Two Dimensional Representations of the Braid Group B3 and its Burau Representation Squared

TL;DR

This paper classifies all two-dimensional irreducible complex representations of the braid group B3 and analyzes the decomposition of the squared Burau representation, providing insights into their structure.

Contribution

It provides an elementary listing of all two-dimensional irreducible representations of B3 and decomposes the square of its Burau representation, a novel analysis in this context.

Findings

Complete classification of 2D irreducible representations of B3

Decomposition of the squared Burau representation

Insights into the structure of B3 representations

Abstract

We list the irreducible two dimensional complex representations of the Braid group B3 in elementary way. Then, we make a decomposition of the square of its irreducible Burau representation.

Equations

$$\sigma_{1}\rightarrow\left[\begin{array}[]{cc}-z&0\\ 1&1\end{array}\right],\sigma_{2}\rightarrow\left[\begin{array}[]{cc}1&z\\ 0&-z\end{array}\right]$$

$$\sigma_{1}\rightarrow\left[\begin{array}[]{cc}-z&0\\ 0&1\end{array}\right],\sigma_{2}\rightarrow\left[\begin{array}[]{cc}\frac{1}{z+1}&-\frac{z}{z+1}\\ -\frac{z^{2}+z+1}{z+1}&-\frac{z^{2}}{z+1}\end{array}\right]$$

$$R(S_3)=\mathbb{Z}[\xi ,\rho ]/({\xi }^2=1,{\rho }^2=\rho +\xi +1,\xi \rho =\rho )$$

Full text

1. Introduction

Artin braid group $B_{n}$, $n\geq 2$, is given by generators $\sigma_{1}$, $\sigma_{2}$, …, $\sigma_{n-1}$ subject to the relations:

\begin{array}[]{l}\left(1\right)\text{ }\sigma_{i}\sigma_{j}=\sigma_{j}\sigma_{i}\text{ for }\left|i-j\right|\geq 2\\ \left(2\right)\text{ }\sigma_{i+1}\sigma_{i}\sigma_{i+1}=\sigma_{i}\sigma_{i+1}\sigma_{i}\text{ for }1\leq i\leq n-2.\end{array}

A quick proof of the fact that these are the minimal relations for the geometric braid group is given by Jun Morita in [3] by induction.

Note that $B_{2}=Z$ is not interesting, on the other hand $B_{3}$ is generated by $\sigma_{1}$, $\sigma_{2}$ subject to a single relation $\sigma_{1}\sigma_{2}\sigma_{1}=\sigma_{2}\sigma_{1}\sigma_{2}$ and it is already very complicated. For $n\geq 3,$ Braid groups are non-abelian infinite groups.

A $k$-dimensional complex representation of $B_{n}$ is a group homomorphism from this group into $GL_{k}(\mathbb{C}),$ the group of $k\times k$ complex invertible matrices. Two representations are isomorphic if they are conjugate by a fixed invertible matrix. We say that a representation is irreducible if it is not isomorphic to direct sum of two lower dimensional representations. All 1-dimensional representations are irreducible by definition. A famous example of a higher dimensional irreducible representation of the braid group $B_{n}$ is its irreducible Burau representation $\beta(z)$, [1].

Edward Formanek’s paper [1] on Braid group representations of low degrees discusses the classification of the complex representations of the Braid group $B_{n}$ for dimensions $\leq n-1.$But, his first and easiest classification result, Theorem 11, seems to list less 2-dimensional representations of $B_{3}$ than we had by solving matrix equations.

In the first part of this note, we classify the complex 2-dimensional representations of $B_{3}$ in the elementary way. Then, in the second part, we will make the decomposition of $\beta(z)\otimes\beta(z)$ hoping to find an analogous of the relation that we have in $R(S_{3}).$ It turned out to be different, not 1+1+2 decomposition but 1+3. The 3-dimensioanal irreducible representation we obtained obeys a Pascal triangle pattern. This kind of patterns first observed in [2]. Higher dimensional versions should be related to higher tensor powers of the Burau representation and shouldn’t be hard to generalize.

2. Two Dimensional Representations of $B_{3}$

Due to the relation $\sigma_{1}\sigma_{2}\sigma_{1}=\sigma_{2}\sigma_{1}\sigma_{2}$, if $\xi:\sigma_{1}\rightarrow z_{1}$, $\sigma_{2}\rightarrow z_{2}$ where $z_{1}$, $z_{2}\in\mathbb{C}^{\ast}$ is a 1-dimensional representation of $B_{3}$, then we must have $z_{1}=z_{2}.$ So for each $z\in\mathbb{C}^{\ast},$ let us denote the representation, $\sigma_{1},$ $\sigma_{2}\rightarrow z$ by $\xi(z).$ All non-isomorphic 1-dimensional representations of $B_{3}$ and also of other higher order braid groups are in this form.

