Thanks! Cool, it makes sense to me how we can make the pullback of L with Xn×R, in n different ways to get n different line bundles, and then tensor them all together. (I actually developed that hypothesis during a car ride earlier today :-p.)
(I’m still not quite sure what the syntax L⊗qi means, but presumably the idea is that there’s an automorphism on 1D vector fields that flips the sign, and we flip the sign of the negative-charge line bundles before tensoring everything together?)
(Also, fwiw, when I said “they’re all isomorphic to C”, I meant that I didn’t expect to figure much out by looking at a single fiber in isolation, and did not mean to imply that there was a canonical isomorphism; it’s clear to me that lacking access to a particular isomorphism is kinda the whole point. That said, I appreciate the pedagogy anyway! I prefer overexplanations to underexplanations whenever my convo-partner is up for generating them.)
The syntax L⊗q means ”L to the tensor power of q”. For q>0, it just means tensoring L with itself q times. For q=0, L⊗q is just the trivial line bundle with total space Y×C (and, yes, all line bundles are isomorphic to the trivial line bundle, but this one just is the trivial bundle… or at least, canonically isomorphic to it). For q<0, we need the notion of a dual vector bundle. Any vector bundle V has a dual V∗, and for a line bundle the dual is also the inverse, in the sense that L⊗L∗ is canonically isomorphic to the trivial bundle. We can then define all negative powers by L⊗q:=(L∗)⊗−q. Notice that non-negative tensor powers are defined for all vector bundles, but negative tensor powers only make sense for line bundles.
It remains to explain what is V∗. But, for our purposes we can take a shortcut. The idea is, for any finite-dimensional complex vector space U with an inner product, there is a canonical isomorphism between U∗ and ¯U, where ¯U is the complex-conjugate space. What is the complex-conjugate space? It is a vector space that (i) has the same set of vectors (ii) has the same addition operation and (iii) has its multiplication-by-scalar operation modified, so that multiplying u by z in ¯U is the same thing as multiplying u by ¯z in U, where ¯z is just the complex number conjugate to z.
Equipped with this observation, we can define the dual of a Hermitian line bundle L to be ¯L, where ¯L is the bundle obtained for L by changing its multiplication-by-scalar mapping in the obvious way.
Thanks! Cool, it makes sense to me how we can make the pullback of L with Xn×R, in n different ways to get n different line bundles, and then tensor them all together. (I actually developed that hypothesis during a car ride earlier today :-p.)
(I’m still not quite sure what the syntax L⊗qi means, but presumably the idea is that there’s an automorphism on 1D vector fields that flips the sign, and we flip the sign of the negative-charge line bundles before tensoring everything together?)
(Also, fwiw, when I said “they’re all isomorphic to C”, I meant that I didn’t expect to figure much out by looking at a single fiber in isolation, and did not mean to imply that there was a canonical isomorphism; it’s clear to me that lacking access to a particular isomorphism is kinda the whole point. That said, I appreciate the pedagogy anyway! I prefer overexplanations to underexplanations whenever my convo-partner is up for generating them.)
Thanks again!
The syntax L⊗q means ”L to the tensor power of q”. For q>0, it just means tensoring L with itself q times. For q=0, L⊗q is just the trivial line bundle with total space Y×C (and, yes, all line bundles are isomorphic to the trivial line bundle, but this one just is the trivial bundle… or at least, canonically isomorphic to it). For q<0, we need the notion of a dual vector bundle. Any vector bundle V has a dual V∗, and for a line bundle the dual is also the inverse, in the sense that L⊗L∗ is canonically isomorphic to the trivial bundle. We can then define all negative powers by L⊗q:=(L∗)⊗−q. Notice that non-negative tensor powers are defined for all vector bundles, but negative tensor powers only make sense for line bundles.
It remains to explain what is V∗. But, for our purposes we can take a shortcut. The idea is, for any finite-dimensional complex vector space U with an inner product, there is a canonical isomorphism between U∗ and ¯U, where ¯U is the complex-conjugate space. What is the complex-conjugate space? It is a vector space that (i) has the same set of vectors (ii) has the same addition operation and (iii) has its multiplication-by-scalar operation modified, so that multiplying u by z in ¯U is the same thing as multiplying u by ¯z in U, where ¯z is just the complex number conjugate to z.
Equipped with this observation, we can define the dual of a Hermitian line bundle L to be ¯L, where ¯L is the bundle obtained for L by changing its multiplication-by-scalar mapping in the obvious way.