Relationship between Levi-Civita symbol and complex/quaternionic numbers

Question

Complex numbers can be represented as matrices (https://en.wikipedia.org/wiki/Complex_number#Matrix_representation_of_complex_numbers). Especially the complex number $-i$ has the matrix representation: $$ \left[ \begin{array}{cc} 0 & 1\\ -1 & 0 \end{array} \right] $$ But this is the same as the matrix representation of the Levi-Civita symbol in two dimensions. Why is the Levi-Civita symbol in two dimensions equal to $-i$ ?

There is also a relationship between quaternions and the Levi-Civita symbol in three dimensions. But the Levi-Civita Symbol in three dimensions can be arranged as a 3x3x3 array. Does this mean quaternions can also be represented as 3x3x3 array ? But up to now I always thought that quaternions are represented by 4x4 matrices. So I'm quite confused.

What is the relationship between complex or quaternionic numbers and the Levi-Civita symbol?

Answer 1

There is a connection.

A cross product of type $(r,d)$ is an $r$-ary alternating multilinear operation on a $d$-dimensional real inner product space which spits out a vector orthogonal to all its inputs and sends orthonormal $r$-frames to unit vectors. Here is the classification of all cross products:

• Degenerate cross products which are identically zero. Occurs when $r> d$.
• Geometric (or co-unary) cross products occur when $r=d-1$. In this case, the output vector is uniquely determined by the properties of the cross product and a choice of orientation. With a choice of coordinates, $X(u_1,\cdots,u_{d-1})$ is the unique $v$ such that $\det(u_1\cdots u_{d-1}w)=\langle v,w\rangle $. Or, with the universal property of $\Lambda$ we speak of a linear map $X:\Lambda^{d-1}\mathbb{R}^d\to\mathbb{R}^d$, and by identifying $\mathbb{R}$ with $\Lambda^1\mathbb{R}^d$ this is the Hodge star operator.
• Unary cross products occur when $r=1$ and $d$ is even. These are induced from multiplication by $i$ in a complex inner product space $V$. Taking any orthonormal basis $B$ for $V$ as a complex vector space, $B\sqcup iB$ is then a basis for $V$ as a real vector space and induces an inner product independent of choice of $B$. (It is $\mathrm{Re}\langle u,v\rangle$ where $\langle\cdot,\cdot\rangle$ is the complex inner product.)
• Binary cross products occur when $r=2$ and $d=0,1,3,7$. These are induced from the normed division algebras $\mathbb{R},\mathbb{C},\mathbb{H},\mathbb{O}$ (the last two being quaternions and octonions). In all four cases, there is an inner product (corresponding to the norm) and the orthogonal complement of the real axis has a cross product $\times$ defined on it corresponding to the multiplication formula $uv=-\langle u,v\rangle+u\times v$.
• Exceptional cross products are any not in the above families. There is only one: a cross product of type $(3,8)$ defined on the octonions $\mathbb{O}$ by the formula $X(a,b,c)=\frac{1}{2}[a(\overline{b}c)-c(\overline{b}a)]$. Why does that formula work? I ask that question too.

I made up the names myself. There is some overlap: types $(2,0)$ and $(2,1)$ are both binary and degenerate, type $(1,2)$ is both unary and geometric, and type $(2,3)$ is both binary and geometric.

Above I used $v_1,\cdots,v_d$ to enumerate $d$ different vectors in $\mathbb{R}^d$. Now let's say I do the physics thing of writing $v^i$ for the $i$th coordinate of a vector $v$. Then the geometric cross product of $u,v,\cdots,w$ (pretend I have $d-1$ different letters) is given in Einstein summation notation by

$$ u^{i_1}v^{i_2}\cdots w^{i_{d-1}}\varepsilon_{i_1\cdots i_{d-1}}.$$

This gives multiplication by $i$ on $\mathbb{C}\cong\mathbb{R}^2$ when $d=2$, and it gives the usual cross product on $\mathbb{R}^3$ when $d=3$, the latter of which is present in quaternion multiplication. Keep in mind the cross product of type $(1,2)$ is multiplication by $i$, it does not represent multiplying two complex numbers together, whereas the cross product of type $(2,3)$ corresponds to multiplication of two quaternions (or more accruately, the imaginary part of the product of two purely imaginary quaternions).

