# Basis vectors and covectors

1. Jul 10, 2006

### masudr

I often think I have fully understood this, then some question comes up in my mind, and I get confused again (which implies I never understood it in the first place).

We have a co-ordinate basis for vectors ${\partial_\mu}.$ I can think of two ways to get a corresponding basis for covectors.

1. Define basis covectors ${dx^\mu}$ such that

$$\partial_\mu dx^\nu = \delta^\nu_\mu.$$

This gives me a problem. Am I allowed to combine vectors and covectors without a metric? This leads me onto the second way...

2. Use the metric so that

$$g(\partial_\mu, dx^\nu) = \delta^\nu_\mu$$

but actually... the metric maps from two vectors to a scalar, am I correct? So perhaps what I mean is

$$g(\partial_\mu, \partial_\nu) = g^{\mu\nu}\partial_\mu \partial_\nu = \partial_\mu dx^\nu = \delta^\nu_\mu$$

which is in fact the same way as "method 1".

I guess my real question is can I define covectors without a metric? This is related to another thread in the forum, but I could do with a quick answer saying if my approach is correct.

Masud.

P.S. It appears that I have largely "thought aloud" in this post, and may well have answered my own question... but I could do with confirmation.

Last edited: Jul 10, 2006
2. Jul 10, 2006

### garrett

Yes, this is correct.

3. Jul 10, 2006

### nrqed

I am learning the subject myself so hopefully someone more knowledgeable will correct me if I am wrong but the way I understand it, "feeding" vectors to covectors does not require a metric.
So the notation $$g(\partial_\mu, dx^\nu) = \delta^\nu_\mu$$
seems incorrect to me. It's simply
$$(\partial_\mu, dx^\nu) = \delta^\nu_\mu$$
A metric allows to map a vector to a covector. And then a sceond vector can be "fed" to that covector, obviously. So what the matric allows is to introduce an inner product between vectors, as in your method 2.

I guess my point is that your first method does not require a metric. The second method does. *IF* there is a metric then the two methods are equivalent. If there is no metric, there will be no inner product between vectors (because no mapping between vectors and covectors) but the first method (with the g removed from the equation) will *still* be defined.

Maybe someone will correct me.

Patrick

4. Jul 10, 2006

### masudr

Patrick, garrett,

Thanks for your reply. So that confirms my suspicion that the metric maps a rank (m, n)-tensor to a rank (m-1, n+1)-tensor (or the other way around), and also maps two vectors to a scalar.

So from what I gather, one can still define covectors without a metric, but the structure is not as rich?

5. Jul 10, 2006

### George Jones

Staff Emeritus
To see how this all works out, I think that a definition is an appropriate place to start.

What is a covector?

6. Jul 10, 2006

### masudr

A map from vectors to scalars?

7. Jul 10, 2006

### George Jones

Staff Emeritus
Right, with linear included.

Doesn't this make

$$\partial_\mu dx^\nu = \delta^\nu_\mu$$

seem a little strange notionally as a definition of covectors?

A map is defined by pinning down how it acts on its domain, so here, we need to say how the $dx^\nu$ act on vectors, and the usual notation writes the mapping first followed by the argument of the mapping.

I'm picking nits, and what you have written in 1. will turn out to be OK, but I prefer starting from the beginning.

8. Jul 10, 2006

### mathwonk

duality is inherently a bit confusing, but inescapable.

if V is a vector space, then the set V* of linear functions from V to the real numbers R is the dual space, and these are called in thats ettinhg covectors. but then the set V** of linear fuinctions from V* to R is dual to the first space. so do we call them co-covectors?

actually there is a pairing VxV*-->R, and in this sense V and V* are symmetrical. i.e. either one can be regarded as vectors and the other as dual to those.

From this point of view, then V** = V quite naturally (in finite dimensions), since both are paired with V.

as an example of a vector acting dually on something else, consider F = the space of all smooth functions defined near a point p, and let v be a vector with foot at p. then the operation of sending a function f to its directional derivative at p in the direction of v, is a function from functions to numbers, i.e. the vectors are dual to functions.

but this operation only depends on the value of the function near p. and if f is locally a product of two functions both vanishing at p, then all the derivatives at p are zero. moreover if two functions differ by a constant then all their derivatives are the same.

so the space of vectors at p is really dual to the space of functions defined near p, and vanishing at p, but where two functions are considered equivalent if their taylor series have the same linear terms.

thus in this sense vectors are dual to germs of functions. thus since a space really has only one dual space, these germs of functions must in some sense be the space of covectors. i.e. anything dual to vectors must somehow be the covectors.

that is why the germ of a function at p, regarded as a covector, i.e. as dfp, is a covector and not a vector.

it goes on and on in many guises but the idea is always the same. to understand variance or covariance (unfortunately the prefix co here is historically used for the wrong one of the two, i.e. for the vectors and not the co vectors) just ask yourself which objects act on which others?

Anyhting that acts on vectors is essentially a covector and anything that acts on functions is essentially a vector.

E.g. curves through p are representatives of vectors (take the tangent or velocity vector at p of the curve to se how they act on functions) and we agree then to call functions defined near p representative of covectors (they pair dually with vectors or curves by differentiating the function in the direction of the curve.)

its just a game when you understand the abstract principle. otherwise it is all symbol pushing (which i refrained from calling a pejorative adjective, only with difficulty).:tongue2:

Last edited: Jul 10, 2006
9. Jul 10, 2006

### masudr

Fair enough, but can I not also say that a vector maps a covector to a scalar? In which case what I wrote is the usual way?

