ManifoldGroupTesting.jl

check_adjoint_action(G, grp_rep, alg_rep, χ, ξ)

The group representation $ρ$ and algebra representation $ρ$ commute with the adjoint action:

\[ρ(χ) ρ(ξ) ρ(χ)^{-1} = ρ(\mathrm{Ad}_χ (ξ))\]

check_adjoint_action_in_alg(G, χ, ξ)

The adjoint action of $χ$ on $ξ$ is an element of $\mathrm{Alg}(G)$:

\[\mathrm{Ad}_{χ}ξ ∈ \mathrm{Alg}(G)\]

check_adjoint_action_lie_bracket(G, χ, ξ1, ξ2)

Adjoint action commutes with Lie bracket.

\[χ [ξ_1, ξ_2] χ^{-1} = [χξ_1χ^{-1}, χξ_2χ^{-1}]\]

check_alg_rep(G, alg_rep, ξ1, ξ2)

The algebra representation $ρ$ is an algebra morphism.

\[ρ([ξ_1, ξ_2]) = [ρ(ξ_1), ρ(ξ_2)]\]

where the latter is a matrix commutator.

check_apply_diff_group_at_id(G, side::GroupActionSide)

The covariant group operation action on itself $α(χ_1, χ_2)$ is either (left side)

\[α(χ_1, χ_2) = χ_1 χ_2\]

or (right side)

\[α(χ_1, χ_2) = χ_2 χ_1^{-1}\]

Now fix $χ_2 = 1$ ($1$ is the identity of $G$) and define $f : G → G$ by $f(χ) := α(χ, 1)$. Since $f(1) = 1$, its differential at identity is a map $\mathrm{Alg}(G) → \mathrm{Alg}(G)$. This map is either

• $I$ (left side)
• $-I$ (right side)

check_exp_ad(G, ξ_1, ξ_2)

$\mathrm{Ad}$ and $\exp$ commute in the sense that

\[{\exp(ξ_1)}ξ_2 \exp(-ξ_1) = \exp([ξ_1, \cdot]) ξ_2\]

where the second exponential is the matrix exponential for the linear operator $ξ ↦ [ξ_1, ξ]$.

check_exp_exp_(G, exp, exp!, χ_, χ, v)

Compare exp and exp!.

check_exp_invariant(G, exp, χ, v, χ_, conv=(LeftAction(), LeftSide()))

The invariant exponential of a Lie group fulfils

\[χ' \exp_{χ}(v) = \exp_{χ'χ}(χ' v)\]

There is a right version which is:

\[\exp_χ(v) χ' = \exp_{χχ'}(v χ')\]

The group exponential commutes with the adjoint action:

\[\exp(χξχ^{-1}) = χ\exp(ξ)χ^{-1}\]

check_exp_lie_point(G, ξ)

The Lie group exponential sends the vector $ξ \in \mathfrak{g}$ in the algebra to an element in the group.

\[\exp(ξ)\in G\]

check_exp_log(G, exp, log, χ1, χ2)

Check the identity

\[\exp_{χ_1}(\log_{χ_1}(χ_2)) = χ_2\]

check_grp_rep_Identity(G)

The representation works at the point Identity(G).

check_grp_rep_compose(G, ρ, χ1, χ2)

The group representation $ρ$ commutes with composition.

\[ ρ(χ_1 χ_2) = ρ(χ_1) ρ(χ_2)\]

where the latter is a matrix multiplication.

check_inv_rep(G, grp_rep, χ)

The group representation $ρ$ commutes with the inverse.

\[ρ(χ)^{-1} = ρ(χ^{-1})\]

check_log_exp(G, log, exp, χ, v)

Check the identity

\[\log_{χ}(\exp_{χ}(v)) = v\]

check_log_log_(G, log, log!, v_, χ1, χ2)

Compare log and log!.

check_zero_Identity(G)

The zero vector at Identity(G) is the same as the zero vector at identity_element(G).

check_action_morphism(α::GroupAction, χ1, χ2, p)

For a group action $α$,

\[α(χ_1, α(χ_2, p)) = α(m(χ_1, χ_2), p)\]

where for a left action

\[m(χ_1, χ_2) = χ_1 χ_2\]

and for a right action

\[m(χ_1, χ_2) = χ_2 χ_1\]

check_apply_diff_linear(A::AbstractGroupAction, χ, p, X, Y, λ=1.)

If the action $α$ is a group action, its differential with respect to the point $p$ is linear. Fix $χ ∈ G$, define $π_χ(p) ≔ α(χ, p)$ and $q ≔ π_{χ}(p)$. We have

\[⟨Dπ_{χ}, X + λY ⟩_p = ⟨Dπ_{χ}, X⟩_q + λ ⟨Dπ_{χ}, Y⟩_q\]

check_apply_morphism_Identity(α::AbstractGroupAction, p)

For any action $α$ and $p$ in the manifold, one has

\[α(1, p) = p\]

where $1$ denotes Identity(G) and $G$ is the action group.

check_switch_action_direction(α::GroupAction, χ, p)

For any group action $α$ and its dual $α'$ obtained by switch_direction one has

\[α(χ,p) = α'(χ^{-1}, p)\]

check_trivial_infinitesimal_action(A::GroupAction, p)

For an action $α$ and a point $p∈ M$, consider the function $f : G → M$ defined as $f(χ) := α(χ, p)$. The tangent map at identity $D := T f(1)$ is a linear map and sends zero to zero:

\[D 0 = 0\]

check_apply_diff_group(
    A::AbstractGroupAction{TAD}, # group action G ⊂ Diff(M)
    χ, # element of G
    ξ, # element of Alg(G)
    p, # element of M
    id_func, # `identity_element` or `Identity`
)

This should hold for any group action $α$ on any manifold. If you define $π_p(χ) := α(χ, p)$ for $χ ∈ G$ and $p ∈ M$, and define, for $ξ ∈ Alg(G)$, $T_R(χ, ξ) := ξχ$ (the right translation), and $T_L(χ, ξ) := χξ$ (the left translation), then we have the identity:

\[⟨Dπ_{p}(χ), T(χ, ξ)⟩ = ⟨Dπ_{α(χ,p)}(1), ξ⟩\]

where, for a left action, $T$ is the right translation, and for a right action, $T$ is the left translation.

check_inv_diff(
  G, # Group
  χ, # group element
  ξ, # Lie algebra element
  side::Manifolds.GroupActionSide,
  )

Test the differential of the inverse on a Lie group G. Denote this inverse by $I(χ) := χ^{-1}$. If the left and right transports are $T_L(χ,ξ) := χξ$ and $T_R(χ,ξ) := ξχ$ respectively, then

\[⟨DI(χ), T_L(χ,ξ)⟩ = -T_R(χ^{-1}, ξ)\]

and

\[⟨DI(χ), T_R(χ,ξ)⟩ = -T_L(χ^{-1}, ξ)\]