# 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 Wedge^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 Wedge^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 (d choose 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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I believe the answer to your question on simplicity is no, e.g. (1,0,0,0) ^ (0,0,1,0) + (0,1,0,0) ^ (0,0,0,1) cannot be written in the form f ^ 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

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

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 euklidean scalar product that consists of products of two 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

if you can get ahold of that.

Short answerts 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 times 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 times e_2 means the product in the Clifford algebra build 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 itime 2). 5. A bivector can be visualized as a survace with a "direction" and a "size", e.g. in three dimensions a part of the x-y-plane plus "clockwise" or "counterclockwise". The vector would then be parallel to the z-axis, it's 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 times 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 times e_2 + e_1 times 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

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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Actually I found the explanation of Baez more attractive than the one of Barrett, the big question is which constraints the 10 bivectors have to obey in order to define uniquely the 4-simplex: the answer of Baez is clear:

What constraints do these 10 bivectors satisfy? They can't be arbitrary! First, for any four triangles that are all the faces of the same tetrahedron, the corresponding bivectors must sum to zero. Second, every bivector must be "simple" - it must be the wedge product of two vectors. Third, whenever two triangles are the faces of the same tetrahedron, the sum of the corresponding bivectors must be simple.

The second just says that each bivector defines a surface. The third is a consequence of [b' is the common edge] (b' ∧ b) + (b' ∧ b") = b' ∧ (b + b") [simple] For the first I'm still looking...

but there is something titillating me: if I have two triangles that lives in the same hypersurface (3D) and that do not share a common edge, the sum of their two bivectors will be simple (because it is a bivector which lives in 3D) so the third condition of Baez is not bijective!!

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To our hosts: Sorry, this thread delevops in a kind of blog thread, maybe we need to think about how to handle this kind of situation (either prevent e.g. the author of the question to post an answer - which would be very restrictive - or allow the topic to spin off in a seperate blog, any other suggestions?)

To Pedro: You need to distinguish the notions "vector space" and "affine space". The bivectors form a vector space. When you speak about disjoint "triangles" you think about them as an affine space. There cannot be a bijection between these two. That is what John (and me too) means when he speaks about a bivector representing a face of a simplex with orientation: A bivector represents the face if you think of the face translated back to the origin. So two different faces of your simplex will be represented by one and the same bivector (meaning in the set of 10 bivectors we spoke about ealier, not all of those need to be different). If you want to get a bijection you have to supplement the bivector with a translation.

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1. First, I'm sorry about the manner I posted my questions, it was my first visit in MO, and I could not register with my OpenID!!! So I work with my cookies whenever I can, I remarked however that the "comment" link under each answer disappeared... Actually I don't know how to proceed in order to develop my question so I post replies.

2. 2.

So two different faces of your simplex will be represented by [...]

I guess you mean "may be"?

[...] meaning in the set of 10 bivectors we spoke about ealier, not all of those need to be different [...]

Really? The 4-simplex will not be degenerate in this case?

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