# Using metric to raise and lower indices

Everything I read on tensors makes it clear that using the metric matrix $g_{ab}$ and its inverse $g^{ab}$ to respectively lower and raise indices of a tensor is very important. As far as I know (and I might be wrong) a tensor is defined to be something whose representation in coordinates $x^\alpha$ and $x'^\alpha$ is related either by a specific combination of partial derivatives of the new coordinates w.r.t. the old coordinates or vice-versa, the combination given by (for [1,1] tensors, for example)

$T'^a_b=\dfrac{\partial x'^c}{\partial x^a } \dfrac{\partial x^b}{\partial x'^d}T^c_d$ where the upper indices are called contravariant and the lower covariant. I understand that given a covariant vector say $x^\mu$, we define $x_\mu:=g_{\mu \nu}x^{\nu}$. My first problem is that I'm not sure why $x_\mu$ defined in this way should be covariant i.e. I'm not sure that this is well defined.

Secondly, if I write something like $V^\alpha$, am I implying that this is contravariant? Say I have a vector that is neither contravariant nor covariant, that is, just a collection of $n$ measurements in a vector $(a_1,a_2,\dots, a_n)$ and I write this as $a_i$. In a general relativity setting, am I implying that this is covariant?

I'm new to general relativity, so it would be really helpful if in addition to the two questions I had, someone could confirm that what I've written otherwise is correct (or tell me that I've got it completely wrong!). Thanks for any help.

Edit: If $g_{ab}$ complies with the tensor notation then in the new coordinates

$g'_{ab}=\dfrac{\partial x^c}{\partial x'^a } \dfrac{\partial x^d}{\partial x'^b}g_{cd}$ and so

$g'_{ab}x'^b=(\dfrac{\partial x^c}{\partial x'^a } \dfrac{\partial x^d}{\partial x'^b}g_{cd})(\dfrac{\partial x'^b}{\partial x^e}x^e$)

and since $\dfrac{\partial x^d}{\partial x'^b } \dfrac{\partial x'^b}{\partial x^e}=4\delta^d_e$ (where $\delta^d_e$=1 iff $d=e$, otherwise $\delta^d_e=0$), the coefficient of 4 because 4 goes from 0 to 3, I get to

$g'_{ab}x'^b=4\dfrac{\partial x^c}{\partial x'^a }g_{cd}\delta^d_e x^e=4\dfrac{\partial x^c}{\partial x'^a }g_{cd}x^d$

I'm quite close but there's a factor of 4 that shouldn't be there?

• If I may say so you have the "wrong" way of thinking about tensors. You should think of tensors as multilinear maps. I recommend reading Wheeler, Thorne, and Misner's "Gravitation". Also you should be reading differential geometry books. Actually, before all that you should read John Baez's delightful introduction here: math.ucr.edu/home/baez/gr/gr.html – Steven Gubkin Jul 26 '14 at 20:49
• @StevenGubkin Why? I can't seem to find a complete introduction to this stuff anywhere. My course introduces it (badly) like this, and I'm sitting the exam sat by that lecturer, so I'm not sure why I should start thinking about it in a completely different way. It's like no author understands that simultaneously shoving a bunch of new concepts in someone's face isn't the best way to learn. – Lammey Jul 26 '14 at 21:02
• Matrix algebra is to linear algebra as tensor algebra is to multilinear algebra. It is very hard to understand matrixes if you do not know vector spaces and linear maps. It is similarly hard to get tensors if you do not know multilinear algebra. For a very lowbrow approach to this stuff, you could look at my course here: ximera.osu.edu/course/kisonecat/m2o2c2/course/activity/welcome. It actually only deals with so called $(0,n)$ tensors. But I think having an understanding of multivariable calculus along these lines will help you. – Steven Gubkin Jul 26 '14 at 21:09
• The book Gravitation really does have most of what I think you want to know though. – Steven Gubkin Jul 26 '14 at 21:09
• @StevenGubkin Thanks for the recommendations - I will take a look at some point but I have an exam on this in ~3 weeks so I can't really afford to be learning things in a way that my course doesn't teach at the moment! – Lammey Jul 26 '14 at 21:43

Hint for part I: Consider how the product $g_{\mu\nu} \; x^\nu$ transforms, since you know how the individual terms in the product transform under a change of basis. You will see that it transforms exactly like $x_\mu$.

For the second part of your question: "a vector that is neither contravariant or covariant" is not a vector (in the sense used in physics). A simple collection of n quantities is not a vector or a co-vector.

