# What is a Vector Multiplet?

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1. Nov 3, 2015

### Emilie.Jung

I have been trying for a while to read a precise definition of a Vector Multiplet (to whom $N=2$ Supergravity theories couple to in $4D$) but was not lucky in finding a self-contained one. The best I got was that on https://en.wikipedia.org/wiki/Supermultiplet though it was on Supermultiplets rather than vector multiplets. It only ;said one thing related to Vector Multpilets and is:
The most commonly used supermultiplets are vector multiplets
which does not really give me what I'm looking for.

So, in short, what is a Vector Multiplet?

2. Nov 4, 2015

### fzero

I would recommend that you try to learn some of the details of SUSY that might help you later on. There are many good references, but a short presentation that covers a number of important topics is Section 2 of http://arxiv.org/abs/hep-th/9701069. You won't master the subject from that, but you will get an idea of what N=2 global SUSY is all about. Supergravity itself quite a bit more complicated, but you have to start somewhere.

To cover your question in relatively simple terms, lets recall how multiplets work for something easier, namely $SU(2)$ spin in quantum mechanics. We have generators $J_z, J_\pm$ and they satisfy the familiar algebra that I won't bother to write down. Multiplets correspond to a spin $j$, where $j(j+1)$ is the eigenvalue of $\vec{J}^2$ and the $2j+1$ states $|j,m_j\rangle$ in a multiplet are labeled by the integer $m_j = -j , \ldots, j$. We can use the raising and lowering operators $J_\pm$ to move from one state to another since. Typically we build states by starting with the highest weight state $|j,j\rangle$ and repeatedly acting with $J_-$ until we get $J_- | j,-j\rangle = 0$. Equivalently we could have started with the lowest weight state and built up the multiplet with $J_+$.

The above procedure is fairly general and an analogous procedure can be used to study what are called representations of (Lie) algebras. In general terms, the states correspond to vectors living in some vector space and the generators of the algebra act like matrices on these vectors. These vector spaces are the representations of the algebra.

With 4D N=1 SUSY, we have fermionic supersymmetry generators $Q_\alpha$ and $\bar{Q}_\dot{\alpha}$, where the indices label two-component spin representations of the Lorentz group. We can pick from the SUSY generators components that act like raising and lowering operators (see those notes above for the details). We assume that we have already labeled the particle states according to the Lorentz group, so we already have spin labels on them. The SUSY operators act on a scalar particle state to give a fermionic state (and vice versa). If we start with a state that is a massless Weyl fermion, $\psi_\alpha$, an appropriate choice of the raising operator will give us a complex scalar $\phi$. Acting again with the raising operator gives zero, since $(\bar{Q}_\dot{\alpha})^2=0$ because of the fermionic nature. So we get a multiplet $(\psi_\alpha, \phi)$ consisting of a 2-component spinor and a complex scalar (2 real degrees of freedom: this is the N=1 chiral multiplet.

If we start with a massless vector state $v^\mu$, it turns out that the raising operator gives a Weyl fermion, $\lambda_\alpha$. Again, $(\bar{Q}_\dot{\alpha})^2=0$, so the multiplet $(v^\mu,\lambda_\alpha)$ just consists of 2 real degrees of freedom of the massless vector and a 2-component spinor: this is the N=1 vector multiplet.

With extended supersymmety, we have additional SUSY generators, so that we add another index $I,J = 1, \ldots, N$ and consider the algebra of . $Q^I_\alpha$ and $\bar{Q}^I_\dot{\alpha}$ . So for N=2, we have $Q^1_\alpha$, $Q^2_\alpha$ and their conjugates. The new ingredient now when we construct multiplets is that we can act on the lowest weight state with $\bar{Q}^1_\dot{\alpha}$, $\bar{Q}^2_\dot{\alpha}$ and $\bar{Q}^1_\dot{\alpha}\bar{Q}^2_\dot{\beta}$ before we get a zero. If we start with a massless Weyl spinor $\psi_\alpha$, we get a pair of complex scalars $\phi^I$ now from acting with the pair of single $Q$s. The term $Q^1Q^2$ gives us a Weyl fermion $\chi_\dot{\alpha}$ with the opposite chirality from the we started with. This is called the N=2 hypermultiplet. In $N=1$ language, it is the sum of a chiral multiplet with an antichiral multiplet.