10. Jul 10, 2006

### mathwonk

now in coordinates some people (i.e. worshippers of 19th century gods) write coordinates for covectors down, and coordinates for vectors up, or vice versa, i mean who cares? forgive me, of course you care, but i have trouble remembering which it is, because there is no reason for either choice. (except that einstein did it a certain way of course).

and it does enable one to evaluate a pairing mindlessly by "contracting" indices which are written oppositely. i.e. if the indices are opposite then one object acts on the other, so together they yield a number. but isn't it nice to know why you can contract tensors of opposite variance?

11. Jul 10, 2006

### mathwonk

now would someone do me a real favor and explain why some natural physical concept like stress, or action, or gravity, anything really, has a natural representation in terms of several naturally paired quantities, and why the pairing between them is physically meaningful?

i.e. in the context of an old science fiction story, theoretically i know how to fix your air conditioner, but I have never seen an actual functioning air conditioner, and it is hot out here.

12. Jul 10, 2006

### George Jones

Staff Emeritus
What I'm going to write is similar to part of what mathwonk wrote in a more general way.

If you have a vector space V, than the dual space V* is defined to the set of linear maps from V to the space of scalars. If scalar multiplication and "vector" addition are defined appropriately, then V* is a vector space, and its elements are sometimes called covectors.

Since V* is itself a vector space, the dual of V*, denoted by V**, can be taken. So, the elements of V** are linear mappings from the vector space V* to the scalars. For finite-dimensional vector spaces, there is a natural (basis-independent) isomorphism between V and V**. Since the elements of V** act on element of V*, this isomorphism allows the elements of V to act on the elements of V*.

It is in this sense that a vector can be considered a mapping from the space of covectors to the scalars.

I agree very much with what mathwonk wrote - It's quite important to get a handle on the abstract ideas involved, becuase this makes it more than just "symbol pushing."

13. Jul 10, 2006

### mathwonk

masudr, yes i think i said precisely that a vector maps a covector to a number. that does not make it a covector however but a co-covector, ie a vector.

or as george said, an element of (V*)*, i.,e. a function on covectors, is an element of V, i.e. a vector.

14. Jul 10, 2006

### mathwonk

my apologies masudr for "answering" QUESTIONS I have not read. i was just saying thigns, that are correct but possibly unrelated to your questions. my excuse is i hate to reqad indicial notation. but maybe i will learn something so here goes:

youe first post:

"We have a co-ordinate basis for vectors I can think of two ways to get a corresponding basis for covectors.

1. Define basis covectors such that

This gives me a problem. Am I allowed to combine vectors and covectors without a metric? This leads me onto the second way...

2. Use the metric so that

but actually... the metric maps from two vectors to a scalar, am I correct? So perhaps what I mean is

which is in fact the same way as "method 1".

I guess my real question is can I define covectors without a metric? This is related to another thread in the forum, but I could do with a quick answer saying if my approach is correct."

ok the first part is correct but not the second.

i.e. you cannot feed covectors into a metric.

so a covector is a linear function on vectors. hence you can define it by giving itsvalues on a basis of vectors. so if d/dxi is a basis vector, then you can define a covector dxj by saying its value on d/dxi is delta(ij).

i.e. that dxj/dxi = delta (ij).

this does not need a metric. i.e. without a metric we have used a basis of vectors to define a basis of covectors. but notice that the whole basis of aectors was needed tod efine evn one of the covectors, since that covector had its values specified separately on each vector.

but now if you want to use a metric somehow, then you can actually do more. i.e. you can define a covector using only one vector. you do not need the whole basis at all.

i.e. suppose v is any vector, then we can dfinea covector from it dv as follows: the value of dv on d/dxi equals the value of the metric

on the pair of vectors <v,d/dxi>.

in particular dxj has value on d/dxi equal to the metric at <d/dxj,d/dxi>.

so we get OOOPS i see now that what you said afterwrads already pointed this out and corrected it yourself! great! that means both 1 and 2 are correct.

now lets progres to alter posts, which are all probably moot by now.

15. Jul 10, 2006

### mathwonk

yeah thats all there is there. sorry for belaboring something you had essentially right to begin with. it always helps to listen to a question before answering it.

16. Jul 10, 2006

### mathwonk

oh yes,

"I guess my real question is can I define covectors without a metric?"

yes.

but it depends what you mean by this.

i.e. we can define what a covector means. and we can take abasis of vectors and use that to define a basis of covectors, all withut a metric, but we cannot take a single vector and use it to define a single covector, without a metric.

17. Jul 10, 2006

### garrett

I like it because it shows off the linearity -- like
$$\left( a^\mu \partial_\mu \right) dx^\nu = a^\nu$$
which isn't obvious when you write the same thing this way, a form as an operator on vectors:
$$dx^\nu ( a )$$

Also, it suggests one consider the vectors and forms as elements of an algebra.

18. Jul 10, 2006

### mathwonk

since either argument can be viewed here as acting on the other, it is impossible to ever sort out which should be which.

i.e. if f is a linear fucntion and v isa vector argument, then f(v) also equals v(f) when v is viewed as an element of V**, acting on functions.

19. Jul 10, 2006

### masudr

good; excellent. many thanks to all.

20. Jul 10, 2006

### mathwonk

another point of view is a row vector dotted with (acting on) a column vector.

if you take the tramnspose of the whole shebang, th column vector becoems the row vector and vice versa, and now the other one is acting on the first one.

but the vaue is the same.

of cousre now because it is just symbols it has lost all emaning.