# I Misleading naming in D'Inverno textbook

1. Apr 6, 2017

### RiccardoVen

I know this may sound quite as an old adagio, but I have some problems, or at least some concerns, about some chapter title in D'Inverno GR textbook. I'm referring to the chapter devoted to Tensor Algebra:

1. Contravariant tensors
2. Covariant and mixed tensors
Within them you can find definitions like:

"A contravariant vector or contravariant tensor of rank (order) 1 is a set of quantities, written..."

similarly for the "covariant tensors".

This is really a painful definition to me, since I've learnt the tensors are invariant objects, i.e. are neither covariant nor contravariant. You can of course get the tensors component related to some basis, but this is a different story. I mean, you can take a vector and expressing it as the linear combination of the basis vectors (coordinate basis) through the contravariant component of it. But nothing prevent us to express the same vector as a linear combination of covariant components, referred to some 1-form basis.
All in all, you can always raise or lower indices using the metric, but these are the components, not the tensor per se. I find quite misleading confusing the object itself with its contravariant/covariant/mixed components...

An I right in claiming this? (It's not of course about judging the D'Inverno textbook, which is one of the many GR superb books I have in my collection, but to confirm/confute my understanding about this topic).

Recently I've also looked through Gouge field, knots and gravity by Baez once more (every time I run through it, I discover new ways of reading the chapters, that's incredible to me) and I see it's presentation less confusing, since classifies vectors and 1-forms by they way they are pulled back or push, forward.

Ricky

2. Apr 6, 2017

### Staff: Mentor

It's not a "confusion", it's a particular choice of definition. I don't have d'Inverno's textbook, but many GR texts (and more generally texts on differential geometry) define tensors according to the transformation properties of their components under changes of coordinates. With that type of definition, the distinction between contravariant and covariant components is crucial, since those terms refer to two different kinds of transformation properties. Those transformation properties are actually independent of any particular choice of coordinates, but it's not necessarily easy to see why.

Only if you have a metric. But the general definition of a tensor does not require a metric and must be usable in cases where there isn't one.

Yes, this is a different choice of definition, which focuses on properties that can be defined without reference to any choice of coordinates.

3. Apr 6, 2017

### RiccardoVen

Thanks PeterDonis, you explain your point very effectively...
So basic, we could define the Fμν (0,2)-tensor as it is, without thinking about a specific coordinate system.
If we are lucky to have a metric, we can raise some indices of it, turning into a different tensor.

Anyhow, I still think Baez's approach is more natural, without coordinates...but they are leading to opposite definitions, actually.
since it's stating the opposite: it's saying the tangent vectors are pushed forward (which makes sense, of course)
but they a re covariant in this case, while f()s on tangent vectors, like 1-forms are actually pulled back (which again makes
sense) but are in this case contravariant.

So, if you use the coordinate dependent definition, vector are contravariant, while if you use the coordinate-free approach,
they are covariant, since are pushed forward (and vice versa for 1-forms).

An I right in claiming this?

It seems to me the coordinate approach is more "physical", since it helps to do real calculations, while
the coordinates-free one is more Math oriented (and Baez is really Math oriented, to me).

4. Apr 7, 2017

### vanhees71

I couldn't agree more! Of course, the problem is that this is the usual physicists' jargon, and it's good to talk in this jargon not to be misunderstood.

From a mathematical point of view it doesn't make sense since indeed tensors are "invariant objects", i.e., they do not depend on the choice of any particular basis (in the case of special relativity; sometimes you find definitions for tensors in a restricted sense, e.g., tensors as invariant objects under rotations in an Euclidean vector space in Newtonian mechanics), and that's why we use them in physics in the first place. What's co- and contravariant are (a) the basis and co-basis vectors and (b) the corresponding components of tensors.

If you have a basis $\boldsymbol{b}_j$ and its co-basis (dual basis) $\boldsymbol{b}^j$ these are co- and contravariant with respect to change of these bases. A vector is invariant and written in terms of its components as
$$\boldsymbol{V}=V^j \boldsymbol{b}_j.$$
For a one-form you have
$$\boldsymbol{L}=L_j \boldsymbol{b}^j.$$
The components of a vector thus transform contravariantly and those of one-forms covariantly such that the objects themselves are basis independent.

The analogous holds for higher-rank tensors of any kind. E.g., for a 2nd-rank tensor you have
$$\boldsymbol{T}=T_{jk} \boldsymbol{b}^j \otimes \boldsymbol{b}^k,$$
etc. etc.

5. Apr 7, 2017

### RiccardoVen

Thinking better to it, I got your point about the metric, so it's clear to me this should work even without a metric.

Anyhow, I've still some problem with this:

It's clear to me distinction between contravariant e covariant components, but I'm making confusions with the fact the way in which these components are transforming, imply the property of the whole tensor, since to being able to assign it a "contravariant" attribute (for instance) a s whole.

Let's take a vector: it has both covariant and contravariant components, depending on the basis you choose.
So, as far as I know it could be either contravariant or covariant using this definition, because it depends from which basis you started.

