
44 2 Connections
The word “local” means that the operator is constructed and calculated
in a certain chart U
α
on M with respect to a certain trivialization of π
−1
U
α
.
Theorem 2.16 The operator pΓ
m
(·, ·) is linear in the second argument.
Indeed, pΓ
m
(ϑ, Y
1
)=Tπ
−1
(Y
1
)
|H
E
(m,ϑ)
− Tπ
−1
(Y
1
)
|H
(m,ϑ)
.Sincetheop-
eration Tπ
−1
and the operation of taking the difference are both linear,
pΓ
m
(ϑ, Y
1
) is linear in Y
1
.
Theorem 2.17 The operator pΓ
m
(·, ·) is linear in the first argument.
To prove Theorem 2.17 we need the following:
Lemma 2.18 Let B : E → E be a map in the vector space E, smooth and
homogeneous with degree 1.ThenB is a linear operator.
Proof. (of Lemma 2.18) Recall that B is homogeneous with degree k if for
any vector X ∈ E and any λ ∈ R we have B(λX)=λ
k
B(X). From the
homogeneity it follows that B(0) = 0.
Since B is smooth, we can expand it by the Taylor formula in a neighbor-
hood of 0 ∈ E up to a certain degree greater than 1. Thus, since B(0) = 0,
B(X)=B
(X)+
1
2
B
(X, X)+... where B
is the first derivative of B at
the origin (recall that B
is a linear operator), B
is the second derivative of
B at the origin (recall that B
is a bilinear operator), etc. On the right-hand
side only B
is homogeneous with degree 1; B
(X, X) is homogeneous with
degree 2 and the other summands have greater degrees of homogeneity. Thus
the left-hand side is homogeneous with degree 1 only if all summands on the
right hand side except B
are equal to zero. Hence B = B
and so it is a
linear operator.
Proof. (of Theorem 2.17) Since by Definition 2.8(i) both H
(m,ϑ)
and H
E
(m,ϑ)
are smooth in ϑ,sotooispΓ
m
(ϑ, Y
1
). We shall show that pΓ
m
(ϑ, Y
1
)is
homogeneous with degree 1 in ϑ so that the statement of Theorem 2.17 will
follow from Lemma 2.18.
The vector HY = Tπ
−1
(Y
1
) ∈ H
(m,ϑ)
is presented as a quadruple in the
form (m, ϑ, Y
1
,Γ
m
(ϑ, Y
1
)). By Definition 2.8(ii) and by formula (2.8)(de-
scribing Ta) the vector Ta(HY )=(m, aϑ, Y
1
,aΓ
m
(ϑ, Y
1
)) belongs to H
(m,aϑ)
.
Using formula (2.3)wegetTπ((m, aϑ, Y
1
,aΓ
m
(ϑ, Y
1
)) = (m, Y
1
). Since Tπ
is one-to-one on H
(m,aϑ)
, it is the unique vector in H
(m,aϑ)
whose image
under Tπ is (m, Y
1
). But the vector HY = Tπ
−1
(Y
1
) ∈ H
(m,aϑ)
,whose
quadruple takes the form (m, aϑ, Y
1
,Γ
m
(aϑ, Y
1
)), also has this property:
Tπ(m, aϑ, Y
1
,Γ
m
(aϑ, Y
1
)) = (m, Y
1
). Hence,
(m, aϑ, Y
1
,aΓ
m
(ϑ, Y
1
)) = (m, aϑ, Y
1
,Γ
m
(aϑ, Y
1
))
and so, since p is linear, pΓ
m
(aϑ, Y
1
)=apΓ
m
(ϑ, Y
1
).