Understanding the Cross Product of Vectors: Perpendicular Properties Explained

[lethe]

Originally posted by lethe

now, i was hoping you would respond to my original question above, which boils down to "what exactly is the definition of an outer product, and why is the vector cross product an outer product?"

mathworld seems to define an outer product as a tensor product (which is what i was claiming), whereas wikipedia thinks the an outer product is the wedge product, which is what someone else (you? i forget now) claimed, and i objected to. it is also what is claimed in the references that saski links.

i guess it can be either, depending on your preference.

 

[matt grime]

Originally posted by lethe
that is an awesome quote. Physicsforums should have a quote of the day feature. this should be it at least once a month.

now, i was hoping you would respond to my original question above, which boils down to "what exactly is the definition of an outer product, and why is the vector cross product an outer product?"

I'm not sure I can adequately answer that.

In generality, given a finite dimenisional vector space V there is the n-fold tensor product, which can saefly be thought of as:

let v_i be a basis of V, then a basis of $$V^{\otimes n}$$ is given by the set of all symbols $$v_{i_1}\otimes\ldots \otimes v_{i_n}$$

( I am, slightly contradictarily, being a gentleman here by not defining that in a basis free way, if you want I can but it is more confusing for the non-algebraist, trust me)

now the permutation group on n elements acts on the n-fold tesnor product by swapping factors in the expresion.

the exterior algebra is that sub-space where swapping the elements is the same as multiplying by -1


eg in the two fold tensor, it is the subspace spanned by expressions like $$u\otimes v - v\otimes u$$

It is not hard to see that in an r dimensional vector space this is only non-zero if n<=r, and it has dimension r choose n.

The element $$u \times v - v\otimes u$$ is labelled u wedge v (can't recall tex for it off hand; it's probably \wedge)

The reason why its so damn easy and confusing at the same time for R^3 is because 3 choose 2 is 3 and is therefore a 3 dimensional vector space. Different groups label these different things all over the place, but the wedge is usually called the exterior product.


Let's stick in the 3-d case: where does the rule ixj=k come from? Note I'm using x for ease of type setting, it is the wedge above.

well, the answer is: there is an exact pairing from between the first and second exterior algebra


basically, there is a unique element in the third exterior algebra's base:
ixjxk corresponding to the invariant element [tex] i\otimes j \otimes k -j\otimes i \otimes k + j\otimes k \otimes i -i\otimes k \otimes j + k\otimes i \otimes j - k\otimes j \otimes i[\tex] in the tensor algebra.


so the paring matches ixj with k and jxk with i and so on.

Is that clear in the slightest?

I think that might answer the question but I'm not sure.

Incidentally the quote is very old, and I can't recall who came up with it. I think the first time I saw it was in some lecture notes by Tom Korner, but it was a quote from someone else.

 

[lethe]

Originally posted by matt grime
In generality, given a finite dimenisional vector space V there is the n-fold tensor product, which can saefly be thought of as:

let v_i be a basis of V, then a basis of $\bigotimes^nV$ is given by the set of all symbols

$$v_{i_1}\otimes\ldots \otimes v_{i_n}$$

( I am, slightly contradictarily, being a gentleman here by not defining that in a basis free way, if you want I can but it is more confusing for the non-algebraist, trust me)

while i am not myself an algebraist, i do have a fair amount of algebra. i am, for example, familiar with the universal construction of a tensor product. as well as the coordinate definition of a tensor product of vector spaces. so whatever.

mostly, i am just looking to establish what is the correct terminology, and why.

now the permutation group on n elements acts on the n-fold tesnor product by swapping factors in the expresion.

the exterior algebra is that sub-space where swapping the elements is the same as multiplying by -1

sure.

eg in the two fold tensor, it is the subspace spanned by expressions like $u\otimes v - v\otimes u$

It is not hard to see that in an r dimensional vector space this is only non-zero if n<=r, and it has dimension r choose n.

The element $u\otimes v - v\otimes u$ is labelled $u\wedge v$ (can't recall tex for it off hand; it's probably \wedge)

The reason why its so damn easy and confusing at the same time for R^3 is because 3 choose 2 is 3 and is therefore a 3 dimensional vector space. Different groups label these different things all over the place, but the wedge is usually called the exterior product.

yes, as i understand it, wedge product and exterior product are synonymous.

