What is a Tensor? Exploring Vectors & Multiplying Tensors

[7777777]

I am not sure if this a right place to ask what is a tensor. I already asked about vectors
in Math section, but I think a tensor has more to do with physics that mathematics,
so I came here.

I am reading A Zee's book Einstein Gravity that students have fear of tensors. I also think
that they suck, are they really necessary? Can't we manage without them?
Zee writes that:" a tensor is something that transforms like a tensor.
But this should not perplex us. A duck is something that quacks like a duck."

I think that vectors are more fundamental than tensors. Vectors don't behave in
multiplication the same way as tensors. You can multiply tensors and the result is
another tensor, but the same does not happen with vectors in multiplication.

 

[DeIdeal]

Tensors do have a lot to do with mathematics.
Why do you think they 'suck'? What's wrong with tensors? I guess they aren't necessary, but you could just as well do electromagnetism like Maxwell did, without vectors and writing out each of the 4*3+3 equations each time. They do make calculations considerably easier, though. And that's a complete understatement.
Vectors are tensors of type (1,0), so no, they are not more fundamental.

 

[WannabeNewton]

You're being horribly misguided by Zee in more ways than one I'm afraid. Get a proper math book. Tensors are absolutely fundamental and you cannot avoid them, not if you want to do GR. You can use spinor fields but that doesn't make things any less complicated.

 

[Nugatory]

I think that vectors are more fundamental than tensors. Vectors don't behave in
multiplication the same way as tensors. You can multiply tensors and the result is
another tensor, but the same does not happen with vectors in multiplication.

Ahhh... vectors ARE tensors. And you need another tensor (the metric tensor) before you can multiply two vectors to get a number out. The only reason that you haven't noticed this already is that the metric tensor is trivial in Cartesian coordinates - applying it is equivalent to multiplying by one.

Once you get to know tensors, you'll like them... honest, you will. They're not as hard as they look, it's basically impossible to do General Relativity without them, and it will really impress people looking over your shoulder when they see you doing tensor calculations.

 

[D H]

Can't we manage without them?

Sure. We can do without vectors, too. Why would you want too? Vectors simplify things greatly. So do tensors.

 

[D H]

Ahhh... vectors ARE tensors.

It's the other way around, at least with regard to the mathematical definition of a vector. There are lots of vectors that aren't tensors. Tensors, on the other hand, satisfy all the requirements of a vector space.

 

[Nugatory]

It's the other way around, at least with regard to the mathematical definition of a vector. There are lots of vectors that aren't tensors. Tensors, on the other hand, satisfy all the requirements of a vector space.

Fair enough... I was assuming that by "vector" OP didn't mean the general mathematical definition.

 

[1MileCrash]

I learned that tensors in some space are objects with n^k entries, where n is the rank of the tensor and k is the dimension of the space, and these objects transform according to a certain "transformation law."

From what I've read, the way I've learned about them is outdated and pretty bad, and that a better understanding can be reached through studying linear algebra in depth.

 

[PSarkar]

Ahhh... vectors ARE tensors. And you need another tensor (the metric tensor) before you can multiply two vectors to get a number out. The only reason that you haven't noticed this already is that the metric tensor is trivial in Cartesian coordinates - applying it is equivalent to multiplying by one.

Once you get to know tensors, you'll like them... honest, you will. They're not as hard as they look, it's basically impossible to do General Relativity without them, and it will really impress people looking over your shoulder when they see you doing tensor calculations.

I learned tensors as the multilinear forms $T: V^n \to \mathbb{R}$ (you can also add in the dual spaces) where $V$ is a vector space over $\mathbb{R}$. So I don't quite see how vectors are tensors (assuming the vector is from the vector space $V$). A tensor of type $(1, 0)$ would just be a linear functional but not a vector. But I suppose you can correspond it to some vector since $V$ and $V^*$ are isomorphic.

 

[1MileCrash]

I learned tensors as the multilinear forms $T: V^n \to \mathbb{R}$ (you can also add in the dual spaces) where $V$ is a vector space over $\mathbb{R}$. So I don't quite see how vectors are tensors (assuming the vector is from the vector space $V$). A tensor of type $(1, 0)$ would just be a linear functional but not a vector. But I suppose you can correspond it to some vector since $V$ and $V^*$ are isomorphic.

It's because there are "ranks" to tensors. A vector can be represented as a 1-dimensional array and so it is a tensor, just like your linear transformation can be represented as a matrix and so it too is a tensor. Even a scalar can be represented as a "0 dimensional array" and so it too is a tensor.

 

[jgens]

There are lots of vectors that aren't tensors.

This is nonsense. If V is a vector space, then the elements of V are tensors of type (1,0) in the tensor algebra of V. So by definition vectors are tensors :tongue:

 

[PSarkar]

It's because there are "ranks" to tensors. A vector can be represented as a 1-dimensional array and so it is a tensor, just like your linear transformation can be represented as a matrix and so it too is a tensor. Even a scalar can be represented as a "0 dimensional array" and so it too is a tensor.

