ is a prime and $r\geq 1$ is an integer number.

Theorem 2.3**.**

(Wedderburn Theorem)* If $|\mathscr{D}|<\infty$ and $\mathscr{D}$ is finite dimensional over a division ring, then $\mathscr{D}\cong\mathbb{F}_{q}$ for some $q=p^{r}$.*

We shall use 2.3 along with a classification of simple rings due to Artin and Wedderburn. Recall that a ring $R$ is called simple, if $R$ has only trivial two-sided ideals. By $M_{n}(\mathscr{D})$ we understand the ring of $n$ by $n$ matrices over $\mathscr{D}$.

Theorem 2.4**.**

(Artin-Wedderburn)* If $R$ is a simple ring, then $R\cong M_{n}(\mathscr{D})$ for a division ring $\mathscr{D}$ and an integer $n\geq 1$.*

Remark 2.5*.*

The ring $M_{n}(\mathscr{D})$ is a domain if and only if $n=1$. For instance, if $n=2$, then the matrices $\left(\begin{smallmatrix}1&0\cr 0&0\end{smallmatrix}\right)$ and $\left(\begin{smallmatrix}0&0\cr 0&1\end{smallmatrix}\right)$ are zero divisors in the ring $M_{2}(\mathscr{D})$.

3. Proof of theorem 1.1

Theorem 1.1 will be proved by contradiction. Namely, we show that existence of a non-trivial two-sided ideal in $O_{\mathbb{K}}$ contradicts 2.3. To begin, let us prove the following lemma.

Lemma 3.1**.**

$O_{\mathbb{K}}$* is a non-commutative Noetherian domain.*

Proof.

Recall that $O_{\mathbb{K}}$ is generated by the zeroes of a non-commutative polynomial $\mathscr{P}(x):=\sum_{i}a_{i}xb_{i}xc_{i}x\dots e_{i}xl_{i}$, where $a_{i},b_{i},c_{i}\dots,e_{i},l_{i}\in O_{\mathbb{L}}$ and $\mathbb{K}$ is a finite dimensional extension of $\mathbb{L}$. By the Hilbert Basis Theorem for non-commutative rings [Amitsur 1970] [1], if $O_{\mathbb{L}}$ is Noetherian, i.e. any ascending chain of the two-sided ideals of $O_{\mathbb{L}}$ stabilizes, then the ring $O_{\mathbb{K}}$ is also Noetherian. Repeating the construction, one arrives at a finite dimensional extension $\mathbb{H}\subset\mathbb{K}$, where $\mathbb{H}$ is the field of quaternions. The ring of the Hurwitz quaternions $O_{\mathbb{H}}$ is known to be Noetherian. Thus $O_{\mathbb{K}}$ is a Noetherian ring. Lemma 3.1 is proved. ∎

Returning to the proof of theorem 1.1, let us assume to the contrary, that $\mathbf{I}$ is a non-trivial two-sided ideal of $O_{\mathbb{K}}$. By lemma 3.1, there exists the maximal two-sided ideal $\mathbf{I}_{\max}$, such that

[TABLE]

Lemma 3.2**.**

The ring $R:=O_{\mathbb{K}}/\mathbf{I}_{\max}$ is a simple domain.

Proof.

The ring $R$ is simple, since $\mathbf{I}_{\max}$ is the maximal two-sided ideal of $O_{\mathbb{K}}$. The ring $R$ is a domain, since $O_{\mathbb{K}}$ is a domain and the homomorphism

[TABLE]

is surjective. ∎

Remark 3.3*.*

It follows from $R\cong O_{\mathbb{K}}/\mathbf{I}_{\max}$, that $|R|<\infty$. Indeed, any non-trivial subgroup of the abelian group $(O_{\mathbb{K}},+)$ has finite index by the Margulis normal subgroup theorem. In particular, the subgroup $(\mathbf{I}_{\max},+)$ has finite index in $(O_{\mathbb{K}},+)$.

To finish the proof of theorem 1.1, we write

[TABLE]

where $\mathscr{D}$ is a division ring, see Theorem 2.4. Since $R$ is a domain, we conclude that $n=1$ in formula (3.3), see remark 2.5. Thus

[TABLE]

But remark 3.3 says that $|R|<\infty$ and by the Wedderburn Theorem one gets $R\cong\mathbb{F}_{q}$ for some $q=p^{r}$. In particular, the homomorphism (3.2) implies that the ring $O_{\mathbb{K}}$ is commutative. Indeed, since $R$ is a commutative ring, one gets $h(xy-yx)=h(x)h(y)-h(y)h(x)=h(x)h(y)-h(x)h(y)=0,$ where [math] is the neutral element of $R$ . In other words, the element $xy-yx$ belongs to the kernel of $h$, which is a two-sided ideal $I_{h}\subset O_{\mathbb{K}}$. If $h$ is not injective, then $I_{h}$ is non-trivial and taking the multiplicative identity $1\in I_{h}$ we obtain a contradiction $h(1)=0$. Thus $h$ is injective and $xy=yx$ for all $x,y\in O_{\mathbb{K}}$, i.e. $O_{\mathbb{K}}$ is a commutative ring. On the other hand, the ring $O_{\mathbb{K}}$ cannot be commutative by an assumption of theorem 1.1. The obtained contradiction completes the proof of theorem 1.1.