# In Grassmann algebra a la Browne, why are vectors dependent if their wedge product vanishes?

I'm reading John Browne's Grassmann Algebra, Vol 1 : Foundations. Early on, he asserts without proof that if $x$ and $y$ are any two vectors in the underlying (real) vector space such that $x \wedge y = 0$, then $x$ and $y$ are linearly dependent. Take the vector space to be $R^3$, say. The result is equivalent to proving that if $e_i, e_j$ are two of the standard basis vectors, then $e_i \wedge e_j \neq 0$.

In the framework of axioms and or constructions that Browne provides, how does one prove that simple fact?

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Can you provide that framework / those axioms, at least in a nutshell? –  Berci May 18 '13 at 22:01
I'm going to guess one of them is that for an $n$-dimensional vector space with basis $v_1,\dots,v_n$, we have $v_1\wedge\dots\wedge v_n\ne 0$. :) –  Ted Shifrin May 18 '13 at 22:04
Well, it's usually defined as a quotient of tensor power. –  Berci May 18 '13 at 22:06
Well if you believe that the $e_i\wedge e_j$, $i\neq j$ are part of the standard basis for the algebra, then you already know they're nonzero... –  rschwieb May 18 '13 at 22:41
@TedShifrin: No, unfortunately, what you guess in included in the axioms in Browne's treatment is not in fact there. Unless there's something much more subtle, I think what's missing is one or another version of what's called Axiom 4 in William Schulz's document cefns.nau.edu/~schulz/grassmann.pdf (pages 50-51). –  murray May 19 '13 at 3:42

The point is that the determinant gives you a non-zero linear function from $\wedge^n V$ to the ground field (which can be arbitrary) when $n=\mathrm{dim}(V)$. So this space is non-zero. Now if you have a basis $e_1,\dots,e_n$ of $V$, then multilinearity and skew-commutativity together imply $e_1 \wedge \cdots \wedge e_n$ spans the top exterior power, and must therefore be non-zero. The result you want follows.

On the other hand, as you point out in the comments below and I confirm, it seems that Browne's axioms are not enough to imply this fact about determinant, and indeed the result you want might fail: any quotient of the Grassmann algebra also satisfies his axioms.

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But that property of determinants is not part of Browne's axioms. Note that he doesn't use the more familiar approach of defining exterior powers in terms of tensor powers, etc., but instead is trying to lay down a set of axioms for wedge product. –  murray May 19 '13 at 3:43
@murray, Well, if you take 2.7-2.18 in Browne's book as the axioms, and assume there are no other relations, then what I wrote is a consequence. That is, linear functions on $\wedge^k V$ are identified with $k$-multilinear alternating functions on $V$. –  S123 May 19 '13 at 15:23
@murray, But it seems to me that Browne's axioms are underdetermined: any quotient of the usual Grassmann algebra also satisfies them. –  S123 May 19 '13 at 15:29
Yeah, underdetermined seems to be the case. –  murray May 19 '13 at 18:43

Mathematicians often use the term "general position" such as in "Assume three points in the plane in general position." That is, the three points are not in a line. Sometimes general position is assumed without explicitly stating it. In any case, in Browne's application any set of distinct symbols representing vectors are taken to be "in general position (up to the dimension of the underlying space)". It's independence unless specified otherwise.

This is stated clearly in the Help page for the ExteriorProduct. "The GrassmannAlgebra and GrassmannCalculus Applications considers any set of distinct 1-element symbols, without further definitions or values, up to the linear dimension of the space, to be independent."

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To assume distinct 1-element symbols automatically represent independent elements seems dangerous, even for CAS purposes. E.g., suppose x1 is a 1-element (i.e., vector). Let x2 = 2 x1. Then x1 and x2 are not independent. –  murray Jun 9 at 19:51