130 M. HUTCHINGS AND C. H. TAUBES, SEIBERG-WITTEN EQUATIONS
Now we come to the point. Consider an arbitrary Spine structure with S+ =
E EB K-^1 E. We can write any connection A on L = K-^1 E^2 as Ao+ 2a where Ao
is the canonical connection on K-^1 and a is a connection on E. Then the formula
for the Dirac operator is the following.
Lemma 4.5. If a is a connection on E, a is a section of E, and (3 is a section of
K-^1 E, then
(4.9) DAo+2a (~) = J2(Baa + B:(3).
(Here Ba n_O,k(X,E) ----t n_O,k+^1 (X ,E) denotes the antiholomorphic part of
\7 a·)
Proof. First consider the canonical Spine structure. By (4.4), the Leibniz rule as
in (4.8), and (4.7), we have
D Ao (a + (3) = D Ao ( (a + ~) · uo)
(d
d(3 + d* (3)
= a+ 2. uo
= h(Ba + B* (3).
(Since (3 is self-dual, cl( d(3) = cl( d* (3).) The case of a general Spin IC structure and
connection a follows formally from this. D
(When Xis Kahler, Ao comes from the Levi-Civita connection, and ( 4.9) follows
more straightforwardly from ( 4.4).)
Exercise 4.6. Prove Theorem 4.l(c).
4.4. Step 2 : Deforming the curvature equation
Since the Seiberg-Witten invariant does not depend on μ , we can choose μ to suit
our convenience. Normally one might think of choosing μ to be small, so as to
perturb the equations and get a smooth moduli space. But we will choose μ to be
very large. We take
μ = rw - iFf 0
where r is a large positive real number.
Let's also rescale by writing '!/J = ft(a,(3). Then the Seiberg-Witten equations
become, by (4.5) and (4.9),
(4.10) Baa= -B:(3,
(4.11)
We will later need to decompose the curvature equation ( 4.11) as follows. We have
Ai@ <C =<Cw EB K EB K-^1.
So (4.11) implies
(4.12)
( 4.13)
po,2 = ra(3
a '