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> [!definition] Definition. ([[tensor of type (p,q)]])
> Let $R$ be a [[commutative ring|commutative]] [[ring]], $M$ a [[free module|free]] [[submodule generated by a subset|finitely generated]] $R$-[[module]], and $M^{*}$ its [[dual vector space|dual module]].
>
A **tensor of type $(p,q)$**, or **$(p,q)$-tensor**, is an element $T$ of the [[tensor product of modules|tensor space]] $ \bigotimes ^{p}_{q} M:={ (M^{*})^{\otimes p} \otimes (M)^{\otimes q} }$
>
We call the $p$-slots **contravariant** and the $q$-slots **covariant**.
^definition
> [!equivalence] Multilinear form equivalence.
> [[when tensors are canonically realized as multilinear forms|Equivalently]], $(p,q)$-tensors are naturally viewed as precisely the [[multilinear map|multilinear forms]] $T: \underbrace{ M^{*} \times \dots \times M^{*} }_{ p \text{ times} } \times \underbrace{ M_{} \times \dots \times M_{} }_{ q \text{ times} } \to R.$
^equivalence
> [!equivalence] Basis equivalence (transformation law).
>
> Fixing a [[basis]] $(e_{i})$ for $M$ and corresponding [[dual basis]] $(e^{i})$ for $M^{*}$, a $(p,q)$-tensor $T$ is determined by its components $T ^{I}{_{J}:= }T^{i_{1}, \dots, i_{p}}{}_{j_{1},\dots,j_{q}}:= T(e^{i_{1}}, \dots, e ^{i_{p}}, e_{j_{1}},\dots, e_{j_{q}})$for $i_{1},\dots,i_{p}, j_{1},\dots,j_{q} =1,\dots,\text{rk }M$.
>
> These components satisfy the following [[transformation law]]: [[matrix of the identity w.r.t. two bases|for]] $(\widetilde{e}_{i})$ another [[basis]], denote by $A$ the [[matrix of the identity w.r.t. two bases|change of basis]] [[matrix]] from $(\widetilde{e}_{i})$ to $(e_{i})$,[^1] $\tilde{e}_{i}=A^{k}{_{i}}\, e_{k}$. [[matrix of the identity w.r.t. two bases|Then]] $(A ^{-1})^{\top}$ [[adjoint|is]] the [[change of basis formula|change]] [[matrix of the identity w.r.t. two bases|of basis]] [[matrix]] from $(\widetilde{e}^{i})$ to $(e^{i})$, $\tilde{e}^{i}=(A ^{-1})^{i}_{}{_{k}}\, e ^{k}$. Letting $\widetilde{T}^{I}{}_{J}$ denote the components of the tensor wrt the [[basis]] $(\widetilde{e}_{i})$, we thus have[^5] $\widetilde{T}^{I}{}_{J}=(A ^{-1})^{i_{1}}{}_{k_{1}} \cdots (A ^{-1})^{i_p}{}_{k _p} A^{\ell_{1}}{}_{j_{1}} \cdots A^{ \ell_{q}}{}_{j_{q}} T^{k_{1} \dots k_{p}}{}_{\ell_{1}\dots \ell_{q}}. (*)$
> Conversely, a multidimensional array of components $[T^{I}{}_{J}]^{I \subset [p]_{}}_{J \subset [q]}$ satisfying the [[transformation law]] $(*)$ determines a well-defined (i.e. basis-independent) [[multilinear map|multilinear form]] $(M^{*})^{p} \times M^{q}$, thus a well-defined $(p,q)$-tensor.
> [!definition] Definition. (Contraction)
> **Coordinate-free.** For $r \in [p]$ and $s \in [q]$, define the **($r,s$)-pairing map** $\begin{align}
> P:\prod^{p}_{q} M &\to \prod ^{p-1}_{q-1} M \\
> (\theta^{1}, \dots, \theta^{p}, v_{1},\dots,v_{q}) & \mapsto \theta^{r}(v_{s}) \, (\theta^{1}, \dots, \widehat{\theta^{r}}, \dots, \theta^{p}, v_{1},\dots, \widehat{v_{s}}, \dots, v_{q}).
