Most statements are confusing and the statement in bold is meaningless. 

The answer is in any textbook on linear algebra and depends on just how abstract one wishes to go. These notes: http://people.maths.ox.ac.uk/hitchin/hitchinnotes/Projective_geometry/Chapter_3_Exterior.pdf

part of a course on projective geometry, http://people.maths.ox.ac.uk/hitchin/

might be useful. 

This is just a rule:

parallel vectors commute, orthogonal vectors anti-commute

Why it is "meaningless"? 

The sentence doesn't make sense, because the exchange operation isn't well-defined. Two parallel vectors, whose orientation is exchanged, do remain parallel; and the cross product of two orthogonal vectors, whose relative orientation is, also, exchanged, doesn't change sign, either. And two vectors needn't be orthogonal for their cross product to change sign-or not-if they're exchanged without, resp. with changing the relative orientation. 

Who is talking about the cross product? It is about some new product, not cross product.

A product  of vectors is a map between the product of vector spaces to some other space, that's defined by certain properties that must just be proved to be consistent. That's all. 

Example

The  outer  product  A X B  for 4-space  

  We recognize  with  Gibbs  two functions  of  the  indeterminate  product  AB  of two vectors,  namely  the  inner  and  outer  products.  The  inner  or scalar  product  is defined  by  

A·B  =  AB  cos  (A,  B),

 where  A  and  B  are  the  magnitudes  of  the  vectors  A  and  B(in R4) and  (A,  B)  is  the  Euclidean  angle  between  the  two  vectors

 The  outer  product  A X B  for 4-space  is no longer  a vector  whose  magnitude  is  the  area  of  the  parallelogram  determined  by  A  and  B  as defined  in R3,  but  rather  a two-dimensional  quantity-a second order tensor W which  can be reduced  to  a form  wherein  all of  its antecedents  and  consequents  are normal  to  the  plane  of  A  and  B  Further,  if  we  wish  the  product  to  conform  to  the  requirement  of 3-space  that W= A X B  = - B X  A,  the  tensor  must  be antisymmetric  We can  therefore  express  the  product  A X B  in  terms  of two  vectors  normal  to  A  and  B.

 A X  B.  =  aÄb  - bÄa.    (Ä=tensor product)

 If  we  further  specify  that

  AB  sin  (A,  B)  =  ab  sin  (a,  b)

 the  definition  is complete.  If  we select  the  vector  b  so  that  it  is orthogonal  to  a,  then  we can  write  the  product in  the  convenient  form:

 A X B  =  e(i3 Äi4  - i4Äi3)AB  sin (A,  B),  

where i3  and  i4  are  mutually  orthogonal  unit  vectors  normal  to  A  and  B,  and  e  has  the  value  +  1 or  -1  depending  upon  the  order  of selection  of  the  vectors  i3  and  i4  .The  above  definition  of  the  outer  product  admits  of immediate  generalization  to  n-space.

Miroslav

Any set V for which a binary operation addidtion +: V×V → V and a binary operation scalar multiplication *: F×V → V (where F is a field) is defined so that some axioms are satisfied, is a vector space.

For some vector spaces, it is possible to define a vector product from V×V to V (so the product of any two vectors is again a vector). For a vector space over a field F, you then get an algebra over the field F if the vector multiplication satisfies some additional axioms. Some well-known examples are:

However, this product is not necessarily unique - that is, for one and the same vector space more than one multiplication can be defined. For example:

Furthermore, for some vector spaces, vector multiplication is not defined. For example, except for some simple cases there is no product defined for m×n matrices with field valued entries when m and n are different numbers. But there are many more examples.

So there is no general answer to your question "how to multiply vectors" because the notion of a vector space is too general.

To conclude, three remarks about your post.

(a) Your formula ab = ½(ab + ba) + ½(ab - ba) does not hold in general, not even when the product ab is defined. E.g. in the context of a vector space over a finite field ZN there simply is no factor ½.

(b) If the vector product is defined, and ½ is an element of the field F, then of course your "antisymmetric part" ½(ab - ba) is a vector.

(c) Your notions of 'parallel' and 'orthogonal' do not necessarily apply to every vector space. If no inner product is defined, then how are two vectors "orthogonal'? E.g. take the space of all arrays {a1, a2, a3, ... } where the an are real numbers (and the array need not to converge). This is a vector space. Then how are two arrays "parallel" or "orthogonal"?

