Produce Scalar from Tensor: What's the Name?

[snoopies622]

As I understand it, for a tensor of any rank I can produce a corresponding scalar in the following way: Create an inverted form of the tensor by lowering its upper indices and raising its lower indices, and then taking the inner product of this tensor and the original one.

My only question is, is there a name for this result?

 

[CompuChip]

Invertibility?
Metric?
Adjointness (or whatever you call "raising/lowering indices", i.e. going from co- to contravariant and vv)

 

[robphy]

Loosely speaking, you can call it the "square-norm" since what you are really doing is analogous to $$g_{ap} A^a A^p$$.
(The quantity $$A_a= g_{ap} A^p$$ is called the metric-dual of $$A^a$$.)

Given $$A^a{}_b{}^{cd}$$, you are forming the scalar using the metric [and its inverse]:
$$g_{ap} g^{bq} g_{cr} g_{ds} A^a{}_b{}^{cd} A^p{}_q{}^{rs}$$

 

[snoopies622]

Thanks, robphy.

I was wondering because in another thread someone mentioned using $$R^{abcd}R_{abcd}$$ (the "Kretschmann scalar"?) in order to show that the event horizon of a Schwarzschild black hole is not a real singularity, and it occurred to me that such a thing might be useful in other circumstances as well, so surely there must be a name for it...

By the way, am I correct in my belief that the "square-norm" of any metric tensor is the number of dimensions of its manifold?

 

[CompuChip]

Yes, you can show that very easily. By definition, $g^{ab} g_{bc} = \delta^a_c$ so if you contract a with c you get $g^{ab} g_{ba} = \sum_{i = 1}^d 1 = d$.

 

[snoopies622]

By definition, $g^{ab} g_{bc} = \delta^a_c$ so if you contract a with c you get $g^{ab} g_{ba} = \sum_{i = 1}^d 1 = d$.

Ah, yes, of course. Thanks, CompuChip.

 

[robphy]

By the way, am I correct in my belief that the "square-norm" of any metric tensor is the number of dimensions of its manifold?

As seen from CompuChip's response, the metric tensor must have an inverse for that calculation. (Note: The metric of a Galilean spacetime is degenerate.)

 

[snoopies622]

..The metric of a Galilean spacetime is degenerate.

I don't know what this means. Are you referring to a Euclidean metric? Minkowskian? And whether or not their matrix representations have inverses? Don't they?

 

[robphy]

I don't know what this means. Are you referring to a Euclidean metric? Minkowskian? And whether or not their matrix representations have inverses? Don't they?

The nondegenerate (i.e. invertible) metric of an n-dimensional Euclidean space has
the diagonal form (+1,+1,...,+1,+1) in rectangular coordinates.

The nondegenerate (i.e. invertible) metric of an n-dimensional Minkowskian spacetime has
the diagonal form (-1,+1,...,+1,+1) in rectangular coordinates.

The degenerate (i.e. non-invertible) metrics of an n-dimensional Galilean spacetime has
the diagonal forms (0,+1,...,+1,+1) [for the spatial metric] and (+1,0,...,0,0) [for the temporal metric] in rectangular coordinates.

 

[snoopies622]

Oh, OK. Thanks.