General expression for geometric product of blades in terms of scalar and exterior products

Question

I am trying to find a general expression for the geometric product of two blades in terms of the scalar and exterior products of vectors. Some preamble to be clear on conventions:

In a geometric algebra on a finite-dimensional real vector space, we have the geometric product between two vectors defined by: $$vw=\langle v,w\rangle + v\wedge w$$ where $\langle v,w\rangle$ is the scalar product, and $v\wedge w$ is the exterior product. A $k$-blade is an exterior product of $k$ vectors, $v_1\wedge \cdots\wedge v_k$. We can extend the scalar product on vectors to a scalar product on $k$-blades by defining it as the following determinant: $$\langle v_1\wedge \cdots\wedge v_k,w_1\wedge \cdots\wedge w_k\rangle = \begin{vmatrix} \langle v_1,w_1\rangle & \cdots & \langle v_1,w_k\rangle \\ \vdots & \ddots & \vdots \\ \langle v_k,w_1\rangle & \cdots & \langle v_k,w_k\rangle \end{vmatrix}$$ (This, of course, is independent of how the blade is factored into vectors.)

Note that because of the associativity of both the geometric product and the exterior product and their commutativity properties, the exterior product of $k$ vectors (a $k$-blade) can be expressed as the totally antisymmetrized geometric product of $k$ vectors: $$v_1\wedge \cdots\wedge v_k=\sum_{\sigma}{1\over k!}(\operatorname{sgn}\sigma)v_{\sigma(1)}\cdots v_{\sigma(k)}$$ where $\sigma$ ranges over all permutations of $1\ldots k$ and $\operatorname{sgn}$ is the sign of the permutation ($+1$ for even and $-1$ for odd).

Now, after playing around with it in my head and on paper for a bit, I am guessing that the geometric product between a $j$-blade $U=u_1\wedge \cdots\wedge u_j$ and a $k$-blade $V=v_1\wedge \cdots\wedge v_k$ can be found by the following rather involved expression: $$UV=\sum_{i=0}^{\min j,k}\sum_{\sigma_i,\tau_i}(\operatorname{sgn}\sigma_i)(\operatorname{sgn}\tau_i)\langle u_{\tau_i(j)}\wedge \cdots\wedge u_{\tau_i(j-i+1)},v_{\sigma_i(1)}\wedge \cdots\wedge v_{\sigma_i(i)}\rangle u_{\tau_i(1)}\wedge \cdots\wedge u_{\tau_i(j-i)}\wedge v_{\sigma_i(i+1)}\wedge \cdots\wedge v_{\sigma_i(k)}$$ where $\sigma_i$ and $\tau_i$ range over all the permutations of $(1\ldots k)$ and $(1\ldots j)$, respectively, which satisfy the following constraints: $$\sigma_i(1)\lt\cdots\lt\sigma_i(i)$$ $$\sigma_i(i+1)\lt\cdots\lt\sigma_i(k)$$ $$\tau_i(1)\lt\cdots\lt\tau_i(j-i)$$ $$\tau_i(j-i+1)\lt\cdots\lt\tau_i(j)$$ Essentially, this expression is the sum over all the ways of factorizing $U=U_{outside}\wedge U_{inside}$ and $V=V_{inside}\wedge V_{outside}$, taking the inner product of $U_{inside}$ (reversed) with $V_{inside}$, and multiplying it to the exterior product of $U_{outside}$ with $V_{outside}$.

However, I don't know how to prove this.

My question, then, is: does anyone know the best way to go about proving this equation, or can someone find a counterexample to it? Thank you!

Answer 1

For a blade $U = a_i \wedge \ldots \ a_n \rightarrow \prod_{k_i} a_{k_i} $, i.e. simply the geometric product in the order the vectors are given.
Let $[k] = \{ k_i \}$ is the index list. So then $$ U V = sgn( [k]) \prod\limits_{k_i < k_j} a_{k_j} $$ where $a$ traverses both $U$ and $V$. Then you simplify all common vectors, which happen to be neighbors in the sorted order. So then the scalar product $U \star V = <UV>_{0}$.

On the other hand $A_k \wedge B_l = A B \quad k+ l \leq dim \quad | \quad 0 \quad k+ l> dim $ therefore your formula can not be completely correct.