Dimensional reduction from $3+1$ to $2+1$ for $\cal{N}=2$ vector superfield

Question

Let the supersymmetry transformations for the chiral multiplet $(z_k,\psi_{kL},f_k)$ be,

$\delta z_k = 2i \bar{\alpha} \psi_{kL}$

$\delta \psi_{kL} = D_\mu z_k \gamma ^\mu \alpha_R + f_k \alpha_L$

$\delta f_k = 2i\bar{\alpha} \gamma^\mu D_{\mu} \psi_{kL} - 2ie\bar{\alpha}\lambda _R z^k$

Similarly let the gauge multiplet $(A_\mu,\lambda,D)$ transform as,

$\delta A_\mu = i \bar{\alpha}\gamma_\mu \lambda$

$\delta \lambda = -\frac{1}{2}F^{4}_{\mu \nu}\gamma_4 ^{\mu \nu} \alpha + D \gamma_5 \alpha $

$\delta D = i \bar{\alpha}\gamma_5 \gamma ^ \mu \partial _ \mu \lambda$

Here in $F^4$ the $\mu$ and $\nu$ go over indices $0,1,2,3$ and $\gamma_4 ^{\mu \nu} = \frac{1}{2} [\gamma_4^\mu, \gamma_4^\nu]$. (these gamma matrices are defined below)

One does something called "dimensional reduction" of these to $2+1$ dimensional spacetime by assuming that the fields are independent of the third spatial coordinate. In $2+1$ dimensional spacetime the gamma matrices are defined as,

$\gamma ^0_3 = -i\sigma^2$, $\gamma^1_3 = \sigma ^3$ and $\gamma^2_3 = \sigma ^1$

In the so called "Majorana representation" the $4-$dimensional Gamma matrices can be written such that $\gamma_4 ^{0\1\2} = \left [ \begin{array}{cc} & \gamma_3 ^{0\1\2} \\ \gamma_3 ^{0\1\2} & \end{array} \right ] $ and $\gamma_4 ^3 = \left [ \begin{array}{cc} I & \\ & -I \end{array} \right ]$

One renames the third component of the gauge field $A_3$ as $C$ and splits the $4-$component fermions into $2-$component spinors in $2+1$ dimensions as,

$\lambda = \left [ \begin{array}{c} \lambda_1 \\ \lambda_2 \end{array} \right ]$

$\alpha = \left [ \begin{array}{c} \alpha_1 \\ \alpha_2 \end{array} \right ]$

One substitutes the above in the first set of supersymmetric transformations and sets to $0$ all derivatives with respect to the $3^{rd}$ spatial direction.

Then one is supposed to get,

$\delta A_\mu = i \bar{\alpha}_a \gamma_\mu \lambda _a$

$\delta \lambda _a = - \frac{\epsilon ^{\mu \nu \rho}F_{\nu \rho}}{2}\gamma _ \mu \alpha_a + \partial _\mu C \gamma ^\mu \alpha ^a + D \alpha ^a$

$\delta C = i\bar{\alpha}^a \lambda _a$

$\delta D = i \bar{\alpha}^a\gamma ^\mu \partial_\mu \lambda_a$

where $a$ runs over $1,2$ and $\alpha^1 = \alpha_2$ and $\alpha^2 = -\alpha_1$

Doing the above I can get the above transformations for all the fields. (Just that I need to use a fact that is true only for these gamma matrices in the $3$-dimensions that $[\gamma_3^\mu,\gamma_3^\nu]=2\epsilon ^{\mu \nu \rho}\gamma_{3\rho}$)

• In this "dimensional reduction" the particular choice of the gamma matrices seemed crucial. Is this true?

• I don't know how but this is supposed to match the transformations of the $\cal{N}=2$ vector multiplet in $2+1$ dimensions which has componets, $A_\mu, \lambda_a,C,D$ (why?)

• I don't know how one derives the field content (as above) of the $\cal{N}=2$ vector multiplet in $2+1$ dimensions.

