# (Lorentz etc) invariant vector fields

(Background: I know some but not much differential geometry, hopefully enough to formulate this post.)

I want to ask about what physicists mean when they say scalar, vector, etc. The answer in differential geometry is something like:

1) Scalar fields are smooth functions on spacetime ($$\phi \in C^{\infty} (\mathcal{M})$$)

2) Vector fields are smooth derivations / smooth sections of the tangent bundle ($$\vec{A} \in \Gamma(T\mathcal{M})$$)

3) One forms and tensors are defined in the usual way as (multi)linear functions satisfying certain properties.

However when we come to studying physics (maybe in QFT) there seems to be an additional requirement:

4) Suppose the map $$\Lambda$$ sends spacetime to itself (eg a boost). Then we additionally require:

4a) Scalars transform like:

$$\phi(X) \mapsto \phi'(X') = \phi(\Lambda^{-1}(X))$$

4b) Vectors transform like:

$$\vec{A}(X) \mapsto \vec{A}'(X') = \Lambda \vec{A}(\Lambda^{-1}(x))$$

... and so on for tensors.

I think the above is not completely correct so my questions are:

Q1) Is this a correct summary of the situation, that a "physical vector field" is a mathematical (differential geometry) vector field plus an invariance condition that is a completely separate condition, or can this be motivated from geometry alone?

Q2) What is the correct formulation of 4? Clearly we don't care about all transformations in $$\text{Aut}(\mathcal{M})$$ but do we fix a subgroup $$G\subset\text{Aut}(\mathcal{M})$$ arbitrarily? Usually for QFT etc it would be the Lorentz group I know, but are there situations where different $$G$$ are considered? Also, my writing of 4(b) doesn't make much sense as it stands unless $$\mathcal{M}$$ is given a vector space structure which I hope isn't necessary.

• Possible duplicates: physics.stackexchange.com/q/32011/2451 , physics.stackexchange.com/q/41211/2451 , physics.stackexchange.com/q/128026/2451 and links therein. – Qmechanic Dec 24 '18 at 14:19
• My personal view is that when dealing with flat space/spacetime the word "scalar" tends to be more restricted, referring to something whose mathematical expression, and not only its value, is the same in all inertial frames. So for example the function $t^2-x^2$ is a scalar in this sense, while $t+x$ is not. – Javier Dec 24 '18 at 16:05
• In other words, something is a scalar/vector/etc if you don't have to specify in which frame its formula is given. But this isn't always the case. – Javier Dec 24 '18 at 16:13
• @Javier Is that a standard definition? Once you've picked a chart $t+x$ does define a smooth map $f:\mathcal{M}\to R$ and whilst it won't have that nice a coordinate representation in all coordinates it will still be a perfectly good scalar field won't it? Your example won't retain its form under the change of chart $t\mapsto t/2$ $x \mapsto x/2$ for example. – jacob1729 Dec 24 '18 at 16:54
• Not sure if this is a standard definition, it's just how I interpret it. The point is that the expression should be invariant under Lorentz transformations (or Galilean or rotations or whatever), not under any coordinate changes. Of course, in GR this requirement goes out the window. – Javier Dec 24 '18 at 18:09

Nevertheless, any manifold with signature $$+---$$ is guaranteed to look locally like Minkowski space, so you can define these transformations locally. If something fits the differential geometry definition of a scalar or vector, then it's guaranteed automatically that when you express its components in a certain basis, and then apply one of these transformations, the components change as required. We do want vectors to behave this way under parity, or else we actually wouldn't call them vectors, we would call them pseudovectors or axial vectors.
• @jacob1729: The components of a vector field transform properly under any diffeomorphism, i.e., any smooth change of coordinates. The transformation law is $v^\mu=\partial_\nu x^\mu v^\nu$. This law for a contravariant vector is an example of the more general tensor transformation laws. – user4552 Dec 24 '18 at 15:22
• Thank you, I think I understand now. So the geometric object stays the same, but its components need to change since they're with respect to some basis (eg the partials $\partial_i$ with respect to local chart coordinates) and that basis changes after doing a diffeomorphism. – jacob1729 Dec 24 '18 at 15:43