> \end{align}$
> By the [[universal property]] of [[tensor product of modules|tensor products]], there is a unique [[linear map|linear map]] $C_{r, s}:\bigotimes ^{p}_{q} M\to \bigotimes ^{p-1}_{q-1}M$ making the [[diagram]] commute:
>
> ```tikz
> \usepackage{tikz-cd}
> \usepackage{amsmath}
> \begin{document}
> % https://tikzcd.yichuanshen.de/#N4Igdg9gJgpgziAXAbVABwnAlgFyxMJZABgBpiBdUkANwEMAbAVxiRAB120AnaAPTQB9AI4ACALIgAvqXSZc+QigCM5KrUYs2nHv2BoAtMqmDgwo1InTZIDNjwEiZZevrNWiDuwBGWAOYQeAC28AIiVjJy9opEqi7UblqenL4BwaH6FqbmxhHqMFB+8ESgAGa8QUhkIDgQSKoa7mwACqYA1qSinDAMDFLWZRVIAEzUtUgAzAmaHl4wAB5YcDhwogCEogDC7Z3dvf3UDHTePc3yDkog3P4AFjgDIOUQlYjV44jDkY9DiA3vE1IKFIgA
> \begin{tikzcd}
> \prod^p_q M \arrow[r, "{P_{r, s}}"] \arrow[d] & \prod^{p-1}_{q-1} M \arrow[d] \\
> \bigotimes^p_q M \arrow[r, "{\exists ! C_{r, s}}"'] & \bigotimes^{p-1}_{q-1} M
> \end{tikzcd}
> \end{document}
> ```
>
> We call $C_{r, s}$ the **$(r, s)$-contraction** or **$(r, s)$-trace** operation. Explicitly, $C_{r, s}$ is defined on [[tensor product of modules|pure tensors]] as $\begin{align}&
> \theta^{1} \otimes\dots \otimes \theta^{p} \otimes v_{1} \otimes \dots \otimes v_{q} \\
> & \xmapsto{C_{r, s}} \theta^{r}(v_{s}) \, \theta^{1} \otimes \dots \otimes \widehat{\theta^{r}} \otimes \dots \otimes \theta^{p} \otimes v_{1} \otimes \dots \otimes \widehat{v_{s}} \otimes \dots \otimes v_{q}
> \end{align}$
> and then linearly extending.
>
>
> **In coordinates.** Fixing [[basis|coordinates]] $(e_{i})$ on $M$, a general element $T \in \bigotimes_{q}^{p}M$ looks like $T(\theta^{1}, \dots, \theta^{p}, v_{1},\dots,v_{q})=T^{I}{}_{J} \, \theta^{I} v_{J} \text{ for }I \subset [p], J \subset [q].$
> Then with $\theta^{r}=\theta^{r}_{i}e^{i}$, $v_{s}=v_{s}^{j}e_{j}$, we have $\theta^{r}(v_{s})=\theta^{r}_{i} v_{s}^{i}$
> and consequently $\begin{align}
> (C_{r,s}T) &(\theta^{1}, \dots, \widehat{\theta^{r}}, \dots, \theta^{p}, v_{1},\dots,\widehat{v_{s}}, \dots, v_{q}) \\
> &= \theta^{r}_{i} v_{s}^{i} \, T ^{\overline{r}}{}_{\overline{s}} \theta^{\overline{r}} v_{\overline{s}}.
> \end{align}$
> Here, the notation $\overline{r}$ and $\overline{s}$ refer to the complements of $r,s$ in $[p]$ and $[q]$ respectively (preserving the order).[^6]
[^6]: In other words, $\theta^{r}_{i} v_{s}^{i} \, T ^{I \setminus r}{}_{J \setminus s} \theta^{I \setminus r} v_{J \setminus s}=T^{i_{1}\dots \widehat{i_{r}} \dots i_{p}}{}_{j_{1} \dots \widehat{j_{s}}\dots j_{q}} \theta^{i_{1} \dots \widehat{i_{r}}\dots i_{p}}v_{j_{1}\dots \widehat{j_{s}}\dots j_{q}}$.
[^5]: This is where the Einstein convention really shines. Without it, we would have to write $\begin{align}
\widetilde{T}^{\,i_{1}\dots i_{p}}{}_{j_{1}\dots j_{q}}
=\sum_{k_{1}=1}^{n}\!\cdots\!\sum_{k_p=1}^{n}
\sum_{\ell_{1}=1}^{n}\!\cdots\!\sum_{\ell_q=1}^{n}
\bigg[\prod_{r=1}^{p} (A^{-1})^{i_r}{}_{k_r}\bigg]
\bigg[\prod_{s=1}^{q} A^{\ell_s}{}_{j_s}\bigg]
\;T^{k_{1}\dots k_{p}}{}_{\ell_{1}\dots \ell_{q}}
\end{align}$
[^1]: Namely, $A=\begin{align} &\begin{matrix} & \widetilde{e}_1 & \dots & \widetilde{e}_i & \dots & \widetilde{e}_n \end{matrix} \\
&\begin{matrix} e_1 \\ \vdots \\ e_n \end{matrix}
\begin{pmatrix} & & \id (\widetilde{e}_{i}) & \\ & & \vdots & \\ & & \id(\tilde{e}_{i}) & &\end{pmatrix} ,\end{align} $
so that the $i$th column $A^:_{ i}$ of $A$ is by definition the scalars needed to write $\widetilde{e}_{i}$ as a [[linear combination]] of the $(e_{k})$, $\tilde{e}_{i}=A ^{k}_{i} e_{k}$.
----
####
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#### References
> [!backlink]
> ```dataview
> TABLE rows.file.link as "Further Reading"
> FROM [[]]
> FLATTEN file.tags as Tag
> WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom"
> GROUP BY Tag
> ```
> [!frontlink]
> ```dataview
> TABLE rows.file.link as "Further Reading"
> FROM outgoing([[]])
> FLATTEN file.tags as Tag
> WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom"
> GROUP BY Tag
> ```