Answer 2

I object to the term "Levi-Civita symbol", and will call them by a more descriptive, and less inaccurate, term "permutation symbol".

Your observation about the permutation symbol in 2-D is a good one. I don't know what form an answer to your question would take. (I would love to hear one!)

Of course, $-i$ is not the only antisymmetric (bi-)linear transformation: $i$ does just as well, the difference being described geometrically as sense of rotation. In the four-dimensional case, there are more transformations that are like rotation. The definition you referred to (in terms of the sign of permutations) specifies individual elements of a matrix representation, which is good for a starting point.

The rotations and the permutation symbol are manifestations of rather different things, geometrically: rotations are simple transformations of the space, whereas the permutation symbol is related to measure (area, volume) of regions in the space. The former can always be represented by square matrices, whereas the complexity of the latter grows much faster — the summation convention was invented to deal with that greater complexity. Only in two dimensions does the permutation symbol coincide with a rotation — but in all dimensions, the two things are related.

The direct generalization from the 2D case to 4D is the four-dimensional permutation symbol, which can be imagined as a $4\times 4\times 4\times 4$ nested block of numbers. It is related to the quaternion units $i$, $j$, and $k$, but the relation takes a little work to describe.

Note that, in contrast to the case of complex numbers, there are two linear transformations associated with a quaternion: one for multiplication on the left by the quaternion, and one for the right, and that they are typically different. Thus, multiplication by a quaternion is not represented by a single matrix, but two.

For the $4\times 4$ matrices representing (with respect to the usual basis for quaternions $\{ 1, i, j, k \}$) quaternion multiplication on the left and right by $i$, $j$, and $k$, write \begin{align*} i_L &= \left[\begin{matrix} 0 & -1 & & \\ 1 & 0 & & \\ & & 0 & -1 \\ & & 1 & 0 \\ \end{matrix}\right] , &j_L &= \left[\begin{matrix} & & -1 & 0 \\ & & 0 & 1 \\ 1 & 0 & & \\ 0 & -1 & & \\ \end{matrix}\right] , &k_L &= \left[\begin{matrix} & & & -1 \\ & & -1 & \\ & 1 & & \\ 1 & & & \\ \end{matrix}\right] ; \\ i_R &= \left[\begin{matrix} 0 & -1 & & \\ 1 & 0 & & \\ & & 0 & 1 \\ & & -1 & 0 \\ \end{matrix}\right] , &j_R &= \left[\begin{matrix} & & -1 & 0 \\ & & 0 & -1 \\ 1 & 0 & & \\ 0 & 1 & & \\ \end{matrix}\right] , &k_R &= \left[\begin{matrix} & & & -1 \\ & & 1 & \\ & -1 & & \\ 1 & & & \\ \end{matrix}\right] . \end{align*}