Marcoen

After a discussion by different people ,my suggestion is to understand the very meaning of torque .Suppose you define torque in 4D ,where you will visualise Similarly the same holds in 2D.

I like all your answers, to be clear!

My question is not precise enough. Intention was not to discuss general vector spaces, just  our "directed line segments" over real numbers. Excuse me!

Parallel vectors are well defined without any metric, for vector a we have 3a as parallel. Any vector b could be written as sum of parallel to vector a (projection) and rejection. 

Anti-symmetric part of new product is not a vector, for example, in Euclidean 3D we have for orthonormal basis vectors (here we use Euclidean metric):

a = e1, b = e2,  ab = (e1e2+e2e1)/2 +  (e1e2-e2e1)/2 =  (e1e2-e2e1)/2 = e1e2.

Now, due to anti-commutativity we have 

e1e2e1e2 = - e1e2e2e1 = -1

It is not a vector. It is a bivector. There are trivectors also, for example, square of  trivector j =  e1e2e3 is -1 and j commutes with all elements (scalars, vectors, bivectors and trivectors, belongs to the center of algebra), so, it is ideal "imaginary unit", with geometric interpretation. In fact, all trivectors are proportional to j. 

Product of - j and (ab - ba)/2 is just axb (cross product). 

With the new product we have algebra (geometric algebra). In Euclidean 2D we have basis of algebra

{1, e1, e2, e1e2,}.

Euclidean 3D with the "new" product (geometric product, due to Clifford) is so rich in structure, it is magic. And any relation can be easily generalized to any dimension. This richness is my motivation to ask: how to multiply vectors?

Miroslav,

Your statement "Anti-symmetric part of new product is not a vector" is simply not true. You can easily see that as follows:

Furthermore it is not clear what you mean by "e1e2e1e2" in "e1e2e1e2 = -1". Do you mean the vector product ((e1*e2)*e1)*e2? Please explain.

Marcoen

Yes, e1e2e1e2 is a vector product. And no, ab is NOT a vector. It is simple to see, noncommutativity gives that. Symmetric part is a scalar, anti-symmetric part squares to negative real number generally.. For example, square a + b and use the fact that square of vector is real number.  It  leads to many geometric objects that are not subject of usual algebraic treatment. Geometric product defines algebra (Clifford, geometric algebra). Objects that are not vectors are bivectors, trivectros ... Bivector, for example, e1e2 defines a 2D subspace, defines unique plane, squares to -1, rotates vectors, describes complex numbers (with a clear geometric interpretations this time), could be unit quaternion, it is used to naturally define magnetic field (or any axial vector), etc. In fact, we obtain mathematics in which subspaces are manipulated.  

Geometric product is a legacy of Clifford and Grassmann. This beautiful and powerful mathematics is forgotten  and replaced by Gibbs products (scalar and cross). 

This is important, because one can name a part of mathematical physics and it is probably contained here, without coordinates, matrices, tensors, ..., powerful and elegant. That is my experience. For example, just a few (almost high-school) steps is needed to obtain powerful formalism of special relativity in 3D, including hard problems of rotations as part of Lorentz transformations. Without 4th time dimension. It would take a hundred of pages to just mention what can be done with this algebra, this time in the unique language and clear geometric interpretation at each single step.  So, problem of vector product is really important.    

Miroslav,

Your posts are confusing.

If you talk about a vector product ab, then ab is a vector. If ab is not a vector, as you state, then you should not call it a vector product.

In your starting post you wrote:

"In 3D (Euclidean) we have for orthonormal basis

eiej + ejei = 2 dij , dij is Kronecker delta symbol.

You see, it is Pauli matrix rule ..."

In that equation, eiej is the inner product, so it is just a number. For the Pauli matrices however we have {si, sj} = sisj + sjsi = 2dij.I where dij is the Kronecker delta and I the identity matrix. So in your equation you have numbers on both sides of the equation, while your "Pauli rule" has matrices (vector products) on both sides of the equation. The one has nothing to do with the other.

I think you are confusing c.q. mixing vector products and inner products.

I'll leave it at this.

Marcoen

Oh, no, it is not a "vector product", it is geometric product (due to Clifford). It is different to say "vector product" or "product of vectors". 