May be you misunderstood my post. A tensor is a multinear form with the field being the real numbers (at least the way I learned it). You only get a matrix or a multidimensional array representation when you pick coordinates but otherwise they are coordinate free just like linear transformations. Furthermore, they are not the same thing as the multidimensional arrays, just like linear transformations are not matrices (The matrices are only a representation when you pick coordinates. You can formalize it as a linear map from the vector space of linear transformations to the vector space of matrices, given some basis). So that's why I was saying that strictly speaking, vectors are not tensors but can be put into a one to one correspondence with tensors of type $(1, 0)$, but even so, such a map is not unique.

May be you learned tensors in a different approach. The approach I am familiar with is the way it is done in Calculus on Manifolds by Michael Spivak. I have also seen an even more general approach in abstract algebra books starting with tensor products of modules and vector spaces.

 

[1MileCrash]

May be you learned tensors in a different approach. The approach I am familiar with is the way it is done in Calculus on Manifolds by Michael Spivak. I have also seen an even more general approach in abstract algebra books starting with tensor products of modules and vector spaces.

I dug up my old book, this is how I learned them:


So it looks like my definition does rely on the idea of coordinate systems.

 

[jgens]

May be you misunderstood my post. A tensor is a multinear form with the field being the real numbers (at least the way I learned it). You only get a matrix or a multidimensional array representation when you pick coordinates but otherwise they are coordinate free just like linear transformations. Furthermore, they are not the same thing as the multidimensional arrays, just like linear transformations are not matrices (The matrices are only a representation when you pick coordinates. You can formalize it as a linear map from the vector space of linear transformations to the vector space of matrices, given some basis). So that's why I was saying that strictly speaking, vectors are not tensors but can be put into a one to one correspondence with tensors of type $(1, 0)$, but even so, such a map is not unique.

May be you learned tensors in a different approach. The approach I am familiar with is the way it is done in Calculus on Manifolds by Michael Spivak. I have also seen an even more general approach in abstract algebra books starting with tensor products of modules and vector spaces.

Before talking definitively about what tensors are perhaps you should read up on those abstract algebra definitions you mention earlier In any case you are consistently getting the type convention for tensors wrong. Tensors of type (1,0) are elements of the vector space while tensors of type (0,1) are elements of the dual space. The way you learned tensors focuses entirely on this dual space perspective since it leads quickly to differential forms, but it is lacking in generality.

 

[PSarkar]

Before talking definitively about what tensors are perhaps you should read up on those abstract algebra definitions you mention earlier In any case you are consistently getting the type convention for tensors wrong. Tensors of type (1,0) are elements of the vector space while tensors of type (0,1) are elements of the dual space. The way you learned tensors focuses entirely on this dual space perspective since it leads quickly to differential forms, but it is lacking in generality.

Ok, I will explicitly state all the details to be clear. In Calculus of Manifolds, tensors are defined as all multilinear forms with the field being the real numbers,
$$T: V^n \to \mathbb{R}$$
The book doesn't go into the dual spaces but including that would look like,
$$T: V^n \times {V^*}^m \to \mathbb{R}$$
where the type is $(n, m)$. So with this definition, a type $(1, 0)$ tensors would be $T: V \to \mathbb{R}$ which is just a linear functional, $T \in V^*$.

I also looked up the approach from abstract algebra (I have not studied this but only saw it before). In this approach tensors are just the elements of $\underbrace{V \otimes \dotsb \otimes V}_n \otimes \underbrace{V^* \otimes \dotsb \otimes V^*}_m$. In this case you are correct since type $(0, 1)$ would mean $T \in V^*$. But some other algebra books define it as the linear maps, (since the above has a natural isomorphism to this one) $\underbrace{V \otimes \dotsb \otimes V}_n \otimes \underbrace{V^* \otimes \dotsb \otimes V^*}_m \to \mathbb{R}$ which in turn has a natural isomorphism to multilinear forms which was the approach in Calculus of Manifolds.

1MileCrash gives even another approach. Since there are multiple approaches to tensors, I guess we are all just arguing about the definition.

Anyway, does anyone know which approach is "the right one" in pure maths. I think it is the algebra way?

 

[jgens]

Ok, I will explicitly state all the details to be clear. In Calculus of Manifolds, tensors are defined as all multilinear forms with the field being the real numbers,
$$T: V^n \to \mathbb{R}$$
The book doesn't go into the dual spaces but including that would look like,
$$T: V^n \times {V^*}^m \to \mathbb{R}$$
where the type is $(n, m)$. So with this definition, a type $(1, 0)$ tensors would be $T: V \to \mathbb{R}$ which is just a linear functional, $T \in V^*$.

Your type convention is still wrong. You can look to Spivak for support but you will not find it there.

