Covariant and Contravariant: What Are the Differences in Differential Geometry?

[StatusX]

gvk, thanks for your replies. I'm only just beginning, so I don't understand all of what you're saying (quadratic forms, for example). But what you said about imaginary numbers seems to make sense. The only thing, though, is that the source I'm reading doesn't mention imaginary numbers. What it does do is define the first christoffel symbol in these two, allegedly equivalent ways:

[p q, r] = \frac{1}{2}\left(g_{qr,p} + g_{rp,q} - g_{pq,r}\right)

[p q, r] = y_{s,pq} \ y_{s,r}

where, eg.,
y_{s,r}=\frac{\partial y_s}{\partial x_r}

Is that second definition always valid?

I guess my central misunderstanding is about what the ambient space is when the metric isn't positive definite. Is it still euclidean space? Does it have this strange metric as well? Or is the metric just a property of a specific manifold?

 

[mathwonk]

gentlemen,

i have made this subject as clear as I can ever do.

best wishes,

mathwonk. but mathmonk might work too.

 

[gvk]

StatusX,
Thank you for these questions!
You just got to the central point of the Differential Geometry.
I guess you are familiar with basics of field theory and quantum mechanics.
If you do, the notion of 'the space without metric' does not shock you. If you don't you need to read some books of this subject to feel more comfortable yourself.
Yes, the second formula of Christoffel symbol is valid for any space (with any metric or without any metric). Why? Because it can be received from
simple requirement that result of differentiation of the vector or covector should be a tensor. It is not normal partial differentiation, it's called 'covariant differentiation'.
It turned out that the covariant differentiation has deep and profound meaning. First discovered by Ricci in 90th of 19 century but only in 30th with the development of quantum mechanics it was understood that this notion is deeper than the metric properties. And this can be the starting point of new approach to diff.geometry which many people try to pursue. (See e.g. the posts of mathmonk)

However, I'll try to explain this notion in simple words. Let we have a vector or covector in one point of space. Assume the space has some (e.g. curvilinear) coordinates. The purpose of Ricci was to receive the vector or covector in another point which is close to the origin one. Oh, yes, for this we need to find differential. But the normal differentiation does not work because the coordinate lines are curved and normal partial derivitive destroys your vector, it won't be the vector anymore (you can check it youself). To create the vector in the neighboring point Ricci turned slightly the vector (or covector) toward the curved line in order the angle between vector and coordinate line be the same. This new vector or covector becomes the real vector or covector. They satisfy the transformation properties. This procedure is called 'parallel transport'. Of course, in eucleadian space where the coordinate systems are stright lines, the covariant differentiation is exactly the same as normal differentiation. The difference between the normal differential and 'covariant differentiation' is defined by Christoffel symbols.
Because 'covariant differentiation' was invented to connect vector in one point with vector in another point, it's called sometimes 'connection'.
I emphasize that the connection does not require any metric properties, it can be received from the following relations:
A) the operation of covariant differentiation is linear,
B) the result of covariant differentiation of the tensor forms a tensor again.

 

[Peterdevis]

Because 'covariant differentiation' was invented to connect vector in one point with vector in another point, it's called sometimes 'connection'.
I emphasize that the connection does not require any metric properties, it can be received from the following relations:
A) the operation of covariant differentiation is linear,
B) the result of covariant differentiation of the tensor forms a tensor again.

This is correct. So you can define a large number of connections on a manifold and each of them implies a distinct notion of covariant differentiation.
When there is a metric you can define one unique connection that is torsion free and metric compatibele:

[p q, r] = \frac{1}{2}\left(g_{qr,p} + g_{rp,q} - g_{pq,r}\right)



Thats the one used in GR

 

[saski]

Hiya StatusX et all

The two expressions for [pq,r] are equivalent, and this is independent of the metric signature.

I had a little trouble interpreting the y-coordinate notation, but I think it makes sense. Assuming so, the formal proof goes as follows.

We start with a set of basis vectors e_i, having components in the global coordinate system:

{e_i}_s = \partial y_s / \partial x_i = y_{s,i}

so that

g_{ij} = {e_i}_s \bullet {e_j}_s

where we using the summation convection over s. Now if we are simply summing over s, that means we're assuming a metric signatire that's all 1's. But we can easily change that assumption later.