In eiej we have geometric product (ie new product we discuss about). Similarly, ab is not an inner product, it is new anti-commutative product, inner product is with dot. So, e1e2 = - e2e1. If we have orthonormal basis in 3D and this (NEW) product of vectors (which is NOT vector) then 2D matrix representation is given with Pauli matrices. Ok? Please, forget usual products of vectors promoted by Gibbs and widely accepted. I am NOT talking about it. "I think you are confusing ..." ? Who is "confusing"? How it is possible to think of eiej + ejei = 2 dij as the inner product relation? It means or 2 = 2 or 0 = 0, so, it is obvious nonsense as a property. I said: NEW product ab, ok? How then e1e2 is an inner product? 

It is easy to say "it is nonsense" or "you are confused", but it would be nice to ask if something is unclear. 

we have Lie product for example in 2real 3 dimension we have cross product which is Lie algebra product associated with SU(2) group.

Dear Miroslave, I think the problem is laying in this phrase: "It is straightforward to show that antisymmetric part is not a vector".

1) In lie Algebra, ab-ba is a vector. In vector space of matrices, also, AB-BA is a vector ( matrix ), ..... 

2) For the vector product, it is not right that it is defined only in 3-dimensional space, it can be generalized to any Euclidean space, it is always the unique vector orthogonal on the plan of the two ones with a specific length and angle.

• Dear Miroslav, I think the most general way is still accepting that you are working in a vector space, mostly R3 . There you have some coordinate system with basis vectors. A vector is expressed with respect to that coordinate system via components xi.and yj .And then you may form all kinds of combinations. A scalar product, a vector, or a tensor. Your vector product is for instance
• xiyjeijk where the last one is the Levi-Civita tensor. But you may form of tensors with more indices as tijkl
• How to do this is in fact group theory.

Dear Hanifa,

1) I see your point, but I am not talking about Lie algebras (funny, they are in geometric algebra), or matrix algebra, it is about "new" product. So, we have properties: 

- usual real vector space properties

+

- noncommutative product ab of vectors a and b, with associativity and distributivity, to be real-like as possible; inner, outer, cross and other possible products would be denoted differently. Important: ab is NOT scalar product. It would be nice if new product allows inverses! (What about inner and cross products?)

In addition, new product of vectors is not necessarily a vector, there is no such condition. a/2 is well defined. Vectors a and 3a are parallel. 

This new product is due to W. Clifford (19. century).

If that is clear, one can proceed and say:

ab = (ab + ba)/2 + (ab - ba)/2,

it is just usual decomposition for non-commutative products, if well defined.

About "straightforward":

What is a2? Let it be real number. Then 

(a + b)2 = a2 + b2 +ab + ba,

so, ab + ba is real number, ie, symmetric part of product is real number (inner product of vectors). To clear this a little, let use special case: 3D Euclidean vector space and orthonormal basis ei in it. General case follows easy. Then (ab + ba)/2 is ordinary scalar product of vectors. For orthogonal vectors we have 

ab = - ba

so it is clear that orthogonal vectors anti-commute. Parallel vectors are proportional and commute.

What about anti-symmetric part of product? Special case is here enough to demonstrate, so, for orthonormal basis vectors in 3D we have, for example:

e1e2 = (e1e2 + e2e1)/2 + (e1e2 - e2e1)/2 = (e1e2 - e1e2)/2  +  (e1e2+ e1e2)/2

 and we see that this new product gives just anti-symmetric part. Square it:

e1e2e1e2 = - e1e2e2e1, due to anti-commutativity and associativity, we see that is is real number, -1 with Euclidean metric. 

So, in Euclidean 3D we have 

eiej + ejei = 2dij,

so, orthogonal unit 3D vectors are represented with Pauli matrices.

What is e1e2? It is not a vector, squares to -1. It is just a new "beast". We call it bivector.

Some simple properties of e1e2:

1) it is common to define reverse of new product ab: ba. Then abba is a real number. So, e1e2e2e1 = 1 (is invertible). It has a magnitude and orientation (e1e2 = - e2e1). It is easy represented with unit oriented square.

2) (e1e2)e1 = - e1e1e2 = - e2,

     e1(e1e2) = e1e1e2 = e2,

so, it rotates vectors.

