Answer by Jonathan Manton for Geometrical meaning of Grassmann Algebra

2012-09-04T07:52:20Z

The key ingredient, in my mind, is to realise that the Grassmann algebra of a $d$-dimensional vector space $V$ is concerned primarily with $d$-dimensional volumes of parallelotopes and that lower-dimensional parallelotopes are merely building blocks for $d$-dimensional parallelotopes. This is explained at length here: http://jmanton.wordpress.com/2012/09/03/introduction-to-the-grassmann-algebra-and-exterior-products/

All volumes are relative volumes. We act as if we do not know what the underlying metric on $V$ is and we only want to make statements such as "this parallelotope is twice as big as that parallelotope" if it is true with respect to all metrics, not just a single metric.

Since the volume of a $d$-dimensional cube equals the $d$-fold product of its side length, it is not unreasonable to hope that the (signed) volume of an (oriented) parallelotope is some sort of product of its side lengths. In fact, the axioms of a "product" of two things essentially agree with the axioms of a bilinear function, and the volume of a parallelotope is indeed given by a multi-linear function of its sides, leading to the standard definition of the exterior algebra in terms of (alternating) multi-linear maps. Regardless, thinking of volume as a product of lengths gives some intuition as to why the wedge product is used to define parallelotopes.

The notation $v_1 \wedge \cdots \wedge v_i$ should be understood to refer to the parallelotope made from the vectors $v_1,\cdots,v_i \in V$. If $i < d = \dim V$ then the "volume" of the parallelotope $v_1 \wedge \cdots \wedge v_i$ is always zero; keep in mind the key point that the Grassmann algebra on $V$ is a priori concerned with $d$-dimensional volume. Lower-dimensional parallelotopes are merely building blocks for top-dimensional parallelotopes. For example, we say $v_1 \wedge \cdots \wedge v_i = w_1 \wedge \cdots \wedge w_i$ if and only if, for all $u_1,\cdots,u_{d-i}$, it is true that $v_1 \wedge \cdots \wedge v_i \wedge u_1 \wedge \cdots \wedge u_{d-i} = w_1 \wedge \cdots \wedge w_i \wedge u_1 \wedge \cdots \wedge u_{d-i}$ where the latter means the (signed) volumes of the two $d$-dimensional parallelotopes are equal (with respect to every possible metric).

The classical results now follow from this. For example, $v_1 \wedge \cdots \wedge v_i = \lambda w_1 \wedge \cdots \wedge w_i$ for some $\lambda$ if and only if, either both sides are zero because they are degenerate parallelotopes, or $\operatorname{span}{v_1,\cdots,v_i} = \operatorname{span}{w_1,\cdots,w_i}$. It is a posteriori acceptable to interpret $v_1 \wedge \cdots \wedge v_i = \lambda w_1 \wedge \cdots \wedge w_i$ as meaning the $i$-dimensional volume of the parallelotope $v_1 \wedge \cdots \wedge v_i$ is $\lambda$ times the $i$-dimensional volume of the parallelotope $w_1 \wedge \cdots \wedge w_i$, but the underlying reason is that they behave the same way when used as building blocks.

The importance of thinking in terms of top-dimensional parallelotopes is that it is otherwise difficult to explain why $v_3 = v_1 + v_2$ does not imply that the length of $v_1$ plus the length of $v_2$ equals the length of $v_3$. In the Grassmann algebra, vectors and lower-dimensional parallelotopes do not have an independent life of their own but are primarily building blocks for top-dimensional parallelotopes. Vector addition in a Grassmann algebra relates to addition of top-dimensional volume, not to lower-dimensional volumes.

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Answer by Jonathan Manton for Geometrical meaning of Grassmann Algebra