So we have

g_{pq} = y_{s,p} y_{s,q}

We can then use the product rule to write:

g_{pq,r} = y_{s,pr} y_{s,q} + y_{s,p} y_{s,qr}

g_{qr,p} = y_{s,qp} y_{s,r} + y_{s,q} y_{s,rp}

g_{rp,q} = y_{s,rq} y_{s,p} + y_{s,r} y_{s,pq}

For derivatives with respect to coordinates, the second derivatives commute: {...}_{,ij} = {...}_{,ji} .

So we rewrite the above as:


g_{pq,r} = y_{s,pr} y_{s,q} + y_{s,p} y_{s,qr}

g_{qr,p} = y_{s,pq} y_{s,r} + y_{s,q} y_{s,pr}

g_{rp,q} = y_{s,qr} y_{s,p} + y_{s,r} y_{s,pq}

Notice how the first term in the first equation matches the second term in the second, and likewise the second in the first matches the first in the third, and the first in the second matches the second in the third.

If we define

[pq,r] = (1/2) { g_{qr,p} + g_{rp,q} - g_{pq,r}}

then terms cancel so that we get:

[pq,r] = y_{s,pq} y_{s,r}

as required.

This derivation works because the original expression for g_{pq} is symmetric in p and q. If we were treating a non-symmetric bilinear function

h_{pq} = A_{st} y_{s,p} y_{t,q}

with a constant A, we wouldn't be able to match terms like that.

Now with a metric whose signature is not {1,1,...}, but say {-1,1,...}, we would have to write it as

g_{pq} = \eta^{st} y_{s,p} y_{t,q}

where \eta^{11} = -1, \eta^{22} = 1, etc, and \eta^{st} = 0 where s != t.

But we can thread the \eta term through the derivation, making each term

\eta^{st} y_{s,-} y_{t,-}

and still match the terms, because \eta is symmetric in s and t.

So, to express [pq,r] for metrics of indefinite signature, one should really write

[pq,r] = (1/2) { g_{qr,p} + g_{rp,q} - g_{pq,r}} = \eta^{st} y_{s,pq} y_{t,r}

 

[StatusX]

Now with a metric whose signature is not {1,1,...}, but say {-1,1,...}, we would have to write it as

g_{pq} = \eta^{st} y_{s,p} y_{t,q}

where \eta^{11} = -1, \eta^{22} = 1, etc, and \eta^{st} = 0 where s != t.

But we can thread the \eta term through the derivation, making each term

\eta^{st} y_{s,-} y_{t,-}

and still match the terms, because \eta is symmetric in s and t.

So, to express [pq,r] for metrics of indefinite signature, one should really write

[pq,r] = (1/2) { g_{qr,p} + g_{rp,q} - g_{pq,r}} = \eta^{st} y_{s,pq} y_{t,r}

Thank you saski, this is what I've been looking for all along. the eta term was never mentioned in my source, that was the source of my confusion.

 

[gvk]

Hiya StatusX et all

The two expressions for [pq,r] are equivalent, and this is independent of the metric signature.

....

Now with a metric whose signature is not {1,1,...}, but say {-1,1,...}, we would have to write it as

g_{pq} = \eta^{st} y_{s,p} y_{t,q}

where \eta^{11} = -1, \eta^{22} = 1, etc, and \eta^{st} = 0 where s != t.

But we can thread the \eta term through the derivation, making each term

\eta^{st} y_{s,-} y_{t,-}

and still match the terms, because \eta is symmetric in s and t.

So, to express [pq,r] for metrics of indefinite signature, one should really write

[pq,r] = (1/2) { g_{qr,p} + g_{rp,q} - g_{pq,r}} = \eta^{st} y_{s,pq} y_{t,r}

Saski,
No, two expressions are not equivalent. First is dependent on the metric. It is 'specific'. The second is not. It is valid for any connection, even without any metric. They will be equivalent if you take the 'specific' connection which called COMPATIBLE with metric (see the post of Peterdevis).
To be compatible with metric the connection \triangledown (covarient differentiation) should satisfy the condition:

\triangledown_r g_{pq} =0.

