# Supersymmetry and non-compact $R$-symmetry group?

The $R$-symmetry for $N$ supercharges is $U(N)$. Is it possible to generalize $R$-symmetry [let's take $U(4)$) to be something like $U(2,2)$ (maybe analogous to Wick rotation of $SO(3,1)$ to $SO(4)$?)]?

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## 1 Answer

Noncompact internal symmetries – and R-symmetry is an internal symmetry (it doesn't transform positions in the spacetime) – are unacceptable in a physical theory because they would lead to negative-norm states.

Consider the $i$-th superpartner of a bosonic particle state, $|i\rangle$, where $i=1,2,\dots,N$. The inner product $\langle i|j\rangle$ of such 1-fermion states has to respect the symmetry. So for $U(M,N)$, it would be ${\rm diag}(+1,+1,\dots,-1,-1,\dots)$ with $M$ plus signs and $N$ minus signs. It would follow that the Hilbert space contains physical states with norms of both signs and the predicted probabilities could be negative, too.

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Thanks Lubos! I'm OK with everything you say but what happens for example in bosonic string worldsheet theory, where you have inertial symmetry $SO(25,1)$ for scalar fields? There you seem to have noncompact inertial symmetry? –  jancore May 1 at 22:08
Dear @jancore, I meant that internal noncompact symmetries that produce states transforming as a linear finite-dimensional representation of the symmetry are unacceptable (for the simple "sign of thenorm" reason I fully described). The internal symmetry you mention (spacetime Lorentz group) doesn't lead to any linear representation. Similarly, SUGRA theories have noncompact groups like $E_{7(7)}$ but they're realized nonlinearly. –  Luboš Motl May 11 at 4:55
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