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▲{{Expert-subject|Mathematics|date=November 2008}}
In a constrained Hamiltonian system, a dynamical quantity is called a '''first class constraint''' if its Poisson bracket with all the other constraints vanishes on the '''constraint surface''' (the surface implicitly defined by the simultaneous vanishing of all the constraints).
==Poisson brackets==
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Suppose we have some constraints
:<math> f_i(x)=0, </math>
for ''n'' smooth functions
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These will only be defined [[chart (topology)|chartwise]] in general. Suppose that everywhere on the constrained set, the ''n'' derivatives of the ''n'' functions are all [[linearly independent]] and also that the [[Poisson bracket]]s
:{ ''f''<sub>''i''</sub>, ''f''<sub>''j''</sub> }
and
:{ ''f''<sub>''i''</sub>, ''H'' }
all vanish on the constrained subspace. This means we can write
:<math>\{f_i,f_j\}=\sum_k c_{ij}^k f_k</math>
for some smooth functions
:''c''<sub>''ij''</sub><sup>''k''</sup>
(there is a theorem showing this) and
:<math>\{f_i,H\}=\sum_j v_i^j f_j</math>
for some smooth functions
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The ordinary [[Poisson bracket]] is only defined over <math>C^{\infty}(M)</math>, the space of smooth functions over ''M''. However, using the connection, we can extend it to the space of smooth sections of ''f'' if we work with the [[algebra bundle]] with the [[graded algebra]] of ''V''-tensors as fibers. Assume also that under this Poisson bracket,
:{ ''f'', ''f'' } = 0
(note that it's not true that
:{ ''g'', ''g'' } = 0
in general for this "extended Poisson bracket" anymore) and
:{ ''f'', ''H'' } = 0
on the submanifold of zeros of ''f'' (If these brackets also happen to be zero everywhere, then we say the constraints close [[off shell]]). It turns out the right invertibility condition and the commutativity of flows conditions are ''independent'' of the choice of connection. So, we can drop the connection provided we are working solely with the restricted subspace.
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Look at the dynamics of a single point particle of mass ''m'' with no internal degrees of freedom moving in a [[pseudo-Riemannian]] spacetime manifold ''S'' with [[metric tensor|metric]] '''g'''. Assume also that the parameter τ describing the trajectory of the particle is arbitrary (i.e. we insist upon [[Parametric_curve#Reparametrization_and_equivalence_relation|reparametrization invariance]]). Then, its [[symplectic space]] is the [[cotangent bundle]] T*S with the canonical symplectic form ω. If we coordinatize ''T'' * ''S'' by its position ''x'' in the base manifold ''S'' and its position within the cotangent space '''p''', then we have a constraint
:''f'' = ''m''<sup>2</sup> −'''g'''(''x'')<sup>−1</sup>('''p''','''p''') = 0.
The Hamiltonian ''H'' is, surprisingly enough, ''H'' = 0. In light of the observation that the Hamiltonian is only defined up to the equivalence class of smooth functions agreeing on the constrained subspace, we can use a new Hamiltonian H'=f instead. Then, we have the interesting case where the Hamiltonian is the same as a constraint! See [[Hamiltonian constraint]] for more details.
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Consider now the case of a [[Yang-Mills theory]] for a real [[simple Lie algebra]] ''L'' (with a [[negative definite]] [[Killing form]] η) [[minimally coupled]] to a real scalar field σ, which transforms as an [[orthogonal representation]] ρ with the underlying vector space ''V'' under ''L'' in (''d'' − 1) + 1 [[Minkowski spacetime]]. For l in ''L'', we write
:ρ(l)[σ]
as
:l[σ]
for simplicity. Let '''A''' be the ''L''-valued [[connection form]] of the theory. Note that the '''A''' here differs from the '''A''' used by physicists by a factor of ''i'' and "g". This agrees with the mathematician's convention. The action ''S'' is given by
:<math>S[\bold{A},\sigma]=\int d^dx \frac{1}{4g^2}\eta((\bold{g}^{-1}\otimes \bold{g}^{-1})(\bold{F},\bold{F}))+\frac{1}{2}\alpha(\bold{g}^{-1}(D\sigma,D\sigma))</math>
where '''g''' is the Minkowski metric, '''F''' is the [[curvature form]]
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no ''i''s or ''g''s!) where the second term is a formal shorthand for pretending the Lie bracket is a commutator, ''D'' is the covariant derivative
:Dσ = dσ − '''A'''[σ]
and α is the orthogonal form for ρ.
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What is the Hamiltonian version of this model? Well, first, we have to split '''A''' noncovariantly into a time component φ and a spatial part <math>\vec{A}</math>. Then, the resulting symplectic space has the conjugate variables σ, π<sub>σ</sub> (taking values in the underlying vector space of <math>\bar{\rho}</math>, the dual rep of ρ), <math>\vec{A}</math>, <math>\vec{\pi}_A</math>, φ and π<sub>φ</sub>. for each spatial point, we have the constraints, π<sub>φ</sub>=0 and the [[Gaussian constraint]]
:<math>\vec{D}\cdot\vec{\pi}_A-\rho'(\pi_\sigma,\sigma)=0</math>
where since ρ is an [[intertwiner]]
:<math>\rho:L\otimes V\rightarrow V</math>,
ρ' is the dualized intertwiner
:<math>\rho':\bar{V}\otimes V\rightarrow L</math>
(L is self-dual via η). The Hamiltonian,
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