The first expression

[pq,r] = (1/2) ( g_{qr,p} + g_{rp,q} - g_{pq,r})

is compatible with metric, and include the signature inside the notation g_{pq}. The metric can be pseudo-Riemannian (not definite), so you don't need to modify the formula to calculate the compatible connection.

 

[saski]

Thanks gvk, I'm alerted. I'll need some time to think this over, before I post back. It's a very important issue.

 

[StatusX]

So let me see if I have this straight. It is always possible to define a christoffel symbol like this:

[pq,r] = y_{s,pq} y_{s,r}

Even if you don't have a metric. But if you do have one, you can alternatively use this:

[pq,r] = \frac{1}{2} (g_{qr,p} + g_{rp,q} - g_{pq,r})

or equivalently:

[pq,r] = \eta^{st} y_{s,pq} y_{t,r}

In GR, they use the second definition. Is this right?

 

[gvk]

StatusX,
What kind of source (book) do you follow in your study of DiffGeom? You don't need the 3rd formula, which contains some incorrection. The second one is perfectly good for caclulation of connection's coefficients in space with indefinite metric. It is OK for GR.
If you don't have some restictions in using the source I can recommend you some good book (of course, from mine point of view).

 

[gvk]

StatusX, Saski
Add one thing about general connection which seems to be important.
If you want to receive the connection from the relations (not involving a metric):
you will get acctualy the Christoffel symbol of the second kind:

\Gamma_{ij}^{k} = \frac{\partial^2 y^m}{\partial x^i \partial x^j} \frac{\partial x^k}{\partial y^m}

You wrote the expressions for the Christoffel symbol of the first kind

\Gamma_{ij},_{k}.

The well known relation between them (rising lowing indices) exists only for connection compatible with a metric. So, you need to correct the first formula too.

 

[StatusX]

OK, I think I understand why I was confused. Either a metric or an ambient space are specified, not both. For example, you can't have a manifold embedded in euclidean space with a minkowski metric. For a general manifold which isn't defined in terms of an ambient space, the metric presents a way of "pretending" there is one. Do I finally have this right?

If so, then does the definition of the christoffel symbol involving y's (ambient coordinates) apply in non-euclidean ambient spaces? Or do you simply use the special metric and not worry about ambient coordinates, just acknowldeging the ambient space isn't euclidean. I apologize if this was so painfully obvious you assumed I realized it, buy my source (which is http://people.hofstra.edu/faculty/Stefan_Waner/RealWorld/pdfs/DiffGeom.pdf by the way, and I'm using it because it's free and I can't afford to buy a book right now) was very confusing in this respect. Thanks a lot for your help so far.

 

[gvk]

OK, I think I understand why I was confused. Either a metric or an ambient space are specified, not both.
For example, you can't have a manifold embedded in euclidean space with a minkowski metric. For a general manifold which isn't defined in terms of an ambient space, the metric presents a way of "pretending" there is one. Do I finally have this right?
If so, then does the definition of the christoffel symbol involving y's (ambient coordinates) apply in non-euclidean ambient spaces? Or do you simply use the special metric and not worry about ambient coordinates, just acknowldeging the ambient space isn't euclidean. I apologize if this was so painfully obvious you assumed I realized it, buy my source (which is http://people.hofstra.edu/faculty/Stefan_Waner/RealWorld/pdfs/DiffGeom.pdf by the way, and I'm using it because it's free and I can't afford to buy a book right now) was very confusing in this respect. Thanks a lot for your help so far.

