# Is the concept of tensor rank useful in physics?

The term 'tensor rank' is sporadically used in the mathematical literature to denote the minimum number of simple terms (i.e. tensor products of vectors) needed to express the tensor. This is analogous (and equivalent) to the usual and prevalent notion of matrix rank in linear algebra; a good example of its use in the literature is Exploring Tensor Rank by Benjamin Weitz. Is this concept useful in physics?

Note that I am not asking about the total number of indices of the tensor, as in e.g. this MathWorld article.

For example, in general relativity, we deal with tensor fields. For a given field, can the rank of a tensor at one point be different that the rank of the tensor at a different point? Is there any physical significance to this?

I can easily construct an example of smooth tensor field over a manifold whose "rank" changes depending on the point. My idea relies upon the following elementary proposition.

I stress that the notion of "rank" used here is that introduced within the original question and not the standard one.

Proposition. Consider a $n$-dimensional real vector space $V$ and let $e_1,e_2 \in V$ be a pair of linearly independent vectors. The "entangled" tensor $$e_1\otimes e_1 + k e_2 \otimes e_2$$ is of "rank" $2$ necessarily if $k\neq 0$, otherwise it is of "rank" 1.

PROOF. Since $e_1$ and $e_2$ are linearly independent there is a basis $\{e_i\}_{i=1,\ldots,n}$ of $V$ containing both them. Assume $k\neq 0$. If the tensor were fo rank 1, i.e., $$e_1\otimes e_1 + k e_2 \otimes e_2= u\otimes v\quad (1)$$ as we have $u= \sum_i u^i e_i$ and $v= \sum_j v^je_j$, (1) would entail: $$e_1\otimes e_1 + k e_2 \otimes e_2= \sum_{i,j=1}^nu^iv^j e_i\otimes e_j\quad$$ and so: $$0= (u^1v^1-1) e_1\otimes e_1 + (u^2v^2-k) e_2\otimes e_2 + u^1v^2 e_1\otimes e_2 + u^2v^1 e_2\otimes e_1 + \sum_{i,j >2} u^iv^j e_i\otimes e_j\qquad (2)$$ Since, in turn, $\{e_i\otimes e_j\}_{i,j=1,\ldots,n}$ is a basis of the space of tensors $V \otimes V$, (2) implies, in particular, that: $$u^1v^1= 1\:, \:\:u^2v^2= k\:,\:\: u^1v^2=u^2v^1=0$$ multiplying together the first two conditions we have: $$u^1v^1u^2v^2 = k$$ whereas the remaining ones entail: $$u^1v^2u^2v^1 = 0$$ which are in contradiction unless $k=0$. QED

So consider a smooth (Hausdorff) manifold of dimension $n$ and a smooth function $\chi: M \to \mathbb R$ which constantly attains the value $1$ in an open set $U\subset M$ and smoothly vanishes before reaching the boundary of another open set $U' \supset U$. Taking $U$ and $U'$ small enough we can always assume that $U'$ is equipped by local coordinates $x^1,\ldots, x^n$. Under these hypotheses define the smooth tensor field $\Xi$ on $M$, vanishing outside $U'$:

$$\Xi(p) := \chi(p)\left(\frac{\partial}{\partial x^1}|_p \otimes \frac{\partial}{\partial x^1}|_p + k(p) \frac{\partial}{\partial x^2}|_p \otimes \frac{\partial}{\partial x^2}|_p\right)$$

where $k: M \to \mathbb R$ is a smooth function that vanishes somewhere in $U$ but not everywhere.

The tensor field $\Xi$ is smooth, well defined over the whole $M$, but it changes its "rank" three times: $0$ outside $U'$, $1$ and $2$ inside $U$, depending on the choice of $k$.

Though this example is completely mathematical, I think that with some further elaboration, some physical meaning could be given to my example, at least when the coordinate system is defined in the whole manifold of dimension $4$ and Lorentzian. The tensor is symmetric and we can assume that $x^1$ is a temporal Minkowskian coordinate so that $\partial_{x^1}$ could define the rest frame of some continuous body and $\Xi$ its stress-energy tensor. Assuming that this system interacts with some external system even the conservation low $\nabla_a\Xi^{ab}= J_{ext}^b$ can be imposed in order to have a non constant function $k$.

• It seems like a very unorthodox use of the term rank. The tensor $u\otimes v$ is not rank one. It seems that you are interchanging "rank one" and "pure tensor". – MBN Mar 8 '14 at 11:52
• I completely agree, I am just using the notion of "rank" presented within the original question. – Valter Moretti Mar 8 '14 at 11:55
• I added a remark to stress the difference. – Valter Moretti Mar 8 '14 at 12:02
• Yes, it was the OP who used the term "rank". The best term would be "generalized Schmidt rank", as for a tensor product of two vector spaces this notion is called "Schmidt rank". (And the linked paper in the question simply generalizes this concept to tensor products of $d$ number of vector spaces.) – Zoltan Zimboras Mar 8 '14 at 13:06
• Right, I forgot what the question said. Sorry about that. – MBN Mar 8 '14 at 15:47

As Emilio Pisanty pointed out in a comment to the answer of Joshphysics, in the physics literature "the rank of a tensor" means something different than how it is used in the nice paper of Benjamin Weitz that you linked in. However, also this type of "tensor rank" appears - at some level - in General Relativity. If you consider the famous Petrov classification of the Weyl tensor, you can (after a brief calculation) observe that: for the cases when the eigenvalue of the eigenbivectors of the Weyl tensor is degenerate, the tensor rank of the Weyl tensor is reduced. (However, note that the Petrov classification is much more - and much finer - than just a "tensor rank" classification of the Weyl tensor.)

