I Can Tensors of Any Rank Be Approximated by Matrices?

[trees and plants]

Hello. Could we approximate a tensor of (p,q) rank with matrices of their elements? I am talking also about the general case of a tensor not only special cases. For example a (2,0) tensor with i, j indices is a matrix of ixj indices. A (3,0) tensor with i,j,k indices I think is k matrices with ixj elements each one of them. So is a tensor of (p,q) rank with upper indices i₁,i₂,...,i_p and lower indices j₁,j₂,...,j_q a set of (i₁xi₂...xi_p)x(j₁xj₂...xj_q-2) matrices of j_q-1xj_q elements? If I am incorrect what did I do wrong? Thank you.

 

[jedishrfu]

What do you mean approximate a tensor?

A tensor of rank 0 is a scalar.

A tensor of rank 1 is a vector that could be viewed as an ordered set of scalars.

A tensor of rank 2 is a matrix that could be viewed as an ordered set of column vectors.

A tensor of rank 3 is a tensor (no particular name) that could be viewed as an ordered set of matrices.

However, thinking this way diminishes the full import of tensors. Tensors have subscript and superscript notation that represent covariant and contravariant attributes that are useful in differential geometry among other things.

https://en.wikipedia.org/wiki/Tensor




...

 

[fresh_42]

A tensor of rank 3 is a tensor (no particular name) that could be viewed as an ordered set of matrices.

You can consider such a tensor as bilinear multiplication. See Strassen's algorithm for example:
https://www.physicsforums.com/insights/what-is-a-tensor/

 

[etotheipi]

A tensor of rank 1 is a vector that could be viewed as an ordered set of scalars.

Tiny quibble... a scalar is, like you point out, a rank-0 tensor, and is by definition coordinate independent. So the components of a vector w.r.t. some basis are not scalars, instead they are just numbers [which of course transform under change of coordinates].

 

[trees and plants]

It is not necessary to approximate a tensor as i say but it may be a goal one has. But other goals are needed as i read about differential geometry and the use of tensors, they study it differently than i say. What i mean by approximation is that the tensor may be represented as the times i said before matrices. Having the components of those matrices will help solve my question in the general case. I think the components and matrices of a (p,q) rank tensor are so many that we need more papers or space to put them. Perhaps my question is wrong.

 

[trees and plants]

Tiny quibble... a scalar is, like you point out, a rank-0 tensor, and is by definition coordinate independent. So the components of a vector w.r.t. some basis are not scalars, instead they are just numbers [which of course transform under change of coordinates].

They can not be functions? The vector with respect to a basis can not be like f=(f₁,...,f_n) where f_i, ∀ i=1,..,n are functions of m variables where x_j are those ∀ j=1,...,m with x_j∈ℝ? They do not obey the transformation law? That is why?

 

[fresh_42]

They can not be functions? The vector with respect to a basis can not be like f=(f₁,...,f_n) where f_i, ∀ i=1,..,n are functions of m variables where x_j are those ∀ j=1,...,m with x_j∈ℝ? They do not obey the transformation law? That is why?

They can and they are. Take a derivative as example: $f(x)=3x^2$ and $\left. \dfrac{d}{dx}\right|_{x=5}f(x)=30$. This is a number, a scalar, and a linear function, or even a covariant derivative. The latter two are the usual interpretation in differential geometry and physics. While the number $30$ is the slope of the tangent at $x=5$ at school, it becomes the linear function: $30=D_5(f)\, : \,v\longrightarrow 30\cdot v$ in differential geometry. This might be a bit confusing, but it is the one dimensional version of what we call e.g. a curvature tensor. So whether the value $30$ is considered a slope, a number, a scalar or a linear function depends on whom you ask, will say: the context.

 

[ChinleShale]

They can not be functions? The vector with respect to a basis can not be like f=(f₁,...,f_n) where f_i, ∀ i=1,..,n are functions of m variables where x_j are those ∀ j=1,...,m with x_j∈ℝ? They do not obey the transformation law? That is why?

