Connection of $\mathcal{O}(n)$ on a toric manifold

Question

The holomorphic line bundle $\mathcal{O}_X(1)$ over a toric manifold $X$, admits a hermitian connection, $A^{(1)}$, whose $U(1)$ gauge transformation in a local patch of the base space is $$ A^{(1)}_Idx^I\rightarrow A^{(1)}_Idx^I-\textrm{ }d\lambda, $$ where $x^I$ are sections of the bundle. On page 61 of https://arxiv.org/abs/hep-th/0005247, it is assumed without explanation that the hermitian connection of $\mathcal{O}_X(-n)$ is just $$ A^{(-n)}=-nA^{(1)} $$ Why is this true? References would be appreciated.

Answer 1

$\mathcal{O}(-n)=\mathcal{O}(-1)^{\otimes n}$ is a tensor product of line bundles, so the connection is simply $n$ times the connection on $\mathcal{O}(-1)$. If that was not obvious, it will be useful if we review the basic ingredients involved here, which will hopefully clear up any confusion.

Consider a principal $G$-bundle $P$ over a manifold $X$. Suppose we have a connection $A$ on $P$, i.e. a locally defined $1$-form valued in the Lie algebra $\mathfrak{g}$ of $G$. Let $\rho:G \to \mathrm{Aut}(V)$ be a representation of $G$ on the vector space $V$, and let $\hat\rho: \mathfrak{g} \to \mathrm{Aut}(V)$ be the induced representation of $\mathfrak{g}$ on $V$.

Let $E_\rho = P \times_\rho V$ be the vector bundle associated to $P$ via the representation $\rho$. The connection $A$ on $P$ defines the covariant derivative acting on sections of $E_\rho$ by $$\nabla \equiv \mathrm{d} + \hat\rho(A),$$ where $\mathrm{d}$ is the differential on $X$.

Suppose $\rho':G\to \mathrm{Aut}(V')$ is another representation of $G$, and let $E_{\rho'}$ denote the associated vector bundle. Then we can form the tensor product bundle $E_\rho\otimes E_{\rho'}$ whose fibers are the tensor products of the fibers of $E_\rho$ and $E_{\rho'}$. The covariant derivative acting on sections of $E_\rho\otimes E_{\rho'}$ is $$\nabla = \mathrm{d} + (\hat\rho \otimes \hat\rho')(A),$$ where $$(\hat\rho\otimes \hat\rho')(A) = \hat\rho(A)\otimes 1 + 1\otimes \hat\rho'(A).$$

Now, consider the simple case $G = \mathrm{U(1)}$. Its irreducible representations are $1$-dimensional and are labeled by an integer $q\in \mathbb{Z}$, $\rho_q :\mathrm{U}(1) \to \mathrm{Aut}(\mathbb{C})$ with $\rho_q(e^{i\theta}) = e^{iq \theta}$. The associated Lie algebra representation is simply multiplication by $q$. The covariant derivative on $E_{\rho_q}$ is therefore simply $\nabla = \mathrm{d} + q A$. The covariant derivative on $E_{\rho_q} \otimes E_{\rho_{q'}}$ is then $\nabla = \mathrm{d} + (q+q')A$, since the tensor product is trivial for these $1$-dimensional representations.

Thus, if $P$ is a $\mathrm{U(1)}$ bundle with connection $A$ and $\mathcal{L}$ is the charge $1$ associated line bundle with covariant derivative $\nabla = \mathrm{d} + A$, then the covariant derivative on $\mathcal{L}^{\otimes n}$ is $\nabla = \mathrm{d} + n A$.

For the original question, $\mathcal{O}(1)$ is a line bundle with covariant derivative $\mathrm{d} + A$. $\mathcal{O}(-1)$ is its dual, with covariant derivative $\mathrm{d} - A$. $\mathcal{O}(-n)$ is by definition the tensor product bundle $\mathcal{O}(-1)^{\otimes n}$, and its covariant derivative is $\mathrm{d} - n A$.