If we start with a massless vector $v^\mu$, acting with each $Q$ gives a pair of Weyl fermions of the same chirality $\lambda^I_\alpha$. . Acting with the other $Q^2$ on $\lambda^1$ gives a complex scalar $\phi$, but $Q^1 Q^2 = Q^1 Q^2$, so we don't get anything new by swapping SUSY flavor indices. This is the N=2 vector multiplet. In N=1 language, it is the sum of a vector multiplet and a chiral multiplet.

Note: I got sloppy with the notation towards the end. I urge you to look at the reference to get a more careful discussion.

3. Nov 4, 2015

### Ben Niehoff

Either Wess & Bagger's book, or Terning's would give you a nice introduction to SUSY and multiplets.

A "vector" multiplet is just what you get when you start with a vector and act with all of your SUSY generators. So it includes the vector and anything it can be transformed into under SUSY. In N=2 theories, this means the multiplet contains a vector, two fermions, and a complex scalar.

4. Nov 4, 2015

### Emilie.Jung

Huge thanks always @fzero !

I am wondering what is exactly meant by a
I can't imagine its physical meaning or representation before its gets acted on by some SUSY operator?

5. Nov 4, 2015

### fzero

We should think of SUSY as an extension of the Poincare group of spacetime symmetries. Then the free particle states are classified by their representation under the Lorentz group and their momentum. Adding SUSY means that we add additional structures to this classification of states, namely the organization of the states of different spins into supermultiplets.

The massless vector is the massless state with spin 1 under the Lorentz group. It doesn't lose any of the meaning that it had before considering SUSY.

6. Nov 4, 2015

### Emilie.Jung

Thank you for your time always. Your understanding and explanation of those advanced ideas is superb. @fzero

7. Nov 4, 2015

### samuelphysics

@fzero , how would those vector multiplets "couple" to supergravity as addressed in the question?

8. Nov 5, 2015

### haushofer

Enjoying the answers of fzero, just wanted to add a nice reference for Sugra: Ortin's Gravity and Strings, and Van Proeyen's book :)

9. Nov 5, 2015

### haushofer

10. Nov 5, 2015

### samuelphysics

@haushofer the link you wrote down is not available. If you may provide the name of the pdf with the author. I also kindly reask @fzero if another look can be taken at my question.

11. Nov 5, 2015

### fzero

Actually deriving the action for N=2 sugra is extremely complicated and I've never gone through all of the details myself, so I won't try to be rigorous. Presumably the texts that haushofer mentioned in post #8 would explain more, or you could look at http://itf.fys.kuleuven.be/~toine/LectParis.pdf by one of those authors.

I will discuss the coupling of the vector multiplet in general terms in order to help motivate the correct result for the bosonic degrees of freedom. So we first need to spell out what fields are involved. First we have the pure supergravity multiplet, sometimes called the Weyl multiplet. This has a spin 2 field $h_{\mu\nu}$, typically defined as a linearization of the metric tensor around some classical background (like the flat metric). Then, for N=2 SUSY, there are two spin 3/2 fermions $\psi^a_{\alpha \mu}$, $a=1,2$, called gravitini. Finally there is a spin 1 field $A^0_\mu$, called the graviphoton.

I will discuss the case where we include $n_V$ abelian vector multiplets. So we have spin 1 abelian gauge fields $A^i_\mu$, $i= 1,\ldots,n_V$, pairs of spin 1/2 fields $\lambda^{ia}_\alpha$ (gauginos), and complex scalars $\phi^i$. In a nonabelian theory, all fields have to be in the same representation as the gauge fields, which is the adjoint representation. In the abelian theory, the gauge fields have zero charge, so the scalars and gauginos have to have zero charge under the abelian group too. This simplifies the interactions a bit.