Outside general relativity, also such ranks have been used in quantum information theory. For a pure states of a bipartite quantum system, the "Schmidt rank" is exactly such a tensor rank that you want. A similar notion has also been generalized to mixed states, or density matrices, where one defines a "Schmidt number", see this link.

Best,Z

• In the mean time Joshphysics's answer (and with it Emilio Pisanty's comment) disappeared. So I have to add in this comment that the in physics literature the "rank of a tensor" means the number of indices of the tensor (to put this in such simple terms). – Zoltan Zimboras Jan 25 '14 at 21:34
• ok, good links. Any idea regarding the change of tensor rank from point to point? I think the for the euclidean and minkowiski metric tensor, the tensor rank is always 1 at every point, since you can always transform to a frame where it's diagnol. Not sure about the tensor rank of ,say, the stress-energy tensor or the einstein tensor. Maybe some similar statement can be made. – yalis Jan 25 '14 at 23:40

Edit. As EmilioPisanty points out below, my answer is irrelevant because I misunderstood which notion of rank the OP was referring to.

Yes, tensor rank is a useful concept in physics.

For instance, it's useful in distinguishing different curvature tensors. The Riemann tensor is rank $4$, the Ricci Tensor is rank $2$, and the ricci scalar is rank $0$, as it's name suggests.

It's also a useful concept because it immediately indicates how the components of the tensor will transform under coordinate transformations. A tensor of rank $k$ will transform with $k$ factors of the jacobian of the coordinate transformation.

The rank of a tensor field does not change from one point to another (at least in standard treatments). In coordinate-free treatments of differential geometry, a tensor field of rank $k$ can be defined as a mapping that associates to each point $p$ on the manifold, a rank $k$ tensor on the tangent space $T_pM$ at point $p$. By this definition, the constant rank of the tensor field is immediate.

• It seems to me that you're misunderstanding the OP's definition of tensor rank. Admittedly your notation is the standard one, but the OP refers to the minimum number of tensor-product simple tensors needed to construct it. For a (standard) rank-2 tensor, for example, this is its matrix rank when you write it as a matrix. – Emilio Pisanty Jan 25 '14 at 20:38
• @EmilioPisanty Well then. I should have read the question more closely; thanks. – joshphysics Jan 25 '14 at 21:09
• +1 because it's critical to call out the difference between two concepts that use the same terminology. – yalis Jan 25 '14 at 23:41

There are already many good answers. Tensors and their decomposition in terms of simple tensors are important to virtually any topic of physics.

I) For starters, functional analysis and operator theory is used in almost all branches of physics. And rank-one operators

$$\tag{1} T~\in~{\cal L}(V;W)~\cong~ V^{*}\otimes W$$

are important objects here. E.g. in quantum information theory, as Zoltan Zimboras mentions in his answer. In particular a density operator is rank one iff it's a pure state. Else it's a mixed state.

II) The list goes on. One can also consider e.g. antisymmetric tensor products, usually called exterior products and denoted with a wedge $\wedge$. E.g. a totally skewsymmetric trivector-field

$$\tag{2} \pi~=~\frac{1}{3!}\pi^{ijk} \partial_{i}\wedge\partial_{j} \wedge\partial_{k} ~\in~\Gamma(\bigwedge{}^{\!3} TM)$$

on a manifold $M$ is traditionally called decomposable  if there exist three vector fields $X,Y,Z \in \Gamma(TM)$ such that

$$\tag{3} \pi~=~X\wedge Y\wedge Z.$$

The name decomposable trivector-field (3) is a bit unfortunate because (3) precisely plays the role of "atomic" building blocks for all trivector-fields. [There is an obvious generalization of (2) and (3) to $n$-multivector-fields.] Next, we can use (2) to form a $3$-bracket

$$\tag{4} \{f,g,h\}~=~\pi^{ijk}~ \partial_{i}f ~\partial_{j}g~ \partial_{k}h, \qquad f,g,h~\in~C^{\infty}(M).$$

Nambu famously investigated such $3$-brackets in 1973, see e.g. Ref. 1 and this Phys.SE answer. Most authors seem to agree that the correct generalization of a Jacobi-like identity for a $3$-bracket is the so-called Filippov fundamental identity (FI). Perhaps surprisingly, one may prove  that the Filippov fundamental identity implies decomposability (3) of a Nambu bracket, i.e. there do not exist non-decomposable Nambu-Poisson brackets.

References:

1. J.A. de Azcarraga and J.M. Izquierdo, $n$-ary algebras: a review with applications, J. Phys. A43 (2010) 293001, arXiv:1005.1028.
• Hmmm. While they are definitely the correct generalization to the exterior algebra, mustn't the 'decomposable' trivectors $X\wedge Y\wedge Z$ necessarily have rank greater than one? Does the notion of the minimum number of decomposable trivectors needed to express a given trivector $\pi\in\Gamma(\bigwedge{}^{\!3} TM)$ have any traction? – Emilio Pisanty Mar 10 '14 at 13:07
• Yes, the notion of rank also exists for totally skewsymmetrized (or symmetrized) tensor products, which often appear in physics along with their un(skew)symmetrized counterparts. Note that the rank calculated in a (skew)symmetrized tensor product in general differs from the rank calculated in the underlying un(skew)symmetrized tensor product. – Qmechanic Mar 10 '14 at 13:47