Let's take another example: the electromagnetic field, is usually depicted in SR with Fμν, so we are used to say: this is a rank-2 covariant tensor (i.e. a 2-form).
But if I take the Lorentz Metric, we can turn it into a mixed tensor, for instance. So, in this case I would say "its component are transforming as a mixed (1,1) tensor".
Am I allowed to say "the electromagnetic field is a mixed (1,) tensor"?
I would rather say : "the electromagnetic field is a rank-2 tensor. As a rank-2 tensor, it could be fully covariant, mixed or fully contravariant".

I know I'm probably wrong in my reasoning, since the majority accept this to say "covariant tensor", but I cannot see where is my pitfall.

Anyhow, I found may places in which people is struggling with this defintion:

http://www.physicspages.com/2013/02/13/contravariant-and-covalent-components-dual-basis-vectors/

"What is rarely made clear in discussions of vectors (and more generally, tensors) is that speaking of a ‘contravariant’ or ‘covariant’ vector isn’t particularly accurate, since all vectors have both types of components. So technically, we should be speaking of the contravariant or covariant components of a vector, rather than the vector itself".

Also the book from Fleisch (A Student's Guide to Vectors and Tensors) is supporting the fact about the invariance, and that
a tensor has either contravariant or covariant componenets.

Thanks for helping me, it's rally a sublte point to me.

6. Apr 7, 2017

### RiccardoVen

I see your point, but I'm really start thinking this jargon is highly misleading. I'm not a professional Phycist nor a professional Math. Anyhow, as you told above, from a Math POV it simply doesn't make sense (may be I'm more Math addict, who knows).

The only thing I feel to be allowed to say is "a vector is an rank-1 tensor (so invariant), which could have either contravariant or covariant components".
On the same page "the electromagnetic field is a rank-2 tensor (so invariant), which is USUALLY represented in its fully covariant components form".

I see your point, thanks vanhees71. Anyhow, a vector could also have covariant components, depending on the basis you choose. Here there's an explnation of this:

http://math.stackexchange.com/quest...variance-and-contravariance-in-tensor-algebra

and in you can find it the same in Fleisch book.

So what disturbes me (but may be I'm wrong) is saying a vector is represented just by its contravariant compoenent only is just to tell half of the medal.

Last edited: Apr 7, 2017
7. Apr 7, 2017

### RiccardoVen

Just to quote Baez from his book (top of page 34):

"We warn the reader, however, that while the vector field ∂μ is covariant, and has its indices dowstairs, physicist often think of a vector field v as being
its component vμ. these have their indices upstairs, so physicist say that vμ are contravariant! This is one of the little differences that makes communication
between the two subjects a bit more difficult".

Of course, this is because the way in which Baez is attributing contravariant and covariant using the way in which they are pushed forward or pulled back.
I don't know exactly why, but this seems more natural to me.

8. Apr 7, 2017

### RiccardoVen

Probably I found the source of my doubts. when I say:

"a vector is an rank-1 tensor (so invariant), which could have either contravariant or covariant components"

it could be a bit foggy, since with this definition, it would be also true:

"a 1-form is an rank-1 tensor (so invariant), which could have either contravariant or covariant components"

So actually this definition would not let me further distinguish between vectors and 1-forms.
The only way is to use some additional criteria, which comes up to be the way in which the component are transforming.

That's said, this doesn't fully make me happy, since a vector still can have either contravariant or covariant components, so calling it "a contravariant vector" it sounds to me as an abuse of language.

That's why the Baez's approach makes to me a bit more sense, since:

"a vector (contravariant vector in the original jargon) is a quantify which is pushed forward by a map between manifolds"

or:

"a 1-form (covariant vector in the original jargon) is a quantify which is pulled back by a map between manifolds".

I think this last two sentences cannot be misinterpreted in any way, leading to precise definitions of what is a vector/1-form.

Last edited: Apr 7, 2017
9. Apr 7, 2017

### vanhees71

The resolution of the quibble is very simple in this case. It's just a slightly more careful language. Instead of saying $V^{\mu}$ is a "contravariant vector" you say $V^{\mu}$ are contravariant vector components.
Another important point is that you can only canonically (i.e., indpendently from the choice of a particular basis) map vectors and one-forms if you have fundamental form ("pseudo-metric" or metric) as in Minkowski or Euclidean vector spaces (or more generally on tangent spaces of differentiable manifolds).

10. Apr 7, 2017

### RiccardoVen

Perfect, this is exactly what I meant. But my original point was about the D'Inverno chapter named "Covariant tensor" or "Contravariant tensor".
However, I see the point, Phyisicist are usually contracting the statement "tensor with contravariant components" in "contravariant tensor".

OK, this is an important point to me. So the former definition based on the coordinates transformation is metric independent, while the Baez's definition would need a metric to be applied. Where can I find, from you point of view, a text book which is more like the Baez one (i.e. focuses on the coordinate free appraoch)? I guess Wald could be a good candidate, may be.