general relativity - Are covariant vectors representable as row vectors and contravariant as column vectors

I would like to know what are the range of validity of the following statement:

Covariant vectors are representable as row vectors. Contravariant vectors are representable as column vectors.

For example we know that the gradient of a function is representable as row vector in ordinary space $ \mathbb{R}^3$

$\nabla f = \left [ \frac{\partial f}{\partial x},\frac{\partial f}{\partial y},\frac{\partial f}{\partial z} \right ]$

and an ordinary vector is a column vector

$ \mathbf{x} = \left[ x_1, x_2, x_3 \right]^T$

I think that this continues to be valid in special relativity (Minkowski metric is flat), but I'm not sure about it in general relativity.

Can you provide me some examples?

Answer

Yes, the statement holds true in general relativity as well. However, as we need to deal with tensors of higher and in particular mixed order, the rules of matrix multiplication (which is where the idea of the representation via row- and column-vectors comes from) are no longer sufficiently powerful:

Instead, the placement of the index determines if we are dealing with a contravariant (upper index) or a covariant (lower index) quantity.

Additionally, by convention an index which occurs in a product in both upper and lower position gets contracted, and equations must hold for all values of free indices.

If the given metric is non-Euclidean (which is already true in special relativity), mapping between co- and contravariant quantities is more involved than simple transposition and the actual values of the components in a given basis can change, eg: $$ p^\mu = (p^0,+\vec p)\\ p_\mu = (p^0,-\vec p) $$ and in general: $$ p_\mu = g_{\mu\nu}p^\nu $$ where $g_{\mu\nu}$ denotes the metric tensor and a sum $\nu=1\dots n$ is implied.