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In a theory with a gauge symmetry, as I understand it, the gauge symmetry is not a symmetry of a system, but rather its a redundancy in description. The procedure goes like this start with some Hilbert space $\mathscr{H}_{total}$, and define some set of linear operators $\{G_i \}$ where $G_i:\mathscr{H}_{total}\to \mathscr{H}_{total} $ these are my gauge transformations. Then mod out by the orbit of the $G_i$s i.e. let $|{\psi}> \sim G_i|\psi>$, and then you get the space of "gauge invariant states" $\mathscr{H}_{gauge} = \mathscr{H}_{total}/\sim$.

Is this new space actually a Hilbert space, or is it just notation that we call it a Hilbert space? Everything I read treats it like it were actually an Hilbert space but it seems like it should just be some manifold not an actual vector space. To see why, take the following stupidly simple example. Consider a single spin 1/2 system so my starting Hilbert space is $\mathscr{H}_{total} = \mathbb{C}^2$. I then define my linear operators to be $\{\mathbb{C}\}$ that is to say for any $\lambda \in \mathbb{C}$ I define $|{\psi}> \sim \lambda|\psi>$. If I now consider my space of invariant states namely $\mathbb{C}^2/\sim$ I dont get a hilbert space, I get $\mathbb{C}P^1$ which is not even a vector space.

So my question is then, in general is $\mathscr{H}_{gauge}$ a hilbert space?

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    $\begingroup$ Your "stupidly simple example" is called the projective Hilbert space and your idea of quotienting out the gauge orbits is wrong: Given the generators of gauge transformations $G_i$, what we declare equivalent is $0 \sim G_i\lvert \psi\rangle$, i.e. we quotient by the image of the infinitesimal gauge transformations, which is a vector space so the quotient is a vector space, too. $\lvert \psi\rangle \sim G_i\lvert \psi\rangle$ would hold for the finite/exponentiated transformations. $\endgroup$
    – ACuriousMind ♦
    Commented Jul 31, 2018 at 20:20
  • $\begingroup$ I understand that its projective Hilbert space, but my issue is not in naming it, it's in whats going on. Also If I have some infinitesimal transformations $G_i$ which generates some macroscopic transformation $\exp[\lambda^iG_i]$ then saying $G_i|ψ⟩ \sim 0$ is the same as saying $\exp[\lambda^iG_i]|ψ⟩ \sim|ψ⟩$. So I think we have the same condition. $\endgroup$
    – YankyL
    Commented Jul 31, 2018 at 20:54
  • $\begingroup$ My point is that "a group of operators obtained by exponentiating the gauge algebra" is much more restrictive than your "some set of linear operators", and means that your example is not an example of the process you are looking at. The algebra of $\mathbb{C}^\times$ as a Lie group is $\mathbb{C}$ and the mapping is the literal exponential from $\mathbb{C}$ to $\mathbb{C}^\times$. So in order to make the projective space an example of a gauge quotient we would need to also declare $\lambda\lvert \psi \rangle \sim 0$ for all $\lambda\in\mathbb{C}$, meaning the result is just a single point. $\endgroup$
    – ACuriousMind ♦
    Commented Jul 31, 2018 at 21:07

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