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The scalar product, as defined in the book "Geometric-algebraic-for-computer-science" ,is directly based on this determinant. Is it possible for this definition to be independent of the determinant? I would like to understand its motives better (of course, I am only at chapter 4)

Takayama_Maria
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$ \newcommand\tr{\mathop{\mathrm{tr}}} $It is simply the scalar part of the geometric product: $$ A*B = \langle AB\rangle_0. $$

We can actually define it without referring to the exterior algebra structure at all. We consider a geometric algebra over a space of dimension $n$. Fix an element $A$ of the geometric algebra. The function $X \mapsto AX$ using the geometric product is linear so its trace is well-defined; call this the trace of $A$ and denote it $\tr(A)$. The normalized trace is gotten by dividing by the dimension of the algebra, and we the get a scalar product $$ A*B = \frac1{2^n}\tr(AB). $$

I claim that the normalized trace is simply the scalar part. To see this, let $e_i$ be an orthonormal vector basis. This generates a basis for the algebra of the form $$ e_{i_1}\dotsb e_{i_k},\quad 1\leq i_1 < \dotsb < i_k \leq n,\quad 0\leq k \leq n $$ where $k = 0$ is defined to give us the identity $1$. The only $E$ of the above form satisfying $E^2 = E$ is $E = 1$ so this is the only component that contributes to the trace. Because $\tr 1 = 2^n$ we get that $\tr A$ is $2^n$ times the $1$-component of $A$, otherwise known as the scalar part.

Nicholas Todoroff
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  • thanks for the answer.I'm confused about the trace form here. What is it? I only know about the matrix trace. – Takayama_Maria Jan 12 '24 at 14:07
  • @Takayama_Maria The trace is definable for any linear map over any vector space. In this case the vector space is the entire geometric algebra and the map is "multiplication on the left by a fixed element". One way to do this is to pick a basis and construct the corresponding matrix $M$ of your map $\phi$ and then take its trace; then because $\mathrm{tr}(UMU^{-1})=\mathrm{tr}(M)$ for any invertible matrix $U$ it follows that $\mathrm{tr}(M)$ only depends on $\phi$ and is independent of the chosen basis. There is also an abstract definition of trace that avoids bases entirely. – Nicholas Todoroff Jan 12 '24 at 14:39