The Picard-Lefschetz theory of complexified Morse functions 1 ...

Complexified Morse functions 59
ι ∗ J = −J,
π(ι(p)) = π(p).
Then the real part Y = E ι is such that for each x ∈ Y \ Crit(π) the symplectic lift
ξ ∈ Tx(E) of ∂t ∈ Tπ(x)(R) (i.e. (Dπ)(ξ) = ∂t and ω(ξ, v) = 0 for all v ∈ Ker(Dπ))
satisfies
ξ ∈ T(Y),
and in fact
ξ = ∇gf/|∇gf |,
where π = F + iH , f = π|Y = F|Y , and g = ωJ (i.e. g(v, w) = ω(v, Jw)). This means
the symplectic transport preserves Y , and coincides with the unit speed gradient flow
of f = F|Y . (Of course, symplectic transport only makes sense on Y \ Crit(π).)
Proof Let p ∈ Y = E ι . We will show that ξp ∈ T(Y). First note that Tp(Y) = Tp(E) ι∗ .
ξ is characterized by: Dπ(ξ) = ∂t and ξ ∈ Ker(Dπ) ω . Let us check that ι∗(ξ) also
satisfies these. First,
Dπ(ι∗ξ) = (ιC)∗(Dπ(ξ)) = (ιC)∗(∂t) = ∂t.
Next, let v ∈ Ker(Dπ). Notice that ι∗(v) ∈ Ker(Dπ) also, by a similar calculation as
above. Thus
ω(ι∗ξ,ι∗v) = −ω(ξ, v) = 0
shows that ι∗ξ ∈ Ker(Dπ) ω .
Now let’s show that ξ = ∇f/|∇f |, where f = F|Y . First we show ξ = ∇F/|∇F|.
Using π∗(JV) = iπ∗(v), it is easy to check that ∇F = ±XH and ∇H = ±XF . Now
observe that
D(π) ω = span{XH, XF};
this follows at once from π −1 (z) = F −1 (a) ∩ (H) −1 (b), where z = a + bi, and
dimension considerations. From this it follows immediately that ∇F/|∇F| ∈ D(π) ω .
Furthermore
Dπ(∇F/|∇F|) = DF(∇F/|∇F|) = 1 · ∂t
because DH(∇F) = DH(±XH) = 0. Thus we have shown
ξ = ∇F/|∇F|.
Now, since we already checked that ξ ∈ T(Y), it follows that in fact ξ = ∇f/|∇f |,
where f = F|Y .