The symbol $\epsilon_{\mu\nu\rho\sigma}$ can be written as a $4 \times 4$ matrix of $4 \times 4$ matrices, so: $$ \left[\begin{matrix} 0 & \left[\begin{matrix} 0 & 0 & 0 & 0\\0 & 0 & 0 & 0\\0 & 0 & 0 & 1\\0 & 0 & -1 & 0 \end{matrix}\right] & \left[\begin{matrix} 0 & 0 & 0 & 0\\0 & 0 & 0 & -1\\0 & 0 & 0 & 0\\0 & 1 & 0 & 0 \end{matrix}\right] & \left[\begin{matrix} 0 & 0 & 0 & 0\\0 & 0 & 1 & 0\\0 & -1 & 0 & 0\\0 & 0 & 0 & 0 \end{matrix}\right]\\ \left[\begin{matrix} 0 & 0 & 0 & 0\\0 & 0 & 0 & 0\\0 & 0 & 0 & -1\\0 & 0 & 1 & 0 \end{matrix}\right] & 0 & \left[\begin{matrix} 0 & 0 & 0 & 1\\0 & 0 & 0 & 0\\0 & 0 & 0 & 0\\-1 & 0 & 0 & 0 \end{matrix}\right] & \left[\begin{matrix} 0 & 0 & -1 & 0\\0 & 0 & 0 & 0\\1 & 0 & 0 & 0\\0 & 0 & 0 & 0 \end{matrix}\right]\\ \left[\begin{matrix} 0 & 0 & 0 & 0\\0 & 0 & 0 & 1\\0 & 0 & 0 & 0\\0 & -1 & 0 & 0 \end{matrix}\right] & \left[\begin{matrix} 0 & 0 & 0 & -1\\0 & 0 & 0 & 0\\0 & 0 & 0 & 0\\1 & 0 & 0 & 0 \end{matrix}\right] & 0 & \left[\begin{matrix} 0 & 1 & 0 & 0\\-1 & 0 & 0 & 0\\0 & 0 & 0 & 0\\0 & 0 & 0 & 0 \end{matrix}\right]\\ \left[\begin{matrix} 0 & 0 & 0 & 0\\0 & 0 & -1 & 0\\0 & 1 & 0 & 0\\0 & 0 & 0 & 0 \end{matrix}\right] & \left[\begin{matrix} 0 & 0 & 1 & 0\\0 & 0 & 0 & 0\\-1 & 0 & 0 & 0\\0 & 0 & 0 & 0 \end{matrix}\right] & \left[\begin{matrix} 0 & -1 & 0 & 0\\1 & 0 & 0 & 0\\0 & 0 & 0 & 0\\0 & 0 & 0 & 0 \end{matrix}\right] & 0 \end{matrix}\right] $$ (I generated this using sympy.) Here, $\mu$ and $\nu$ index the containing matrix, while $\rho$ and $\sigma$ index the inner matrices, both in row-column order.

In this form, the matrices for multiplication by quaternion units can be read off in terms of blocks of the matrix for the permutation symbol: \begin{eqnarray*} i_L &=\quad \epsilon_{4 3 \rho \sigma} + \epsilon_{3 1 \rho \sigma} \,, \qquad i_R &=\quad \epsilon_{4 3 \rho \sigma} + \epsilon_{1 2 \rho \sigma} \,, \\ j_L &=\quad \epsilon_{2 4 \rho \sigma} + \epsilon_{2 1 \rho \sigma} \,, \qquad j_R &=\quad \epsilon_{2 4 \rho \sigma} + \epsilon_{1 3 \rho \sigma} \,, \\ k_L &=\quad \epsilon_{3 2 \rho \sigma} + \epsilon_{4 1 \rho \sigma} \,, \qquad k_R &=\quad \epsilon_{3 2 \rho \sigma} + \epsilon_{1 4 \rho \sigma} \,. \end{eqnarray*}

To express the permutation symbol in terms of the quaternion units, arrange the six matrices to form two antisymmetric $4 \times 4$ matrices of matrices, so: $$ U_L = \left[\begin{matrix} 0 & -i_L & -j_L & -k_L \\ i_L & 0 & -k_L & j_L \\ j_L & k_L & 0 & -i_L \\ k_L & -j_L & i_L & 0 \end{matrix}\right] , \quad U_R = \left[\begin{matrix} 0 & -i_R & -j_R & -k_R \\ i_R & 0 & k_R & -j_R \\ j_R & -k_R & 0 & i_R \\ k_R & j_R & -i_R & 0 \end{matrix}\right] . $$

From here, it is a matter of verification that $$ \epsilon_{\mu\nu\rho\sigma} = \frac{1}{2} ( U_L - U_R ) . $$

I am at a loss to answer "why", beyond a simple verification, but I expect that a succinct, convincing proof, based on symmetries, is achievable.