Let's start out by thinking about what sort of renormalizable terms we can have for the bosonic fields in the vector multiplets in 4D, ignoring gravity for the time being. We have kinetic terms for the scalars, which take the form $\partial_\mu \phi^i\partial^\mu\bar{\phi}_i$. Similarly we have the Maxwell term for the gauge fields $(F_{\mu\nu}^i)^2$. Additionally, we could consider the term $\delta_{ij} \epsilon^{\mu\nu\rho\sigma} F^i_{\mu\nu} F^j_{\rho\sigma}$. For the abelian theory here, these terms vanish, but similar terms will be important momentarily.

Now, consider the case where we turn on gravitational interactions. The first modification is that we would restore the metric in the sums over spacetime indices, as well as include the appropriate covariant derivative in the definition of the field strengths $F^i_{\mu\nu}$. From the point of view of effective field theory, the interactions will generate nonrenormalizable terms in the effective action, which are allowed to couple the scalars to the gauge fields (including the gravitphoton). These terms are restricted by gauge-invariance and covariance, but some allowed terms take the form
$$g(\phi,\bar{\phi})_{i\bar{i}} \partial_\mu \phi^i\partial^\mu\bar{\phi}^{\bar{i}}, ~~P_{IJ} (\phi,\bar{\phi}) F^I_{\mu\nu}F^{J\mu\nu}, ~~Q_{IJ} (\phi,\bar{\phi})\epsilon^{\mu\nu\rho\sigma} F^I_{\mu\nu} F^J_{\rho\sigma}.$$
Here I have introduced the notation $I,J=0,1,\ldots, n_V$, so that the graviphoton is included. Further terms that we could include are a scalar potential $V(\phi,\bar{\phi})$, as well as the nonminimal coupling to the curvature $R N_{i\bar{i}} \phi^i\bar{\phi}^\bar{i}$.

So far we haven't included any SUSY constraints. It would take some complicated discussion of N=2 superspace to really explain these, but I will just state the result. The functions appearing in each of the terms above can be shown to derive from a single holomorphic function $F(\phi^I)$, called the prepotential. If we define the quantities
$$F_{I_1\cdots I_n} = \frac{\partial}{\partial \phi^{I_1}} \cdots \frac{\partial}{\partial \phi^{I_n}} F(\phi),$$
then the Kahler potential for the scalars is
$$K(\phi,\bar{\phi}) = i \phi^I \bar{F}_I (\bar{\phi}) - i \bar{\phi}^I F_I(\phi)$$
and the target space metric is
$$g_{I\bar{I}} = \frac{\partial}{\partial \phi^{I}}\frac{\partial}{\partial \bar{\phi}^{\bar{I}} }K.$$
The quantities $P$ and $Q$ can also be written down in terms of $F_{IJ}$ and $\bar{F}_{IJ}$, but, since I would have to introduce the self-dual and anti-self-dual components of the gauge field strengths to write them in a nice way, I would refer you to the references for exact formulas.

12. Nov 6, 2015

### samuelphysics

Thank you for the answer @fzero !

1)So, the thing is that we first take our theory's vector multiplet, then we see what is present in this multiplet then try to make terms out of the constituents of the multiplet and put them in a Lagrangian? That is how coupling goes? Why should they couple to start with(this might sound as a crazy question but why, I cannot comprehend why coupling takes place or rather doesn't)?

2)Now if the procedure is like what I understood, then where is the term of the gauginos? We only saw that of the scalars and the gauge fields.

3)Lastly, are the N=2 SUSY vector multiplets the same thing as SUGRA N=2 vector multiplets?

