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I am looking for some theorems about matrix representations of Clifford algebras.
Let $a \in G_{p,q,r}$, where $p$ elements square to $1$, $q$ to $-1$, and $r$ to $0$, that is $a=\sum_{i=1}^{2^{p+q+r}}a^{(i)}g_i$, where $g_i$ denotes $i$'th basis of clifford algebra $G_{p,q,r}$ arising from orthonormal basis. Then we can represent $a$ in form of matrix of real numbers in at least two different ways as following:

Let operator $V()$ denote vector form of a clifford number, that is if $x=\sum_{i=1}^{2^{p+q+r}}x^{(i)} g_i$, then $V(x)=[x^{(1)},x^{(2)},x^{(3)},...,x^{(2^{p+q+r})}]^{T}$. Take a geometric product of two Clifford numbers $a,b \in G_{p,q,r}$, and calculate the necessary matrix of real numbers $C$ such that $CV(b)=V(ab)$. The matrix $C$ is the first representation. The second representation can be obtained by calculating matrix $D$ such that $DV(b)=V(ba)$. In other words $C$ and $D$ are matrices for the linear transformations induced by left and right (respectively) multiplication by $a$.

For sake of generalizing below observations to algebras where $r$ is nonzero, from now on we assume that if $g_i^{2}=0$ then associated $a^{(i)}$ is equal to zero. I am looking for proofs of the following observations about these representations:

(1) det($C$)=det($D$), and eigenvalues of $C$ equal to eigenvalues of $D$.
(2) Each unique nonzero element of matrices $C$ and $D$ is either always symmetric, always antisymmetric, or always has a zero symmetry. Moreover, $a^{(i)}$ is symmetric if and only if $g_i^{2}=1$, antisymmetric iff $g_i^{2}=-1$, and has zero symmetry iff $g_i^{2}=0$.

I should explain what I mean by (2). For example take algebra $G_{1,1,0}$. Then we have the following matrices $C$ and $D$ for it: \begin{align} C=\left[ \begin{matrix} a^{(1)}& a^{(2)}& -a^{(3)}& a^{(4)}\\ a^{(2)}& a^{(1)}& -a^{(4)}& a^{(3)}\\ a^{(3)}& -a^{(4)}& a^{(1)}& a^{(2)}\\ a^{(4)}& -a^{(3)}& a^{(2)}& a^{(1)} \end{matrix}\right] \end{align}

\begin{align} D=\left[ \begin{matrix} a^{(1)}& a^{(2)}& -a^{(3)}& a^{(4)}\\ a^{(2)}& a^{(1)}& a^{(4)}& -a^{(3)}\\ a^{(3)}& a^{(4)}& a^{(1)}& -a^{(2)}\\ a^{(4)}& a^{(3)}& -a^{(2)}& a^{(1)} \end{matrix}\right] \end{align}

It can be seen that:
if $C_{ij}=\pm a^{(1)}$, then $C_{ji}=\pm C_{ij}$. $\qquad$ ($a^{(1)}$ is symmetric, $e_0^2=1$)
if $C_{ij}=\pm a^{(2)}$, then $C_{ji}\pm C_{ij}$ $\qquad$ ($a^{(2)}$ is symmetric, $e_1^2=1$)
if $C_{ij}=\pm a^{(3)}$, then $C_{ji}=\mp C_{ij}$ $\qquad$ ($a^{(3)}$ is antisymmetric,$e_2^2=-1$)
if $C_{ij}=\pm a^{(4)}$, then $C_{ji}=\pm C_{ij}$ $\qquad$ ($a^{(4)}$ is symmetric,$e_{12}^2=1$)

The same is true for $D$ also.
Here is another example, this time for $G_{0,1,1}$: \begin{align} C=\left[ \begin{matrix} a^{(1)}& -a^{(2)}& 0& 0\\ a^{(2)}& a^{(1)}& 0& 0\\ a^{(3)}& a^{(4)}& a^{(1)}& -a^{(2)}\\ a^{(4)}& -a^{(3)}& a^{(2)}& a^{(1)}\\ \end{matrix}\right] \end{align} \begin{align} D=\left[ \begin{matrix} a^{(1)}& -a^{(2)}& 0& 0\\ a^{(2)}& a^{(1)}& 0& 0\\ a^{(3)}& -a^{(4)}& a^{(1)}& a^{(2)}\\ a^{(4)}& a^{(3)}& -a^{(2)}& a^{(1)}\\ \end{matrix}\right] \end{align}

if $C_{ij}=\pm a^{(1)}$, then $C_{ji}=\pm C_{ij}$. $\qquad$ ($a^{(1)}$ is symmetric, $e_0^2=1$)
if $C_{ij}=\pm a^{(2)}$, then $C_{ji}\mp C_{ij}$ $\qquad$ ($a^{(2)}$ is antisymmetric, $e_1^2=-1$)
if $C_{ij}=\pm a^{(3)}$, then $C_{ji}=0$ $\qquad$ ($a^{(3)}$ is zero-symmetric,$e_2^2=0$)
if $C_{ij}=\pm a^{(4)}$, then $C_{ji}=0$ $\qquad$ ($a^{(4)}$ is zero-symmetric,$e_{12}^2=0$)

The above observations (1) and (2) are just my ideas after observing many of these $C$ and $D$ matrices for different algebras, so they may be false. Ideally i would like to prove or disprove each of them.

