# Are there Gauge fields that are not 4-vectors?

In my understanding Gauge fields are fields that have some kind of redundancy, i.e. a transformation that does not change the physical state. As far as I can see all the Gauge fields in the Standard Model have 4-vector representations.

Are there any Gauge fields that are not 4-vectors? Maybe even fields with a lower dimensional representation?

• It's not very clear to me how you would like to define a gauge field. The action $S=\int d^4x (\partial \phi)^2$ has a ''gauge'' redundancy ($\phi \mapsto \phi+a$). Do you regard this as a gauge field, or not? I think you need to fix a definition. – Danu Jun 26 '16 at 15:53
• Isn't a gauge field a section of a vector bundle that is associated to a principal bundle? If so, none of the gauge fields are 4-vectors. – Bence Racskó Jun 26 '16 at 15:56
• @Uldreth In the usual mathematical treatment, they are connections (or, equivalently, connection 1-forms) on principal bundles, yes. This is partly why I'm asking Numrok to be more precise. – Danu Jun 26 '16 at 15:58
• @Danu well to be fair, the question does clearly specify a definition, albeit not one that corresponds with your (or most people's) predefined notion of what a gauge field is. – David Z Jun 26 '16 at 16:49
• @DavidZ I am looking for something more specific/rigorous than "some kind of redundancy". If that's all to go by, my first comment gives an example of a scalar "gauge field". – Danu Jun 26 '16 at 16:50

Right now, I have no idea about what you actually mean by gauge fields, or what anyone ever means by gauge fields. According to Danu, the gauge field is the connection itself, while if I remember well, Bleecker defines gauge fields as sections of those vector bundles that are associated to principal bundles.

However, I can, I think, give a definitive answer to the redundancy issue, which you allude to in your comments.

In gauge theories (and that will be my definition of a gauge field), you usually have some kind of fields that are defined throuout spacetime, but are not actual tensor fields, as per the usual differential geometry definition of tensor fields.

Instead, you have some sort of vector space $E$, and separate copies of $E$ for every point of spacetime, eg., if $p$ is a point of spacetime, $E_p$ is a separate copy of $E$ that is "located at" $p$.

These $E_p$ spaces are strung together in such ways, that they are all identical, but you cannot tell that which element of one space corresponds to that of another. For example, if $v\in E_p$, and $w\in E_q$, you have no canonical way of saying that $w$ is the same thing as $v$ (unless we are talking about the zero vectors). This structure is called a vector bundle.

The redundancy is that, like usual in physics, we tend to use components to describe stuff, so if $\psi(p)$ is a section of the vector bundle, eg. it is a field that maps to a point $p$ an element of $E_p$, and $e_A(p)$ is a local basis field of the vector bundle, then you can decompose $\psi$ as $$\psi(p)=\psi^A(p)e_A(p),$$ where the index $A$ is often called an "internal index" (and its range does not have to be the same as the spacetime manifold's dimension).

Now, what matters is $\psi$, not the components of $\psi$, so if you make a change of basis $e_A(p)\mapsto g_A(p)=M^B_{\ \ A}(p)e_B(p)$ for some invertible matrix $M$, the new components $\tilde{\psi}^A(p)=(M^{-1})^A_{\ \ B}(p)\psi^B(p)$ will describe the same physics as the old components.

As it usually happens, there is some additional structure imposed on the $E_p$s, which select some preferred bases, like there is a sense of "orthonormality" in each $E_p$, and you want only to use "orthonormal bases", so usually the matrices $M$ are not just invertible matrices, but are elements of some more restrictive group, but the point remains - that there are physically equivalent choice of bases, which are switched around by elements of a Lie-group. This is the redundancy.

In this sense, none of the gauge fields are 4-vectors, unless the internal vector bundle is the tangent bundle itself. Which is true in general relativity, BUT in GR we do not have any dynamical vector fields in the theory.

If by gauge fields, we mean the connection forms, those are local Lie-algebra valued 1-forms $A_\mu^i$ (where $i$ is a Lie-agebra index). These objects are always Lie-algebra valued 1-forms, therefore, as long as spacetime is 4 dimensional, they will always be (Lie-algebra valued) "covariant 4-vectors".

I hope this is what you were curious about, since I have a hard time understanding your question.

If the graviton exists it's not a 4 vector, but a tensor.

Well, the main point is that Yang-Mills theory is just one out of many gauge theories, cf. e.g. this Phys.SE post. E.g. SUGRA is a gauge theory. In fact, theoretically there are relativistic gauge theories with gauge fields transforming in virtually any possible representation of the Lorentz group. Whether they are realized in Nature is another story.