# Contracted multiplication of tensors

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.

1. Technically, you can only contract a covariant slot with a contravariant slot, which amounts to using the canonical pairing $$T^*M\otimes TM\longrightarrow\mathbb{R}\ .$$
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 $$Cont_{12}(e_i\otimes e_j\otimes e_k) = (e_i\cdot e_j)e_k = \delta_{ij}e_k\ .$$