Contact Geometry of second order I - Dept. Math, Hokkaido Univ ...

3. Symbols of Second Order Equations.
In view of the Bäcklund Theorem, we will discuss the symbols of second order equations
as the first invariants in the Contact Geometry of Second Order.
3.1. Symbol Algebras. Let R be a submanifold of L(J) satisfying the following condition:
(R.0) p : R → J ; submersion,
where p = π |R and π : L(J) → J is the projection. This condition implies that the system
of equations R of second order contains no equations of first order. We have two differential
systems C 1 = ∂E and C 2 = E on L(J). We denote by D 1 and D 2 those differential
systems on R obtained by restricting C 1 and C 2 to R. Moreover we denote by the same
symbols those 1-forms obtained by restricting the defining 1-forms {ϖ, ϖ1, · · · , ϖn} of
the canonical system E to R, where ϖ = dz − ∑ n
i=1 pidxi, and ϖi = dpi − ∑ n
j=1 pijdxj
(i = 1, . . . , n). Then it follows from (R.0) that these 1-forms are independent at each
point on R and that
D 1 = {ϖ = 0}, D 2 = {ϖ = ϖ1 = · · · = ϖn = 0}.
Thus D 1 and D 2 are subbundles of T (R) such that ∂D 2 ⊂ D 1 . Hence subbundles D 2 , D 1
and T (R) define a filtration on R. Namely, putting D −1 = D 2 , D −2 = D 1 , D p = T (R)
for p ≦ −3, we have
[D p , D q ] ⊂ D p+q
for p, q < 0,
where D p = Γ(D p ).
Now we define the Symbol Algebra s(v) of R at v ∈ R by
s(v) = s−3(v) ⊕ s−2(v) ⊕ s−1(v),
where s−3(v) = Tv(R)/D 1 (v), s−2(v) = D 1 (v)/D 2 (v) and s−1(v) = D 2 (v). The bracket
operation in s(v) is defined, similarly as in §2.3, as follows: For X ∈ sp(v) and Y ∈ sq(v),
let us take ˜ X ∈ D p and ˜ Y ∈ D q such that X = πp(( ˜ X)v) and Y = πq(( ˜ Y )v),where
πp : D p (v) → sp(v) is the projection. Then the bracket product is defined by
[X, Y ] = πp+q([ ˜ X, ˜ Y ]v) ∈ sp+q(v).
The bracket product [X, Y ] is well-defined for X ∈ sp(v) and Y ∈ sq(v), i.e., is independent
of the choice of ˜ X and ˜ Y . In fact, in our case, this can be shown as follows: The defining 1forms
ϖ, ϖ1, . . . , ϖn for D 1 and D 2 actually define a basis {A} of s−3(v) and {B1, . . . , Bn}
of s−2(v) such that ϖ( Ã) = 1, π−3( Ã) = A, ϖi( ˜ Bj) = δ i j, π−2( ˜ Bi) = Bi and ˜ Bi ∈ D 1 (v).
Then, for X1, X2 ∈ s−1(v) = D 2 (v), we calculate
dϖi(X1, X2) = ˜ X1(ϖi( ˜ X2)) − ˜ X2(ϖi( ˜ X1)) − ϖi([ ˜ X1, ˜ X2]) = −ϖi([ ˜ X1, ˜ X2]).
Thus, putting βi = −dϖi(X1, X2), we get
[X1, X2] = β1B1 + · · · + βnBn ∈ s−2(v).
For X ∈ s−1(v) and Y ∈ s−2(v), we calculate
dϖ(X, ˜ Yv) = ˜ Xv(ϖ( ˜ Y )) − ˜ Yv(ϖ( ˜ X)) − ϖ([ ˜ X, ˜ Y ]v) = −ϖ([ ˜ X, ˜ Y ]v).
Similarly we have dϖ(X1, X2) = 0 for X1, X2 ∈ s−1(v). Thus dϖ(X, ˜ Yv) depends only on
X ∈ s−1(v) and Y ∈ s−2(v). Hence, putting α = −dϖ(X, ˜ Yv), we have
[X, Y ] = αA ∈ s−3(v).
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