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3 fixed signs

1.) Yes; the extension to pure k-covectors is given by the skew Liebniz identity $$\nabla \left(X_1 \wedge \cdots \wedge X_k\right) = \sum_k (-1)^{k+1} X_1 \wedge \cdots \wedge \nabla X_k \wedge \cdots \wedge X_k$$ Extend to all of $\bigwedge^k T^*M$ by linearity.

2.) Yes again. This is easiest to see by thinking about parallel transport. Given a curve $\gamma$ from $p$ to $q$ and an orthogonal frame $\Psi_p, \Psi_q$ at each point, parallel transport along $\gamma$ determines a unique orthogonal transformation $O$ such that $$P^\nabla_\gamma \Psi_p = O \cdot \Psi_q$$ This means that any invariants of the group $O(n)$ are parallel, such as the volume form.

2 vector --> covector

1.) Yes; the extension to pure k-vectors k-covectors is given by the skew Liebniz identity $$\nabla \left(X_1 \wedge \cdots \wedge X_k\right) = \sum_k (-1)^{k+1} X_1 \wedge \cdots \wedge \nabla X_k \wedge \cdots \wedge X_k$$ Extend to all of $\bigwedge^k T^*M$ by linearity.

2.) Yes again. This is easiest to see by thinking about parallel transport. Given a curve $\gamma$ from $p$ to $q$ and an orthogonal frame $\Psi_p, \Psi_q$ at each point, parallel transport along $\gamma$ determines a unique orthogonal transformation $O$ such that $$P^\nabla_\gamma \Psi_p = O \cdot \Psi_q$$ This means that any invariants of the group $O(n)$ are parallel, such as the volume form.

1

1.) Yes; the extension to pure k-vectors is given by the skew Liebniz identity $$\nabla \left(X_1 \wedge \cdots \wedge X_k\right) = \sum_k (-1)^{k+1} X_1 \wedge \cdots \wedge \nabla X_k \wedge \cdots \wedge X_k$$ Extend to all of $\bigwedge^k T^*M$ by linearity.

2.) Yes again. This is easiest to see by thinking about parallel transport. Given a curve $\gamma$ from $p$ to $q$ and an orthogonal frame $\Psi_p, \Psi_q$ at each point, parallel transport along $\gamma$ determines a unique orthogonal transformation $O$ such that $$P^\nabla_\gamma \Psi_p = O \cdot \Psi_q$$ This means that any invariants of the group $O(n)$ are parallel, such as the volume form.