§ 4. Differential Manifolds
229
velocity vector –, provided, once again, of not letting classical, but misleading
images mystify us.
Any tangent vector at a ∈ X can be obtained in this manner: if h corresponds to the vector h(ϕ) ∈ R
d in the chart (U, ϕ), it suffices to choose the
map
ϕ [μ(t)] = ϕ(a) + th(ϕ)
as μ. In particular, the vectors a i (ξ) of the basis for X
(a) correspond to the
trajectories t → ϕ(a) + te i in R
d , where (e i ) is the canonical basis. Denoting
the inverse map of ϕ by f : ϕ(U ) −→ U so that, for all ξ ∈ ϕ(U ), f (ξ) is
the point of X whose coordinates in the chart considered are precisely the
canonical coordinates ξ
i of the point ξ, the corresponding curve μ is obviously
the map
t −→ f
ξ
1 , . . . , ξ
i + t, . . . , ξ
d
,
where ξ = ϕ(a). The basis for X
(a) associated to the chart considered is
obtained by calculating the tangent vectors to these supposed “ curvilinear
coordinate axes ” at t = 0.
Suppose, for example, that X is an n-dimensional Cartesian space E,
and hence isomorphic but not identical to R
n . The simplest charts for E
are obtained by choosing a basis (a i ) for E and by associating to each x =
ξ
i a i ∈ E the point ϕ(x) = ξ
i e i of R
n . If (b α ) is another basis for E, we get
the formulas b α = g
i
α a i with an invertible matrix (g
i
α ) ; to calculate ψ(x) for
the new chart, set x = η
α b α , and so, by definition, ψ(x) = η
α e α . Then, as
ξ
i = g
i
α η
α , formula (1) can be applied with ρ
i
α (η) = g
i
α . Then consider a
vector h ∈ E
(x), where x is an arbitrary point of E. It has corresponding
vectors
h(ϕ) = h
i (ϕ)e i , h(ψ) = h
α (ψ)e α
in the charts (E, ϕ) and (E, ψ) thus constructed. The general formula (3)
shows that h
i (ϕ) = g
i
α h
α (ψ), which means that
h
i (ϕ)a i = h
α (ψ)g
i
α a i = h
α (ψ)b α .
So E
(x) can be canonically identified with the vector space E itself, by
the map h → h
i (ϕ)a i which, in conformity with Italian mechanics, does not
depend on the chosen basis.
Conversely, the simplest among the curves μ running through some x ∈ E
are the lines t → x + th, where h ∈ E is given. Hence, each h ∈ E canonically
defines an element of E
(x). It would be extremely surprising if the map
E −→ E
(x) thus defined was not the inverse of the one obtained by using
the bases for E; we leave it to the reader to check this. It is for good reason
229
velocity vector –, provided, once again, of not letting classical, but misleading
images mystify us.
Any tangent vector at a ∈ X can be obtained in this manner: if h corresponds to the vector h(ϕ) ∈ R
d in the chart (U, ϕ), it suffices to choose the
map
ϕ [μ(t)] = ϕ(a) + th(ϕ)
as μ. In particular, the vectors a i (ξ) of the basis for X
(a) correspond to the
trajectories t → ϕ(a) + te i in R
d , where (e i ) is the canonical basis. Denoting
the inverse map of ϕ by f : ϕ(U ) −→ U so that, for all ξ ∈ ϕ(U ), f (ξ) is
the point of X whose coordinates in the chart considered are precisely the
canonical coordinates ξ
i of the point ξ, the corresponding curve μ is obviously
the map
t −→ f
ξ
1 , . . . , ξ
i + t, . . . , ξ
d
,
where ξ = ϕ(a). The basis for X
(a) associated to the chart considered is
obtained by calculating the tangent vectors to these supposed “ curvilinear
coordinate axes ” at t = 0.
Suppose, for example, that X is an n-dimensional Cartesian space E,
and hence isomorphic but not identical to R
n . The simplest charts for E
are obtained by choosing a basis (a i ) for E and by associating to each x =
ξ
i a i ∈ E the point ϕ(x) = ξ
i e i of R
n . If (b α ) is another basis for E, we get
the formulas b α = g
i
α a i with an invertible matrix (g
i
α ) ; to calculate ψ(x) for
the new chart, set x = η
α b α , and so, by definition, ψ(x) = η
α e α . Then, as
ξ
i = g
i
α η
α , formula (1) can be applied with ρ
i
α (η) = g
i
α . Then consider a
vector h ∈ E
(x), where x is an arbitrary point of E. It has corresponding
vectors
h(ϕ) = h
i (ϕ)e i , h(ψ) = h
α (ψ)e α
in the charts (E, ϕ) and (E, ψ) thus constructed. The general formula (3)
shows that h
i (ϕ) = g
i
α h
α (ψ), which means that
h
i (ϕ)a i = h
α (ψ)g
i
α a i = h
α (ψ)b α .
So E
(x) can be canonically identified with the vector space E itself, by
the map h → h
i (ϕ)a i which, in conformity with Italian mechanics, does not
depend on the chosen basis.
Conversely, the simplest among the curves μ running through some x ∈ E
are the lines t → x + th, where h ∈ E is given. Hence, each h ∈ E canonically
defines an element of E
(x). It would be extremely surprising if the map
E −→ E
(x) thus defined was not the inverse of the one obtained by using
the bases for E; we leave it to the reader to check this. It is for good reason