What is Exterior algebra: Definition and 15 Discussions
In mathematics, the exterior product or wedge product of vectors is an algebraic construction used in geometry to study areas, volumes, and their higher-dimensional analogues. The exterior product of two vectors
u
{\displaystyle u}
and
v
{\displaystyle v}
, denoted by
u
∧
v
{\displaystyle u\wedge v}
, is called a bivector and lives in a space called the exterior square, a vector space that is distinct from the original space of vectors. The magnitude of
u
∧
v
{\displaystyle u\wedge v}
can be interpreted as the area of the parallelogram with sides
u
{\displaystyle u}
and
v
{\displaystyle v}
, which in three dimensions can also be computed using the cross product of the two vectors. More generally, all parallel plane surfaces with the same orientation and area have the same bivector as a measure of their oriented area. Like the cross product, the exterior product is anticommutative, meaning that
u
∧
v
=
−
(
v
∧
u
)
{\displaystyle u\wedge v=-(v\wedge u)}
for all vectors
u
{\displaystyle u}
and
v
{\displaystyle v}
, but, unlike the cross product, the exterior product is associative.
When regarded in this manner, the exterior product of two vectors is called a 2-blade. More generally, the exterior product of any number k of vectors can be defined and is sometimes called a k-blade. It lives in a space known as the k-th exterior power. The magnitude of the resulting k-blade is the volume of the k-dimensional parallelotope whose edges are the given vectors, just as the magnitude of the scalar triple product of vectors in three dimensions gives the volume of the parallelepiped generated by those vectors.
The exterior algebra, or Grassmann algebra after Hermann Grassmann, is the algebraic system whose product is the exterior product. The exterior algebra provides an algebraic setting in which to answer geometric questions. For instance, blades have a concrete geometric interpretation, and objects in the exterior algebra can be manipulated according to a set of unambiguous rules. The exterior algebra contains objects that are not only k-blades, but sums of k-blades; such a sum is called a k-vector. The k-blades, because they are simple products of vectors, are called the simple elements of the algebra. The rank of any k-vector is defined to be the smallest number of simple elements of which it is a sum. The exterior product extends to the full exterior algebra, so that it makes sense to multiply any two elements of the algebra. Equipped with this product, the exterior algebra is an associative algebra, which means that
α
∧
(
β
∧
γ
)
=
(
α
∧
β
)
∧
γ
{\displaystyle \alpha \wedge (\beta \wedge \gamma )=(\alpha \wedge \beta )\wedge \gamma }
for any elements
α
,
β
,
γ
{\displaystyle \alpha ,\beta ,\gamma }
. The k-vectors have degree k, meaning that they are sums of products of k vectors. When elements of different degrees are multiplied, the degrees add like multiplication of polynomials. This means that the exterior algebra is a graded algebra.
The definition of the exterior algebra makes sense for spaces not just of geometric vectors, but of other vector-like objects such as vector fields or functions. In full generality, the exterior algebra can be defined for modules over a commutative ring, and for other structures of interest in abstract algebra. It is one of these more general constructions where the exterior algebra finds one of its most important applications, where it appears as the algebra of differential forms that is fundamental in areas that use differential geometry. The exterior algebra also has many algebraic properties that make it a convenient tool in algebra itself. The association of the exterior algebra to a vector space is a type of functor on vector spaces, which means that it is compatible in a certain way with linear transformations of vector spaces. The exterior algebra is one example of a bialgebra, meaning that its dual space also possesses a product, and this dual product is compatible with the exterior product. This dual algebra is precisely the algebra of alternating multilinear forms, and the pairing between the exterior algebra and its dual is given by the interior product.
Do we really need concept of cross product at all? I always believed cross product to be sort of simplification of exterior product concept tailored for the 3D case. However, recently I encountered the following sentence «...but, unlike the cross product, the exterior product is associative»...
Maxwell's equations in differential form notation appeared as a motivating example in a mathematical physics book I'm reading. However, being a mathematical physics book it doesn't delve much into the physical aspects of the problem. It deduces the equations by setting dF equal to zero and d(*F)...
(A1−A2,B1−B2,C1−C2)∧(A1,B1,C1)(A1−A2,B1−B2,C1−C2)∧(A1,B1,C1)
##=((A1−A2)∗B1−(B1−B2)∗A1)∗(\hat x \wedge \hat y)+((C1−C2)∗A1−(A1−A2)∗C1)∗(\hat z \wedge \hat x)+((B1−B2)∗C1−(C1−C2)∗B1)∗(\hat y \wedge \hat z)##
Is this the correct exterior product?
If we seek a bijection $$\wedge^p V \to \wedge^{n-p} V$$ for some inner product space ##V##, we might think of starting with the unit ##n##-vector and removing dimensions associated with the original vector in ##\wedge^p V ##. Might this be expressed as a sequence of steps by some binary...
No question this time. Just a simple THANK YOU
For almost two years years now, I have been struggling to learn: differential forms, exterior algebra, calculus on manifolds, Lie Algebra, Lie Groups.
My math background was very deficient: I am a 55 year old retired (a good life) professor of...
Hello
I am a mechanical engineer who is teaching himself the math of exterior algebra and differential forms. It is not easy for me and I have had many SIMPLE stumbling blocks due to my not respecting algebra.
May I ask for help on some simple aspects? (Please be patient with me.)
My...
Hi,
I am currently reading about differential forms in "Introduction to Smooth Manifolds" by J. M. Lee, and I was wondering exactly how you define the wedge product on the exterior algebra \Lambda^*(V) = \oplus_{k=0}^n\Lambda^k(V) of a vector space V. I understand how the wedge product is...
Hello,
In R^3, the surface of the parallelogram determined by two vectors u and v is given by the norm of the cross product of u and v. For my research, I have to know if this can be generalized in the following manner:
Let e_1,..,e_n be the canonical basis of R^n, and Ext_k be the exterior...
I'm reading Marsden's vector calculus. In the chapter of differential forms, it mentions the wedge product satisfies the laws:
dy^dx=-dxdy.
and for a 0-form f, f^w=fw.
Does it have formal derivation?
hope someone can give me a hint or even a link.
I understand that there is a way to find a basis \{e_1,...,e_n\} of a vector space V such that a 2-vector A can be expressed as
A = e_1\wedge e_2 + e_3\wedge e_4 + ...+e_{2r-1}\wedge e_{2r}
where 2r is denoted as the rank of A. However the way that I know to prove this seems sort of...
Can someone please thoroughly explain how the determinant comes from the wedge product? I'm only in Cal 3 and Linear at the moment. I'm somewhat trying to learn more about the Wedge Product in Exterior Algebra to understand the determinant on a more fundamental basis. A thorough website or...
Unfortunately there seems to be a misprint in the paper I'm reading which is an introduction to clifford algebra, it says:(I highlighted in red possible misprint, either one of them has to be true misprint if you know what I mean)
The Clifford algebra C(V) is isomorphic to the tensor algebra...
Does anyone have any good refreshers/tutorials for exterior algebra? I need to reacquaint myself with differentials and wedge products specifically. Thanks.