Next, we want to determine all possible forms of irreducible 2-dimensional representations.

Theorem 1. All irreducible 2-dimensional representations of $B_{3}$ up to tensor with 1-dimansional representations are in the form

(i) \sigma_{1}\rightarrow\left[\begin{array}[]{cc}-z&0\\ 0&1\end{array}\right],\sigma_{2}\rightarrow\left[\begin{array}[]{cc}\frac{1}{z+1}&f\\ g&-\frac{z^{2}}{z+1}\end{array}\right] where $fg=\dfrac{z(z^{2}+z+1)}{(z+1)^{2}}$, $z\neq 0,-1$ and $z^{2}+z+1\neq 0,$

or in the form

(ii) \sigma_{1}\rightarrow\left[\begin{array}[]{cc}1&z\\ 0&1\end{array}\right],\sigma_{2}\rightarrow\left[\begin{array}[]{cc}e&z(e-1)^{2}\\ -\frac{1}{z}&2-e\end{array}\right] where $z\neq 0.$

Proof.** **Let us now consider a 2-dimensional representation of $B_{3}$ given by $\sigma_{1}\rightarrow A$, $\sigma_{2}\rightarrow B$ where $A$, $B\in GL_{2}(\mathbb{C}).$

Since we are interested in isomorphism classes of 2-dimensional representations, we can diagonalize $A$ if it is diagonalizable or if not, we can put it in Jordan form. So, we have two cases.

(i) We assume that $A$ is diagonal and also we can tensor with a 1-dimensional representation to assume that one entry of the diagonal is 1.

So, let us assume that A=\left[\begin{array}[]{cc}-z&0\\ 0&1\end{array}\right]  and that B=\left[\begin{array}[]{cc}e&f\\ g&h\end{array}\right] where $z\neq 0$. The choice of $-z$ in the matrix $A$ will be clarified in the next section.

Due to the relation $\sigma_{1}\sigma_{2}\sigma_{1}=\sigma_{2}\sigma_{1}\sigma_{2},$ we must have $ABA=BAB$ and thus we can solve for $e,h$ and the product $fg$ to get the result as expressed above.

Since if $z^{2}+z+1\neq 0$ the given matrices do not have common eigenvectors for their same eigenvalues, we can not find one dimensional subrepresentation. So, the representations given in this case are all irreducible.

(ii) We assume that $A$ is in Jordan from and also we can tensor with a 1-dimensional representation to assume that diagonal entries are 1.

So let A=\left[\begin{array}[]{cc}1&z\\ 0&1\end{array}\right]  and that B=\left[\begin{array}[]{cc}e&f\\ g&h\end{array}\right] where $z\neq 0$. Due to the relation $\sigma_{1}\sigma_{2}\sigma_{1}=\sigma_{2}\sigma_{1}\sigma_{2},$ we must have $ABA=BAB$ and thus we find that $f=z(e-1)^{2},$ $g=-\dfrac{1}{z}$ and $h=2-e$.

Since we have one eigenvector in this case, we can not split these representations. So, all representations in this case are irreducible too.

Edward Formanek’s paper [1] also discusses the condition $z^{2}+z+1\neq 0$ related to Burau representation. He generalizes this condition for all $n$ in Lemma 6 of the paper. His main tool is pseudoreflections.

3. Burau Representation of $B_{3}$ Squared

The irreducible Burau representation of $B_{3}$ is given by

[TABLE]

where $z\in\mathbb{C}^{\ast}$ is a parameter. We will denote this representation by $\beta(z).$ Hence, $\beta(z):B_{3}\rightarrow GL_{2}(\mathbb{C}).$

**Proposition 2. **The irreducible Burau representation of $B_{3}$ is isomorphic to the representation, $z\neq 1,$

[TABLE]

*Proof. *The matrix A=\left[\begin{array}[]{cc}-z&0\\ 1&1\end{array}\right] is diagonalizable by P=\left[\begin{array}[]{cc}-(z+1)&0\\ 1&1\end{array}\right] to P^{-1}AP=\left[\begin{array}[]{cc}-z&0\\ 0&1\end{array}\right]. And then for B=\left[\begin{array}[]{cc}1&z\\ 0&-z\end{array}\right], we get P^{-1}BP=\left[\begin{array}[]{cc}\frac{1}{z+1}&-\frac{z}{z+1}\\ -\frac{z^{2}+z+1}{z+1}&-\frac{z^{2}}{z+1}\end{array}\right] as required.

