My basic understanding is that a field theory consists of symmetry groups, a space $S$ that the symmetry groups act on and of fields defined on that space $S$. In other words, the space $S$ is the domain of the fields. I have understood that the space $S$ can be a rather complex construct, usually a tensor product of the Minkowski space and other spaces that the gauge groups act on.

We define the Lagrangian as a composition of the fields and require it to be invariant under the symmetry group transformations of the domain space $S$.

Then, for example, a Lorentz transformation turns the scalar field $\phi(x_\mu)$ into $\phi(L_\mu^\nu x^\mu)$. The codomain, the space of field values, is part of $S$, it has no symmetry groups defined on it and it is not being transformed.

I would then expect that a field $A^\nu(x_\mu)$ is transformed into $A^\nu(L^\alpha_\mu x_\alpha)$.

However, applying a the Lorentz Symmetry transformation to, e.g. the Lagrangian of QED requires me to do something like: $L^\nu_\beta A^\beta(L^\alpha_\mu x_\alpha)$.

Now, suddenly, the codomain of the fields is subject to transformations just like the field domain $S$.

It has always confused me that the codomain of the fields is also transformed. I thought the symmetry groups are defined on the domain of the fields?

Can it maybe be seen in a way that only the field domain is being transformed? Maybe we can re-define the vector fields a scalar fields on a bigger domain?

I will be very grateful is someone can finally clear this up for me.

  • $\begingroup$ This has nothing to do with QFT. That's just how a vector field transforms. $\endgroup$ – ACuriousMind Aug 20 '15 at 15:53
  • $\begingroup$ I wonder if I formulated my question so badly, because I don't feel that you've addressed it at all. $\endgroup$ – Konstantin Schubert Aug 20 '15 at 16:24
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    $\begingroup$ Either I have not understood your question, or you have not understood my comment ;) Let's try again: A vector field takes values in the tangent spaces at the points. Applying a Lorentz (or any other) transformation not only changes the coordinate description ($x\mapsto x' = Lx$), but also changes the basis for the tangent space ($\partial_x \mapsto L\partial_{x'}$, since the tangent space is spanned by the derivatives, and so $A\mapsto L^{-1} A$). $\endgroup$ – ACuriousMind Aug 20 '15 at 16:30
  • $\begingroup$ I mean, my question was basically: I was always told that the symmetry operations are defined on the domains of the fields. But for vector fields we are suddenly transforming the codomain as well. How is this possible? Are the symmetry groups defined on both domains? $\endgroup$ – Konstantin Schubert Aug 20 '15 at 16:31
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    $\begingroup$ Depending on the confusion, I would recommend reading either Wald's GR book (first chapter on basic differential geometry), or Weinberg QFT vol 1 chapters 2 and 5. The important thing is that the symmetry group acts on the fields primarily (the action on the base manifold can be thought of as a transformation of the fields). Any consistent transformation law is allowed. $\endgroup$ – TotallyRhombus Aug 20 '15 at 16:32

First, terminology: Symmetry groups are not "defined on domains". Symmetry groups exist in the abstract, and they are then represented on certain spaces.

If we have a spacetime manifold $\mathcal{M}$, then the fields are functions $$ f : \mathcal{M} \to V$$ where $V$ is some vector space upon which a representation $\rho : \mathrm{SO}(1,3)\to \mathrm{GL}(V)$ of the Lorentz group exists. Why has there to exist such a representation? Because we need a rule how the fields transform. In principle you could imagine an uglier transformation law than a linear representation, but it turns out this is never really needed, at least for global symmetries. But, no matter how, you have to define how the fields transform (even if you define them to not transform at all, i.e. choose $V$ as the trivial representation $\mathbb{R}$, where every transformation is just the identity).