• From the above it is claimed that the following forms a lagrangian respecting the above symmetry and is what gets called the ``supersymmetric Chern-Simons theory"

$L = \frac{\kappa}{2} (\epsilon ^{\mu \nu \rho} A_\mu \partial _\nu A_\rho - i \bar{\lambda_a}\lambda_a + 2CD)$

But to establish the above claim I need to assume two results,

• First is a property which again seems typical of $3$ dimensional gamma matrices that, $\gamma_3^{\mu \dagger} \gamma_3 ^0 = -\gamma_3^0\gamma_3^\mu$

• Secondly I need to assume the following variation for the fermion self-coupling,

  $\delta (\bar{\lambda} \lambda) = 2\delta (\bar{\lambda}) \lambda$

It is not clear to me as to why should the above hold.

• A crucial difference between the above and the usual Abelian supergauge theory is that the kinetic term for the fermion is missing. How does one understand this?

Answer 1

Dear Anirbit, this is a very large number of explicit and implicit, somewhat elementary questions embedded into a copy of a rather technical derivation that doesn't seem to be important for your real "primary" questions at all. It's hard to see what your questions really are because you seem to be hiding them in the convoluted formalism. But I see the following ones:

The particular form of gamma matrices is crucial, isn't it?

No, the particular form of gamma matrices is never crucial for physics. Of course, if you want to agree with particular values of components or matrix entries that someone else has derived, you need to use the same convention for the gamma matrices (and other things). But if you want to derive a physical insight, you may choose any basis of the gamma matrices. The derivations will be related by a trivial conjugation of matrices, or sign flips if one chooses different conventions for the metric etc.

Sorry, this is really not a question about representations of 3D supersymmetry algebras. It is a basic question about the difference between physics and social conventions. My answer would be identical in any other context where we talk about conventions. Everyone who has heard or read at least 10 minutes of a lecture about gamma matrices must know that there are different ways to represent them and all of them may be used to do physics.

How one derives the field content of a supermultiplet?

One takes a state of the multiplet and acts with it by raising and/or lowering supersymmetry generators - those that increase or decrease some $U(1)$ charges by $\pm 1/2$. Some of them are annihilated by the supersymmetry generators. One obtains the spins of all the states in the multiplet. Then one may also attempt to guess Lorentz-covariant fields whose physical polarizations match the desired spectrum of physical particles.

In $d=2+1$, the vector multiplet of $N=2$ algebra - which has 4 supercharges just like $N=1$ in $d=3+1$ - is just the dimensional reduction of the $N=1$ $d=4$ multiplet into 3 dimensions. So the vector multiplet becomes one vector and one scalar - one of the transverse component becomes a scalar - while the Majorana spinor becomes a pair of spinors in 3 dimensions.

The chiral multiplet in $d=4$ $N=1$ gets directly reduced to three dimensions. In four dimensions, one has a complex scalar and a Majorana spinor. In three dimensions, this gets reorganized as two real scalars and two real two-component spinors - therefore the index $k$ of those fields at the beginning of your text.

How one decomposes multiplets under dimensional reduction?

First, the supersymmetries themselves transform as spinors, so one decomposes spinors e.g. of $SO(2,1)$ under the $SO(1,1)$ subgroup. If one removes a dimension from spacetime, the higher-dimensional spinor always becomes the lower-dimensional spinor or their pair. Similarly, for a decomposition of a spinor of $SO(d)$ under a maximal $SO(d_1)\times SO(d_2)$ group, a spinor of the former becomes a tensor product of the spinors of the latter smaller groups, or the direct sum of two such tensor products (one has to do a simple exercise concerning the chirality and/or reality of the spinors).

It's hard to understand from your text whether you are familiar with these basic facts about spinors or not i.e. whether you want these things to be explained or whether it is just a waste of time because you know them, or whether you are familiar with the very fact that most of these "decomposition" and "dimensional reduction" questions are about decompositions of representations of Lie groups and supergroups under their subgroups. If you're not, you shouldn't have started with obscure technical problems such as $N=2$ vector multiplets in three dimensions; you should have asked what is the procedure to dimensionally reduce a theory. But one should focus at such a question - one question at a time. It just becomes extremely confusing if a physics SE question is composed out of 10 such things.

