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In Dan Freed's notes Exercise 9.30 he outlines the proof of the existence criterion which is that there exists $\tilde{c} \in H^2(M;\mathbb{Z})$ such that $2\tilde{c} = c_1(E)$. His approach is to pass to the determinant line bundle Det$(E) \to M$ of a complex vector bundle and work on $c_1(\operatorname{Det}(E))$ via its equality to $c_1(E)$. I actually do not appreciate the benefit of passing to determinant line bundles and in particular I have no idea about (1) how to prove the first Chern class (step ii) there and (2) how to define the suggested Lie group homomorphism (step iii) there. Could somebody please help me?

By the way I have asked a question on the same topic but for a different approach given in Cohen's notes on the topology of the fiber bundles. If you have some idea about it please do me a favor drop it there!

Community
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PhysicsMath
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1 Answers1

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Here are several hints.

For step (ii), the splitting principle allows you to prove the identity $c_1(E) = c_1(\operatorname{Det} E)$ under the additional assumption that $E = L_1 \oplus \cdots \oplus L_n$. It might be useful to know that $$\operatorname{Det}(L_1 \oplus \cdots \oplus L_n) \cong L_1 \otimes \cdots \otimes L_n.$$ This follows from induction and the fact that $\Lambda^n (V \oplus W) \cong \bigoplus_{i + j = n} \Lambda^i(V) \otimes \Lambda^j(W)$.

For step (iii), the existence of a continuous map $\tilde{U}(1) \to \mathbb{T}$ follows from the lifting criterion for covering spaces. However, to show that you have a Lie group homomorphism requires a little bit more work. First, you need to pick to the lift that sends the identity to the identity. Once you do, the uniqueness of lifts will show that it is a group homomorphism. Smoothness can checked locally, and all the covering maps are local diffeomorphisms while $\det$ is polynomial and hence smooth. So indeed we have a homomorphism of Lie groups.

Finally, the reason why you want to consider the determinant bundle is that it's a line bundle, and for line bundles $L_1, L_2$ we have the formula $c_1(L_1 \otimes L_2) = c_1(L_1) + c_1(L_2)$. The argument in step (iv) is to show that using (iii), one can produce a line bundle $\sqrt{\operatorname{Det} E}$ whose tensor square is $\operatorname{Det}(E)$. Using the formula for the first Chern class of tensor products of line products, this completes the proof of the proposition.

JHF
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  • great answer! Especially the two key ideas: (1) $\operatorname{Det}(L_1 \oplus \cdots \oplus L_n) \cong L_1 \otimes \cdots \otimes L_n$ and (2) $\sqrt{\operatorname{Det} E}$ whose tensor square is $\operatorname{Det}(E)$ which make the benefit of using determinant line bundle more than obvious! I wish Freed could have included something similar to what you wrote here! By the way what I see here is that you cross out a few sentences. I am not sure whether it is because you want to emphasize that is your guess. If not could you polish it so that we can make it the official answer? – PhysicsMath Jul 29 '16 at 00:07
  • to show that it is a Lie group homomorphism you mentioned that as long as I make sure that the lift sends the identity to the identity the uniqueness of lifts will show that it is a group homomorphism. What theorem is that? Could you give a reference? – PhysicsMath Jul 29 '16 at 00:13
  • I'll remove the crossed out part if you want. For the group homomorphism, I'm referring to the argument that if $\phi: \tilde{U}(n) \to \mathbb{T}$ is the lift, then there are two lifts $\tilde{U}(n) \times \tilde{U}(n) \to \mathbb{T}$ corresponding to $(x,y) \mapsto \phi(xy)$ and $(x,y) \mapsto \phi(x)\phi(y)$. If $\phi$ sends the identity $e$ to the identity that these two maps agree at the point $(e,e)$, so must be the same map. This is a variant of the argument used in the proof of theorem 7.7 in John Lee's Introduction to Smooth Manifolds. – JHF Jul 29 '16 at 00:49
  • correct me if I am wrong. You mention that the existence of a continuous map $\tilde{U}(1)→\mathbb{T}$ follows from the lifting criterion for covering spaces. But as far as I can recall and by what is said in the Wikipedia the lifting property refers to maps of the opposite direction i.e. from $\mathbb{T} \to U(n)$ to $\mathbb{T} \to \tilde{U}(n)$. Are you not referring to this lifting property of the covering $\tilde{U}(n) \to U(n)$? – PhysicsMath Jul 29 '16 at 00:59
  • Apply the lifting criterion to the double cover $\mathbb{T} \rightarrow \mathbb{T}$ and the map $\tilde{U}(n) \rightarrow U(n) \rightarrow \mathbb{T}$ instead. – JHF Jul 29 '16 at 02:04
  • I see. So it is applied to $\mathbb{T} \to \mathbb{T}$ instead of $\tilde{U}(n) \to U(n)$. Got you JHF! – PhysicsMath Jul 30 '16 at 18:24