Edit: deleted observations which I have proofs of now, namely that $CD=DC$ and $C^{T}D=DC^{T}$. These can be proved by using the associativity of Clifford algebra. Though my proof of $C^{T}D=DC^{T}$ still relies on some parts of observation (2) being true.

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Cross-posted on MO: mathoverflow.net/questions/118466/… –  Alex B. Jan 9 '13 at 18:19
    
If you explain $C$ and $D$ by saying that they are matrices for the linear transformations induced by left and right (respectively) multiplication by $a$, it will be a little easier to read. I spent a few minutes convinced you meant something more complicated, but it seems like that's not the case. –  rschwieb Jan 9 '13 at 19:10
    
@rschwieb Thanks for noting, I fixed it. –  Sunny88 Jan 9 '13 at 19:49
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1 Answer 1

up vote 2 down vote accepted

Let's take the exmaple of $G_{1,1,0}$ as you did and let $e_1, e_2$ be an orthonormal basis for $V$ with the first squaring to 1 and the second squaring to -1.

Let's look at $C$ and $D$ corresponding to $e_1$:

$$C=\begin{bmatrix}0&1&0&0\\1&0&0&0\\0&0&0&1\\0&0&1&0\\\end{bmatrix} D=\begin{bmatrix}0&1&0&0\\1&0&0&0\\0&0&0&-1\\0&0&-1&0\\\end{bmatrix}$$

Similarly for $e_2$:

$$C=\begin{bmatrix}0&0&-1&0\\0&0&0&1\\1&0&0&0\\0&-1&0&0\\\end{bmatrix}D=\begin{bmatrix}0&0&1&0\\0&0&0&1\\-1&0&0&0\\0&-1&0&0\\\end{bmatrix}$$

and for $e_1e_2$:

$$C=\begin{bmatrix}0&0&0&1\\0&0&-1&0\\0&-1&0&0\\1&0&0&0\\\end{bmatrix}D=\begin{bmatrix}0&0&0&1\\0&0&1&0\\0&1&0&0\\1&0&0&0\\\end{bmatrix}$$

And of course for the identity, both matrices are the identity matrices.

In retrospect, it's easy to see now that because the product of two basis elements is $\pm$ another basis element, or else it is zero, all of these sorts of matrices will be partly "permutation," and sometimes they will have zero columns/rows where the products are zero. Finally, they "don't overlap". At any rate, that explains their symmetry antisymmetry properties that you were observing.

Since every other matrix you are interested in is a linear combination of these, I think this, roughly speaking, explains the symmetries you are seeing.

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Anyhow, I think you will find it useful to look at what the basis elements do, rather than generic elements of the algebra. –  rschwieb Jan 9 '13 at 19:36
    
Sorry, I don't understand "At any rate, that explains their symmetry antisymmetry properties that you were observing.". We can see from the matrix that this is true for $G_{1,1,0}$, but how can i generalize to $G_{p,q,r}$? –  Sunny88 Jan 9 '13 at 19:46
    
@Sunny88 Since each of the basis matrices has the symmetry property you described, and they don't overlap on nonzero entries, any linear combination of them is going to have the same symmetry properties. (That's clear, right?) You're right, my solution doesn't rigorously explain it for all $G_{ijk}$, but I think this is the right idea. All of the symmetry properties you were describing are present here stem from the anticommutativity of the basis elements and their square being in $\{\pm 1,0\}$. The only thing can't see now is why, in general, the basis matrices don't "overlap." –  rschwieb Jan 9 '13 at 19:57
    
@Sunny88 For nondegenerate metrics, I think you can argue that no overlap occurs because of the orderly way that the basis elements act on the basis elements. For example, if $e_ib=e_jb$ for a basis elt $b$, then $e_i=e_j$ (very strong!) Moreover, $e_ib=-e_jb$ for $b^2\neq0$ gives a contradiction since the basis elements are linearly independent. This amounts to the fact that the matrices for $e_i$ and $e_j$ won't have overlapping nonzero entries. (continued in next comment) –  rschwieb Jan 9 '13 at 20:45
    
@Sunny88 If the metric is degenerate then we have to look at how the $e_i$ such that $e_i^2=0$ move the basis. Upon examination, we see that they annihilate half the basis elements, and "translate" the other half onto the annihilated elements. Precisely as last time, if $b^2\neq 0$, then $e_ib=e_jb$ implies $i=j$. I guess the last case is to consider what to do when $b^2=0$, or in the extreme case when we are working in the exterior algebra :) The solution to this is probably obvious... –  rschwieb Jan 9 '13 at 20:50
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