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In the definition of pre-quantization of representation $f\to \hat{f}$, (here $\hat{f}$ is Hermitian operator)of $C^{\infty}(M)$ on $L^2(M,L,\mu)$ where $\mu$ is Hermitian form, suppose that there exists some $(L,\mu,\Delta)$, such that $\Delta$ has the curvature $\omega$. Let given $(L,\mu)$, then why the choices of $\Delta$, should be parametrized by $\frac{H^1(M,\mathbb{R})}{H^1(M,\mathbb{Z})}$.?

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This is a general fact in differential cohomology, which in degree two classifies Hermitian line bundles with connection.

Differential cohomology has an exact sequence $$ 0 \to \frac{H^{k-1}(M,\mathbb{R})}{H^{k-1}(M,\mathbb{Z})} \to \hat H^k(M,\mathbb{Z}) \to H^k(M,\mathbb{Z}) \times_{H^k(M,\mathbb{R})} \Omega^k_{cl}(M). $$

In your case, the pair $((L,\mu),\omega)$ is an element in the group on the right hand side, and the possible prequantizations are those elements in the middle mapping to that element.

EDIT: The third arrow in above sequence sends a Hermitian line bundle with connection $(L,\mu,\Delta)$ to the pair $(c_1(L),R^{\Delta})$ consisting of the first Chern class of $L$ and the curvature of $\Delta$. Two choices of $\Delta$ lead to two different elements $(L,\mu,\Delta)$ and $(L,\mu,\Delta')$ in $\hat H^2(M,\mathbb{Z})$. By your assumption that $\Delta$ and $\Delta'$ have the same curvature, the difference between the two elements lies in the kernel of the third arrow, and hence in the image of the second.

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Konrad Waldorf@ I need a little bit explanation about your answer. Why choices of $\Delta$ should be parametrized by $\frac{H^1(M,\mathbb{R})}{H^1(M,\mathbb{Z})}$. – baba ab dad Jul 3 '13 at 22:08
Please specify: are you asking (A) Why is above sequence exact, or (B) Why does exactness answer the question? – Konrad Waldorf Jul 4 '13 at 7:59
I am asking about B) – baba ab dad Jul 4 '13 at 8:26
I'll edit the question and add an explanation. – Konrad Waldorf Jul 4 '13 at 20:31

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