# Invariant of supersymmetry?

Given two vectors in 3D superspace $$(x_1^\mu,\theta_1^\alpha,\overline{\theta}_1^\alpha)$$ and $$(x_2^\mu,\theta_2^\alpha,\overline{\theta}_2^\alpha)$$ I am trying to find a polynomial invariant under supersymmetry.

e.g. Something like:

$$I_{12} = (x_1-x_2)_\mu(x_1-x_2)^\mu + \theta_1 \gamma_\mu (x_1-x_2)^\mu\overline{\theta_2}+...$$

The first term is the length-squared invariant of normal Euclidean space.

Basically the conditions I want are:

(1) First term must be the $$|x_1-x_2|^2$$

(2) Must be invariant under simultaneous supersymmetry transformations $$Q^\alpha$$ and $$\overline{Q}^\beta$$ of both pairs of coordinates simultaneously, (and Poincare transformations).

(3) The reason is, just to see if it can be done.

It seems like it shouldn't be too hard but I'm getting stuck trying to make it supersymmetric in both $$Q$$ and $$\overline{Q}$$.

If that is not possible, a 2D or 1D superspace invariant would be useful.

Edit: I think I was basically on the right path but I had a sign wrong on my $$\overline{Q}$$ generator!

(I think there may be more than one invariant that works)

Edit I will explain why I think this is interesting. If you take the Cayley-Menger determinant of the invariants between enough points and then raise it to a big enough power you get zero. This is an interesting fact about superspace.

• See how hard this is with trivial superspace; the gamma matrices are a red herring. Nov 24 '20 at 20:30
• I think in the Wikipedia article they may have the sign wrong for the $i$ in $Q^\dagger$. Nov 24 '20 at 20:40

There is a geometrical approach to that which is viewing the superspace as a coset

$$\frac{\text{d=3 super-Poincarè}}{SO(3)}$$

where your coordinates parametrize the coset as

$$g(x,\theta,\bar\theta)=e^{x^{i}P_{i}+\theta^{\alpha}q_{\alpha}+\bar\theta^{\alpha}\bar q_{\alpha}}$$

Now you compute the left invariant currents

$$g^{-1}dg = (dx_i + (\bar\theta\gamma_{i}d\theta)+c.c.)P_i + \dots$$

This means that we have the object

$$d\pi_i=dx_i + (\bar\theta\gamma_{i}d\theta)+c.c.$$

invariant under left translations, and so supersymmetry. Now we integrate

$$\pi_{i}(x^{i}_{2},\theta^{\alpha}_{2},\bar\theta^{\dot\alpha}_{2}|x^{i}_{1},\theta^{\alpha}_{1},\bar\theta^{\dot\alpha}_{1})=\int_{(x^{i}_{1},\theta^{\alpha}_{1},\bar\theta^{\dot\alpha}_{1})}^{(x^{i}_{2},\theta^{\alpha}_{2},\bar\theta^{\dot\alpha}_{2})} d\pi_i$$

This integral is understood as

$$\int_{(x^{i}_{1},\theta^{\alpha}_{1},\bar\theta^{\dot\alpha}_{1})}^{(x^{i}_{2},\theta^{\alpha}_{2},\bar\theta^{\dot\alpha}_{2})} d\pi_i =\sum_{k=1}^{N} (x^{i}_{k}-x^{i}_{k-1}) + \prod_{k=1}^{N}(\bar\theta_{k}\gamma^{i}(\theta_{k}-\theta_{k-1}))+c.c$$

for $$N\rightarrow\infty$$. We obtain $$x_{12}^i=(x_2-x_{1})^{i}+(\theta_2\gamma^{i}\bar\theta_{1})-(\bar\theta_1\gamma^{i}\theta_{2})$$

which is invariant under global supersymmetry. If you also want it to be a scalar just do

$$g_{ij}x_{12}^ix_{12}^j$$

• Sounds sensible, but I'm not able to do the integration myself. Dec 4 '20 at 14:41