# Relationship between Levi-Civita symbol and complex/quaternionic numbers

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?

• "Can be represented as" is pretty mushy. We often represent $a + bi$ as $M = \begin{bmatrix} a & -b \\ b & a \end{bmatrix}$, but there's an equally good representation where we use $M^t$ instead of $M$. Which one's better? Depends whether you like to read horizontally or vertically, I guess. I suspect that if you clear up your notion of "can be represented," you'll either find that your question -- whatever it may really be -- is answered or that it's merely coincidence. – John Hughes Jan 4 '17 at 22:01
• 3D levi-civita is 3x3x3 because it takes two 3D vectors and spits out a 3D vector. If you wanted to, you could represent quaternion multiplication as contraction with a 4x4x4 tensor in the same way, but that's different from representing individual quaternions by 4x4 matrices. In any case, the 3D cross product may be represented by a 3x3x3 tensor, namely the 3D levi-civita symbol, and the 3D cross product is only a component of quaternion multiplication. – arctic tern Jan 8 '17 at 1:40

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).