# Vectors - are the definitions truly equivalent?

1. Jul 30, 2014

### 1MileCrash

There are two "formal" definitions of vectors (and tensors in general) which I've learned.

The first is what I consider the "better" definition, one I learned in linear algebra. We call a set X a vector space over a field F whenever that set has properly defined operations of scalar multiplication and addition, and follows certain properties. The elements of X are then called vectors.

The second definition, the one I learned as a physics major, is the one I consider the "pain in the ***" definition, that is, it is a vector is a "group of numbers" that transform a certain way (transformation law) into a new coordinate system.

I've always wondered if these definitions were, in fact, equivalent and I've seen it mentioned that they are.

However, what makes me doubt that they are really equivalent definitions is the emergence of "pseudovector" in the case of the second definition. These are vector-like objects which do not transform correctly upon inversion.

My questions are:
Consider a vector that meets the first definition; does it meet the second?

Consider a vector that meets the second definition. Does it meet the first?

2. Jul 30, 2014

### jbunniii

 I had never heard of pseudovectors or any definition of "vector" that refers to transformations. But I found this Wiki entry which gave some insight. As usual the physicists decided to do things a bit differently.

http://en.wikipedia.org/wiki/Pseudovector

I'm not sure I fully understand the "definition" above (for one thing, what is the difference between "displacement vector" and "vector"?), but already it's clear that this is not equivalent to the usual definition of a vector space, for a couple of reasons:

(1) In a general vector space, it's not necessarily possible to express a vector as an array of finitely many numbers/coordinates. This is only possible in a finite-dimensional vector space. Many interesting vector spaces have infinite dimensions, for example, the space of all real-valued sequences, or the space of all real-valued functions defined on $\mathbb{R}$.

(2) In a general vector space, there need not be a notion of "rotation" or "angle" - this requires some additional structure such as an inner product.

Hopefully someone who knows something about tensors (i.e. not me :tongue: ) can give some more insight here.

Last edited: Jul 30, 2014
3. Jul 30, 2014

### HallsofIvy

The first definition is the "Linear Algebra" definition- it is more general than the second or "physics definition". That is mainly because the first definition allows infinite dimensional vector spaces. For example, if X is the set of all polynomials, with scalar multiplication defined to be multiplying every coefficient by that number and vector addition defined to be adding coefficients of the same power of the variable, then X is a vector space. It is NOT "group of numbers" because there is no upper bound on the degree of the polynomial- it is "infinite dimensional".

On the other hand, if V is a finite dimensional vector space, using the first definition, then we can choose a basis for the vector space, write every vector as in terms of those basis vectors, $v= a_1v_1+ a_2v_2+ \cdot\cdot\cdot+ a_nv_n$, and then identify vector v with the "group of numbers" $(a_1, a_2, \cdot\cdot\cdot, a_n)$. If we were to select a different basis for the vector space, we would have a different set of numbers: $v= b_1u_1+ b_2u_2+ \cdot\cdot\cdot+ b_nu_n$. However, the set of numbers $(b_1, b_2, \cdot\cdot\cdot, b_n$ and $(a_1, a_2, \cdot\cdot\cdot a_n)$ are connected- each is a matrix times the other. That matrix multiplication is the "transformation law" you refer to.

Going the other way, given a "group of numbers", $(a_1, a_2, \cdot\cdot\cdot, a_n)$, with the usual scalar multiplication and addition, does satisfy all of the requirements for a "vector space".

Now, you say that a "pseudovector" does not transform correctly upon inversion. What do you mean by "inversion"? Does that even correspond to an allowable operation in a vector space?

4. Jul 30, 2014

If you know something about differential forms I give you some citations of Frankel's book:

"....Half of the "forms", "vectors," and "scalars" that occur in physics are in fact "pseudo-objects" that make sense only when an orientation prescribed...for example the magnetic field pseudovector B is perhaps the most famous example..."

"Definition: A pseudo-p-form $\alpha$ on a vector space E assigns, for each orientation $\circ$ of E, an exterior p-form $\alpha_{\circ}$ such that if the orientation is reversed the exterior form is replaced by it's negative $\alpha_{-\circ}=-\alpha_{\circ}$.....Similarly we can define define pseudovectors, pseudoscalars and so on, pseudo always reffering to a change of sign with a change of orientation"

Maybe that helps !

Greets

5. Jul 30, 2014

### AlephZero

I think the motivation for the different approaches is that mathematicians want to consider the properties of vectors in an abstract sense, but physicists and engineers want to do calculations, which in practice means representing a finite-dimensional vector in some definite (sually orthogonal) basis, or "coordinate system".

The physics/engineering definition of vectors (and similarly of tensors) focuses on the transformations required to do the calculations, when the underlying field is $\mathbb{R}$ or $\mathbb{C}$, and the underlying vector space corresponds to n-dimensional Euclidean geometry.

