Let $(X,\omega)$ be a compact K"ahler manifold of dimension $n$, and let $D=\sum_{i=1}^r D_i$ be a simple normal crossing divisor on it, i.e., a divisor with smooth components $D_i$ intersecting each other transversally in $X$. Let $\mathcal{V}$ be a locally free coherent sheaf on $X$ and let $$\nabla:\mathcal{V}\to \Omega^1_X(\log D)\otimes\mathcal{V}$$ be a $\mathbb{C}$-linear map satisfying \begin{align}\nabla(f\cdot e)=f\cdot\nabla e+df\otimes e.\end{align} One defines $$\nabla_a:\Omega^a_X(\log D)\otimes \mathcal{V}\to \Omega^{a+1}_X(\log D)\otimes \mathcal{V}$$ by the rule $$\nabla_a (\omega\otimes e)=d\omega\otimes e+(-1)^a \omega\wedge \nabla e.$$ We assume that $\nabla_{a+1}\circ\nabla_a=0$ for all $a$. Such $\nabla$ will be called an integrable logarithmic connection along $D$, or just a connection.
My Question is: Given a connection $\nabla$ in $\mathcal V$, can it induce a connection in its dual bundle $\mathcal V^*$?
Here is my thought: Locally, $\nabla$ can be written as the form $\nabla=d+\omega$, where $\omega$ is a holomorphic section of $\Omega_{X}^1(\log D)\otimes\text{End}(\mathcal{V})$. For $\mathcal{V}$ and its dual bundle $\mathcal{V}^*$, the dual pairing $$\langle\,,\,\rangle:\mathcal{V}_x^*\times\mathcal{V}_x\longrightarrow\mathbb{C}$$ induces a dual pairing $$\langle\,,\,\rangle:A^0(\mathcal{V}^*)\times A^0(\mathcal{V})\rightarrow A^0.$$ Given a connection $\nabla$ in $\mathcal V$, we define a connection, also denoted by $\nabla$, in $\mathcal V^{*}$ by the following formula: $$ d(\xi, \eta)=\langle \nabla \xi, \eta\rangle+\langle\xi, \nabla \eta\rangle \text { for } \xi \in H^{0}(\mathcal V), \eta \in H^{0}\left(\mathcal V^{*}\right). $$ Given a local frame ficld $s=\left(s_{1}, \ldots, s_{r}\right)$ of $\mathcal V$ over an open set $U,$ let $t=$ $\left(t^{1}, \cdots, t^{r}\right)$ be the local frame field of $\mathcal V^{*}$ dual to $s$ so that $$ \left\langle t^{i}, s_ j\right\rangle=\delta^{i}_j, \text { or }\left\langle t, s\right\rangle=I_{r}, $$ where $s$ is considered as a row vector and $t$ as a column vector. If $\omega=(\omega_{j}^{i})$ denotes the connection form of $\nabla \text { with respect to } s \text { so that }$ $$ \nabla s_{i}=\sum s_{j} \omega_{i}^{j} \qquad \text{or} \,\,\nabla s=s \omega{,}$$ then $$ \nabla t_{j}=-\sum \omega_{i}^{j}t^j \qquad \text{or} \,\,\nabla t= -\omega t{.}$$This follows from $$ 0=d \delta_{j}^{i}=\left\langle \nabla t^{i}, s_{j}\right\rangle+\left(t^{i}, \nabla s_{j}\right)=\left\langle \nabla t^{i}, s_{j}\right\rangle+\omega_{j}^{i}. $$
In short, $\omega_{\mathcal V^*}=-\omega_{\mathcal V}$.
Am I right? Any advice and suggestion will be appreciated. Thanks a lot.
