simple tensor

The tensor product (http://planetmath.org/TensorProduct) tensor product$U\otimes V$ of two vector spaces $U$ and $V$ isanother vector space which is characterised by being universalfor bilinear maps on $U\times V$. As part of this package,there is an operation $\otimes$ on vectors such that$𝐮\otimes 𝐯\in U\otimes V$ for all$𝐮\in U$ and $𝐯\in V$, and the primary subjectof this article is the image of that operation.

Definition 1.

The element $𝐰\in U\otimes V$ is said to be asimple tensor if there exist $𝐮\in U$ and $𝐯\in V$ such that $𝐰=𝐮\otimes 𝐯$.

More generally, the element $𝐰\in W=U_1\otimes \mathrm{\cdots }\otimes U_k$ is said to be a simple tensor (with respect to thedecomposition $U_1\otimes \mathrm{\cdots }\otimes U_k$ of $W$) if thereexist ${𝐮}_i\in U_i$ for $i=1,\mathrm{\dots },k$ such that$𝐰={𝐮}_1\otimes \mathrm{\cdots }\otimes {𝐮}_k$.

For this definition to be interesting, there must also be tensorswhich are not simple, and indeed most tensors aren’t. In order toillustrate why, it is convenient to consider the tensor product oftwo finite-dimensional vector spaces $U={𝒦}^m$ and $V={𝒦}^n$over some field $𝒦$. In this case one can let $U\otimes V={𝒦}^{m\times n}$ (the vector space of $m\times n$ matrices), since${𝒦}^{m\times n}$ is isomorphic to any generic construction of$U\otimes V$ and the tensor product of two spaces is anyway onlydefined up to isomorphism. Furthermore considering elements of $U$and $V$ as column vectors, the tensor product of vectors can bedefined through

$$𝐮\otimes 𝐯=𝐮\cdot {𝐯}^{\mathrm{T}}$$

where $\cdot$ denotes the product of two matrices (in this case an$m\times 1$ matrix by a $1\times n$ matrix). As a very concreteexample of this,

One reason the simple tensors in $U\otimes V$ cannot exhaust thisspace (provded $m,n⩾2$) is that there are essentiallyonly $m+n-1$ degrees of freedom in the choice of a simple tensor, but$mn$ dimensions (http://planetmath.org/Dimension2) in the space $U\otimes V$as a whole. Hence

$${𝒦}^m\otimes {𝒦}^n\ne \{𝐮\otimes 𝐯\text{<span class="ltx\_rule" style="width:1px;background:black;display:inline-block;"> </span>}𝐮\in {𝒦}^m,𝐯\in {𝒦}^n\}\mathit{\hspace{1em}\hspace{1em}}\text{when }m,n⩾2\text{.}$$

How can one to understand the non-simple tensors, then? In general,they are finite sums of simple tensors. One way to see this is fromthe theorem that $U\otimes V$ has a basis consisting of products ofpairs of basis vectors.

Theorem 2 (tensor product basis (http://planetmath.org/TensorProductBasis)).

Let $U$ and $V$ be vector spaces over $\mathrm{K}$ with bases${\mathrm{\{}{\mathrm{e}}_i\mathrm{\}}}_{i\mathrm{\in }I}$ and ${\mathrm{\{}{\mathrm{f}}_j\mathrm{\}}}_{j\mathrm{\in }J}$respectively. Then ${\mathrm{\{}{\mathrm{e}}_i\mathrm{\otimes }{\mathrm{f}}_j\mathrm{\}}}_{\mathrm{(}i\mathrm{,}j\mathrm{)}\mathrm{\in }I\mathrm{\times }J}$ is a basis for $U\mathrm{\otimes }V$.

Expressing some arbitrary $𝐰\in U\otimes V$ as a linearcombination

$$𝐰=\sum _{r=1}^n{\lambda }_r({𝐞}_{i_r}\otimes {𝐟}_{j_r})$$

with respect to such a basis immediately produces the decomposition

$$𝐰=\sum _{r=1}^n({\lambda }_r{𝐞}_{i_r})\otimes {𝐟}_{j_r}$$

as a sum of simple tensors, but this decomposition is often farfrom optimally short.Let ${𝐞}_1=\left(\begin{array}{c}\hfill 1\hfill \\ \hfill 0\hfill \end{array}\right)\in {𝒦}^2$ and ${𝐞}_2=\left(\begin{array}{c}\hfill 0\hfill \\ \hfill 1\hfill \end{array}\right)\in {𝒦}^2$. The tensor is notsimple, but as it happens the tensor is simple. In general itis not trivial to find the simplest way of expressing a tensor asa sum of simple tensors, so there is a name for the length of theshortest such sum.

Definition 3.

The rank of a tensor $𝐰$ is the smallest naturalnumber $n$ such that $𝐰={𝐰}_1+\mathrm{\cdots }+{𝐰}_n$for some set of $n$ simple tensors ${𝐰}_1$, …, ${𝐰}_n$.

In particular, the zero tensor has rank $0$, and all other simpletensors have rank $1$.

• Warning.  There is an entirely different concept which is also called‘the rank of a tensor (http://planetmath.org/Tensor)’,namely the number of components (factors) in the tensorproduct forming the space in which the tensor lives. This latter‘rank’ concept does not generalise‘rank of a matrix (http://planetmath.org/RankLinearMapping)’.The ‘rank’ of Definition 3 doesgeneralise ‘rank of a matrix’. (It also generalisesrank of a quadratic form (http://planetmath.org/Rank5).)

one way

One area where the distinction between simple and non-simple tensorsis particularly important is in Quantum Mechanics, because the statespace of a pair of quantum systems is in general the tensor productof the state spaces of the component systems. When the combined stateis a simple tensor $𝐰=𝐮\otimes 𝐯$, then thatstate can be understood as though one subsystem has state $𝐮$and the other state $𝐯$, but when the combined state $𝐰$is a non-simple tensor ${𝐮}_1\otimes {𝐯}_1+{𝐮}_2\otimes {𝐯}_2$ then the full system cannot beunderstood by considering the two subsystems in isolation, even ifthere is no apparent interaction between them. This situation isoften described by saying that the two subsystems areentangled, or using phrases such as “either $U$ is in state${𝐮}_1$ and $V$ is in state ${𝐯}_1$, or else $U$ is in state${𝐮}_2$ and $V$ is in state ${𝐯}_2$.”Entanglement is an important part of that which makes quantum systemsdifferent from probabilistic classical systems. The physicalinterpretations are often mind-boggling, but the mathematical meaningis no more mysterious than ‘non-simple tensor’.

Entanglement can also be a useful concept for understanding pure mathematics.One reason that the comultiplication $\mathrm{\Delta }:C⟶C\otimes C$of a coalgebra $C$ cannot simply be replaced in the definition by twomaps ${\mathrm{\Delta }}_L,{\mathrm{\Delta }}_R:C⟶C$ that compute the ‘left’and ‘right’ parts of $\mathrm{\Delta }$ is that value of $\mathrm{\Delta }$ may beentangled, in which case one left part ${\mathrm{\Delta }}_L(c)$ and one rightpart ${\mathrm{\Delta }}_R(c)$ cannot fully encode $\mathrm{\Delta }(c)$.