# I Using the metric to raise and lower indices

1. Mar 10, 2016

### Kevin McHugh

Let me see if I understand this correctly. Using the metric to raise an index converts a vector into a one form and lowering the index converts a one form into a vector. The contraction on the indices is the dot product between the two. Am I correct so far?

If so, here is my question. What is the physical significance to these operations? Why is it done, or more precisely, when is it necessary to do so? I'm trying to understand the logic behind the operation. As always, TIA for your insight.

Regards, Kevin

2. Mar 10, 2016

### Orodruin

Staff Emeritus
The usual notation is to use upper indices for the components of a vector (tangent vector) and lower indices for the components of a one-form (covector).

The metric can be used to convert a tangent vector into a one-form and without it there is no way of assigning properties such as length to a vector (or angles between vectors). By definition, one-forms are linear maps from the set of tangent vectors to scalars. The metric (invertibly) assigns a one-form to any given tangent vector.

Physically, tangent vectors describe displacements (think velocities and accelerations) while one-forms are more akin to level curves and naturally describe gradients. Only when you have a metric are their interpretations intertwined.

3. Mar 10, 2016

### Twigg

Not quite. Using the metric tensor to lower the index of some vector's components gives you the components of the corresponding one-form, and raising the index converts components of a one-form to the components of a vector.

Each contraction of an index tells you to sum up like you would if you were doing a dot product, yes.

These are purely mathematical operations without any physical significance, though we apply them a lot. Raising or lowering components of a vector/one-form using the metric tensor is just how you compute coefficients when you use the metric to go from a vector to its unique dual one-form, or from a one-form to its unique dual vector. "Dual" sounds fancy but it's just a way of distinguishing between vectors and one-forms (e.g., a projection operator) when they don't always have the same coefficients (which they don't on manifolds). For a familiar example of where it's necessary, take a velocity vector in $\mathbb{R^{3}}$ and transform it into a basis of tangent vectors in spherical coordinates, then evaluate $\vec{v} \cdot \bigtriangledown$ (remember to use the scale factors for del in spherical coordinates!). Suddenly tangent vectors and directional derivatives don't look the same any more. Try it.

4. Mar 10, 2016

### Kevin McHugh

Thanks Twigg, that helps. So when the the alpha index is lowered on the Riemann tensor, it is to find the components of the dual vector? Also, when the first index is raised on the Cristoffel symbol, we find the components of the dual one form? Or is that necessary to determine the Riemann tensor?

5. Mar 10, 2016

### Orodruin

Staff Emeritus
First of all, people will not understand (we may be able to guess, but that is besides the point) when you refer to "the alpha index". There is no a priori naming convention for any indices and you can call them anything you wish.

The Riemann tensor is not a vector, it is a type (1,3) tensor and needs to be treated as such. Lowering the contravariant index results in a type (0,4) tensor. Technically, this is a multilinear map taking four tangent vectors to a scalar. Geometrically, the Riemann tensor relates to the change in a vector after parallel transport around a small loop. This is why it is a type (1,3) tensor, which maps three tangent vectors to a single tangent vector. The three vectors it takes is the directions of the displacements and the parallel transported vector. The resulting vector is proportional to the change after parallel transport around a small loop defined by the displacement directions.

The Christoffel symbols are not tensors, nor do they have the right number of components to be tangent vectors or one forms. They are defined by an affine connection and give you a prescription for how nearby vectors should be compared. They essentially correspond to the components of the directional derivatives of the coordinate basis.

6. Mar 10, 2016

### robphy

If you take for granted the metric, then you could argue that it is a mathematical operation.
However, relativity, quantum mechanics, etc... tells us that we shouldn't take the metric for granted.
Without a metric, you can't raise and lower indices....
Without an invertible metric, you might be able to lower indices but not raise.
What metric we have (or make use of) will dictate how a vector maps [in a metric-dependent way] to a particular one-form.
That is to say, using another metric (say Minkowskian instead of Euclidean) leads to a different one-form

7. Mar 10, 2016

### Twigg

You're totally right, robphy. I've been spending too much time in the topology forum.