Yes, you are almost right, exept for the case that metric is specified in addition to the ambient space (coordinate system).
I wrote the Christoffel symbol which depend only on transformation rules from y's to x's and does not involve any metric. Of course, it's far away from 'perfect' definition, but it can give you a sense that Chistoffel symbols and, hence, connection can be defined without metric. The transformation's rule of Christoffel symbol may serve as a general definition for this notion itself:

\Gamma_{ij}^{k} = \frac{\partial x^k}{\partial y^m} ( \Gamma_{ls}^{m} \frac{\partial y^l}{\partial x^i} \frac{\partial y^s}{\partial x^j}+ \frac{\partial^2 y^m}{\partial x^i \partial x^j})

It is the same kind as the definition of vector or covector through the transformation rules. And here the same relation to the controvesial 'visualization' problem. Here, the metric is the only condition to visualize the connection as a 'parallel transport'.
I think, your source gives you the level of knowledge up to 20-th. In my opinion, the best book of such type is the 'classical' book Levi-Chivita, Absolute Differential Calculus. Blackie&Son Limited London Glasgo 1927. It include all detail about metric Riemmannian and Minkowski space and GR and
written in clear 'nonformal' language.
I would also recommend the book Dubrovin, Novikov, Fomenko,
Modern Geometry-Methods and Applications, Part I,II,III (Universitext)
Springer-Verlag (1990). The part I would be enough for first reading. It is written in 'nonformal' language for physicists. All books you can borrow from a library.

 

[makris]

Even though I am posting this quite late, I have decided to post it hopping that I will save some people's time.
I have found the most lucid explanation on the covariant - contravariant issue in the book
"Tensor Analysis and Continuum Mechanics" by Wilhelm Flugge. (pages 2-7).

The explanation there assumes only the dot product of two vectors and the work (Force* distance) as prerequisites. No knowledge about manifolds or relativity is needed. If you have access to the book (via a library probably...) it will help you a lot.

 

[gvk]

Makris,
Please explain what did you learn from "Tensor Analysis and Continuum Mechanics" by Wilhelm Flugge about covariant vs. contravariant.

 

[makris]

This is a reply to the message by gvk.

Sorry I am replying this late...
The book by Wilhelm Flugge gives an introduction to tensors assuming minimum prerequisites.
It explains very well the following (pages 2-8 or a little bit more):

1.)Covariant and contravariant base vectors.
2.)Covariant and contravariant components of a vector and a second order tensor.
3.) Metric tensor.

I tried myself to read several book on tensors but there were too mathematical for me. I found the most lucid explanation on the book I mentioned. I undestood the consepts immediately despite the fact that I am neither mathematician nor physisist. It also helped me understand more advance books on the subject. From this point of view I suggested reading this part of the book. The book contains additional topics like Christoffels symbols etc but in subsequent chapters. However, I have not compared these topics with simillar ones written in other books to give my opinion.

Makris

 

[StatusX]

Just wondering, is the covariant/contravariant distinction the same as the one between bra and ket vectors in quantum mechanics?

 

[jcsd]

Yes, exactly the same distinction.

 

[Haelfix]

I always remember the distinction between the two b/c they are exactly the opposite of the logical way of assigning them.

 

[mathwonk]

Here is what i think haelfix means: if F(X) is a functor, then F is "covariant" if for any map f:X-->Y we get a map f*:F(X)-->F(Y).

On the other hand F is "contravariant" if for any map f:X-->Y we get a map f*:F(Y)-->F(X).

i.e. covariant means "same direction", and contravariant means "opposite direction".


Thus tangent spaces are covariant, since if f:X-->Y, then for each p in X we get

dfp:Tp(X)-->Tf(p)Y.

On the other hand the dual of a tangent space, i.e. the cotangent space T*(X), is contravariant, since when f:X-->Y, then we get dfp*:T(Y)*-->T(X)*.


Unfortunately this is the opposite convention from that adopted for some historical reason in diifferential geometry, and hence in physics. I.e. those subjects use the words backwards.

Does this help anyone? Sadly to me, many of the books recommended here discuss these terms purely with regard to whether the indices used to describe them are "up" or "down". Anyone wishing to become expert is advised to try to get beyond that type of mechanical understanding of the topic.

The intelligent student will not be misled by discussions which seem easier because they omit explanation of the concepts involved and discuss only the notation. Remember, one can go from an understanding of the concepts to a grasp of the notation, but not the other way.

 

[ObsessiveMathsFreak]

Thread Necromancy!

I'd like to bump this old thread to reopen discussion on the idea of covariance and contravariance.