We notice that the Burau representation of $B_{3}$ can be obtained by taking $f=-\frac{z}{z+1}$ in Theorem 1 (i).

$S_{3},$ the symmetric group on 3 letters is generated by cycles $s_{1}$ and $s_{2}$ subject to the relations $s_{1}^{2}=s_{2}^{2}=1$ and $s_{1}s_{2}s_{1}=s_{2}s_{1}s_{2}.$

We have the natural projection homomorphism $\pi:B_{3}\rightarrow S_{3}$ which sends $\sigma_{1}$ to $s_{1}$ and $\sigma_{2}$ to $s_{2}$. Note that we have these natural homomorphisms $\pi:B_{n}\rightarrow S_{n}$ for each $n$ and the kernel of these homomorphisms are called pure braid groups.

The complex representation ring of $S_{3}$ is given by

[TABLE]

where $\xi$ is the 1-dimensional representation given by $s_{1}\rightarrow-1,$ $s_{2}\rightarrow-1$ and is called the sign representation and $\rho$ is the 2-dimensional representation given by s_{1}\rightarrow\left[\begin{array}[]{cc}-1&0\\ 1&1\end{array}\right],s_{2}\rightarrow\left[\begin{array}[]{cc}1&1\\ 0&-1\end{array}\right] which is called the standard representation.

The standard representation of $S_{3}$ is obtained from Burau representation of $B_{3}$by taking $z=1.$ In other words, $\rho=\pi\circ\beta(1).$ Inspired of this connection, we suspect that $\beta(z)\otimes\beta(z)$ might satisfy a similar relation.

**Proposition 3. Let $z\neq 0,-1.$ **Then $\beta(z)^{2}=\xi(-z)+\mu(z)$ where $\mu(z)$ is a 3-dimensional representation which is irreducible if $z^{3}\neq 1$ and it is given by

\sigma_{1}\rightarrow\left[\begin{array}[]{ccc}1&0&0\\ 0&-z&0\\ 0&0&z^{2}\end{array}\right], \sigma_{2}\rightarrow\left[\begin{array}[]{ccc}\frac{z^{4}}{\left(z+1\right)^{2}}&\frac{z^{2}}{\left(z+1\right)^{2}}\left(z^{2}+z+1\right)&\frac{1}{\left(z+1\right)^{2}}\left(z^{2}+z+1\right)^{2}\\ 2\frac{z^{3}}{\left(z+1\right)^{2}}&z\frac{z^{2}+1}{\left(z+1\right)^{2}}&-\frac{2}{\left(z+1\right)^{2}}\left(z^{2}+z+1\right)\\ \frac{z^{2}}{\left(z+1\right)^{2}}&-\frac{z}{\left(z+1\right)^{2}}&\frac{1}{\left(z+1\right)^{2}}\end{array}\right]

*Proof. *First, we calculate

$A=\beta(z)\otimes\beta(z)(\sigma_{1})=$ \left[\begin{array}[]{cc}-z&0\\ 1&1\end{array}\right]\otimes\left[\begin{array}[]{cc}-z&0\\ 1&1\end{array}\right]=\left[\begin{array}[]{cccc}z^{2}&0&0&0\\ -z&-z&0&0\\ -z&0&-z&0\\ 1&1&1&1\end{array}\right] and that

$B=\beta(z)\otimes\beta(z)(\sigma_{2})=$ \left[\begin{array}[]{cc}1&z\\ 0&-z\end{array}\right]\otimes\left[\begin{array}[]{cc}1&z\\ 0&-z\end{array}\right]=\left[\begin{array}[]{cccc}1&z&z&z^{2}\\ 0&-z&0&-z^{2}\\ 0&0&-z&-z^{2}\\ 0&0&0&z^{2}\end{array}\right].

Now, $A$ is diagonalizable with the matrix P=\left[\begin{array}[]{cccc}0&0&0&z^{2}+2z+1\\ 0&-z-1&-1&-z-1\\ 0&0&1&-z-1\\ 1&1&0&1\end{array}\right] and

$P^{-1}AP=$ \left[\begin{array}[]{cccc}1&0&0&0\\ 0&-z&0&0\\ 0&0&-z&0\\ 0&0&0&z^{2}\end{array}\right].