Let's stick in the 3-d case: where does the rule $\imath\times\jmath=k$ come from? Note I'm using x for ease of type setting, it is the wedge above.

well, the answer is: there is an exact pairing from between the first and second exterior algebra

sure, because $\binom{3}{1}$ and $\binom{3}{2}$ are equal, there is an isomorphism between the two given by the Hodge dual (which depends on the orientation and the inner product we give to the vector space)

basically, there is a unique element in the third exterior algebra's base:
ixjxk corresponding to the invariant element $\hat{\imath}\otimes\hat{\jmath} \otimes \hat{k} -\hat{\jmath}\otimes \hat{\imath}\otimes\hat{k} + \hat{\jmath}\otimes\hat{k}\otimes\hat{\imath} -\hat{\imath}\otimes \hat{k}\otimes\hat{\jmath} + \hat{k}\otimes\hat{\imath}\otimes\hat{\jmath} - \hat{k}\otimes\hat{\jmath} \otimes \hat{\imath}$ in the tensor algebra.

but this exactly illustrates my point of why the exterior algebra is not the cross product algebra. first of all, the exterior algebra is associative, and the cross product algebra is not. it follows that in the cross product algebra,

$$(\hat{\imath}\times\hat{\jmath})\times\hat{k}=\hat{\imath}\times(\hat{\jmath}\times\hat{k})=0$$

whereas in the exterior algebra, you get the invariant element $e_1\wedge e_2\wedge e_3\neq 0$.

if we assign a product on the exterior algebra $e_i\times e_j=\star(e_i\wedge e_j)$, then we get an explicit correspondence with the vector cross product algebra in R³

so the paring matches ixj with k and jxk with i and so on.

Is that clear in the slightest?

I think that might answer the question but I'm not sure.

well, this isn't really my question. i know what the exterior algebra is and how it relates to the cross product.

i guess all i really want to know is: what does the term "outer product" mean to you? some people are saying "outer product"="exterior product", i think saski might have even said "outer product = Clifford product (geometric product)", but i guess we ruled that out.

on the other hand, i think "outer product"="tensor product"

i really just wanted you to weigh in with your opinion of what the term outer product should mean.

 

[matt grime]

Oh, ok an opinion, erm, well i never use the phrase outer product, but would take it to be the exterior product. sorry, i thought you wanted to see where the link with tensors is. sorry for telling you stuff you already know.

as for ixjxk=0 but i^j^k not being zero, well, I'm not making the identification between (R^3,x) as an algebra and $$\wedge^{\bullet}(R^3)$$ that would be required for that contradiction to make sense. One is a 3 dimensional ungraded algebra, the other is an 8 dimensional graded algebra.

 

[lethe]

Originally posted by matt grime
Oh, ok an opinion, erm, well i never use the phrase outer product, but would take it to be the exterior product. know.

OK. well, that's that then. perhaps i will endeavor to just never use that phrase as well. too much ambiguity.

as for ixjxk=0 but i^j^k not being zero, well, I'm not making the identification between (R^3,x) as an algebra and $$\wedge^{\bullet}(R^3)$$ that would be required for that contradiction to make sense. One is a 3 dimensional ungraded algebra, the other is an 8 dimensional graded algebra.

right, but there is an identification you can make, i am not quite sure how you make it explicit. i think it is what you were describing here, right?:

Originally posted by matt grime

Let's stick in the 3-d case: where does the rule ixj=k come from? Note I'm using x for ease of type setting, it is the wedge above.

well, the answer is: there is an exact pairing from between the first and second exterior algebra

so if you consider the R³ subspace of $\bigwedge \mathfrak{R}^3$, and identify the space of bivectors with their Hodge duals, then this becomes isomorphic to R³ with the cross product.

i know this is true, is there a nicer way to describe it though?

 

[matt grime]

take the two 3-d bits in the exterior algebra, the vectors and 'bivectors' and mod out be the relation v~*v (its hodge star dual, I mean) just considered as a vector space. See if the product descends to the quotient? Which might just be another way of saying what you just said. That seems the slickest way to do it. But that isn't the same as nice.

 

[BlackBaron]

I'm not 100% sure of this, but I believe that, historically vector products (and vectors) actually originated with quaternions.
I'm not telling the how quaternions originate (for that, look here http://www-groups.dcs.st-and.ac.uk/~history/Mathematicians/Hamilton.html), only how vector came from them.

Quaternions may be taken to have a real (scalar) part and a vector part, so in that notation, they are written like this: $$a=(a_o,\vec{a})$$, where $$\vec{a}=a_1\hat{i}+a_2\hat{j}+a_3\hat{k}$$.
Then, the product of 2 quaternions, can be written like this: $$ab=(a_ob_o-\vec{a}\cdot\vec{b},a_o\vec{b}+b_o\vec{a}+\vec{a}\times\vec{b})$$.
From what I read, phycisist (and mathematics) originally used quaternions instead of vectors, but then they noticed that they could just drop the scalar part.
Notice now what we get if we multiply 2 scalar-less quaternions:
$$ab=(0,\vec^{a})(0,\vec^{b})=(-\vec{a}\cdot\vec{b},\vec{a}\times\vec{b})$$
See, the scalar part is just minus the dot product and the vector part is the cross product. And so (I've been told), vectors and their products were born.

I've also been told that Maxwell original work on electromagnetism was done using quaternions, I've been trying really hard to confirm that (looking for a copy of Maxwell original work) but I haven't had any luck, can anyone here help me with that?

P.S. To find more about quaternions, look here http://mathworld.wolfram.com/Quaternion.html

*Edit: I had wrong the first link above (about the history of Hamilton and quaternions), but now is fixed.

