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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

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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.

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  • $\begingroup$ Very nicely explained. Thanks for the clarification! $\endgroup$ Commented Feb 7, 2014 at 13:05

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