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Let $X$ be a manifold, $L$ be a complex line bundle over $X$, and $L^{*}$ be the associated principal bundle. Suppose $\alpha$ is a connection form on $L^{*}$, with associated connection $D$ on $L$. Suppose we wish to define a Hermitian product $\langle,\rangle$ on $L$ compatible with $D$ in the sense that for every pair of sections $s, s'$ and vector field $v$, we have $v\langle s,s'\rangle=\langle D_{v}s,s'\rangle+\langle s,D_{v}s'\rangle$. According to Sniatycki's book 'Geometric Quantization,' such a pairing exists iff $2\pi i(\alpha - \overline{\alpha})$ is exact. However, I cannot figure out a proof of this, and he does not provide one.

Does anyone know where I can find a proof of this, or could anyone provide the proof here?

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Please use \langle and \rangle instead of < and >, this makes things much more readable. – MTS Sep 20 '11 at 0:50
up vote 1 down vote accepted

Wanted to clarify one potentially confusing difference between Śniatycki's and Prof. Figueroa-O'Farrill's notation: $\beta$ is actually the local representative of the connection $2\pi i~\alpha$ with respect to the non-vanishing local section $\sigma$, i.e., its pullback under $\sigma$ $$ \beta = 2\pi i~\sigma^*\alpha~. $$ Then Prof. Figueroa-O'Farrill's equation $ Ds = (df_s + \beta f_s) \sigma $ is equivalent to Śniatycki's equation (3.17), $ \nabla_X\sigma = 2\pi i~\sigma^*\alpha(X)~\sigma. $

The function $h$ mentioned above is the logarithm of (the positive-valued) $\langle\sigma,\sigma\rangle$ $$ h = \ln\langle\sigma,\sigma\rangle = \sigma^{*}g, $$ where $g$ is the function $p\mapsto\ln\langle p,p\rangle$ on the non-vanishing part of the line bundle $L$.

Hence $dh=\beta+\overline{\beta}$ becomes $ d(\sigma^* g) = 2\pi i(\sigma^* \alpha - \overline{\sigma^* \alpha})~, $ or, since $\sigma$ is arbitrary and pullback commutes with exterior derivative $$ dg = 2\pi i(\alpha-\overline{\alpha}). $$

Conversely, if such a $g$ exists, the quadratic form $e^g$ (necessarily positive since $g$ is necessarily real) will give your desired pairing via the polarization identity, and this pairing will be unique up to a positive constant for connected base space.

The geometric interpretation of this condition is as follows: if $p\in L-\textrm{zero section}, u\in \mathbb{R}\simeq\mathfrak{u}(1)$, and $\eta_u(p) = \frac{d}{dt}e^{2\pi i ut}p\big\vert_{t=0}$ is the fiber $U(1)$-generator at $p$ (which by definition satisfies $\alpha(\eta_u(p))=u$ in Śniatycki's conventions), then $dg = 2\pi i(\alpha-\overline{\alpha})$ gives $$ \eta_u(p)g = 0 $$ or in other words $g(e^{2\pi i u}p) = g(p)$, as would be expected if $g$ is to equal $\ln\langle\cdot,\cdot\rangle$ for some pairing. Correspondingly, the fiber radial generators are given by $\eta_{-iv},~v\in\mathbb{R}$, and $$ \eta_{-iv}(p)g = 4\pi v $$ or $g(e^{2\pi v}p) = 4\pi v+g(p)$ for $p\in L-\textrm{zero section}, v\in\mathbb{R}$, again as would be expected.

The best place to learn the full theory is still Kostant's original "Quantization and Unitary Representations", published in Lecture Notes in Mathematics 170. It can be tough going at times, though.

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I'm not sure about your conventions and I don't have Sniatycki's book handy. However the following should be easily adaptable to your conventions.

Choose a local trivialisation of the line bundle. This is the same as choosing a local nowhere-vanishing section $\sigma$. Then any other section $s$ can be written as $f_s \sigma$, for some complex-valued function $f_s$. Let $\langle \sigma,\sigma\rangle = e^h$, for some real function $h$. Then the hermitian inner product of two sections $s,s'$ is given in that local trivialisation by $$ \langle s,s'\rangle = e^h \overline{f_s} f_{s'}~. $$

The covariant derivative of a section $s$ is given by $$ Ds = (df_s + \beta f_s) \sigma~, $$ for some connection 1-form $\beta$. (My $\beta$ is your $\alpha$ in conventions such that $2=\pi=i=1$ in the immortal words of Warren Siegel.)

The compatibility condition says that $$ d\langle s,s'\rangle = \langle Ds, s'\rangle + \langle s, Ds'\rangle~. $$ Using the definitions above one easily finds that this is equivalent to $$ \beta + \overline{\beta} = dh~. $$

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