A 2-blade is a simplebivector. Sums of 2-blades are also bivectors, but not always simple. A 2-blade may be expressed as the wedge product of two vectors a and b:
A 3-blade is a simple trivector, that is, it may be expressed as the wedge product of three vectors a, b, and c:
The highest grade element in a space is called a pseudoscalar, and in a space of dimension n is an n-blade.[4]
In a vector space of dimension n, there are k(n − k) + 1 dimensions of freedom in choosing a k-blade for 0 ≤ k ≤ n, of which one dimension is an overall scaling multiplier.[5]
A vector subspace of finite dimension k may be represented by the k-blade formed as a wedge product of all the elements of a basis for that subspace.[6] Indeed, a k-blade is naturally equivalent to a k-subspace, up to a scalar factor. When the space is endowed with a volume form (an alternating k-multilinear scalar-valued function), such a k-blade may be normalized to take unit value, making the correspondence unique up to a sign.
^For Grassmannians (including the result about dimension) a good book is: Griffiths, Phillip; Harris, Joseph (1994), Principles of algebraic geometry, Wiley Classics Library, New York: John Wiley & Sons, ISBN978-0-471-05059-9, MR1288523. The proof of the dimensionality is actually straightforward. Take the exterior product of k vectors and perform elementary column operations on these (factoring the pivots out) until the top k × k block are elementary basis vectors of . The wedge product is then parametrized by the product of the pivots and the lower k × (n − k) block. Compare also with the dimension of a Grassmannian, k(n − k), in which the scalar multiplier is eliminated.