If one has vector fields $X_{i}$ in some neighborhood that form a basis in each tangent space then any vector field in this neighborhood can be expressed as a linear combination of these basis fields. The coefficients of this linear combination will be functions on the neighborhood. So if $V=∑_{i}a_{i}X_{i}$ then the $a_{i}$ are functions(scalars). If one changes basis $V$ will be written using different functions. The same thing holds for any tensor.

If one has a vector field $V$ and a 1 form $ω$ then the value $ω(V)$ of $ω$ on $V$ is a function(scalar). Given the basis $X_{i}$ one has an n-tuple of 1 forms $ω_{i}$ that give the $i$'th coefficient of any vector field in its linear expansion with respect to the basis fields. So for an arbitrary vector $V$, $V=∑_{i}ω_{i}(V)X_{i}$. The $ω_{i}(V)$ are functions for each vector field $V$.

 

[LCSphysicist]

What do you mean approximate a tensor?

A tensor of rank 0 is a scalar.

A tensor of rank 1 is a vector that could be viewed as an ordered set of scalars.

A tensor of rank 2 is a matrix that could be viewed as an ordered set of column vectors.

A tensor of rank 3 is a tensor (no particular name) that could be viewed as an ordered set of matrices.

However, thinking this way diminishes the full import of tensors. Tensors have subscript and superscript notation that represent covariant and contravariant attributes that are useful in differential geometry among other things.

https://en.wikipedia.org/wiki/Tensor




...

"Tensor is a thing that transform like a tensor" is definitely my favorite definition

 

[jedishrfu]

Yeah, recursive definitions often leave oneself in the dust.

 

[Stephen Tashi]

So the components of a vector w.r.t. some basis are not scalars, instead they are just numbers [which of course transform under change of coordinates].

Yes, but human endeavors permit contradictions, so an elementary physics text may speak of "resolving a vector into components" and draw the components as vectors. For example, the gravitational force on a mass resting on inclined plane can be represented a "component" perpendicular to the surface of the plane and a "component" parallel to the surface of the plane.

 

[ChinleShale]

Tiny quibble... a scalar is, like you point out, a rank-0 tensor, and is by definition coordinate independent. So the components of a vector w.r.t. some basis are not scalars, instead they are just numbers [which of course transform under change of coordinates].

Actually the components are scalars. When you transform the tensor to a new coordinate system the new components are different scalars. Scalars are just functions defined on the manifold. Components of a vector field in a given coordinate system are functions.

For instance the components of a vector field(contravariant tensor of rank 1) in a given coordinate system form a n-tuple of functions on the coordinate domain. The most primitive examples are the coordinate functions $x_{i}$ themselves which are the projections of the coordinate chart onto the standard axes in Euclidean space.

 

[etotheipi]

I thought, that $\phi$ is a scalar if for any two coordinate systems $C$ and $C'$, $$\phi'(x') = \phi(x)$$that means, a scalar is invariant under a change of coordinate system. That definition is not satisfied by the components of a vector with respect to some basis, though.

 

[ChinleShale]

I thought, that $\phi$ is a scalar if for any two coordinate systems $C$ and $C'$, $$\phi'(x') = \phi(x)$$that means, a scalar is invariant under a change of coordinate system. That definition is not satisfied by the components of a vector with respect to some basis, though.

The components in any given coordinate system are scalars. When one changes coordinates one gets a new set of scalars. If one considers all components in all coordinate systems together as a single object then one has the underlying tensor. But that is different.

 

[etotheipi]

I think I see what you mean, thanks.

 

[ChinleShale]

I think I see what you mean, thanks. I delete!

No need to delete. Your comment leads to clarification of ideas. The index definition of tensors is not easy to master.

 

[wrobel]

Hello. Could we approximate a tensor of (p,q) rank with matrices of their elements? I am talking also about the general case of a tensor not only special cases. For example a (2,0) tensor with i, j indices is a matrix of ixj indices. A (3,0) tensor with i,j,k indices I think is k matrice

Yes you can consider tensor by means of matrices
that is completely useless by itself for understanding what tensor is

 

[trees and plants]

Yes you can consider tensor by means of matrices
that is completely useless by itself for understanding what tensor is

Could someone answer the following?Why do we use tensors without referring to matrices and vectors and their theories? Could a tensor be generalised to another object and what would that generalisation allow in mathematics and in physics?

