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I am trying to understand the relationship between the wedge product and linear subspace. Let $e_1,\cdots, e_4$ be the standard basis of $\mathbb{R}^4$. The wedge product $$(e_1+2e_2)\wedge (3e_1+e_3+e_4) $$ can be thought of as the oriented linear subspace generated by the vectors $e_1+2e_2$ and $3e_1+e_3+e_4$.

Upon expanding we get $$e_{13}+e_{14}-6e_{12}+2e_{23}+2e_{24},$$ where $e_{ij}=e_i\wedge e_j$. Each $e_{ij}$ can be thought of as the linear subspace spanned by $e_i,e_j$. But what role do the coefficients $1,1,-6,2,2$ play?

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I don't see how it is useful to think of a bivector as a linear subspace of the original space. For example, $\{e_1,e_2\}$ and $\{2e_1,e_2\}$ span the same subspace, but $e_1\wedge e_2$ and $2e_1\wedge e_2$ are different bivectors -- one is twice the size of the other. How would identifying them with subspaces work? –  Henning Makholm Nov 27 '11 at 22:23
    
The identification is not one to one. In your example, the two wedge products correspond to same linear subspace. –  TCL Nov 27 '11 at 22:31
    
$v_1\wedge v_2$ is a basis for a linear subspace of $\wedge^2 \mathbb R^4$, not for a subspace of $\mathbb R^4$ itself. –  Grumpy Parsnip Nov 27 '11 at 22:33
    
From a book by Frank Morgan: The oriented m-planes through the origin in $\mathbb{R}^n$ are in one-to-one correspondence with the unit, simple m-vectors in $\wedge^m\mathbb{R}^n$. –  TCL Nov 27 '11 at 22:46
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@TCL: Ah, "simple" is the operative word. –  Grumpy Parsnip Nov 27 '11 at 23:04

2 Answers 2

As you mention in the comments, there is a 1-1 correspondence between subspaces and unit simple $m$-vectors in $\wedge^m\mathbb R^n$. When you expand out $v_1\wedge v_2$ in a basis, the result is no longer a simple $2$-vector but a linear combination of simple $2$-vectors, so you should not try to interpret the coefficients of this expansion as having geometric meaning.

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These coefficients are called Plucker coordinates. There are many ways of thinking about them; the following is perhaps the most geometric.

Take your linear space (the span of $e_1+2e_2$ and $3e_1+e_3+e_4$) and project it onto $\mathrm{Span}(e_1, e_2)$. Then this linear map reduces area by a factor of $$\frac{6}{\sqrt{1^2+1^2+6^2+2^2+2^2+0^2}}.$$ If you project onto $\mathrm{Span}(e_1, e_3)$ instead, then the numerator changes to $1$ (the coefficient of $e_1 \wedge e_3$) and the denominator stays the same. The sign of the Plucker coordinate tells you whether the projection is orientation preserving or reversing.

Note: I am using the inner product on $\mathbb{R}^4$ to measure areas. Without this, I could still make sense of a ratio of two Plucker coordiantes, but there wouldn't be a good way to interpret one Plucker coordinate by itself.

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