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2

Take for example $\mathcal{N}=2$ supersymmetry. The algebra, including a central charge $Z$, is given by $$\{Q_\alpha^a,Q^\dagger_{\dot{\alpha}b}\}=2\sigma^\mu_{\alpha\dot{\alpha}}P_\mu \delta^a_b$$ $$\{Q_\alpha^a,Q_\beta^b\}=2^{3/2}\epsilon_{\alpha\beta}\epsilon^{ab}Z$$ $$\{Q^\dagger_{\dot{\alpha} a},Q^\dagger_{\dot{\beta} ... 1 Supersymmetry is broken by electroweak symmetry breaking, by so-called F-terms. There must be another, separate source of supersymmetry breaking, because the EWSB would result in light supersymmetric particles, with sum rules for their masses, e.g.$$ m_{e_L}^2 + m_{e_R}^2 = m_e^2. This is excluded experimentally. With exact supersymmetry, EW symmetry is ... 1 The point is that the SUSY algebra, \begin{align} & \left\{ Q _\alpha ,Q _\beta \right\} = \left\{ \bar{Q} _{\dot{\alpha}} , \bar{Q} _{\dot{\beta}} \right\} = 0 \\ & \left\{ Q _\alpha , \bar{Q} _{\dot{\beta}} \right\} = 2 \sigma ^\mu _{ \alpha \dot{\beta} }P ^\mu \end{align} is invariant under multiplication of  Q _\alpha  by a phase, ... 0 The consequence if the vacuum (i.e. the VEV of the scalar field) was not SU(2)_L\times U(1)_Y charged is that it will not break this symmetry and also the consequence if the vacuum was SU(3) charged is that it will break this symmetry which is believed to be an exact symmetry in the Standard Model (SM). More precisely, for your first question, check my ... 0 I will present the simplest example of beta functions arising in string theory, specifically within bosonic string theory. The states transform in the 24 \otimes 24 representation of SO(24), equivalent to three irreducible representations; schematically,(\mathrm{traceless \;symmetric} )\otimes (\mathrm{antisymmetric}) \otimes (\mathrm{singlet})$$To ... 1 The answer is yes; this can be shown by evaluating the variation of the action. The action consists of three terms, we will consider them separately:$$\delta_\epsilon K(\Phi,\bar{\Phi})=\frac{\partial K(\Phi,\bar{\Phi})}{\partial\Phi^i}\delta_\epsilon \Phi^i+\frac{\partial K(\Phi,\bar{\Phi})}{\partial\bar{\Phi}_i}\delta_\epsilon\bar{\Phi}_i=i\epsilon K_i ...

1

Explanation for the electromagnetism aspect: If the vacuum carried charge under some generator, that would mean that the generator would not annihilate the vacuum. That would mean that even if such a generator corresponds to a symmetry of the theory, the vacuum however is not symmetric under that operation. Then the gauge boson corresponding to this ...

0

Well, as regards the second part of the first question, "and QCD color neutral", the if the vacuum state of any Standard Model field had a colour charge surely it would be in opposition to the Confinement property of QCD that we observe. That is, we never see colour charge in an experiment, we only ever see colour singlet (white) field states. I'm not sure ...

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Well, the Logic is reversed compared to the one you are implicitly using. It is the direction of the Higgs'vev that's defining what we call electric charge. In practice SU(2)xU(1) is broken to a certain U(1) that we can always choose to point in a certain direction, and accordingly assign the electric charges afterwards. Since the Higgs can't carry color it ...

2

In principle, $\mathcal{N}$ gives you the number of supercharges in your theory. There are, however, cases with more than one irreducible (pseudo-)real spinor representations. If you have $N$ charges in one and $N'$ charges in the other representation, you can denote the total number of charges as $\mathcal{N}=(N,N')$ in order to emphasize the difference. ...

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I happen upon this old thread now. Maybe it is still worth giving an update, and more of an answer. The latest account (as of the time of this writing) of the conjectural statement in question here appears as Conjecture 1.17 in Stephan Stolz, Peter Teichner, Supersymmetric field theories and generalized cohomology in H. Sati, U. Schreiber (eds.) ...

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It looks like this loophole is not explicitly discussed in the "axioms", but it is mentioned in the paragraph before equation (2) which I copy here: A symmetry transformation is said to be an internal symmetry transformation if it commutes with P. This implies that it acts only on particle-type indices, and has no matrix elements between particles ...

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