# Bivectors/Trivectors: How to imagine?

A bivector is a wedge product of two independent vectors which can be imagined as a parallelogram and a trivector is a wedge product of three independent vectors which can be imagined as a paralleopiped. So am I correct in thinking that this parallelogram and paralleopiped is the bivector and the trivector itself, respectively? (With its magnitude as the area and the volume, respectively?).

If so how can you imagine the orientation of a bivector? (its even more harder to imagine the orientation of a trivector!).

robphy
Homework Helper
Gold Member
A bivector is a wedge product of two independent vectors which can be imagined as a parallelogram and a trivector is a wedge product of three independent vectors which can be imagined as a paralleopiped. So am I correct in thinking that this parallelogram and paralleopiped is the bivector and the trivector itself, respectively? (With its magnitude as the area and the volume, respectively?).

If so how can you imagine the orientation of a bivector? (its even more harder to imagine the orientation of a trivector!).

Those figures are accurate geometrical representations of those multivectors.
Orientation is specified by indicating an ordering of the vectors, which you could draw in as a directed-arc from the first vector to the other vector... or as a directed-helix. Look at Burke's "Applied Differential Geometry" or Schouten's "Tensor Analysis for Physicists".

I'm working on a paper and computer program to visualize tensors [metrics, differential forms, and multivectors]. A while back I visualized differential forms in electromagnetism using VRML http://www.phy.syr.edu/courses/vrml/electromagnetism/ [Broken] ... but since no one uses VRML any more, I need to revise it. Check out the references.

You might be interested in this thread.

Last edited by a moderator:
I was reading this article recently and it says that bivectors need not be paralellograms; in fact they can be of any shape! Here's the exact quote:

We should remark that the shape of the plane segment representing a bivector is not relevant. Thus the representation of a bivector as the outer product of two vectors (one could say ‘factorisation’) is highly non-unique. The plane segment need not even be a parallelogram. It may be any flat figure with an orientation.

Here's the original article:

Last edited by a moderator:
robphy
Homework Helper
Gold Member
Yes, that's correct. They need not be parallelograms... just any similarly oriented directed area. ..and, I believe, this is restricted to simple bivectors (those that can be written as the wedge of two vectors). For visualization purposes, it might be useful to choose a shape that conveys some of the symmetries of the problem.

Jancewicz is one of the authors in my references.

BTW, thanks for those programs robphy. I'll check em out this weekend.

Chris Hillman
Visualizing multivectors

Hi, Swapnil,

Just thought I'd chime in with a more elaborate answer.

First, note that k-multivector is to vector as k-multivector field is to vector field. That is, you are asking about exterior algebra (due to Hermann Grassmann), not exterior calculus (due to Elie Cartan).

Swapnil said:
So am I correct in thinking that this parallelogram and paralleopiped is the bivector and the trivector itself, respectively? (With its magnitude as the area and the volume, respectively?).

The answer I'd give to your question is that a simple k-multivector in the vector space R^n represents a k-dimensional subspace of R^n together with an oriented k-volume element. As Cartan realized, this is just what we need to capture the notion of "a surface element of a k-dimensional submanifold". In exterior calculus, when we integrate a k-form over a k-dimensional surface, we automatically account for the Jacobian determinant factor familiar from the change of variables formula for a multidimensional integral as treated in vector calculus.

As you know, the key algebraic rule for exterior algebra is that the wedge product is antisymmetric: $\vec{e}_1 \wedge \vec{e}_2 = -\vec{e}_2 \wedge \vec{e}_1$ and so on. Thus for the vector space R^2 with basis $\vec{e}_1, \; \vec{e}_2$, the wedge product of two vectors (a simple bivector) is
$$\left( a \, \vec{e}_1 + b \, \vec{e}_2 \right) \wedge \left( p \, \vec{e}_1 + q \, \vec{e}_2 \right) = \left( a q - b p \right) \; \vec{e}_1 \wedge \vec{e}_2$$
where I expanded out the product using the antisymmetry and simplified.

As a bit of thought shows, in R^2 the exterior algebra, considered as a four dimensional real algebra, is the direct sum of a one dimensional space (the zero-vectors or real constants), a two dimensional space (the vectors), and a one dimensional space (the bivectors), so our bivector is completely determined by the value of the determinant. In this sense, we have infinitely many representations of a given simple bivector in terms of parallelograms.

Similarly, you can decompose the exterior algebra on R^n into subspaces whose dimension is given by binomial coefficients; the first few are the space of constants, the space of vectors, the space of bivectors, the space of trivectors, and so on. The series is symmetric and the last subspace is the space of n-multivectors $\omega = a \, \vec{e}_1 \wedge \vec{e}_2 \wedge \dots \vec{e}_n$, which is one dimensional. Any n-fold wedge product of n vectors in R^n is thus completely characterized by an n by n determinant. Thus, in a sense exterior algebra naturally gives rise to determinants and offers a geometric interpretation of determinants in terms of volume factors. Indeed, in exterior calculus, n-multivectors give rise to n-forms on n-dimensional manifolds, or "volume forms".

