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My understanding of a simple k-vector is that it is the wedge product of k vectors.

Also, two simple k-vectors are the same, when their magnitude, attitude and orientation match. Now my question is, could I just define a simple k-vector in this way? Meaning "a simple k-vector is an equivalence class of ordered k-tuple of vectors. Two tuples are equivalent if their attitudes and orientation match, and if the parallelograms they span attain the same magnitude."

Especially a reference where this is stated would be greatly appreciated! I feel like I have read something like this somewhere, but I cannot find it anymore.

I am only using k-vectors for something I am writing for university and having to explain the wedge product would deviate from my topic a little. That's why I am trying to avoid this definition.

waldyrious
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dontknow3
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  • Can you give an example of two ordered tuples that are equivalent but not equal? – John Douma Nov 15 '23 at 17:06
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    You could define the equivalence for two sets of $k$ linearly independent vectors in terms of them (i) having the same span and (ii) the linear transformation taking one set to the other having determinant $1$. – blargoner Nov 15 '23 at 18:51
  • @JohnDouma yes. From what I understand the wedge product of [v1,v2,v3,v4] in that order would be the same as [v2,v1,v4,v3] for any four linearly independent vectors. – dontknow3 Nov 17 '23 at 09:10
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    @blargoner Thank you for your contributions to the discussion. Your remarks on the determinant were quite helpful :) – dontknow3 Nov 17 '23 at 09:11

2 Answers2

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Yes, this works, though I'm not sure you'll find any reference.

(I assume we're talking about the Euclidean case. This fails in general since we can have nonzero $k$-vectors which have a "magnitude" of zero.)

What you mean by orientation of a $k$-vector is two things: the subspace represented by the $k$-vector and the "orientation" as described by an ordered basis. Define the (squared) $k$-vector magnitude $M(v_1,\dotsc,v_k)$ of vectors $v_1,\dotsc,v_k$ by $$ M(v_1,\dotsc,v_k) = \det\Bigl(v_i\cdot v_j\Bigr)_{i,j=1}^k. $$ For example, with $k=3$ we have $$ M(v_1,v_2,v_3) = \det\begin{pmatrix} v_1\cdot v_1&v_1\cdot v_2&v_1\cdot v_3 \\ v_2\cdot v_1&v_2\cdot v_2&v_2\cdot v_3 \\ v_3\cdot v_1&v_3\cdot v_2&v_3\cdot v_3 \end{pmatrix}. $$

The theorem you are interested in can be stated as follows:

  • Suppose $k\ne 0$. Define a relation $\sim$ between $k$-tuples of linearly independent vectors as follows: $$ (v_1,\dotsc,v_k)\sim(w_1,\dotsc,w_k) \iff \Bigl[\mathrm{span}\{v_1,\dotsc,v_k\} = \mathrm{span}\{w_1,\dotsc,w_k\}\Bigr]\text{ and }\Bigl[\exists T : S\to S.\; \det T > 0\text{ and }(Tv_1,\dotsc,Tv_k) = (w_1,\dotsc,w_k)\Bigr]\text{ and }\Bigl[M(v_1,\dotsc,v_k) = M(w_1,\dotsc,w_k)\Bigr] $$ where $S = \mathrm{span}\{v_1,\dotsc,v_k\}$. Then $(v_1,\dotsc,v_k) \sim (w_1,\dotsc,w_k)$ iff $v_1\wedge\dotsb\wedge v_k = w_1\wedge\dotsb\wedge w_k$.

The restriction to linearly independent tuples is crucial (though we could add the stipulation that any two linearly dependent tuples are equivalent).

I will sketch a proof that equivalence of tuples implies equality of $k$-vectors. First note that $$ \mathrm{span}\{v_1,\dotsc,v_k\} = \{v \;:\; v\wedge v_1\wedge\dotsb\wedge v_k = 0\}. $$ Additionally, for any $k$-vector $X$, if $v\wedge X = 0$ then we can factor $X = v\wedge Y$ for some $(k-1)$-vector $Y$. Inducting on $k$, this shows that there is a scalar $\alpha$ such that $$ v_1\wedge\dotsb\wedge v_k = \alpha w_1\wedge\dotsb\wedge w_k. \tag{$*$} $$ Let $$ S = \mathrm{span}\{v_1,\dotsc,v_k\} = \mathrm{span}\{w_1,\dotsc,w_k\}. $$ The subalgebra generated by $S$ of the exterior algebra is naturally isomorphic to the exterior algebra of $S$, so conflate the two and consider the outermorphism of $T_\wedge$ of $T$. Then $$ w_1\wedge\dotsb\wedge w_k = T_\wedge(v_1\wedge\dotsb\wedge v_k) = (\det T)v_1\wedge\dotsb\wedge v_k = \alpha(\det T)w_1\wedge\dotsb\wedge w_k $$ so $1 = \alpha(\det T)$ and thus $\alpha > 0$. Finally, take the magnitude of both sides of ($*$) to get $$ M(v_1,\dotsc,v_k) = \alpha^2M(w_1,\dotsc,w_k) \implies 1 = \alpha^2. $$ Because of the Euclidean assumption we know that these magnitudes are nonzero, so $\alpha = 1$ and this completes the proof.

Nicholas Todoroff
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  • Do we really need the Euclidean structure for the equivalence relation, or just to formalize that notion of magnitude? In your proof since $\alpha=1$, it follows that $\det T=1$, so can't we just use that? – blargoner Nov 15 '23 at 23:06
  • @blargoner If we want to involve the magnitude then yes we need the Euclidean structure to rule out the magnitudes being zero (otherwise you cannot conclude that $\alpha^2 = 1$, only that $\alpha = 1/\det T$). We could ignore the magnitude and just assume $\det T = 1$, but I think that is against the spirit of the question. – Nicholas Todoroff Nov 15 '23 at 23:42
  • I hear you re: spirit -- sorry, I wasn't suggesting totally ignoring the magnitude concept but rather, if we just assume $\det T=1$ in the equivalence relation, we get that the magnitudes agree in the Euclidean case for free. – blargoner Nov 16 '23 at 00:09
  • @blargoner What I'm saying is that if we use $\det T = 1$ in the definition of $\sim$ then the use of the magnitude is redundant---we can remove it entirely, whereas OP wanted to capture equality of $k$-vectors as subspace+orientation+magnitude. So instead I opted for $\det T > 0$ which ensures that $v_1,\dotsc,v_k$ has the same orientation as $w_1,\dotsc,w_k$ without saying anything about scaling, and then scaling is taken care of by constraining the magnitudes; and under this definition $\sim$ fails to capture the equality of arbitrary $k$-vectors when we use an indefinite metric. – Nicholas Todoroff Nov 16 '23 at 02:38
  • Thank you! This helped me a lot. – dontknow3 Nov 17 '23 at 09:08
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I think the reference you are looking for is my article, The Wedge Product and Analytic Geometry, in the August-September 2008 edition of The American Mathematical Monthly, pp. 623-644. The idea there is to define the wedge product of k given vectors as an equivalence class representing an oriented parallelpiped. This is meant to extend the elementary concept of a vector as an equivalence class of directed line segments. You can find a copy of it at https://www.mdeetaylor.com/wp-content/uploads/2017/06/wedge_monthly.pdf.

Michael D. Taylor
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