# Raising and Lowering indices of tensor

Why we use metric tensors $g$ to raise or lower indices of tensors, why not using other (invertible) order-2 tensors to do the job?

Given a finite-dimensional vector space $V$ over some field, there is no natural isomorphism with $V^*$, its dual vector space. If $V$ is equipped with an inner product, this can be used to define an isomorphism between $V$ and $V^*$ by sending any linear form $\omega\in V^*$ to the unique vector $v_\omega \in V$ that is such that $$\omega(u) = (v_\omega,u),\qquad\forall u\in V.$$ Given a basis $\{e_k\}_{k=1,\ldots,n}$ of $V$, this isomorphisms allows to define the so-called dual basis of $V^*$ by taking vectors $e^k$ that satisfy to $$(e^i,e_k)=\delta^i_k$$ and mapping them back to $V^*$.
These ideas can be generalised "fibre-wise" to manifolds. Hence on a Riemannian manifold $(M,g)$, the metric $g$ provides an isomorphism between the cotangent bundle $T^*M$ and the tangent bundle $TM$. So given a basis of $T_pM$, there is a dual basis on $T^*_pM$, and the way a vector of $T_pM$ is mapped to the corresponding 1-form in $T^*_pM$ in terms of component w.r.t. these basis is through the components of the metric tensor. In a coordinate-free approach you would have that, given a 1-form $\omega\in T^*M$, there is a vector field $v_\omega\in TM$ such that $$\omega(u) = g(v_\omega,u),\qquad\forall u\in TM.$$ Locally, when referring to a chart $(U,\phi)$, you can stick to the coordinate basis $\frac{\partial}{\partial x^1},\ldots,\frac{\partial}{\partial x^n}$ of $TM$ on $U$, whose dual basis is denoted by $\text dx^1,\ldots,\text dx^n$, and with the property that $$\text dx^i\left(\frac{\partial}{\partial x^k}\right) = \delta^i_k.$$