To begin on the wrong foot, I think that covariance and contravariance are in fact red herrings in the study of tensors, but before I go any further, I'd like to explain myself through equations. A word of warning first. I won't be using superscripts and subscripts to denote contravariant or covariant components. In fact, I'm going to avoid using the terms contravariant and covariant at all, for reasons which I hope will become clearer later

I'm going to talk about rank 1 tensors first, i.e. vectors. I'll distinguish the two types of vector by use of lower case for the first type and uppercase for the second. So the basis vectors, and components for the first type will be denoted by \mathbf{e}_u^i and a_u^i, and for the second type by \mathbf{E}_u^i and A_u^i. Here the subscript "u" denotes the coordinate system, and the superscript i is denoting the index.

Any vector \mathbf{w} can be represented by either coordinate system. Using the Einstein summation convention;

\mathbf{w} = a_u^i \mathbf{e}_u^i = A_u^i \mathbf{E}_u^i.

OK so far these two vector types seem to simply be two different basis for the vector space. Now we distinguish them by how their representations change under a change of coordinates.

If we make a change of coordinates from the system "u" to the system "v", We must change both the basis vectors and the components to obtain the representations of \mathbf{w} in the new coordinate system.

The first type of vector transforms in the following way.

\mathbf{e}_v^i = \frac{\partial u^j}{\partial v^i}\mathbf{e}_u^j
a_v^i = \frac{\partial v^i}{\partial u^j}a_u^j

And the second type of vector transform in, in some sense, an opposite way.
\mathbf{E}_v^i=\frac{\partial v^i}{\partial u^j} \mathbf{E}_u^j
A_v^i = \frac{\partial u^j}{\partial v^i} A_u^j

As you can see, whichever transformation you decribe as covariant or contravariant, for each type of vector, its components transform one way and its basis vectors transform the other way. This is I think a big part of the confusion between the two terms, and I think, the primary reason for their inappropriateness as decriptive terms. Depending on your point of view, one type is contravariant and the other covariant, but this depends on whether you are speaking from a component viewpoint or a basis vector viewpoint.

More later.

 

[ObsessiveMathsFreak]

At this time, I'd like to drop the terms contravariant and covariant, and instead simply concentrate on these two types of transforming vectors. I would like, if I may, to denote the lower case type as a tangent vector, and, if I may be so bold, denote the upper case type as a cotangent vector. I'm not entirely sure about the appropriateness of the nomenclature here, but I will press on.

So represented by tangent vectors
\mathbf{w} = a_u^i \mathbf{e}_u^i.
And by cotangent vectors
\mathbf{w} = A_u^i \mathbf{E}_u^i.

So, if we make a change of cooridinate system from the system \mathbf{u}=(u^1,u^2,\ldots,\u^n) to the system \mathbf{v}=(v^1,v^2,\ldots,\v^n), both the basis vectors and components of the tangent and cotangent vectors change.

Lets denote the Jacobian matrix for the transformation from u to v by J with
J_{ij} = \frac{\partial v^i}{\partial u^j}
And the inverse Jacobian matrix J^{-1} is then given by
J_{ij} = \frac{\partial u^i}{\partial v^j}

To keep things from becoming totally abstract (this may already be a lost cause), I'd like to deal specifically with 3d vectors in 3d space. The following arguments do extend to higher, and lower, dimensions.

First, consider the tangent basis vectors\mathbf{e}_u^1 , \mathbf{e}_u^2, \mathbf{e}_u^3. These are vectors in their own right, which have a representation in carteasian coordinates. For the sake of argument, let's represent the carteasian coordinates basis vectors in carteasian space as column vectors. For example
\mathbf{e}_u^1=\left[\begin{array}{c}a\\b\\c\end{array}\right], or something.

