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I’m trying to figure out what $Con{{t}_{12}}({{\mathbf{e}}_{\mathbf{i}}}\otimes {{\mathbf{e}}_{\mathbf{j}}}\otimes {{\mathbf{e}}_{\mathbf{k}}})$ simplifies to. Here $Con{{t}_{ij}}\mathbf{T}=\mathbf{T(}{{\mathbf{x}}_{\mathbf{1}}},...,{{\mathbf{x}}_{\mathbf{i-1}}},{{\mathbf{e}}_{\mathbf{k}}},{{\mathbf{x}}_{\mathbf{i+1}}},...,{{\mathbf{x}}_{\mathbf{j-1}}},{{\mathbf{e}}_{\mathbf{k}}},{{\mathbf{x}}_{\mathbf{j+1}}},...,{{\mathbf{x}}_{\mathbf{m}}}\mathbf{)}$ is the definition of the contracted multiplication of an m-th order tensor T (for example, $\mathbf{u(x)}={{u}_{i}}{{x}_{i}}=Con{{t}_{23}}(\mathbf{A}\otimes \mathbf{v}(\mathbf{x}\mathbf{,y}\mathbf{,z}))=\mathbf{A}(\mathbf{x}\mathbf{,}{{\mathbf{e}}_{\mathbf{j}}})\mathbf{v}({{\mathbf{e}}_{\mathbf{j}}})={{x}_{i}}{{A}_{ij}}{{v}_{j}}$ , so $Con{{t}_{23}}\mathbf{A}\otimes \mathbf{v}(\mathbf{x}\mathbf{,y}\mathbf{,z})={{A}_{ij}}{{v}_{j}}$ when ${{A}_{ij}}$ is second order and ${{v}_{k}}$ first order tensors and $\mathbf{x}\mathbf{,y}\mathbf{,z}$ are the vector arguments), and ${{e}_{i}},\ {{e}_{j}},\ {{e}_{k}}$ are the standard basis vectors on${{\mathbb{R}}^{n}}$. I know that $\mathbf{u(x)}=\mathbf{u}\bullet \mathbf{x}={{u}_{i}}{{x}_{i}}=Con{{t}_{23}}(\mathbf{a}\otimes \mathbf{b}\otimes \mathbf{c})=\mathbf{a}\otimes \mathbf{b}\otimes \mathbf{c}\ (\mathbf{x}\mathbf{,}{{\mathbf{e}}_{\mathbf{j}}}\mathbf{,}{{\mathbf{e}}_{\mathbf{j}}})={{x}_{i}}\mathbf{a(}{{\mathbf{e}}_{\mathbf{i}}}\mathbf{)b(}{{\mathbf{e}}_{\mathbf{j}}}\mathbf{)c(}{{\mathbf{e}}_{\mathbf{j}}}\mathbf{)}={{x}_{i}}{{a}_{i}}{{b}_{j}}{{c}_{j}}$

which implies ${{[Con{{t}_{23}}(\mathbf{a}\otimes \mathbf{b}\otimes \mathbf{c})]}_{i}}={{u}_{i}}={{a}_{i}}{{b}_{j}}{{c}_{j}}={{a}_{i}}\mathbf{b}\bullet \mathbf{c}=(\mathbf{a}\otimes \mathbf{b}){{c}_{j}}$ for the first order tensors $\mathbf{a},\ \mathbf{b}$ and $\mathbf{c}$ ($\mathbf{a(}{{\mathbf{e}}_{\mathbf{i}}}\mathbf{)}=\mathbf{a}\bullet {{\mathbf{e}}_{\mathbf{i}}}$ is the first order tensor a acting on ${{\mathbf{e}}_{\mathbf{i}}}$). I tried the same strategy with my problem and got $\mathbf{u(x)}=\mathbf{u}\bullet \mathbf{x}={{u}_{k}}{{x}_{k}}=Con{{t}_{12}}({{\mathbf{e}}_{\mathbf{i}}}\otimes {{\mathbf{e}}_{\mathbf{j}}}\otimes {{\mathbf{e}}_{\mathbf{k}}})={{\mathbf{e}}_{\mathbf{i}}}\otimes {{\mathbf{e}}_{\mathbf{j}}}\otimes {{\mathbf{e}}_{\mathbf{k}}}\ ({{\mathbf{e}}_{L}}\mathbf{,}{{\mathbf{e}}_{L}},\mathbf{x})={{\mathbf{e}}_{\mathbf{i}}}\mathbf{(}{{\mathbf{e}}_{L}}\mathbf{)}{{\mathbf{e}}_{\mathbf{j}}}\mathbf{(}{{\mathbf{e}}_{L}}\mathbf{)}{{\mathbf{e}}_{\mathbf{k}}}\mathbf{(x)}={{x}_{k}}{{\delta }_{iL}}{{\delta }_{jL}}={{x}_{k}}{{\delta }_{ij}}$
which gives ${{[Con{{t}_{12}}({{\mathbf{e}}_{\mathbf{i}}}\otimes {{\mathbf{e}}_{\mathbf{j}}}\otimes {{\mathbf{e}}_{\mathbf{k}}})]}_{k}}={{u}_{k}}={{\delta }_{ij}}$ where ${{\delta }_{ij}}$ is the Kronecker delta. If I now multiply both sides by ${{\mathbf{e}}_{k}}$ (can I do this?) I get $\mathbf{u}(\mathbf{x})=Con{{t}_{12}}({{\mathbf{e}}_{\mathbf{i}}}\otimes {{\mathbf{e}}_{\mathbf{j}}}\otimes {{\mathbf{e}}_{\mathbf{k}}})={{\mathbf{e}}_{k}}{{u}_{k}}={{\mathbf{e}}_{k}}{{\delta }_{ij}}$ as my answer. Is this correct? Is my method correct? I’m pretty sure $Con{{t}_{23}}({{\mathbf{e}}_{\mathbf{i}}}\otimes {{\mathbf{e}}_{\mathbf{j}}}\otimes {{\mathbf{e}}_{\mathbf{k}}})={{\mathbf{e}}_{\mathbf{i}}}{{\delta }_{jk}}$ as this was used in my textbook for divergence of a tensor, but they only applied the recipe $\mathbf{a}(\mathbf{b}\bullet \mathbf{c})=(\mathbf{a}\otimes \mathbf{b})\mathbf{c}$. How do I solve my problem without using this recipe? My textbook is very brief about this.

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First let's clear a couple of things up.

  1. Technically, you can only contract a covariant slot with a contravariant slot, which amounts to using the canonical pairing $$T^*M\otimes TM\longrightarrow\mathbb{R}\ .$$
  2. However, in the presence of a metric, you can use it to contract two covariant (or two contravariant) slots, which amounts either to using the metric, or equivalently to apply the isomorphism between the tangent and cotangent bundles induced by it on one of the slots and then using the natural pairing.

I gather that you are working on $\mathbb{R}^n$ (which is its own tangent space) with the metric given by the dot product. Thus, we have $$Cont_{12}(e_i\otimes e_j\otimes e_k) = (e_i\cdot e_j)e_k = \delta_{ij}e_k\ .$$

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