Ways to compute inner products of tensors

Question

One way to evaluate the following sums is combining Table and Sum:

$u_{abcd} = \sum_{e=1}^3 v_{aeb}w_{ced}$

$q_{ab} = \sum_{d,e=1}^3 v_{d e a}w_{deb}$

It will look like

v = Table[Times[i, j, k], {i, 3}, {j, 3}, {k, 3}]
w = Table[Times[i+2, j-1, k+1], {i, 3}, {j, 3}, {k, 3}]

u = Table[Sum[v[[a, e, b]] w[[c, e, d]], {e, 3}], {a, 3}, {b, 3}, {c, 3}, {d,  3}]
q = Table[Sum[v[[d, e, a]] w[[d, e, b]], {d, 3}, {e, 3}], {a, 3}, {b, 3}]

Is there a more elegant way to evaluate sums like these?

Answer 1

If we transpose the indices of $v$ and $w$ so that $v'_{abe} = v_{aeb}$ and $w'_{ecd} = w_{ced}$, then we can compute $u = v' \cdot w'$:

u === Transpose[v, {1, 3, 2}] . Transpose[w, {2, 1, 3}]
(* True *)

We can use a similar trick to compute $q$ if we reorder $v_{dea} \to v''_{ade}$, except that this time the $d$ and $e$ indices in $v_{ade}''$ and $w_{deb}$ must treated as if they comprised a single index to be contracted. Flatten can do this for us:

q === Flatten[v, {{3}, {1, 2}}] . Flatten[w, {{1, 2}, {3}}]
(* True *)

For details about the Flatten syntax used here, see Flatten command: matrix as second argument.

Answer 2

Incorporating b.gatessucks's recommendation:

u === Outer[v[[#1, All, #2]].w[[#3, All, #4]] &, ##] & @@ Range@{3, 3, 3, 3}

True

Answer 3

In Mathematica version 9, you can do these kinds of things much more naturally. Here are the two quantities that you wanted:

u = TensorContract[TensorProduct[v, w], {2, 5}];

q = TensorContract[TensorProduct[v, w], {{1, 4}, {2, 5}}];

The contraction is performed on the tensor product in which the first three indices belong to the factor v and the last three indices label w. Therefore, the indices corresponding to $e$ in your sum for u are in slots {2, 5}, and the two summations for q run over slots {1, 4} (for the variable $d$) and {2,5} (as in v).