# Square vs non-square tensors?

In mathematics, tensors are objects that operates on vector space. In physics or engineering, tensors usually operates on one vector space and its dual space:

$V^{*} \times V^{*} \times V^{*} \times \dots \times V \times V \times V \times \dots$

leading to a notation where $n$ is the number of contravariant indexes, and $m$ the number of covariant indexes. And that's fine. For order-2 tensors, it leads to a square matrix notation, for order-3 it leads to cubic multimatrix notations...

But mathematically, I think that a tensor can operates over different spaces likes: $U \times V \times W \times \dots$ with $U$, $V$ and $W$ vector spaces that may have different numbers of dimensions leading to non-square tensors.

I have several questions:

1. Are the previous statements true?
2. Do you confirm that covariance and contravariance does not have any sense for non-square tensors?
3. As tensors that operates on one vector space and its dual space are a sort of special case of non-square tensors, do they have a particular name? (for example like a ring is a semiring with the additional property that its additive group is an Abelian group)
4. Is there any use of non-square tensors in physics or engineering?

The word tensor is often abused. Firstly, a tensor is simply an element of the tensor product of some vector spaces or bimodules or something. In this sense, of course there are non-square tensors. For example an element of $V\otimes_k W$ would be called a tensor, for any $k$-vector spaces $V$ and $W$. But the words covariant and contravariant don't have any meaning here.

Secondly (and this is more closely aligned with the topic of your question), tensor might also mean a tensor (in the first sense above) valued function on a manifold. For example, let $T_p M$ denote the tangent space to a smooth manifold at the point $p\in M$. A tensor can mean a choice of element $Z_p\in T_pM\otimes \cdots \otimes T_pM \otimes (T_pM)^\ast\otimes \cdots\otimes (T_pM)^\ast$ for each point $p\in M$, which depends differentiably on $p$. For example vector fields are tensors in this sense. The words covariant and contravariant have their origins here in how the coordinates of $Z$ behave with respect to coordinate transformations on $M$.

For a "non-square" tensor of this type, one especially important example is the second fundamental form. If $M^k$ is a Riemannian manifold isometrically immeresed in some Riemannian manifold $N^{k+n}$, then the second fundamental form is roughly this: for a point $p\in M$ and a pair of tangent vectors $v,w\in T_pM\subset T_pN$, there is a normal vector $S_p(v,w)\in (T_pM)^\perp$ which is something like a second derivative (hence measures curvature). Since $S_p$ chews on two tangent vectors and spits out a normal vector, we can think of $S_p$ as an element of $(T_pM)^\perp\otimes (T_pM)^\ast\otimes (T_pM)^\ast$. This $S$ is a very important non-square tensor (dimensions $n\times k\times k$)!

For a more precise response to your questions:

1. Not really. Tensors don't really act on anything. However $\operatorname{End}(V)\cong V\otimes V^\ast$, so operators can be thought of tensors, but not usually vice versa. A tensor usually just means an element of a tensor product of vector spaces (mathematician) or as a tensor valued function (physicist).

2. I would say this is right. Without any context there's no reason to call an element of $V\otimes W^\ast$ a tensor of type $(1,1)$ or $(2,0)$, or whatever. These notions are undefined in general.

3. No.

4. Yes! See above.

• Thank you very much. As I am not friendly at all with the second fundamental form, could you explain me what is exactly the number of dimension over each index? – Vincent Dec 31 '13 at 0:50
• Are you sure of what you are saying because in some documents, authors seem to say that the second fundamental form is a tensor of type (2, 0) (see section 17.3 of this document: maths-people.anu.edu.au/~andrews/DG/DG_chap17.pdf, or page 171 of this document : win.tue.nl/~rvhassel/Onderwijs/Tensor-ConTeX-Bib/…) – Vincent Dec 31 '13 at 1:04
• The second fundamental form can also refer to a type $(2,1)$ tensor like I have above (I learned from Do Carmo's Riemannian geometry book). The two are closely related. However, I did make a mistake in describing this tensor! It might be distracting to give a full-on definition, but here goes: suppose a manifold $M^k$ is embedded in $\mathbb{R}^n$. Extend a tangent vector $w\in T_pM\subset T_p\mathbb{R}^n$ to a vector field $W$ on all of $\mathbb{R}^n$. Define $S(v,w)$ to be the normal component of the derivative of $W$ in the direction of $v$. This does not depend on the extension $W$. – Matt Hogancamp Dec 31 '13 at 1:39
• The usual second fundamental form (along a normal vector $\eta$) is the quadratic (not bilinear!) form $\langle \eta, S(v,v)\rangle_N$, where $S$ is like I have above. I'm getting all this from chapter 6 of Do Carmo. Oops, I just noticed that he uses the letter B where I have used S. – Matt Hogancamp Dec 31 '13 at 1:54

Yes, a tensor is just a multilineal function which operates in vector spaces. But in the context of differential geometry (which is closely related to physics), where one has tensor and vector fields (note the fundamental difference: in physics, a tensor is often understood as a tensor field, a field that varies smoothly which assigns a tensor to every point of the manifold), a difference between contravariant and covariant indexes has to be made: imagine you want to make a change of coordinates (for whatever reason: you need to make a rotation, a simmetry, or something like that). Then, depending on what is the origin of the vector (it may arise as a gradient of some scalar function, or as the time derivative of some parametrization) and what is the kind of change of coordinates you want to do, you'll find that the coordinates expression of certain parts of the tensor change in one way or another. Is then when the difference becomes apparent.

But if you're not dealing with tensor fields but with tensors, then there's no difference, because a dual vector is a vector too, and tensors act on vector spaces.

I don't know if the tensors $T_1^1$ have a special name. They can be interpreted in a number of different ways (as square matrices too).

As for the last question, yes, there are non-square tensor (fields) in mathematical physics. Maybe the most famous example is the Riemann curvature tensor, which is fundamental in general relativity.

• Maybe the term non-square tensor was not well-chosen then. The Riemann curvature tensor in this denomination IS a square tensor (by square I mean non-hyperrectangle: same dimension over each index). – Vincent Dec 31 '13 at 0:09