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Take two k-vectors $x = \bigwedge_{i = 1}^k \vec{x}_i$, $y = \bigwedge_{i = 1}^k \vec{y}_i$ in $\Lambda^k(\mathbb{R}^n)$. Define the inner product between them as: $$ \langle x, y \rangle = \det(\langle \vec{x}_i, \vec{y}_j\rangle) \equiv \star(x\wedge\star y). $$ Now we let both x and y transform under a rotor $x \mapsto RxR^{-1}$, $y \mapsto RyR^{-1}$. Does $\langle x, y \rangle$ stay constant?

Half Attempted proof: $$ \langle RxR^{-1}, RyR^{-1} \rangle $$ $$ \overset{?}{=} \det(\langle A \vec{x}_i, A\vec{y}_j\rangle)$$ where $A$ is a unitary rotation matrix representing the rotor 'distributed' over the constituent vectors. Not 100% sure if this is valid or possible. If it is then $$ \langle A \vec{x}_i, A \vec{y}_j \rangle = \langle \vec{x}_i, \vec{y}_j\rangle$$

meaning each component of the Gram matrix is the same and the inner product stays constant.

If the inner product does stay constant, is there a simpler proof than the one I attempted? If it doesn't stay constant, is there a proof of that?

iglizworks
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  • Related: https://math.stackexchange.com/questions/3198338/inner-product-structure-on-geometric-algebra – mr_e_man Aug 01 '22 at 17:21

1 Answers1

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$ \newcommand\form[1]{\langle#1\rangle} \newcommand\rev\widetilde $

The inner product between multivectors $X, Y$ is given in geometric algebra by the scalar product: $$ \form{X, Y} = \form{\rev XY}_0, \tag{$*$} $$ where $\rev X$ is the reversal of $X$ and $\form{\cdot}_0$ is the scalar projection. The scalar product is symmetric, from which it follows that $$ \form{\rev{(RX\rev R)}RY\rev R}_0 = \form{R\rev X\rev RRY\rev R}_0 = \form{R\rev XY\rev R}_0 = \form{\rev XY\rev RR}_0 = \form{\rev XY}_0, $$ where we've used the fact that $R^{-1} = \rev R$. Alternatively, we could also say $$ \form{R\rev XY\rev R}_0 = R\form{\rev XY}_0\rev R = \form{\rev XY}_0R\rev R = \form{\rev XY}_0 $$ since $x \mapsto Rx\rev R$ is grade-preserving and $\form{\cdot}_0$ is a scalar.

Assuming by $\star$ you mean the Hodge star defined by $$ x\wedge(\star y) = \form{x, y}I $$ where $I$ is some chosen unit pseudoscalar, your equation $\form{x, y} = \star(x\wedge\star y)$ is incorrect for arbitrary metrics (though does work for some metric such as Euclidean metrics) and should instead be $$ \form{x, y} = \star^{-1}(x\wedge\star y), $$ where $\star^{-1}$ is the inverse of the Hodge star.

I don't know if there is a straightforward way to prove ($*$) that isn't tedious or overtly abstract, but if you're willing to accept that $$ \star X = \rev XI,\quad \star^{-1}X = \rev{XI^{-1}}, $$ then we see $$ \form{x, y} = \star^{-1}(x\wedge\star y) = (x\wedge(\rev yI)I^{-1})^{\rev{\hphantom X}} = \rev{I^{-1}} (\rev Iy)\wedge\rev x = \rev{I^{-1}}\form{\rev Iy\rev x}_n = \rev{I^{-1}}\rev I\form{y\rev x}_0 = \form{\rev xy}_0 $$ where $\form{\cdot}_n$ is pseudoscalar projection.

Nicholas Todoroff
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  • Yup I see now that the second formula for the inner product is incorrect. I was trying to be clever and stumbled a bit, though the idea was there. Your proof makes sense though. Thank you! – iglizworks Aug 02 '22 at 02:34
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    @iglizworks No problem! As an aside, I just noticed that your "half-attempted" proof also does work: because conjugation by a rotor is grade-preserving, $$R(A_r\wedge B_s)R^{-1} = \langle RA_rB_sR^{-1}\rangle_{r+s} = \langle RA_rR^{-1}RB_sR^{-1}\rangle_{r+s} = (RA_rR^{-1})\wedge(RB_sR^{-1}) $$ for $A_r$ an $r$-vector and $B_s$ an $s$-vector. Applying this to the definition of $\langle\cdot,\cdot\rangle$ and using the fact that rotor conjugation is a vector isometry gives the desired result. – Nicholas Todoroff Aug 02 '22 at 03:35