In geometric algebra, a blade is a generalization of the concept of scalars and vectors to include simple bivectors, trivectors, etc. Specifically, a k-blade is any object that can be expressed as the exterior product (informally wedge product) of k vectors, and is of grade k.

In detail:[1]

$\mathbf{a} \cdot \mathbf{b} .$
• A 1-blade is a vector. Every vector is simple.
• A 2-blade is a simple bivector. Linear combinations of 2-blades also are bivectors, but need not be simple, and are hence not necessarily 2-blades. A 2-blade may be expressed as the wedge product of two vectors a and b:
$\mathbf{a} \wedge \mathbf{b} .$
• A 3-blade is a simple trivector, that is, it may expressed as the wedge product of three vectors a, b, and c:
$\mathbf{a} \wedge \mathbf{b} \wedge \mathbf{c}.$
• In a space of dimension n, a blade of grade n − 1 is called a pseudovector.[2]
• The highest grade element in a space is called a pseudoscalar, and in a space of dimension n is an n-blade.[3]
• In a space of dimension n, there are k(nk) + 1 dimensions of freedom in choosing a k-blade, of which one dimension is an overall scaling multiplier.[4]

In a n-dimensional spaces, there are blades of grade 0 through n. 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.[5]

## Examples

For example, in 2-dimensional space scalars are described as 0-blades, vectors are 1-blades, and area elements are 2-blades known as pseudoscalars, in that they are one-dimensional objects distinct from regular scalars.

In three-dimensional space, 0-blades are again scalars and 1-blades are three-dimensional vectors, but in three-dimensions, areas have an orientation, so while 2-blades are area elements, they are oriented. 3-blades (trivectors) represent volume elements and in three-dimensional space, these are scalar-like—i.e., 3-blades in three-dimensions form a one-dimensional vector space.

4. ^ 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, ISBN 978-0-471-05059-9, MR 1288523. The proof of the dimensionality is actually straightforward. Take k vectors and wedge them together $v_1\wedge\cdots\wedge v_k$ and perform elementary column operations on these (factoring the pivots out) until the top k × k block are elementary basis vectors of $\mathbb{F}^k$. The wedge product is then parametrized by the product of the pivots and the lower k × (nk) block.