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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!

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  • $\begingroup$ You have not defined the geometric product between two blades (only between two vectors), so I guess solving the question will require a look at the literature. My impression is that if you look at the right text (some students of Rota have written a lot about this), you might find the formula in there... $\endgroup$ – darij grinberg Oct 3 '17 at 19:26
  • $\begingroup$ I have edited the question to give more information; the geometric product is associative and blades can be expressed as antisymmetrized geometric products of vectors. $\endgroup$ – Matt Dickau Oct 13 '17 at 18:39
  • $\begingroup$ I don't think you have given an actual definition of the geometric product on blades -- other than your formula, which would be a valid definition if you didn't want to prove it! $\endgroup$ – darij grinberg Oct 13 '17 at 18:42
  • $\begingroup$ darijgrinberg, I've expressed a blade as a sum of geometric products of vectors. The geometric product is associative. This specifies the geometric product of blades. $\endgroup$ – Matt Dickau Oct 13 '17 at 18:52
  • $\begingroup$ Ah! That's a fairly indirect definition, but it does uniquely specify the product. I regret I have neither a reference nor a written-up proof to offer, but here is a roadmap that should take you there if you try hard enough: (1) Define the geometric product via your formula (which should be correct, up to sign). Prove that it is associative. (This becomes easy if you know Hopf superalgebras, because then your formula says something like $UV = \sum_{(U)} \sum_{(V)} \left< U_{(1)}, V_{(1)} \right> U_{(2)} \wedge V_{(2)}$ in the exterior algebra (which is a Hopf superalgebra), ... $\endgroup$ – darij grinberg Oct 13 '17 at 19:42
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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.

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  • $\begingroup$ This... doesn't answer my question, and isn't even completely correct. The outer product is only equal to the geometric product if all the vectors are mutually orthogonal. $\endgroup$ – Matt Dickau Nov 27 '17 at 2:07
  • $\begingroup$ In most sources blades are interpreted as geometric products of orthogonal vectors. If you mean something else you should state it. $\endgroup$ – user48672 Nov 28 '17 at 21:01
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    $\begingroup$ I did state what a blade was in my preamble to the question. $\endgroup$ – Matt Dickau Dec 15 '17 at 23:34

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