# Bivectors in 3 and 4 dimensions

The big questions behind are:

1. Is a bivector a two-form?
2. Why a bivector is simply a vector in 3 dimensions?
3. How to distinguish between vectors and bivectors in 3D?
4. Why all bivectors are not vectors in 4D?
5. How to imagine the dual of a bivector?
6. How can a bivector be non-simple?
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What is a bivector for you? Is it just v wedge w for v, w, vectors? If so, then aren't most of your questions answered by the observation that 3 choose 2 is 3, but 4 choose 2 isn't 4? – Kevin Buzzard Nov 4 '09 at 11:09
Yes but no :-) actually I don't even know if we can speak of bivectors in 3D! And I'm talking about 4D precisely because I'm studying how to describe a 4-simplex by a set of 10 bivectors – Pedro Nov 4 '09 at 13:43

A bivector is an element of $\bigwedge^2 V$, so it is dual to a $2$-form on $V$. You can think of a bi-vector as a tiny piece of area.

If $V$ is three dimensional and comes with an inner product, then one can choose an isomorphism between $V$ and $\bigwedge^2 V$ which commutes with all the orthogonal maps for your inner product. In elementary math, this is the map which we call the cross product. This is not quite unique; you have to decide whether to use the left-hand-rule or the right-hand-rule to take cross products.

In my opinion, the best way to learn to distinguish between vector and bivectors is to get in the habit of not identifying $V$ and $V^*$. One way to do this is to work with an inner product given by an arbitrary symmetric matrix $g$ and keep the matrix $g$ in all your computations, rather than changing to an orthonormal basis.

A quicker way which I find useful is to think about whether the quantity in question has a natural direction, or has a sign ambiguity which comes from some arbitrary convention. For example, the normal vector to an orientated surface in $3$ space is going to be a bivector, because we need to decide whether the orientation circles the normal to the left or the right.

Writing down a bi-vector in $d$ dimensions takes $\binom{d}{2}$ coordinates. So, for $d = 4$, we need $6$ coordinates and we can't fit them into a single vector. I'm guessing that "simple" means a wedge of two vectors. So $e_1 \wedge e_2 + e_3 \wedge e_4$ is not simple. Once we get up into higher than $3$ dimensions, there is nothing that prevents this, so it can happen.

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Metaquestion to our hosts: Is there another way to post a follow-up question than postion it as an answer to the first one? To Pedro: Ok, now I get what you mean by simple bivector. Your question is answered in John Baez' blog:

http://math.ucr.edu/home/baez/week120.html

(Just do a text search for "bivector").

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When I posted a follow up question, I made it its own question, linked from the new question to the old (in the body of the question) and from the old question to the new (in a comment on the question.) See mathoverflow.net/questions/3289/… . I would agree that it would be nice if there were an automated way to do this. – David Speyer Nov 4 '09 at 16:09

I believe the answer to your question on simplicity is no, e.g. $(1,0,0,0)\wedge (0,0,1,0) + (0,1,0,0)\wedge (0,0,0,1)$ cannot be written in the form $f\wedge g$.

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An easy way to verify that is to observe that if W = x ∧ y, then W ∧ W = 0. This is not satisfied by your example, so it can't be simple. – Darsh Ranjan Nov 4 '09 at 15:22

One possible interpretation of the question uses Clifford algebras: A bivector could be defined as an element of the Clifford algebra of the $n$-dimensional real vector space with the Euclidean scalar product that consists of products of two orthogonal elements.

Have a look at the Wikipedia entry and the book

Lounesto, Pertti (2001), Clifford algebras and spinors, Cambridge: Cambridge University Press, ISBN 978-0-521-00551-7 (MR)

if you can get ahold of that.

Short answers to the questions would then be:

1. Strictly speaking, no, bivectors and two-forms live in different algebraic objects, but there is a canonical isomorphism of vector spaces.

2. In three dimensions there is a canonical isomorphism between the two, e.g., $e_1 \wedge e_2$ (bivector) is taken to $e_3$ (vector). The $e_1$ etc. are the elements of the canonical basis of your vector space, $e_1 \wedge e_2$ means the product in the Clifford algebra built from that (as mentioned above).

3. If you write bivectors and vectors explicitly as elements of the Clifford algebra I mention above, the difference is manifest.

4. In 4-dim there is no canonical isomorphism; this works in 3-dim only (see item 2).

5. A bivector can be visualized as a surface with a "direction" and a "size", e.g., in three dimensions a part of the $xy$-plane plus "clockwise" or "counterclockwise". The vector would then be parallel to the $z$-axis, its length equal to the size of the bivector. It's pretty easy to draw, but hard to describe with words. But if you write the bivector as, e.g., $e_1 \wedge e_2$, you get your vector by the familiar cross product of $e_1$ and $e_2$.

6. I do not know what "non-simple" means in this context, but maybe you think of elements like $e_1 \wedge e_2 + e_1 \wedge e_3$. That would be a bivector that consists of two elementary bivectors.

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Actually I want to avoid this abstract picture of bivectors as elements of a Clifford algebra. I prefer a more intuitive approach. Simplicity means of the form b = f ∧ g – Pedro Nov 4 '09 at 14:41
Probably for (6) you mean your vector to be something like $e_1 \wedge e_2 + e_3 \wedge e_4$, since $e_1 \wedge e_2 + e_1 \wedge e_3 = e_1 \wedge (e_2 + e_3)$ is simple? – L Spice Jun 30 at 17:54

Tim's answer was mostly good, but I have to point out one thing: the isomorphism between bivectors and vectors in dimension 3 is not canonical; it depends on choosing something like a basis (as Tim did) or a Riemannian metric.

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Thank you for all your answers, now I can ask the true question: A 4-simplex (in R^4) determines a set of 10 surfaces (true ones i.e: 2-simplices) so it determines also a set of 10 bivectors, for a 4-simplex to be uniquely determined by a set of 10 bivectors (up to parallel translation and inversion through the origin) the later must satisfy some conditions, among these conditions we find: (2) Each bivector is simple, i.e. of the form b = f ∧ g. (Are there bivectors which are not simple? The simplicity condition is encoded in their own definition!!!)

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Regarding the simple bivectors question, see the last paragraph of David Speyer's answer. One can take linear combinations of bivectors, but this does not preserve simplicity. – S. Carnahan Nov 4 '09 at 15:32

Ok so condition (2) simply says that each simple bivector (by its simplicity) will define two vectors thus will define a 2D surface, thus we have constructed the 10 surfaces of the future 4-simplex by this requirement.

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I'm not sure if I understand you, but I'll try :-)

You are handed a set of 10 bivectors that live in four (real) dimensions. You want to find out if there is a simplex and an orientation on each of it's faces, such that the surfaces of the simplex are represented by your bivectors.

A necessary condition is that each of the bivectors is simple. Why? Answer: In 4 dimensions it could actually happen that some of the bivectors are not simple. That cannot happen e.g. in 3 dimensions. A bivector that is not simple does not represent a surface, therefore the condition is necessary.

But this condition alone is not sufficient to ensure that a set of 10 bivectors represent faces of a simplex. There are more (see my link to the blog of John Baez).

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