Clearly, $g_{\mu\nu}$, being covariant components transform as

$$g^{'}_{\alpha\beta} = \frac{\partial x^{\mu}}{\partial x'^{\alpha}}\frac{\partial x^{\nu}}{\partial x'^{\beta}}g_{\mu\nu}$$

The contravariant components $x^{\nu}$ transform the other way $$x^{'\alpha} = \frac{\partial x^{\alpha}}{\partial x'^{\nu}}x^{\nu}$$

So that $$g^{'}_{\alpha\beta} x^{'\alpha} = \frac{\partial x^{\mu}}{\partial x'^{\alpha}}\frac{\partial x^{\nu}}{\partial x'^{\beta}}\frac{\partial x'^{\alpha}}{\partial x^{\rho}} g_{\mu\nu} x^{\rho} \\ = \delta^{\mu}_{\rho} \frac{\partial x^{\nu}}{\partial x'^{\beta}} g_{\mu\nu} x^{\rho} \\ = \left( \frac{\partial x^{\nu}}{\partial x'^{\beta}} \right) g_{\rho\nu} x^{\rho}$$

The bracketed term in the last equation shows clearly that the product transforms as covariant components do, i.e. like $x_\mu$ No extra factors of 4 anywhere.

Also, as Steven Gubkin says, this is not a good way to think about tensors at all. The reason is that we were talking about components of the tensors (in a basis) all the time, rather than the tensors themselves. This is a historical artifact of how tensors were used in physics (especially GR).

To see why you won't have a 4 or 3, think about the term $$\frac{\partial x^{\mu}}{\partial x'^{\alpha}} \frac{\partial x'^{\alpha}}{\partial x^{\rho}}$$

If you expand the sum you'll have
$$\frac{\partial x^{\mu}}{\partial x'^{1}} \frac{\partial x'^{1}}{\partial x^{\rho}} + \ldots$$ But if you remember the chain rule, you need to add all 4 (or 3) terms to replace this sum by $\delta^{\mu}_{\rho}$ $$\frac{\partial x^{\mu}}{\partial x^{\rho}}$$

• I actually don't know how $g_{\mu \nu}$ transforms, maybe that's what I'm missing? – Lammey Jul 26 '14 at 20:47
• I've edited the question to highlight what's gone wrong for me – Lammey Jul 26 '14 at 21:40
• Thanks for the edit. I don't see why there is no factor of four. You just cancel the $\partial x'^\alpha$'s off but isn't there a sum over them? Like for example if I write $\frac{x^\alpha}{x^\alpha}$ surely this is 4 and not 1? – Lammey Jul 27 '14 at 1:04
• Oh I see, thanks a lot! – Lammey Jul 27 '14 at 1:09
• @Lammey You're welcome. – user_of_math Jul 27 '14 at 1:10

The terms "contravariant" and "covariant" refer to how vectors change when you move from one coordinate system to another. So just declaring a vector like $V$ or $[2,2]$ does not make it contra or covariant. It's what you do with it that matters.

For example, let's say there is coordinate system representing the width and length of furniture in meters. A vector $[2,2]$ in this coordinate system could mean two different things:

Firstly, it could be the dimensions of a piece of furniture: 2 meters by 2 meters. That would be a contravariant vector. The reason is that if you transform into a feet coordinate system, the numbers $[2,2]$ would get bigger $[6.28,6.28]$ while the measurements (aka basis vectors) get smaller (feet are smaller than meters). Since the numbers vary contrary to the bases, it's a contravariant vector.

But $[2,2]$ could also represent a function for computing the perimeter of the furniture. If you had a desk that was 1.5m x 3m, you can find the perimeter by multiplying by $[2,2]$. Eg the perimeter is $[2,2] * [1.5,3] = 2*1.5 + 2*3 = 9$. To transform that function into feet (so you get the same answer), the vector $[2,2]$ would have to get smaller $[0.6,0.6]$. Since the numbers co-vary with the bases, it's a covariant vector.

So you have to know what the vector does and how it transforms in order to determine whether it's contravariant or covariant.

For the summing $\frac{\partial x^d}{\partial x'^b}$ is a 4x4 matrix. Multiplying any matrix by its inverse results in the identity matrix, which is the same as the Kronecker Delta $\delta$. Each entry in the result will be a sum with 4 parts, but they'll cancel out to leave either 1 or 0.

• That's a good example to introduce the concept. – Leonardo Castro Nov 10 '16 at 11:03