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Why is the projective symmetry group (PSG) called projective?

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As discussed by Prof.Wen in the context of the quantum orders of spin liquids, PSG is defined as all the transformations that leave the mean-field ansatz invariant, IGG is the so-called invariant gauge group formed by all the gauge-transformations that leave the mean-field ansatz invariant, and SG denotes the usual symmetry group (e.g., lattice space symmetry, time-reversal symmetry, etc), and these groups are related as follows SG=PSG/IGG, where SG can be viewed as the quotient group.

However, in math, the name of projective group is usually referred to the quotient group, like the so-called projective special unitary group $PSU(2)=SU(2)/Z_2$, and here $PSU(2)$ is in fact the group $SO(3)$.

So physically why we call the PSG projective rather than the SG? Thank you very much.

This post imported from StackExchange Physics at 2016-02-26 10:00 (UTC), posted by SE-user Kai Li
asked Jun 1, 2014 in Mathematics by Kai Li (975 points) [ no revision ]

1 Answer

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It depends on what group you consider the starting point, and that depends on context.

One context is the mathematical one (forget all you know about spins etc) where we start with the vector space $\mathbb C^2$ and the natural action of $SU(2)$ on it. If we then look at the projective vector space $\mathbb C^2/\mathbb C^* = \mathbb CP^1$, then the action of $SU(2)$ is given by $PSU(2) = SU(2)/ \mathbb Z_2$. Since this group now acts on the projective vector space, we can call it a projective group.

In the physical context--in the case of rotation symmetry--our starting group is not $SU(2)$ but rather $SO(3)$. The way this acts on the Hilbert space of a spin $1/2$ particle is by a nice linear representation $\rho: SO(3) \to \mathbb CP^1$ (this is in fact the identity map, since as you point out $\mathbb CP^1 \cong SO(3)$!). However physicists don't really like to think about projective Hilbert spaces, and so we prefer to think of our symmetry as acting on the linear Hilbert space: $\tilde \rho : SO(3) \to \mathbb C^2$. However, it turns out that the best you can do is that $\rho$ is a projective representation (which means the group structure is only respected up to a complex scalar). Hence you can say we traded in the projective Hilbert space for a projective group action. Again, physicist don't like thinking about projective representations, and so instead we use the linear representation of the covering group. Indeed, if we first extend $SO(3)$ to $SU(2)$ then it can act linearly on the linear space $\mathbb C^2$.

I am sure you know that that is why we use $SU(2)$ instead of $SO(3)$, but I wanted to go through the reasoning explicitly, to demonstrate that our original symmetry group is $SO(3)$, but since using that induces ''projectiveness'' down the road (either on the space or in the way it acts), we instead use the extended symmetry group $SU(2)$, which we can call the projective symmetry group of our system since it encodes all the projective realizations of our original symmetry group.


In conclusion: it's not that one name is better than the other, and you are correct in noticing that they are not talking about the same thing, it's just that it depends on the context when you tag the label ''projective''. In the former case we call it the projective group since it is the way in which the original group acts on the projective vector space. In the latter case we might call the extended symmetry group projective because its (linear) representations correspond to all projective representationsn of our original symmetry group.

This post imported from StackExchange Physics at 2016-02-26 10:00 (UTC), posted by SE-user Ruben Verresen
answered Dec 1, 2015 by Ruben Verresen (205 points) [ no revision ]

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