2
$\begingroup$

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?

$\endgroup$
  • 3
    $\begingroup$ "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. $\endgroup$ – John Hughes Jan 4 '17 at 22:01
  • $\begingroup$ 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. $\endgroup$ – arctic tern Jan 8 '17 at 1:40
3
+25
$\begingroup$

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).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.