Using this column vector representation, we can create a 3x3 matrix out of the tangent basis vectors.

e_u = \left[\begin{array}{ccc}<br /> \mathbf{e}_u^1 &amp; \mathbf{e}_u^2 &amp; \mathbf{e}_u^3 \end{array}\right]

Using this shorthand, we can express the tangent basis vector transformation rules as


\left[\begin{array}{ccc}<br /> \mathbf{e}_v^1 &amp; \mathbf{e}_v^2 &amp; \mathbf{e}_v^3 \end{array}\right] = \left[\begin{array}{ccc}<br /> \mathbf{e}_u^1 &amp; \mathbf{e}_u^2 &amp; \mathbf{e}_u^3 \end{array}\right] J^{-1}

Or if you like

e_v = e_u J^{-1}

Also, let's express the components of the tangent vector a_u^1\ ,\ a_u^2\ ,\ a_u^3 as a column vector
\mathbf{a_u}=\left[\begin{array}{c}a_u^1\\a_u^2\\a_u^3\end{array}\right]

Using this notation, the transformation rule for the tangent vector components can be expressed as
\left[\begin{array}{c}a_v^1\\a_v^2\\a_v^3\end{array}\right]= J \left[\begin{array}{c}a_u^1\\a_u^2\\a_u^3\end{array}\right]

Or if you like

\mathbf{a_v} = J \mathbf{a_u}

Next, I'd like to do something similar for the cotangent vectors, except this time I'll do things with row vectors, instead of column vectors.

So let's represent the carteasian coordinates of cotangent basis vectors as a row vector, For example:
\mathbf{E}_u^1=\left[\begin{array}{ccc} a &amp; b &amp; c \end{array}\right], or something.

Again, form a matrix, but this time the basis vectors are rows, not columns.

E_u = \left[\begin{array}{c}\mathbf{E}_u^1\\\mathbf{E}_u^2\\\mathbf{E}_u^3\end{array}\right]

So in this way we can represent the transformation rule for cotangent basis vectors as
\left[\begin{array}{c}\mathbf{E}_v^1\\\mathbf{E}_v^2\\\mathbf{E}_v^3\end{array}\right]=J \left[\begin{array}{c}\mathbf{E}_u^1\\\mathbf{E}_u^2\\\mathbf{E}_u^3\end{array}\right]

Or if you like

E_v = J E_u

And finally, let's express the components of the cotangent vector A_u^1\ ,\ A_u^2\ ,\ A_u^3 as a row vector
\mathbf{A_u} = \left[\begin{array}{c}A_u^1 \\ A_u^2 \\ A_u^3 \end{array}\right]

And so the transformation rule for cotangent components can be expressed as
\left[\begin{array}{c}A_v^1 \\ A_v^2 \\ A_v^3 \end{array}\right] = \left[\begin{array}{c}A_u^1 \\ A_u^2 \\ A_u^3 \end{array}\right] J^{-1}
Or if you like
\mathbf{A_v} = \mathbf{A_v} J^{-1}

So summing up, we have the tangent vector components as columns, and the cotangent vector components as rows, with the tangent and cotangent basis matrices, and the four transformations as follows,

\left[\begin{array}{ccc}<br /> \mathbf{e}_v^1 &amp; \mathbf{e}_v^2 &amp; \mathbf{e}_v^3 \end{array}\right] = \left[\begin{array}{ccc}<br /> \mathbf{e}_u^1 &amp; \mathbf{e}_u^2 &amp; \mathbf{e}_u^3 \end{array}\right] J^{-1}

\left[\begin{array}{c}a_v^1\\a_v^2\\a_v^3\end{array}\right]= J \left[\begin{array}{c}a_u^1\\a_u^2\\a_u^3\end{array}\right]

\left[\begin{array}{c}\mathbf{E}_v^1\\\mathbf{E}_v^2\\\mathbf{E}_v^3\end{array}\right]=J \left[\begin{array}{c}\mathbf{E}_u^1\\\mathbf{E}_u^2\\\mathbf{E}_u^3\end{array}\right]

\left[\begin{array}{c}A_v^1 \\ A_v^2 \\ A_v^3 \end{array}\right] = \left[\begin{array}{c}A_u^1 \\ A_u^2 \\ A_u^3 \end{array}\right] J^{-1}

More later, weh I get to my main point.

