# Why are vectors in the $\left(\frac{1}{2}, \frac{1}{2}\right)$ representation of the Lorentz Group?

I just finished reading the excellent book that is "Physics from Symmetry" by Jakob Schwichtenberg.

I am left with a doubt. In the book he starts with a group theoretical introduction and clearly shows how we can see the Lorentz group as two copies of the $$SU(2)$$ group and derives all the Lagrangian and equation of motion from the Lorentz representations of the group and the mathematical properties of the vectors these transformations act upon. This description allows us to use Spin (Intended as the sum of the two casimirs of the two different copies of $$SU(2)$$ that compose the Lorentz Group) to distinguish the representations and that Spin turns out to be the actual physical Spin of the field described by objects that transform under that representation which by the way I found extremely cool.

He defines 4-Vectors as being objects that transform according to the spin 1 $$\left(\frac{1}{2}, \frac{1}{2}\right)$$ representation of the Lorentz Group and that lines up with the fact that object that are described by Vector fields (i.e. objects obeying the Proca equation of motion) are Bosons.

But shouldn't it be possible to construct boson fields from objects that transform according to the $$(1,0)$$ and $$(0,1)$$ representations?

I honestly haven't still tried to see what they would look like but I don't see any reason why not.

They would still have Spin 1, they would still be 4 components objects but that would introduce left and right handed versions of the bosons (which I don't know if it's acceptable but could be interesting if it is).

Is there something I am missing about this? and, if not has anyone tried to describe particles and their interactions through fields of this kinds?

• I'm not sure what you're asking here. It is a fact that the ordinary definition of 4-vectors transforms in the $(1/2,1/2)$ representation. In terms of vectors and forms, the $(1,0)$ and $(0,1)$ representations are the (anti-)self-dual 2-forms, hence $(1,0)\oplus (0,1)$ is a 2-form, not a vector. Who's saying it's not possible that these are bosons? – ACuriousMind Jul 20 at 22:55
• the question is, "what is inside these representations and is it possible to describe left and right chiral bosons from them?" – Defcon97 Jul 20 at 22:57
• – Qmechanic Jul 20 at 23:00

The $$(1,0)$$ and $$(0,1)$$ representations are not four-dimensional, but three-dimensional. Together - as the direct sum $$(1,0)\oplus (0,1)$$ - they form the 2-forms, i.e. antisymmetric tensors, on $$\mathbb{R}^4$$, as opposed to vectors. The parts of the sum are the self-dual and anti-self-dual parts of the 2-form under the Hodge dual.
So this is, for example, the representation the electromagnetic field strength tensor transforms in. It's certainly a bosonic object. This would also be the representation relevant to a gauge theory of rank two - where the gauge field is not a 1-form or vector $$A_\mu$$ but a 2-form $$B_{\mu\nu}$$.
• ok that's what i was looking for. I still don't get why they should be 3 and not 4 dimensional, the 1 representation of $SU(2)$ is 3 dimensional and the 0 representation is 1 dimensional thus when you take the cartesian product it looks like you get 4 degrees of freedom. – Defcon97 Jul 20 at 23:05
• @Defcon97 The $(m,n)$ representation is the tensor product of the $m$ and the $n$ representation of $\mathfrak{su}(2)$, not the Cartesian product. – ACuriousMind Jul 20 at 23:22