And then P^{-1}BP=\allowbreak\allowbreak\left[\begin{array}[]{cccc}\frac{z^{4}}{\left(z+1\right)^{2}}&\frac{z^{2}}{\left(z+1\right)^{2}}\left(z^{2}+z+1\right)&0&\frac{1}{\left(z+1\right)^{2}}\left(z^{2}+z+1\right)^{2}\\ 2\frac{z^{3}}{\left(z+1\right)^{2}}&z\frac{z^{2}+1}{\left(z+1\right)^{2}}&0&-\frac{2}{\left(z+1\right)^{2}}\left(z^{2}+z+1\right)\\ -\frac{z^{3}}{z+1}&-\frac{z}{z+1}\left(z^{2}+z+1\right)&-z&\frac{1}{z+1}\left(z^{2}+z+1\right)\\ \frac{z^{2}}{\left(z+1\right)^{2}}&-\frac{z}{\left(z+1\right)^{2}}&0&\frac{1}{\left(z+1\right)^{2}}\end{array}\right]\allowbreak.

Let $\mathbb{C}^{4}=Span\left\{e_{1},e_{2},e_{3},e_{4}\right\}.$ From the calculations above, we observe that $Span\left\{e_{3}\right\}$ is an invariant subspace of the representation $\beta(z)\otimes\beta(z)$ with eigenvalue $-z.$ So $\beta(z)\otimes\beta(z)$ is reducible. Hence, we have $\beta(z)\otimes\beta(z)=\xi(-z)\oplus\mu(z)$ where $\xi(-z)$ is the mentioned 1-dimensional subrepresentation and

$\mu(z):$

\sigma_{1}\rightarrow C=\left[\begin{array}[]{ccc}1&0&0\\ 0&-z&0\\ 0&0&z^{2}\end{array}\right],

\sigma_{2}\rightarrow D=\left[\begin{array}[]{ccc}\frac{z^{4}}{\left(z+1\right)^{2}}&\frac{z^{2}}{\left(z+1\right)^{2}}\left(z^{2}+z+1\right)&\frac{1}{\left(z+1\right)^{2}}\left(z^{2}+z+1\right)^{2}\\ 2\frac{z^{3}}{\left(z+1\right)^{2}}&z\frac{z^{2}+1}{\left(z+1\right)^{2}}&-\frac{2}{\left(z+1\right)^{2}}\left(z^{2}+z+1\right)\\ \frac{z^{2}}{\left(z+1\right)^{2}}&-\frac{z}{\left(z+1\right)^{2}}&\frac{1}{\left(z+1\right)^{2}}\end{array}\right]

is a 3-dimensional subrepresentation.

The matrix $D$ has eigenvectors \left[\begin{array}[]{c}1\\ -2\\ 1\end{array}\right] for the eigenvalue $1,$ \left[\begin{array}[]{c}-\frac{1}{z}\left(z^{2}+z+1\right)\\ \frac{1}{z}\left(z^{2}+1\right)\\ 1\end{array}\right] for the eigenvalue -z,\allowbreak\left[\begin{array}[]{c}\frac{1}{z^{2}}\left(z^{4}+2z^{3}+3z^{2}+2z+1\right)\\ \frac{1}{z}\left(2z^{2}+2z+2\right)\\ 1\end{array}\right] for the eigenvalue $z^{2}\allowbreak.$ Whereas, eigenvectors of $C$ are the standard basis vectors of $\mathbb{C}^{3}$. If $z\neq 1$ and $z^{2}+z+1\neq 0,$ which means that $z^{3}\neq 1,$ there is no common eigenvectors, and we deduce that these eigenvectors do not produce a 1-dimensional subrepresentation. Therefore, $\mu(z)$ is irreducible.

Remark 4. For $z=1,$ the eigenvectors of $D$ corresponding to 1 are \left[\begin{array}[]{c}1\\ -2\\ 1\end{array}\right] and \left[\begin{array}[]{c}9\\ 6\\ 1\end{array}\right] and thus \left[\begin{array}[]{c}3\\ 0\\ 1\end{array}\right] also is an eigenvector of $D$. \left[\begin{array}[]{c}3\\ 0\\ 1\end{array}\right] is also an eigenvector of $C$ for $z=1$ corresponding to the eigenvalue 1. This vector spans the trivial 1-dimensional subrepresentation $\xi(1)$ which corresponds to 1 in the decomposition $\rho^{2}=\rho+\xi+1$ which occurred in $R(S_{3})$.

**Remark 5. **If we choose basis vectors, $\left\{u_{1}=e_{1},\text{ }u_{2}=e_{2}+e_{3},\text{ }u_{3}=e_{4}\right\},$ we can easily show that the representation $\mu(z)$ is isomorphic to the representation \sigma_{1}\rightarrow\left[\begin{array}[]{ccc}z^{2}&0&0\\ -z&-z&0\\ 1&2&1\end{array}\right],

\sigma_{2}\rightarrow D=\left[\begin{array}[]{ccc}1&2z&z^{2}\\ 0&-z&-z^{2}\\ 0&0&z^{2}\end{array}\right]. We observe a Pascal triangle pattern in matrices. These patterens also mentioned in [2].