Einstein notation and nabla I have a problem with Einstein's summation convention. If I now have two vectors $v^i$ and $w^j$, is $v^i w^j$ a tensor? I always thought that if the index is at the top, then the vector is written vertically, so $K^{n \times 1}$. And if the index is at the bottom, then the vector is written horizontally, so $K^{1 \times n}$. So by matrix multiplication at $v_i w^j$ you have the scalar product of two vectors,
\begin{align}
 (v_1, \ldots,v_n) \cdot \left(\begin{array} (w_1 \\ \vdots \\ w_n \end{array}\right) = v_1w_1 + v_2w_2 + \ldots 
\end{align}
and at $v^i w_j$ a Tensor,
\begin{align}
 \left(\begin{array} (v_1 \\ \vdots \\ v_n \end{array}\right) \cdot (w_1, \ldots,w_n) = \left(\begin{array}( v_1w_1 & v_1w_2 & \ldots & v_1w_n \\ v_2w_1 & \ddots \\ \ \ \ \vdots \end{array}\right)
\end{align}
But with $v^i w^j$ this does not work. 
\begin{align}
 \left(\begin{array} (v_1 \\ \vdots \\ v_n \end{array}\right) \cdot \left(\begin{array} (w_1 \\ \vdots \\ w_n \end{array}\right) \overset{???}{=} v_1 \cdot \left(\begin{array} (w_1 \\ \vdots \\ w_n \end{array}\right) + v_2 \cdot \left(\begin{array} (w_1 \\ \vdots \\ w_n \end{array}\right) + \ldots
\end{align}
I also wonder what $\nabla^\alpha f$ is. I know that $\nabla_\alpha f$ is the gradient.

If $V$ is a real vector space with an inner product $g$, basis $\{e_i\}_{1 \leq i \leq n}$, and $v, w \in V$, then $g(v, w) = g_{ij} v^{i} w^{j}$, where the $v^{i}$ are the components of $v$ (so that $v = v^{i} e_i$, and the same holds analagously for $w$) and $g_{ij} = g(e_i, e_j)$. But the expression $v^{i} w^{j}$, without any further context, just means "multiply the $i$-th component of $v$ by the $j$-th component of $w$".
And the notation $\nabla^{\alpha} f$ stands for $g^{\alpha \beta} \nabla_{\beta} f. $