13. Nov 6, 2015

### fzero

My method is just a simple method to motivate what sort of terms appear in the Lagrangian for the interacting theory. The rigorous method explained in the references (such as the van Proeyen lectures that I linked above) ensures that the Lagrangian is actually locally N=2 supersymmetric.

Let me review. Starting from the fields in the vector multplet, we start with the free field theory terms. If we had a nonabelian gauge theory, then we would have gauge interactions between the fields in the vector multiplet (that I ignored). Next is to add gravitational interactions. All physical fields in the theory have some energy associated with corresponding particle states, so they must interact gravitationally. For the abelian theory, the gravitational interaction is the only fundamental interaction.

So coupling between fields must take place because of the nature of gravity.

I didn't want to get involved with the fermions since there are many terms to deal with. For example, if you start with the free field theory kinetic term $\bar{\lambda}_{aI} \gamma^\mu \partial_\mu \lambda^{aI}$, after turning on gravity we get a term like $g_{IJ} \bar{\lambda}_{a}^I \gamma^\mu D_\mu \lambda^{aJ}$. If you want to see some additional terms, check out eq (4.81) in the van Proeyen lectures (he uses $\Omega_i^I$ for the gauginos).

The physical fields are the same. However, it is customary to add various auxiliary fields in supermultiplets in order to make SUSY manifest. Dealing with local supersymmetry (SUGRA) generally requires many more auxiliary fields than global SUSY, so the equations for SUGRA will be more complicated. But after imposing the various constraints and equations of motion, the remaining physical degrees of freedom are the same in both cases.

14. Nov 6, 2015

### haushofer

Introduction to Sugra, by Henning Samtleben. The most accesible first exposure I know of :)

15. Nov 6, 2015

### haushofer

By the way, an efficient way of describing matter couplings in Sugra is given by the 'superconformal tensor calculus', discussed by Van Proeyen. Essentially it uses gauged superconformal symmetry to couple matter to ordinary sugra (the gauge fields of the special conf.transfo's and dilatations are dependent and are Stuckelberg fields respectively, leaving you for the bosonic fields with the vielbein and spin connection; this spin connection is dependent).

Last edited: Nov 6, 2015
16. Nov 6, 2015

### samuelphysics

Let me see if I got this right. You say particles must interact gravitationally (here you do not mean in abelian theories? Because the sentence that follows talks that in abeliean ones, grav. interaction is fundamental.)

Last sentence says couplings take place because of nature of gravity, is this in Abelian, non-Abelian or both?

17. Nov 6, 2015

### samuelphysics

Right, I would like if you can elaborate (in a not-necessary-detailed way) on why superconformality plays this role in helping us couple sugra to matter? I don't know about superconformality but I know about CT and their relation to sclae-invariance? So, does scale-invariance play any role here?

Last edited: Nov 6, 2015
18. Nov 6, 2015

### fzero

What I meant was that in the abelian theory there is no gauge interaction between the scalar and vector field. In the nonabelian theory there is a tree-level interaction of the form $A \phi \partial \phi$. In either theory, once we include gravity, there are tree-level interaction terms involving the graviton, which have the form $h \partial \phi\partial\phi$, $h \partial A \partial A$ (plus others).

19. Nov 6, 2015

### MacRudi

So you think that Gravitons have the same properties as Gluons? What is about the quadropol?

20. Nov 6, 2015

### fzero

I'm not sure why you conclude that from what I've written. In particular, the gravitational interaction terms I've written down don't mix the scalar with the vector. The gravitational interactions also have an extra derivative compared to the gauge interaction, which is precisely because the graviton has spin 2 vs the spin 1 of the vector field. Spin 2 is also why the multipole solutions of the classical wave equation for the graviton start with the quadrupole.

That said, there are gross properties that the graviton would have in common with the gluon. They are both massless. They both are associated with a gauge symmetry, so there are certain parallels in the discussion of their wave equation or their quantum theory. However, there are clear differences, as above.