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I am reading through some material about $4$-vectors. And came across the following for which an explanation woud be greatly appreciated.

The index for $\partial_\alpha$ can be raised giving $\partial^\alpha$ because spacetime is flat.

Perhaps I should provide some context. It is dealing with EM fields and essentially referring to $$F_{\alpha\beta}=\partial_\alpha A_\beta-\partial_\beta A_\alpha$$ implies that $$F^{\alpha\beta}=\partial^\alpha A^\beta-\partial^\beta A^\alpha$$

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On a smooth manifold, there is a priori no coordinate-free relationship between vectors and covectors (one forms). A choice of coordinates $\{x^\alpha\}$ induces a basis $\partial_\alpha = \frac{\partial}{\partial x^\alpha}$ of the tangent space at each point covered by the coordinates, and a dual basis $dx^\alpha$ of the cotangent space at each point.

The presence of a pseudo-Riemannian metric $g$($=g_{\alpha\beta}$ in coordinates) induces a coordinate-free identification between vectors and covectors. A vector field $W$ is identified with the one-form $\omega$ defined by pairing $\omega(\cdot) = g(W,\cdot)$. If we choose a local basis for the tangent bundle, the coordinates of $\omega$ are found by "raising the indices" of $W$. This is shorthand for multiplication by the inverse of the metric: $\omega^\alpha = g^{\alpha\beta}W_\beta$. If $e_\alpha$ is local orthonormal frame field, the metric is the identity matrix (up to some signs) $\delta_{\alpha\beta}$ and the dual frame field is obtained by $e^\beta = \delta^{\alpha\beta}e_\alpha$.

In this case, Minkowski spacetime is flat, so orthonormal frames are integrable to coordinates. Therefore, one obtains the metric dual to coordinates by raising indices, i.e., multiplying by the inverse of the metric, which is the identity matrix.

One last note: I seem to recall a common convention distinguishing between Roman ($i,j$) indices and Greek ($\alpha,\beta$) indices which nicely hides confusion about signs in the Lorentz metric. It's something like $\partial_\alpha = \partial_i$ for $\alpha,i=0,1,2,3$ but $\partial^\alpha = \partial^i$ for $\alpha,i = 1,2,3$ and $\partial^\alpha = -\partial^i$ for $\alpha = i = 0$.

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Thank you, Neal. –  tom Jan 9 '13 at 23:42

$\partial_\alpha$ is a vector, actually $\{\partial_\alpha \}$ form a basis for the tangent space, and if we have a metric, $\partial^\alpha$ it is its natural dual. With the physicist notation this means that you just raise the index. In Minkowski space you can have the whole space covered by a single chart so you can do it globally.

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Thank you, inquisitor. –  tom Jan 9 '13 at 23:42

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