 

[fresh_42]

Could someone answer the following? Why do we use tensors without referring to matrices and vectors and their theories?

Because they are something different. Matrices and vectors are certain tensors, but not vice versa. A tensor can be seen as a multilinear function, in the same sense as a vector can be seen as a linear function and a matrix as a bilinear function.

Could a tensor be generalised to another object ...

What does that mean? A tensor is already a generalization in the sense that it has the so called universal property.

... and what would that generalisation allow in mathematics and in physics?

It allows to consider multilinear structures such as e.g. Graßman algebras or Lie algebras as quotient space of tensor algebras.

 

[etotheipi]

Could someone answer the following?Why do we use tensors without referring to matrices and vectors and their theories? Could a tensor be generalised to another object and what would that generalisation allow in mathematics and in physics?

Maybe I misunderstand the question but the matrix representations of a tensor I believe just help to do actual calculations given a choice of coordinates. For example, you can construct a homomorphism from the group of rotations in 3D space to the group of 3x3 orthogonal matrices w/ unit determinant, e.g. something like $R^{\mathbf{n}}_{\theta} \mapsto \rho(R^{\mathbf{n}}_{\theta})$ requiring that $\rho(R^{\mathbf{n}}_{\theta_1}) \rho(R^{\mathbf{n}}_{\theta_2}) = \rho(R^{\mathbf{n}}_{\theta_1 + \theta_2})$, so now you can instead work with the coordinate representations of the vectors in $V$.

Nonetheless the original abstract tensor $R^{\mathbf{n}}_{\theta}$ is still just a rank 2 tensor which can be viewed naturally as a map from $V \rightarrow V$ (or instead $V \times V \rightarrow \mathbb{R}$), which has nothing to do with matrices a priori.

 

[trees and plants]

Nonetheless the original abstract tensor $R^{\mathbf{n}}_{\theta}$ is still just a rank 2 tensor which can be viewed naturally as a map from $V \rightarrow V$ (or instead $V \times V \rightarrow \mathbb{R}$), which has nothing to do with matrices a priori.

So if we have a rank 4 tensor $R_{ijkl}$ (like the Riemann curvature tensor), it can be viewed as a map from $V\times V \times V \rightarrow V$ or $V \times V \times V \times V \rightarrow \mathbb{R}$) ?In this case (sorry for the question) what is V?

 

[fresh_42]

So if we have a rank 4 tensor $R_{ijkl}$ (like the Riemann curvature tensor), it can be viewed as a map from $V\times V \times V \rightarrow V$ or $V \times V \times V \times V \rightarrow \mathbb{R}$) ?In this case (sorry for the question) what is V?

A vector space. And it can also be seen as with dual vector spaces $V^*$ instead. It all depends on the context and the purpose.

 

[trees and plants]

A vector space. And it can also be seen as with dual vector spaces $V^*$ instead. It all depends on the context and the purpose.

This means that a Riemannian manifold is a vector space and a topological space? Is it also a group? So it is a Lie group?

 

[fresh_42]

This means that a Riemannian manifold is a vector space ...

No.

... and a topological space?...

Yes.

Is it also a group? ...

Normally not, but it can be a group, e.g. ...

... So it is a Lie group?

Yes. Every Lie group is a differentiable manifold, but not vice versa.

 

[trees and plants]

So if we could generalise the multilinearity and axioms of vector space in order to generalise we could get a generalisation of Riemannian manifolds or differentiable manifolds?What else would allow in that structure for calculus to be applied?Or is it just that ok?

 

[wrobel]

why do not you read a textbook?
for example
Lectures on Differential Geometry
Shlomo Sternberg

 

[trees and plants]

Another question i have is what do tensor relations, like equations express? Mathematically or physically. Could someone explain?