3) Vector have inverses, for vector a it is just a/a2. For a vector in Euclidean 2D

v = xe1 + ye2 we have 

e1(xe1 + ye2) = x + e1e2y,

so, x + e1e2y is a complex number. What is ve1? Complex numbers are in geometric algebra, but all complex numbers theory now has clear geometric meaning and extends to any dimension. In fact, one can use any plane in any dimension, define unit bivector in it and define complex numbers. There are so many "imaginary units".

4) e1e2 defines unique plane spanned by vectors e1 and e2, How? 

Let me use symbol ^ for outer product (anti-symmetric part of geometric product), usually we use a "wedge" symbol. Then e1e2 = e1^e2 and we can find all vectors x 

x^(e1^e2) = 0.

Solution is x = r e1 + s e2, r and s are reals. For example, e1^(e1^e2) = e1^e1^e2=0, due to anti-commutativity.  Any plane in any dimension is well defined with some unit bivector. 

It is something new! There are trivectors, generally k-vectors, all of them define subspaces and have a lot of properties. 

And one example to think about:

In quantum mechanics we use Pauli matrices. But we can use orthonormal unit vectors basis as well, in fact, it is better. Geometric interpretation is clear at each step. 

For example, we could have a product of Pauli matrices (s is short for usual sigma) s1s2. It is diagonal matrix with i and - i on the diagonal. Ok, this is a simple one. My question is: what geometric properties we "see" in such a matrix? Or any matrix? But with e1e2 geometric meaning is clear. As a "side effect" we don't need square of -1, we have bivectors instead, with clear geometric meaning. 

Benefits are numerous, so, it would be wise to rethink way we are multiplying vectors. Grassmann and W. Clifford are pioneers, we should listen to them. 

Dear Peter,

not just R3, any R(p,q,r).

Coordinates, matrices, tensors, quaternions, complex numbers, ... all that "stuff" is not subject of geometric algebra. Of course, it is always possible to invoke coordinates, or relate results, coordinates are necessary when checking experimental data, but with geometric product forget about coordinates. It works and it is powerful.

About cross product:

1) How to define it in 2D? With bivectors there is no problem, in fact, simple bivector defines a plane. Poor 2D "people" cannot define a cross product. But can easily define bivector and that beast will do all we need. One can imagine it as little "whirligig" in the plane, in any dimension. Why then bother with "vector orthogonal to the plane"?

2) I know about attempts to define cross product in higher dimensions, In 4D let define e1xe2. It should be a vector orthogonal to the plane, but which one? Is it e3 or e4, or some linear combination? Ok, if one succeeds to define it uniquely, question is why bother? Bivector e1e2 defines plane and all you need: you can rotate objects, define complex numbers in the plane, etc. In 3D, cross product is just proportional to bivector, so, it is easy to change formulas.

3) There is one more obstacle with cross product: you need a right hand rule. For example, due to right hand rule there is a problem with electromagnetism: use straight wire with current, plot one magnetic field vector B, place mirror near it and what you see? There is no symmetry. In geometric algebra all axial vectors are replaced with bivectors. It is an oriented area (shape is irrelevant). Now, mirror image is symmetric! A right hand rule is unnecessary. Please, give me another way to do that! 

Furthermore, Maxwell equations in geometric algebra are just nablaF = J. Ok, it is possible to do it without geometric algebra, but, here "nabla" is vector-like quantity and have an inverse via Green functions. This is not a "just another formalism", it is something really new and important, as I see it.

4) One more thing, what is e1 in 3D? A polar vector, isn't it? What is e2xe3? An axial vector. So, e1 is polar and axial vector? Or not? In geometric algebra it is clear what is e1e2 , an bivector, new kind of geometric objects. It is Grassmann who noticed that there are new geometric objects and we can naturally define them. My point is that we really need geometric objects that are not vectors and algebra to deal with them. We don't need a cross product! Ever! Even if one succeeds to define it in 2D. 

