Tensor Einstein summation notation I have two tensors $A^i$ and $B_j$ with components $(2,3,4)$ and $(1,2,3)$ respectively.
What is the difference between $A^i B_i$ and $A^i B_j$? 
Is it just:
$A^i B_i = 2+6+12 = 20$
$A^i B_j =$
$
        \left(\begin{matrix}
        2 \\
        3 \\
        4 \\
        \end{matrix}\right)
$
$
        \left(\begin{matrix}
        1 &2 & 3\\
        \end{matrix}\right)
$
$
        \left(\begin{matrix}
        2 & 4 & 6 \\
        3 & 6 & 9 \\
        4 & 8 & 12 \\
        \end{matrix}\right)
$?
Thanks
 A: When you take $A^i B_j$, without any repeated indices, then, indeed, you're forming a $(1,1)$-tensor $C^i_{\;j} = A^i B_j$ with matrix
$$
 \begin{pmatrix} 2\\3\\4\end{pmatrix}\begin{pmatrix}1&2&3\end{pmatrix} = \begin{pmatrix}2&4&6\\3&6&9\\4&8&12\end{pmatrix}.
$$
When you repeat an index, so that it appears both as a superscript and as a subscript, then you sum over that index; in particular, if $T^i_{\;j}$ is a $(1,1)$-tensor, then $T^i_{\;i}$ is precisely the trace of the corresponding matrix. Hence, in this case
$$
 A^i B_i = C^i_{\;i} = 2 + 6 + 12 = 20.
$$
On the other hand, given the repeated index, $A^i B_i = B_i A^i$ can also be interpreted as the linear functional (or covector) $B_j$ acting on the vector $A^i$ to yield a scalar:
$$
 B_i A^i = \begin{pmatrix}1&2&3\end{pmatrix}\begin{pmatrix} 2\\3\\4\end{pmatrix} = 1 \cdot 2 + 2 \cdot 3 + 3 \cdot 4 = 20.
$$
The nice thing about this notation, however, is that this interpretation is entirely consistent with the first interpretation---the whole system of subscripts (corresponding to covectors) and superscripts (corresponding to vectors) guarantees this. Again, the essential point is that you sum over an index that appears both as a subscript and as a superscript; this is a process called contraction over that index.