The scalars - singlets - remain scalars - singlets. Vectors are decomposed to vectors plus the singlets coming from the reduced directions of the space. When one knows that, it's easy to determine what names you used for those fields.

Why Chern-Simons theory

Because the Chern-Simons action may be written using a 3D gauge field and it is gauge-invariant and Lorentz-invariant. However, that's not the only action that one can write down for a gauge field. The Yang-Mills action may also be supersymmetrized. So it's not true that the $N=2$ SUSY implies that the gauge field has to be the Chern-Simons one (which has no physical polarizations since the theory is topological). But the source you are reading simply did study this theory, so it made whatever assumptions are necessary to pinpoint the particular Lagrangian.

Hermitian conjugation of gamma matrices

At some point, you seem to be puzzled by the relation that you incorrectly claim to "only hold for 3D matrices" that the Hermitian conjugation is equivalent to (minus) the conjugation by the temporal gamma matrix.

This actually holds in any dimension and any (natural!) convention in which the gamma matrices are chosen Hermitian or anti-Hermitian for temporal and spatial directions, respectively. (This choice is needed because the individual matrices must square to plus identity or minus identity matrix, depending on their being spacelike or timelike, because it follows from the anticommutator relations of gamma matrices and the signature of the metric.) It's not hard to see why. The anti-Hermitean ones change the sign by the conjugation, which are exactly those that anti-commute with $\gamma_0$: note that $\gamma_0$ commutes with itself but anticommutes with others, so the conjugation by $\gamma_0$ changes the sign of the "others".

Again, you shouldn't study obscure supermultiplets in obscure dimensions if you don't understand what gamma matrices do under the Hermitian conjugation. This fact is a basic component of any quantum field theory course. In fact, it is normally discussed in the context of the (single-particle) Dirac equation even before the students start with quantum field theory. The usual pedagogical treatment may have a $d=4$ bias, for obvious reasons, but it's true that gamma matrices only become worth their name if they can be used above $d=2$ or $d=3$, too: knowing gamma matrices in $d=2$ or $d=3$ only means not to know gamma matrices at all; $d=4$ is already complicated enough so that one could guess how it generalizes to any dimension.

You must have missed this whole thing in some way. Again, I recommend you some basic course on quantum field theory because it covers all those questions or, less naturally, an equivalent mathematical introduction to representations of Lie groups.

The variation of the fermions

The other "mysterious" formula for the variation of the fermionic bilinears is just the Leibniz rule for the derivative of a product (now, I would love to believe that you know how to differentiate a product, but this question of yours - in which you didn't do a single step to reduce the question to a more elementary one, i.e. in which you didn't really apply even the Leibniz rule - reduced my certainty even about this point), combined with the fact that the spinors in 3 dimensions are real so that the Dirac conjugation effectively contains "transposition" only and the two terms from the Leibniz rule are equal, producing the factor of two. Again, it's not a big deal; it's just a symmetry of the inner product of two real vectors (spinors) in this case. The only thing one could worry about is the relative sign but because $\bar\lambda \lambda$ doesn't vanish for a fermionic $\lambda$, because they're contracted with an antisymmetric $\epsilon_{\alpha\beta}$ of $SL(2,R)=Spin(2,1)$, it follows that its variation can't be zero, either. So the relative sign has to be plus and the first term from the Leibniz rule has to be doubled rather than canceled.

It's very hard to answer those questions meaningfully because it's a stream of relatively clearly irrelevant technicalities that have nothing to do with the "primary" things you seem to be confused by; combined with your opinions and guesses that are almost always wrong (like guesses that one is not allowed to use different bases); and typographic errors that make a technical discussion impossible (for example, there is no "backslash 1" or "backslash 2" in TeX).