There are two operations involved in the definition of W: pullback and tensor product.
Pullback is defined for arbitrary bundles. Given a mapping f:X→Y (these X and Y are arbitrary manifolds, not the specific ones from before) and a bundle B over Y with total space TotB and projection mapping prB:Tot(B)→Y, the pullback of B w.r.t. f (denoted f∗B) is the bundle over X with total space Tot(B)×YX and the obvious projection mapping. I remind that Tot(B)×YX is the fibre product, i.e. the submanifold of Tot(B)×X defined by prB(t)=f(x). Notice that the fibre of f∗B over any x∈X is canonically isomorphic to the fibre of B over f(x). The word “canonical” means that there is a particular isomorphism that we obtain from the construction.
It is easy enough to see that the pullback of a vector bundle is a vector bundle, the pullback of a line bundle is a line bundle, and the pullback of a Hermitian vector bundle is a Hermitian vector bundle.
Tensor product is an operation over vector bundles. There are different ways to define it, corresponding to the different ways to define a tensor product of vector spaces. Specifically for line bundles there is the following shortcut definition. Let L1 and L2 be line bundles over X. Then, the total space of L1⊗L2 is the quotient of L1×XL2 by the equivalence relation given by: (v1,v2)∼(w1,w2) iff v1⊗v2=w1⊗w2. Here, I regard v1,w1 as vectors in the vector space which is the corresponding fibre fo L1 and similarly for v2,w2 and L2. The quotient of a manifold by an equivalence relation is not always a manifold, but in this case it is.
I notice that you wrote “a particular fiber is isomorphic to C”. Your error here is, it doesn’t matter what it’s isomorphic to, you should still think of it as an abstract vector space. So, if e.g.V1 and V2 are 1-dimensional vector spaces, then V1⊗V2 is yet another “new” vector space. Yes, they are all isomorphic, but they are not canonically isomorphic.
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.
There are two operations involved in the definition of W: pullback and tensor product.
Pullback is defined for arbitrary bundles. Given a mapping f:X→Y (these X and Y are arbitrary manifolds, not the specific ones from before) and a bundle B over Y with total space TotB and projection mapping prB:Tot(B)→Y, the pullback of B w.r.t. f (denoted f∗B) is the bundle over X with total space Tot(B)×YX and the obvious projection mapping. I remind that Tot(B)×YX is the fibre product, i.e. the submanifold of Tot(B)×X defined by prB(t)=f(x). Notice that the fibre of f∗B over any x∈X is canonically isomorphic to the fibre of B over f(x). The word “canonical” means that there is a particular isomorphism that we obtain from the construction.
It is easy enough to see that the pullback of a vector bundle is a vector bundle, the pullback of a line bundle is a line bundle, and the pullback of a Hermitian vector bundle is a Hermitian vector bundle.
Tensor product is an operation over vector bundles. There are different ways to define it, corresponding to the different ways to define a tensor product of vector spaces. Specifically for line bundles there is the following shortcut definition. Let L1 and L2 be line bundles over X. Then, the total space of L1⊗L2 is the quotient of L1×XL2 by the equivalence relation given by: (v1,v2)∼(w1,w2) iff v1⊗v2=w1⊗w2. Here, I regard v1,w1 as vectors in the vector space which is the corresponding fibre fo L1 and similarly for v2,w2 and L2. The quotient of a manifold by an equivalence relation is not always a manifold, but in this case it is.
I notice that you wrote “a particular fiber is isomorphic to C”. Your error here is, it doesn’t matter what it’s isomorphic to, you should still think of it as an abstract vector space. So, if e.g.V1 and V2 are 1-dimensional vector spaces, then V1⊗V2 is yet another “new” vector space. Yes, they are all isomorphic, but they are not canonically isomorphic.
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.