# Conceptual question: invariant tensors, raising and lowering indices

1. Aug 5, 2013

### physicus

1. The problem statement, all variables and given/known data

When we raise and lower indices of vectors and tensors (in representations of any groups) we always use tensors which are invariant under the corresponding transformations, e.g. we use the Minkoski metric in representations of the Lorentz group ($\Lambda^\mu{}_\rho\Lambda^\nu{}_\sigma\eta_{\mu\nu}=\eta_{\rho\sigma}$), for an $SL(2,\mathbb{C})$ representation (spinor representation) we use the antisymmetric $\epsilon_{ab}$ ($S^a{}_c S^b{}_d \epsilon_{ab}=\epsilon_{cd}$).

The question is why we require this invariance of the tensor.

2. Relevant equations

3. The attempt at a solution

For example, instead of the Minkoski metric one could choose the metric $\delta_{\mu\nu}$ which we define to be $diag(1,1,...,1)$ in some coordinate system. If we define it to transform under Lorentz transformations covariantly according to its two antifundamental indices, i.e. $\delta'_{\mu\nu}=\Lambda_\mu{}^\rho\Lambda_\nu{}^\sigma \delta_{\rho\sigma}$, then the inner product $\delta_{\mu\nu}v^\mu w^\nu$ is invariant, just like the inner product defined with the Minkowski metric. The only difference is, that the contraction symbol $\delta_{\mu\nu}$ does transform, while $\eta_{\mu\nu}$ is invariant.

So what makes the invariance of the tensor we use to raise and lower indices so crucial?

2. Aug 5, 2013

### Vic Sandler

Any equation that varies under Lorentz transformations fails to satisfy the requirement that the physics look alike to all inertial observers.

3. Aug 5, 2013

### physicus

That does not really answer my question. Also, equations don't have to be Lorentz invariant. It is sufficient if they are covariant (e.g. both sides of the Einstein equations transform as (0,2) tensors and are therefore not Lorentz invariant). The thing is that both $\delta_{\rho\sigma}v^\rho w^\sigma$ and $\eta_{\rho\sigma}v^\rho w^\sigma$ are Lorentz invariant, when $\delta_{\mu\nu}$ is defined as specified above.

It is obvious that it is neater to use $\eta_{\mu\nu}$ to raise/lower indices. But why is invariance (rather than just covariance) always required for the object that is used to raise/lower indices?

4. Aug 5, 2013

### WannabeNewton

$\eta_{ab}$ and $\delta_{ab}$ define completely different musical isomorphisms. The latter does not map you from vectors to covectors nor from covectors to vectors in Minkowski space-time.

5. Aug 5, 2013

### Avodyne

Your delta is different in different frames. Thus any expression involving delta is not frame independent.

6. Aug 5, 2013

### physicus

Thank you two! Ok, I agree that for a certain metric space there is a canonical isomorphism between its tangent space and the cotangent space, which is defined by the metric.

However, there are more isomorphisms. I can simply map an arbitrary basis of the tangent space to an arbitrary basis of the cotangent space. Any non-degenerate (non-singular) bilinear form defines such an isomorphism. Why can I not use another bilinear form and not the metric to define the raising and lowering of indices? Is it possible to explicitly show that this breaks, e.g. Lorentz invariance.

(You said, for example, that the choice of my delta is frame dependent (which I am aware of). Why does that mean any expression involving it is frame dependent? It transforms covariantly (i.e. nice enough, as a (0,2) tensor), such that when I contract its indices with two vectors I get Lorentz invariants, in my opinion.)

Your answers have made me aware that I am not sure about even more basic things. When we study representations of the Lorentz group we do that in Minkowski space, i.e. a 4-dimensional vector space equipped with a certain metric. When I studied representation theory in maths we never spoke about the metric in the representation vector space, only the dimension was important. I suppose we select the Minkowski metric for our 4-dimensional real representation, since it is invariant under the representation matrices, i.e. we choose the metric such that Lorentz transformations are isometries. Why exactly do we do that?

Similarly, for the Weyl spinor representation we choose the metric such that our 2 dim. vector space is symplectic. Then the spinor transformations are isometries of the representation space.

Is that correct or is the logic completely twisted?

As said above, even in the case we have a metric space I don't understand why the canonical isomorphism defined by the metric is disitinct from any other isomorphism.

I am very thankful for any explanation.

7. Aug 5, 2013

### WannabeNewton

Hi physicus! Your logic is not twisted in any way at all, as far as I can tell. Certainly one can define non-canonical isomorphisms between the cotangent and tangent spaces. However, there is a subtle difference between general covariance (which is what you are speaking of) and special covariance (which is what we are seeking). Instead of trying to explain it myself, let me just provide passages from Wald's GR text (read the third one in particular):
http://postimg.org/gallery/1kq5s3e0/7bf35aee/

8. Aug 5, 2013

### physicus

Thanks, that helped a lot. However, I have questions remaining. Special covariance requires equations to be covariant under certain transformations (i.e. the isometries) even after plugging in the explicit form of the metric. That means there are components that behave nicely under isometries even though they are not generally covariant (because under general transformations the above components might "mix").

So my definition of $\delta_{\mu\nu}$ to lower indices $v_\mu=\delta_{\mu\nu}v^\nu$ is not compatible with special covariance because if I plug in a coordinate expression for my "metric" the equation is not Lorentz invariant any more (because $\delta$ cannot transform under Lorentz the way I forced it upon it after plugging in a coordinate form)? But why do I have to plug it in, I don't really say it is the metric? Do I also violate general covariance, because according to Wald all physical laws can only depend on one space-related quantity, which is the metric? But what I wrote was not really a law, was it?

9. Aug 6, 2013

### Oxvillian

I think the choice of special 2nd-rank tensor to perform the raising/lowering operation depends on the intended PHYSICAL APPLICATION of all this differential geometry.

The choice of $g_{ab}$ is appropriate for relativity, because $g_{ab}$ is what connects with the physics - it's what we use to calculate spacetime intervals. Contractions with the metric tensor therefore crop up all over the place in calculations. Certainly you could define a "different" raising and lowering using your $\delta_{ab}$, and it would be a totally well-defined mathematical operation. Just not a terribly useful one.

On the other hand if you're just calculating distances in Euclidean space, then $\delta_{ab}$ would absolutely be the right thing to use for your raising & lowering.

Similar comments for the symplectic 2-form in Hamiltonian mechanics.