Dear Miroslav:

There are many vector spaces which allow a multiplication; these are called algebras. E.g. euclidean 3-space with the vector product is a Lie algebra (the Lie algebra of the Lie group SO(3)). To make sense of your question you should pose more assumptions. One strong assumption is that the algebra is \it{normed}, that is there is a norm |.| (which must be euclidean as it turns out) with  |ab| = |a||b|. Then there is an easy answer, given by Hurwitz 1898: The only solutions are the complex numbers, the quaternions, and the octonions, see the adjacent file for an elementary and short proof. The common vector product on R^3 is the imaginary part of the quaternionic multiplication, and there is an analogue on R^7, the imaginary part of the octonionic multiplication, and there is no other such product in any dimension.

Does this help?

Best regards

Jost

Hi Miroslav,

As you see I and others were led astray by your use of the term 'vector product'. But I see now that you mean the geometric product. You can define the geometric product of vectors a and b as

ab = a·b + a^b

so ab is a multivector, which is the sum of a scalar (the dot product a·b) and a bivector (the wedge product a^b). And then you get the formula of your original post ab = ½(ab + ba) + ½(ab - ba).

You can then also define a generalized geometric product AB for multivectors A and B. The product is then again a multivector.

I and many others agree with you that this has some interesting applications e.g. in mathematical physics. But as I see it the gain is still questionable. On the one hand you do get a shorter expression for e.g. the Maxwell equations, but on the other hand the definition of the framework within which these shorter equations are formulated is more complicated.

Furthermore, I'm confused as to what your question now is.

Marcoen

Dear Macroen,

thank you, but I am using geometric algebra a lot and know such a things. 

Why question?

1) To see what other people think, it is interesting!

2) In physics we need to multiply vectors. So, how to do that? Imagine 2D world and 2D physicists. They see that force can produce rotation of body. How to define quantity to formulate law? Effect is proportional to force magnitude and some distance. But there is a problem of direction. So, what to do? I suspect that they wouldn't formulate 3D theory for that (however, they might). Non-commutative product of vectors would be natural thing to start with. If one accepts this, then geometric algebra is inevitable.

3) There is many things against products of vectors, especially against the cross product.

4) Geometric algebra is not just the matter of simplicity (or complexity). For example, my point on Maxwell equation is not "simplicity", I  said that geometric algebra is not unique in that, my point was about existence of "operator" inverse. 

5) As I see it, geometric algebra has promising potential  to be universal language of mathematical physics.

Most important, I enjoy this discussion and like all comments!

Best ...

Dear Jost,

it helps, as I said, I enjoy all comments.

True, I am aware of things you said, but here is the Clifford algebra in question. With explicit geometric interpretation we name it geometric algebra (due to Cifford himself). 

What I understand is that the linear space theory is a fruit of study of system of forces, which naturally follow the addition rule and multiplication by scalar. So I feel that if one can find how forces are multiplied, then we may get to a product of vector, as well.

Dear Rajiv,

 we have motivation to add forces if they act on the same body, but why to multiply them if we are in the linear theory? Do you have any idea about it?

I think that there is quite enough  motivation for vector multiplication as I discussed here. 

I like simple examples, so consider 2D world and force that acts on body and rotates it. We all know solution, we use the cross product rxF, but there is a problem in 2D, there is no a "third dimension". But in geometric algebra we have outer product r^F. It is anti-symmetric (as anti-symmetric part of geometric product) and has a module just rFsin(angle), there is no need for third dimension and it is easy to generalize formulas to any dimension. For example, in 3D we have just r^F again and we obtain quantity proportional to cross product (it is a dual of cross product). In 4D nothing changes. Outer product is defined as usual and problem of "vectors orthogonal to plane" disappears. There is no need for it. Bivectors (outer product of two vectors) are all we need: uniquely define a plane, generate rotations (so powerful), act as a complex numbers,  have direct connection to Lie groups and Lie algebras, in 3D we can treat them as quaternions (with the geometric interpretation this time), they are constituents of spinors, ..., list is really long. 

With the geometric product distinction between "polar" and "axial" vectors disappears. In fact, axial vectors are not vectors at all, they are bivectors (recall that in the language of coordinates and indices we can treat them as the second rank tensors).

There is a beautiful Zen story "Empty your cup of tea". My question here is on that: "forget" all we know about multiplication of vectors and rethink it again. It is fun.

https://www.springer.com/gp/book/9783030017552

Marcoen J.T.F. Cabbolet

" the vector product is defined, so the products ab and ba are both vectors "

ab and ba are not vectors, they are quaternions.