1.2. Measures and forms [014V]
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1.2. Measures and forms
Any finite-dimensional real vector space comes equipped with a Lebesgue (or Haar) measure , uniquely defined up to a multiplicative constant. Any lattice allows us to normalize by .
To any top-dimensional differential form on a manifold is associated a positive measure on . For example, if is a lattice as above, is a basis of the dual lattice, then is Lebesgue measure on normalized by .
If is a complex manifold of dimension , and is a section of , that is, a holomorphic -form, we define as the positive measure
The normalization is chosen so that the measure associated to the form on is Lebesgue measure on .
This construction induces a natural bijection between smooth metrics on the canonical bundle and (smooth, positive) volume forms on , which associates to a smooth metric on the volume form locally defined by
for any local section of . If is another metric on , then
where is the usual exponential of the smooth function . This can be used to make sense of as a positive measure for any (possibly singular) metric on . Similarly, is a volume form for every metric on , .
Now assume is a pair in the sense of the Minimal Model Program, i.e. is a normal complex space and is a (not necessarily effective) -Weil divisor on such that
is a -line bundle. Denote by the canonical singular metric on , viewed as a -line bundle. If is smooth metric on the -line bundle , then is a smooth metric on , and is thus a volume form on .33 3 Here and in what follows, we write for the complement of the support of a (not necessarily reduced) divisor in a complex space .
A pair is subklt if for some (or, equivalently, any) log resolution of , the unique -divisor such that and has coefficients . The pair is klt if is further effective.
Lemma 1.1.
For any smooth metric on , is subklt if and only if the measure has locally finite mass near each point of .
Proof.
With the above notation it is immediate to check that
We are thus reduced to a log smooth pair , i.e. is smooth and has snc support, and the proof is then trivial. ∎