I am having trouble in distinguishing the difference between two types of multiplication. Basically if we have $X^{\mu}=(ct,\textbf{x})^{\mu}$, what is the difference between
$$\textbf{j}_{\mu}~X^{\mu}$$
and
$$\textbf{j}^{\mu}~X^{\mu} $$
where $\textbf{j}$ is the four-current.
If you work with Einstein summation notation, the first one, $j_\mu X^\mu$ , is actually the scalar product of these two 4-vectors. This scalar product works with the helps of the metric tensor $\eta_{\mu \nu}$ in the following way \begin{align} j_\mu X^\mu = j^\mu X_\mu = j^\mu \eta_{\mu \nu} X^\nu \, . \end{align} The metric tensor tells you how distances in spacetime are measured. In standard Minkowski space ,this is $(\eta_{\mu \nu}) = diag(1,-1,-1,-1)$.
(Note that Minskowski metric only holds for flat spacetimes. Curved spacetimes of general relativity are described by different metric tensors, in fact, Einstein's field equations are differential equations for metric tensors.)
So you can see from the equation above, that co- and contravariant vectors are related by the metric tensor as \begin{align} X_\mu = \eta_{\mu \nu} X^\nu \, . \end{align} This is the reason why it is often said that \begin{align} x^\mu = \begin{pmatrix} ct \\ \vec{x} \end{pmatrix} \, , x_\mu = \begin{pmatrix} ct \\ -\vec{x} \end{pmatrix} \, , \end{align} which is a little sloppy in notation. Think of them as vectors and covectors.
The object $j^\mu X^\mu$, on the other hand, is not a sum but only a product of components depending on the value of $\mu$, since Einstein summation only sums if there are "the same indices bottom and top".
