I believe you are confused because you are mixing up related but slightly different quantities.

**Yes, a partial derivative is a vector and yes, a vector is an object with an upper index.**

The above statement may seem contradictory, but in fact it is not for the following reason. A vector is an abstract quantity that is an element of a "vector space". In this case, the vector space that is being discussed is the tangent space. On a vector space, one can choose a basis, any basis. Once a basis has been chosen, any other vector in the vector space can be described by simply prescribing a set of numbers. For instance in ${\mathbb R}^2$ (rather the corresponding affine space), one can choose a basis of vectors as ${\hat x}$ and ${\hat y}$. Once this has been done any other vector can be described simply by 2 numbers. For instance the numbers $(1,2)$ really implies that we are talking about the vector ${\hat x} + 2 {\hat y}$.

How does the discussion above apply here?

On the tangent space, a natural choice of basis are the set of partial derivatives $\partial_\mu = \{ \partial_0 , \partial_1 , \partial_2 , \partial_3 \}$ (assuming we are in $M_4$. Each partial derivative is in itself a vector.

Now, once this basis has been chosen, every other vector can be described by a set of 4 numbers $v^\mu = (v^0 , v^1 , v^2 , v^3)$ which corresponds to the vector $v^\mu \partial_\mu$. It is this sense, that the bold statement above is true. Often, since the basis of partial derivatives is obvious, one simply describes a vector as an object with an upper index $v^\mu$.

Next, let us discuss co-vectors (quantities with a lower index). These are lements of the dual vector space (which is the space of linear functions on the vector space) of the tangent space. Given the partial derivative basis on the tangent space, one then has a natural basis in the cotangent space denoted by $dx^\mu = \{ dx^0 , dx^1 , dx^2 , dx^3 \}$. Note that each differential itself is a covector. This natural basis is defined by the relation $dx^\mu (\partial_\nu ) = \delta^\mu_\nu$. As before, once this natural basis has been chosen, any element of the cotangent space can be described by 4 numbers, namely $v_\mu = \{ v_0, v_1 , v_2 , v_3 \}$ which corresponds to the covector $v_\mu dx^\mu$.

In **summary**, $\partial_\mu$ for each $\mu$ corresponds to a 4-dimensional vector whereas $v^\mu$ for each $\mu$ corresponds to 4 components of a single vector. Similarly, $dx^\mu$ for each $\mu$ corresponds to a 4-dimensional covector whereas $v_\mu$ for each $\mu$ corresponds to 4 components of a single covector.

PS 1 - Sometimes people like to use bases other than $\partial_\mu$ and $dx^\mu$ on the tangent and cotangent spaces respectively. These are known as non-coordinate bases.

PS 2 - Just to be clear, $\partial_\mu$ is a vector, but $\partial_\mu F$ is a function

This post imported from StackExchange Physics at 2014-11-11 14:51 (UTC), posted by SE-user Prahar