Kevin, if you're still reading, please disregard what I said about the metric being a mathematical operator with no physical significance. That mindset doesn't teach you how to use the metric tensor. The physical significance of the metric tensor is that wherever you stand on an n-dimensional Riemannian manifold, the metric tensor gives you a ruler and a protractor that are specially calibrated for your present location, allowing you to measure the length of a tangent vector passing through your position and the angle between a pair of tangent vectors passing through your position, respectively. If you can imagine looking at hyperplanar level sets of any tangent vector's incremental length as measured along the ruler, where each surface element's area equals the speed of the original vector, then you've constructed dual one-forms using the metric. "Lowering the index" of the vector is how you calculate the components of the covector you built from the components of the metric tensor and the components of the original vector.

8. Mar 11, 2016

### Kevin McHugh

Orodruin, what is the purpose of lowering the contravariant index on the Riemann tensor? Why is the first covariant index raised in the Christoffel symbol?

I was under the impression that the first slot in the Riemann tensor was for a one form and the first index is the index of the components of the one form. Then the second slot is for a tangent vector, the third slot is for the relative acceleration vector, and the last slot another tangent vector. So that indicies are for the components of the tangent vector, relative acceleration vector, and tangent vector respectively. Is this correct?

9. Mar 11, 2016

### Orodruin

Staff Emeritus
It is very difficult to really understand your intended meaning if you do not write out explicitly what you refer to and instead describe it in words. I hope you are aware that you can actually use LaTeX commands in this forum (see https://www.physicsforums.com/help/latexhelp/).

10. Mar 11, 2016

### Kevin McHugh

Being new to the forum I was not aware. Thanks for that. However the question remains, what is the purpose of Rαβγδ → Rαβγδ, or Γθφφ → Γθφφ ?

11. Mar 11, 2016

### robphy

When the contravariant index of Riemann is lowered, some symmetries of the lowered-index Riemann are revealed.

12. Mar 11, 2016

### Kevin McHugh

Thanks Rob, how about the Christoffel symbol?

13. Mar 11, 2016

### robphy

14. Mar 11, 2016

### robphy

https://www.physicsforums.com/threa...a-covariant-or-contravariant-quantity.666861/

It can be argued (as Burke [referenced in those threads] does)
that some physical quantities are more naturally described as
covectors (index-down) rather than vectors (index-up).
Similarly other quantities are born naturally with some particular index structure.
With this collection of structures---say, without appealing to a metric,
what operations are possible? Such operations are independent of a metric.

When additional structures [like a metric] are included, then the metric allows
more operations... introducing more symmetries. In some sense, one can
blur the distinction between a vector and its (metric-dual) covector.
This blurring, however, may obscure the physics.

A related example compares the electric and magnetic vector fields.
They seem to be both 3-element vectors... although one is a pseudovector. (Indeed, one never adds them together.)
If we imagine physics with a different dimensionality of space, will they still have the same number of elements [i.e. independent components]?
By asking such questions, one may realize that their equality of the number of elements is a numerical accident of three dimensions...
and that the magnetic field is structurally different from the electric field (and should probably be drawn differently).

15. Mar 13, 2016

### stevendaryl

Staff Emeritus
The Christoffel symbol $\Gamma^\alpha_{\beta \gamma}$ has the indices raised and lowered the way they are because of its typical use in performing covariant derivatives: The geodesic equation for a slower-than-light geodesic obeys:

$\frac{d^2 x^\alpha}{d\tau^2} + \Gamma^\alpha_{\beta \gamma} \frac{dx^\beta}{d\tau} \frac{dx^\gamma}{d\tau} = 0$

The point of the raised and lowered indices is so that you can tell what type of object the expression produces. If there is one "free" index (one that isn't summed over) that is raised, then the result is a vector.

The version with lowered first index, $\Gamma_{\alpha \beta \gamma}$, is typically not used very often, as far as I know.