 

[ObsessiveMathsFreak]

Just looking at those four transformations again, in shorthand.

e_v = e_u J^{-1}
\mathbf{a_v} = J \mathbf{a_u}

E_v = J E_u
\mathbf{A_v} = \mathbf{A_v} J^{-1}

Now, I don't know about you, But I'm very tempted to refer to the transformation involving the jacobian as covariant, and those involving the inverse jacobian as contravariant. Perhaps I'm backwards here, but my point is that these transformations are opposite. With good reason too, as now for a tangent vector

\mathbf{w} = e_v \mathbf{a_v} = e_u J^{-1}J \mathbf{a_u} = e_u \mathbf{a_u}
and for a cotangent vector
\mathbf{w} = \mathbf{A_v} E_v = \mathbf{A_v} J^{-1}J E_u = \mathbf{A_u} E_u

So you can see the reason for when if the components transform either covariantly or contravariantly, the basis vectors must transform oppositely, so that those jacobians cancel.

My main point is, I think that reffering to vectors as being either tangent or cotangent is a lot more appropriate as referring to them as being contravariant or covariant. Contravariant and covariant depend on whether you are talking about the basis vectors, or the components. Tangent and cotangent refer to the vector as a whole object. I'm not entirely sure how this idea would extend to higher ranked tensors, but for vectors at least, I find thinking this way to be far less confusing than thinking of contra and covariance.

The question is obviously raised on what exactly the tangent and cotangent vectors are. Well, my understanding of them comes from the basis vectors. The tangent basis vectors are just the regular tangent space basis for a manifold. i.e.

\mathbf{e}_u^i \equiv \frac{\partial \mathbf{x}}{\partial u^i} \equiv \frac{\partial }{\partial u^i}
Where \mathbf{x}=\mathbf{x}(u^1,u^2,\ldots,u^n) is a point in cartesian space.

The cotangent basis vectors, to my understanding, are in fact the gradients of the coordinates in cartesian space.

\mathbf{E}_u^i \equiv \nabla u^i \equiv grad(u^i)
Where u^i=u^i(\mathbf{x}) is the inverse function of the coordinate mapping into cartesian space..

In other words, \mathbf{E}_u^i is normal to the level surfaces of u^i in cartesiann space. I think. I wish I had a good diagram here.

What are people's thoughts on all this?

 

[Chris Hillman]

At this time, I'd like to drop the terms contravariant and covariant, and instead simply concentrate on these two types of transforming vectors. I would like, if I may, to denote the lower case type as a tangent vector, and, if I may be so bold, denote the upper case type as a cotangent vector. I'm not entirely sure about the appropriateness of thttps://www.physicsforums.com/editpost.php?do=editpost&p=1205732he nomenclature here, but I will press on.

The standard terms you seek would be a vector field and its dual covector field.

A covector is a (simple) one-form, i.e. a real-valued function on vector fields. Given one vector field on a smooth manifold, you can extend this to a basis of vector fields on the manifold, and then the dual one-form takes value unity on the original vector field and zero on all the others.

To be precise, you are discussing a coordinate vector field \partial_w and its dual covector dw, which is the exterior derivative of the coordinate w (that is, a monotonic function on our manifold). Since the exterior derivative of a function is dual to a vector field called the gradient of the function in ordinary vector calculus, we can say this: a coordinate is just a monotonic function; it is associated with a unique vector field (the gradient) and a unique covector field (the dual of the gradient). Just as you said!

To make a coordinate system on some open neighbhood N of a d-manifold, you choose d coordinates on N, such that the corresponding gradients are never parallel. The coordinate vector fields are then also called a holonomic basis, meaning that their Lie brackets always vanish. Then, the Riemann curvature tensor is obtained immediately as the difference between iterated covariant derivatives performed in either order. Also, in this case, given an arbitrary vector field \vec{X}, by applying the various coordinate covector fields we can pick off the "components" of \vec{X} with respect to the given coordinate basis. If we have a frame field (an orthonormal set of vector fields) expressed in terms of our coordinate basis, we can then convert these to the components wrt the frame field. These are sometimes called the "physical components" since in specific scenarios they correspond, in principle, to measurable quantities.