Since the Lorentz symmetry is not an internal symmetry, however, but one of spacetime, we have certain types of fields that naturally appear:

  • Scalars: $\phi : \mathcal{M} \to \mathbb{R}$. They are invariant - the value of $\phi$ at $p\in\mathcal{M}$ is just a constant number. If we describe the point $p$ by some coordinates $x^\mu\in\mathbb{R}^4$, then we write $$ \phi(x^\mu) \mapsto \phi({L^\nu}_\mu x^\mu)$$

  • Vectors: The tangent vectors are where a vector field naturally takes values in - to every point $p\in\mathcal{M}$, there is the space spanned by the derivatives of the coordinates $\frac{\partial}{\partial x^0},\frac{\partial}{\partial x^1},\frac{\partial}{\partial x^2},\frac{\partial}{\partial x^3}$. If we change coordinates as $x'(x)$, these transform by the chain rule as $$ \frac{\partial}{\partial x^\mu}\mapsto\frac{\partial}{\partial x^\nu}\frac{\partial x^\nu}{\partial x'^\mu}$$ and if we had a vector $v\in T_p\mathcal{M}$ before as $v = v^\mu\frac{\partial}{\partial x^\mu}$, then it now it's $v'^\mu\frac{\partial}{\partial x'^\mu}$ with $v'^\mu = \frac{\partial x'^\mu}{\partial x^\nu}v^\nu$.

  • Tensors: These take values in the tensor product of the tangent space $\otimes_{i = 1}^n T_p \mathcal{M}$ and have components $T^{\mu_1\dots\mu_n}$, where every index transforms as in the vector case

For a coordinate change $x'(x) = \Lambda x,\Lambda\in\mathrm{SO}(1,3)$ that is a Lorentz transformation, we have $\frac{\partial x'^\mu}{\partial x^\nu} = {\Lambda^\mu}_\nu$, and we get the transformation laws for scalars and vectors you have written.

Lastly, because your comment alluded to such a relation, there is no real relation between the electromagnetic four-potential $A$ being a vector field and it being a gauge field - there could, in principle, be vector fields which aren't gauge fields (precisely the massive ones, in fact), and gauge fields which aren't vector fields, but tensor fields (e.g. the Kalb-Ramond field).

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    $\begingroup$ minor typo: as of the writing of this comment, the above answer gives the transformation rule for 1-forms rather than vectors. To satisfy the chain rule (viewing the vector as a differential operator on scalars), $v'^\mu=\frac{\partial x'^\mu}{\partial x^\nu} v^\nu$ (because when $x'=\Lambda x$, $\phi(x)\rightarrow \phi'(x')=\phi(\Lambda^{-1}x')$). $\endgroup$ – TotallyRhombus Aug 20 '15 at 21:31
  • $\begingroup$ You are saying that scalar fields are invariant, but then you still define a transformation for them : $\phi(x^\mu) \mapsto \phi({L^\nu}_\mu x^\mu)$. Why is that? $\endgroup$ – Konstantin Schubert Aug 20 '15 at 22:21
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    $\begingroup$ @Konstantin: It's not a transformation of the manifold as such - you are just changing the coordinate system (rotating the basis vectors, essentially). $\phi(p)$ for an abstract point on the manifold doesn't transform at all, but when you write $\phi(x\^mu)$, you have chosen some special coordinate system $x^\mu$ - which you then can change by choosing another coordinate system related to it by a Lorentz transformation. The other symmetry group doesn't do that because they are not related to such coordinate choices on spacetime itself. $\endgroup$ – ACuriousMind Aug 20 '15 at 22:53
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    $\begingroup$ @Konstantin: Didn't I show why the vectors transform under a coordinate charge? Whenever you change the coordinates, you also change the basis of the tangent space, so the transformation "on $\mathcal{M}$" is inseparable from that "on $V$". Conversely, changing the basis of the tangent space means changing the coordinates. $\endgroup$ – ACuriousMind Aug 21 '15 at 14:50
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    $\begingroup$ @Konstantin: If the vector fields wouldn't take values in the tangent space, they would just be a bunch of scalars, not a vector on the spacetime manifold in the geometric sense. $\endgroup$ – ACuriousMind Aug 22 '15 at 13:38

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