Lectures on String Theory

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It is divergent since the integrand depends on x − y only. Also one sees that the
group of translations
x → x + a, y → y + a
leaves the measure invariant.
x
Orbits x−y=const
y
x+y=0
gauge slice
intersect each orbit
at one point
Fig. 4. Divergence caused by a presence of the translational symmetry. One
integrates over the orbits of the gauge group while both the measure and
the integrand are translation invariant.
Let us split the coordinates (x, y) as
( x − y
(x, y) = , y − x )
} 2 {{ 2 }
x−y
2 + y−x
2 =0 +
( x + y
, x + y )
} 2 {{ 2 }
shift by (a,a),
a= x+y
2
This suggests to introduce new coordinates “along” the gauge orbit and “orthogonal”
to it:
(x, y) → (u, v) u = x − y, v = x + y,
i.e. x = u+v and y = u−v . Then the integral takes the form
2 2
Z = 1 ∫ +∞
(∫ +∞
) √ ∫ π +∞
e −u2 du dv = d(x + y) = √ π
2 −∞ −∞
2 −∞
} {{ }
√ π
∫ +∞
−∞
da .
} {{ }
volume
This example illustrates the basic idea to define the path integral – one has to divide
the original Z by the infinite volume of a symmetry group. Thus, our discussion
suggests that a proper definition of the path integral in string theory should be
∫
1
Z =
Dh αβ (σ, τ)DX µ (σ, τ)e iSp[X,h] ,
V Diff V Weyl
where we divided over the infinite volumes of the reparametrization and Weyl groups.