Simple k-multivectors are k-fold wedge products of vectors. For example, a typical simple bivector in R^3 is
$$\left( a \, \vec{e}_1 + b \, \vec{e}_2 + c \, \vec{e}_3 \right) \wedge \left( p \, \vec{e}_1 + q \, \vec{e}_2 + r \, \vec{e}_3 \right) = (aq-bp) \, \vec{e}_1 \wedge \vec{e}_2 + (br - cq) \, \vec{e}_2 \wedge \vec{e}_3 + (cp-ar) \, \vec{e}_3 \wedge \vec{e}_1$$
The subspace of bivectors in R^3 is three dimensional, since $\vec{e}_1 \wedge \vec{e}_2, \; \vec{e}_2 \wedge \vec{e}_3, \; \vec{e}_3 \wedge \vec{e}_1$ forms a basis for this subspace. Note that 1+3+3+1=8, the dimension of the exterior algebra over R^3.

It probably will not be clear why not every k-multivector need be simple (expressible as a k-fold wedge product of vectors). To understand this, you need to know a bit about Grassmann-Plucker sygygies. The simplest case arises when we consider the six dimensional subspace of bivectors in R^4. Then if we write a generic simple bivector
$$\left( x^1 \, \vec{e}_1 + x^2 \, \vec{e}_2 + x^3 \, \vec{e}_3 + x^4 \, \vec{e}_4 \right) \wedge \left( y^1 \, \vec{e}_1 + y^2 \, \vec{e}_2 + y^3 \, \vec{e}_3 + y^4 \, \vec{e}_4 \right)$$
as a linear combination of the six basis bivectors,
$$u_{12} \, \vec{e}_1 \wedge \vec{e}_2 + u_{13} \, \vec{e}_1 \wedge \vec{e}_3 + \dots u_{34} \, \vec{e}_3 \wedge \vec{e}_4$$
where $u_{12} = x^1 \, y^2 - x^2 \, y^1$ and so on, then the Grassmann-Plucker syzygy is the (rather sophisticated!) "high school algebra" identity
$$u_{12} \, u_{34} - u_{13} \, u_{24} + u_{23} \, u_{14} = 0$$
This tells us that the six components of a simple bivector in R^4 are not rationally independent: any one can be determined from the other five using the syzygy. Thus, not every bivector in R^4 is simple; the set of simple bivectors forms a kind of hypersurface (an algebraic variety, not a linear subspace!) in the six dimensional subspace of bivectors. Sometimes every k-multivector is simple, for example bivectors in R^3 or n-multivectors in R^n, but these cases are exceptional.

For another example, try trivectors in R^5. (If you are very good at high school algebra, you should find three independent syzygies, not all obviously guessed. Note that the syzygies give algebraic relations among the components, but when there are several syzygies, these may themselves be related by an algebraic identity. This suggests the notion of a kind of sequence which keeps track of relations among coefficients, relations among these relations, relations among the previous relations, and so on.)

The textbook by Harris, Algebraic Geometry: A First Course, provides a wealth of information on some very interesting topics closely related to topics discussed above, including Grassmann manifolds, Stiefel manifolds, and Plucker embeddings of Grassmannians in higher dimensional projective spaces. In addition, if you have followed the This Week postings of John Baez, you may sense (correctly!) that this stuff is related to the Schubert calculus, which concerns incidence relations among k-flats in (complex) projective n-space. I note too that Grassmann and Stiefel manifolds are classic examples of homogeneous spaces (coset spaces for a closed subgroup of a Lie group). The textbook An Introduction To Differentiable Manifolds And Riemannian Geometry by Boothby offers a good introduction to homogeneous spaces.

The textbook by Spivak, Calculus on Manifolds offers a very good introduction to exterior algebra and exterior calculus. The more elementary approach in the classic textbook by Birkhoff and Mac Lane, Modern Algebra, and in the classic monograph by Flanders, Differential Forms with Applications to the Physical Sciences, are well worth studying, in addition to the books already mentioned.

Last edited:
Hurkyl
Staff Emeritus
Gold Member
A bivector is a wedge product of two independent vectors which can be imagined as a parallelogram and a trivector is a wedge product of three independent vectors which can be imagined as a paralleopiped. So am I correct in thinking that this parallelogram and paralleopiped is the bivector and the trivector itself, respectively? (With its magnitude as the area and the volume, respectively?).

If so how can you imagine the orientation of a bivector? (its even more harder to imagine the orientation of a trivector!).
You can always imagine them as algebraic objects!

mathwonk