 

[matt grime]

Vectors were used before Hamilton came up with Quaternions. A quaternion is a 4-vector, I'm not sure where this idea that it is a scalar plus a vector came from (how does one even add scalars and vectors?) but you're not the first person to have said it even in this forum. It is no more a vector plus a scalar than any n vector is a sum of n scalars. Perhaps it is because people think the i,j,k are the unit vectors in R^3?

Hamilton came up with them in order to find an extension of degree greater than 2 of the Real numbers (ie a vector space over R with a Field structure on it) he attempted first to make a degree three extension and it took him a while to figure out it can't be done (it can be for 2,4,8,16 and no others)

 

[BlackBaron]

First, a quaternion is NOT a 4-vector, preety much like a complex is not a 2-vector, because, for example, you can't divide two 4-vectors (anyway, I don't think this is the place to discuss that).
I didn't say a quaternion is a vector sadded to a scalar and I didn't say either that quternion units i, j, k are equal to vector basis i, j, k.
What I said, is that you may look a quaternion as a pair (scalar, vector) (pretty much like a comlex number can be looked as a pair of real numbers). That is just notation, and it's useful, for example, as it makes really easy to compute the product of 2 quaternions using the well known vector products.

It may be right that the stroy is not real and that the use of that notation originated it, as I said, I'm not 100% sure about it and I haven't got the chance to confirm it with serious sources.

You may want to give a look at this link:
http://www.rtis.com/nat/user/jfullerton/school/math251/cproduct.htm

Among other things, it says that Hamilton was the first to use the word vector, based on the Latin veher, "to draw".

 

[matt grime]

Quaternions do form a vector space of dimension 4 over the real numbers, a quaternion is a 4 vector. Quaternions are a (4-d) division algebra (over R). If you wish to say it behaves LIKE a pair (u,v) with u a scalar and v a 3-vector, then fine, but all you are doing is defining a 4-vector there anyway. An n vector is just an n-tuple of elements of a field with an addition and scalar multiplication on them.

Just because not every vector space of dimension A has the property that some object X does, does not mean X cannot be an A-dimensional vector space. The set of L^2 functions on the circle is an infinite dimensional vector space, yet and arbitrary infinite dimensional vector space doesn't have an integral defined on it.


Seeing as a+bi+cj+dk is a quaternion, I was assuming when you spoke of a scalar and a vector part that the a was the scalar part, and bi+cj+dk was the vector, that is where my misunderstanding of the adding scalar to vector came from - just notational issues.


A complex number is a 2-vector over R (C is a two dimensional real space, that happens to also be a field, it was in attempting to make a 3 -dim real space that was also a field that Hamilton realized you had to drop the commutativity requirements and make it 4-d)

As every n-dimensional vector space (over R) is abstractly isomorphic to every other n dimensional vector space (over R) for n in N, one can define a division algebra structure (any element is invertible) on any real 4 dimensional vector space, simply because Quaternionic structure preseves the norm. One may also do this to real vector spaces of dimension 2 (iso to C) 4 (quaternions) 8 (octonions) and 16 (can't remember the name). It might not be very useful to do this.



Hamilton may have been the first to use the word vector, however their use has been implicit for many centuries. Another instance formal group theory wasn't developed until the start of the 20th century, approx, but Galois used groups in the 1830s.

 

[BlackBaron]

Before I forget, 16D are called sedenions.

Well, yes, you're right, that's the definition of a n-vector and according to it, a quaternion is indeed a 4-vector, it just has some extra properties (so it's not just a 4-vector (just in case, I know you never say a quaternion was just a 4-vector, I'm just putting that to leave it clear)).

Coming to think of it, since the parallelogram law is so intuitive, geometrical vector addition may be as old as euclidian geometry, and there's no doubt they were somehow implicitly used when analitical geometry was created around the XVII century.
Vectors use must also be implicit in Newton's work which dates from the XVII century too.
But was there a formal theory of vectors before Hamilton?
What about the dot and the cross products?

 

[matt grime]

As always in mathematical history, it is very hard to decide at what point these abstract, disparate elements came together to form a theory as we now recognize it. I am prepared to state categorically that vectors came before quaternions and defend that statement in all its interpretations since Hamilton came up with quaternions after '10 years' of trying to define a (non-degenerate) multiplication on a 3-d real vector space. St Andrews University has a website about the history of maths that has plenty of details about these things. A an geometer, my defintion of vector spaces (though they weren't called that) were known before Hamilton - Analytic geometry for instance predates Hamiton, however, as physical quantities such as velocity, I'd be prepared to accept Hamilton as the father of that interpretation. Swicthing hats, as an algebraist, he wouldn't have understood our vector spaces over fields of non-zero characteristic, as such fields weren't known to him; all groups to 19th century mathematicians were at best concrete permutation groups, and fields as we now know them weren't defined until such time as abstract groups were developed. Inner products? Pass.

Octonions are now exciting the interest of theoretical physicists, perhaps in the same way quaternions did in the 19th Century.