Of course, physicists and engineers had been successfully doing the calculations for a few centuries before mathematicians decided to make the subject more complicated

6. Jul 30, 2014

### micromass

7. Jul 30, 2014

### Matterwave

To add onto Aleph's point, mathematicians tend to like to make concrete things abstract (vector is a set with over a field along with the vector properties), whereas physicists like to stick to the concrete examples (vector is an arrow, or an object with magnitude and direction, or an n-tuple of numbers) . The difference lies in the ultimate goal differences between a mathematician and a physicist.

To a mathematician, making a concept abstract makes it possible to describe a much broader set of mathematical objects in the same way. This was hinted at in Ivy's post. Because mathematicians decided to define a vector in such an abstract way, they can analyze objects that we wouldn't normally consider vectors (e.g. functions spaces) using vector analysis. In this way a mathematician can find mathematical insight into objects that he wouldn't be able easily get by analyzing it otherwise.

To a physicist, we want to stick to the concrete examples because that helps us deal with the physical problems much easier and allows us physical insight into a problem. If we think of a velocity vector as an arrow in space, for example, we can much more quickly figure out what is going on physically than if we consider a velocity vector as an abstract object which, combined with the other velocity vectors (and the real numbers), obey the vector space postulates.

Because the "physicist's definition" originated first (as is the case for many mathematical objects because the "physicist's definition" is more likely to be the more intuitive one and less abstract one), it is usually found to be a subset of the "mathematician's definition". As is the case here.

8. Jul 30, 2014

### micromass

Moderator note: Please keep this thread on topic. This is not supposed to be a thread on physicists and mathematicians and who do it better. For any such discussions, see the link in my previous reply.

9. Jul 30, 2014

### Fredrik

Staff Emeritus
Any n-tuple of real numbers is a vector in the sense of linear algebra, but to get a vector (or pseudovector) in the sense of the awful definition, you need something that associates an n-tuple of real numbers with each coordinate system. It's the function that makes that association that should be called a tensor/pseudotensor, not one specific n-tuple.

10. Jul 30, 2014

### micromass

So rigorously, what would be the definition of a pseudovector?

11. Jul 30, 2014

### Fredrik

Staff Emeritus
Inversion is just the map $x\mapsto -x$. Physics books often say that a vector transforms according to the tensor transformation law under O(3) transformations, while a pseudovector transforms that way under SO(3) transformations, and does something else (flips the sign?) under inversion.

12. Jul 30, 2014

### Fredrik

Staff Emeritus
I don't think I've ever seen a rigorous version of the awful definitions (of "vector" and "pseudovector"), but something like this should work:

A function $V:O(3)\to\mathbb R^3$ is said to be a vector (under O(3) transformations) if
$$V(R)_i=\sum_{j=1}^3 R_{ij}V(I)_j$$ for all $i\in\{1,2,3\}$ and all $R\in O(3)$.

A function $V:O(3)\to\mathbb R^3$ is said to be a pseudovector (under O(3) transformations) if
$$V(R)_i=(\det R) \sum_{j=1}^3 R_{ij}V(I)_j$$ for all $i\in\{1,2,3\}$ and all $R\in O(3)$.

13. Jul 30, 2014

I already posted a definition of pseudoforms.......

14. Jul 30, 2014

### micromass

Ah yes, I missed that. Which book by Frankel is this exactly?

15. Jul 30, 2014

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16. Jul 30, 2014

### micromass

Last edited by a moderator: May 6, 2017
17. Aug 2, 2014

### 1MileCrash

From what I understand, under inversion, a pseudovector will "stay the same" (or flip the sign an additional time if you prefer to say it that way).

Suppose O and P are coordinate systems such that by the transformation law, a vector V in O would be expressed as -V in P.

Then for vectors A, B in O, A x B does not transform according to the transformation law because:

A x B = - B x A = - A x -B

If A x B were to be a vector by the poopy definition, it would have to be the case that -A x -B = -(A x B).

Right?

18. Aug 3, 2014

### Fredrik

Staff Emeritus
This sounds right to me, assuming that the transformation flips the sign of all elements of $\mathbb R^3$. $A\times B$ is only one element of $\mathbb R^3$, but if we also require that for all O(3) transformations R, the "transformed" $A\times B$ is $RA\times RB$, then we have associated an element of $\mathbb R^3$ with all coordinate systems (of the relevant kind), and we can check the transformation properties of this association. You suggested (if I understand you correctly) that we consider R=-I, where I is the identity. We have
$$(A\times B)'=(-IA\times -IB)=(-A)\times(-B)= A\times B,$$ and therefore
$$(A\times B)'_i =(A\times B)_i=I_{ij}(A\times B)_j =\det(-I)\big((-I)_{ij}(A\times B)_j\big) \neq(-I)_{ij}(A\times B)_j.$$ So the association defined above certainly doesn't define a "vector" under O(3) transformations, but it may define a pseudovector under O(3) transformations. To know for sure, we would have to check what happens with an arbitrary O(3) transformation instead of just -I.