You might be interested in some very recent posts by myself in other threads (past week) in which I mentioned the Coll-Morales classification of coordinate charts on Lorentzian manifolds, and a very recent post by myself briefly discussing various types of derivatives which appear in differential geometry.

 

[mathwonk]

obsessmath, your post, although correct in detail, isquite consistent with the usual terminology. i.e. coordinates of vectors ARE dual vectors. since they are scalar valued linear functions on vectors.

that is why the coordinates transform contravariantly, and the basis vectors themselves transform covariantly. i.e. a choice of (covariant) basis vectors at each point also determines a dual choice of (contravariant) basic covectors at each point, namely the linear coordinate functions for that basis.

 

[mathwonk]

you have put your finger on the fundamental confusion for many people, i.e. they confuse vectors with coordinates of vectors. the coordinates are functions on vectors, not vectors themselves.

it is hard to appreciate this when we have been told, misleadingly, all our life that a vector is an n - tuple of numbers. you seem to be doing this yourself however.

think about it physically, swing a rock around a string and let go, in one second its two positions determine an arrow of velocity, not a sequence of numbers. the numbers are assigned to the arrow as a vehicle of measurement, hence are functions on the arrow.

i.e. the coordinates are dual vectors, or covectors.

 

[mathwonk]

here is one waY TO SE THAT COORDINATES ArE NOT a natural property of vectors. choose just one vector at a point. what are its coordinates? they are not determiend. you need to choose a whole basis of vectors before you get coordinates for even one vector.

a coordinate system determines a basis fo vectors at each point nd also a bsis of coordinate functions, but a coordinate system is something different from a vector. i.e. coordinates are not just another way to view vectors.let me say you are very intelligent to notice this duality. but it is still confusing.

 

[mathwonk]

look at this example: suppose we are given a single vector at each point, i.e. a vector field. then we do not have a basis, hence no coordinates, and cannot take the coordinate point of view, without artificially introducing coordinates.

we wish to understand thigns as intrinsically as possible, hence must use exclusively the covariant point of view here.

simialrly if we have a single one form given naturally, we have only one covector at each point, hence again no dual basis, hence must intrinsically study it by the contravariant point of view.

coordinates are artficial and obscure the diference betwen covariance and contravariance. the intrinsic properties are the phenomena with real physical meaning, hence should be kept distinct for best understanding.

 

[ObsessiveMathsFreak]

Let me see if I understand. You are saying that a coordinate system, say in 3d, defines not only basis tangent vectors \mathbf{e}_1 ,\mathbf{e}_2,\mathbf{e}_3, but basis cotangent vectors or one-forms \acute{E}_1,\acute{E}_2,\acute{E}_3, and that the representation of a vector in that coordinate system is
\mathbf{v} = \acute{E}_1(\mathbf{v})\mathbf{e}_1+\acute{E}_2(\mathbf{v})\mathbf{e}_2+\acute{E}_3(\mathbf{v})\mathbf{e}_3

Now, the basis one forms transform with the jacobian under a change of coordinates , \mathbf{a_v} = J \mathbf{a_u}, and the basis tangent vectors transform with the inverse Jacobian , e_v = e_u J^{-1}.

So the basis one forms are covariant, and the basis tangent vectors are contravariant? Is that the right way around? And if so, then all one forms in the cotangent space are covariant and all vectors in the tangent space are contravariant? Is it correct to say that?

I understand that a vector is an object quite independant of its coordinates, or basis, or what have you. Obviously it is something that exists independantly of coordinate systems, or even our ability to create coordinate systems.

But then, what is all this talk about contravariant vectors and covariant vectors? Under a change of coordinates, the vector itself does not change at all. Only its representation changes. So the vector itself is neither contravariant nor covariant? Only its representations are? Or, are "normal" vectors always considered to be tangent vectors that transform contravariantly?

Would it be correct to say that what are being called covariant vectors are actually one forms, and what are being called contravariant vectors are actually (tangent) vectors? If so, there would be no such thing as a covariant vector, there would only be covariant forms?

A lot of questions here, but I think the fog is clearing.

 

[mathwonk]

clearing fog is what we live for. more power to you my friend,