I also looked up the approach from abstract algebra (I have not studied this but only saw it before). In this approach tensors are just the elements of $\underbrace{V \otimes \dotsb \otimes V}_n \otimes \underbrace{V^* \otimes \dotsb \otimes V^*}_m$. In this case you are correct since type $(0, 1)$ would mean $T \in V^*$. But some other algebra books define it as the linear maps, (since the above has a natural isomorphism to this one) $\underbrace{V \otimes \dotsb \otimes V}_n \otimes \underbrace{V^* \otimes \dotsb \otimes V^*}_m \to \mathbb{R}$ which in turn has a natural isomorphism to multilinear forms which was the approach in Calculus of Manifolds.

Both definitions are equivalent. You have a modified type convention for the second way of dealing with tensors, which for whatever reason, you still have not grasped.

Edit: For clarity, whenever you deal with the multilinear map definition of tensors, the type convention is as follows. If T:V×...×V×V^*×...×V^*→ℝ is a multilinear map, then T has type (n,m) if and only if n V^*s appear in the product and m Vs appear in the product.

Anyway, does anyone know which approach is "the right one" in pure maths. I think it is the algebra way?

All of them are equivalent. Which one is best depends on the context. I am not a fan of doing everything with multilinear maps since it introduces an unnecessary double dualization, but geometers seem to prefer it for whatever reason.

 

[wotanub]

Zee writes that:" a tensor is something that transforms like a tensor.
But this should not perplex us. A duck is something that quacks like a duck."

This is ironic since this is a good definition in the same way that "a vector is something that transforms like a vector." Instead of rejecting this as some sort of circular logic it should beg the question "what does it mean to transform like a tensor?"

It allows you to move from "a vector is a magnitude and direction" to "a vector is an element of a vector space, a mathematical construct with characteristic defining maps and are defined by certain axioms."

 

[Jhenrique]

Tensors (of rank 2, i.e. one nxn matrix) analogously as vector (that have three components, three scalars) have three components too, three vectos as component, that, in its turn, each vector have three components, three scalars, therefore, a tensor have 9 scalars components. The Cauchy stress tensor illustrates very well this concept. But, as a matrix is a tensor and a tensor is a matrix, there is a lot interpretation for a tensor, becase there is a lot interpretation for a matrix.

 

[PSarkar]

Your type convention is still wrong. You can look to Spivak for support but you will not find it there.



Both definitions are equivalent. You have a modified type convention for the second way of dealing with tensors, which for whatever reason, you still have not grasped.

Edit: For clarity, whenever you deal with the multilinear map definition of tensors, the type convention is as follows. If T:V×...×V×V^*×...×V^*→ℝ is a multilinear map, then T has type (n,m) if and only if n V^*s appear in the product and m Vs appear in the product.



All of them are equivalent. Which one is best depends on the context. I am not a fan of doing everything with multilinear maps since it introduces an unnecessary double dualization, but geometers seem to prefer it for whatever reason.

I see my mistake now. I kept corresponding the $V$'s in the multilinear map definition to the $V$'s in the algebraic definition when I should have corresponded the $V^*$'s to the $V$'s in the algebraic definition

Yes all of them are equivalent but what about tensors as multidimensional arrays where it is written with the indices? That is quite different especially since you have some chosen coordinates.

 

[jgens]

Yes all of them are equivalent but what about tensors as multidimensional arrays where it is written with the indices? That is quite different especially since you have some chosen coordinates.

I think the multidimensional array definition is actually the oldest and it too is equivalent to the rest. There are ways of making that definition essentially coordinate independent as well. It just takes a little work.

 

[Chronos]

The simplest description of a tensor I ever heard is a tensor is a matrix of matrices. It looks simple and elegant, but, that is an illusion.

 

[WannabeNewton]

...and it will really impress people looking over your shoulder when they see you doing tensor calculations.

Hehe

 

[HallsofIvy]

Quote by PSarkar

Yes all of them are equivalent but what about tensors as multidimensional arrays where it is written with the indices? That is quite different especially since you have some chosen coordinates.

Quote by jgens
I think the multidimensional array definition is actually the oldest and it too is equivalent to the rest. There are ways of making that definition essentially coordinate independent as well. It just takes a little work.

Quote by Chronos
The simplest description of a tensor I ever heard is a tensor is a matrix of matrices. It looks simple and elegant, but, that is an illusion.

No. Arrays or matrices (or "matrices of matrices") are ways of representing tensors, in a given coordinate system, not tensor themselves, any more than "(1, 1, 3)" is a vector. In order to have a tensor, you must have either a "coordinate free" definition or a way of finding its matrix representation in any coordinate system.

 

[jgens]

No. Arrays or matrices (or "matrices of matrices") are ways of representing tensors, in a given coordinate system, not tensor themselves, any more than "(1, 1, 3)" is a vector. In order to have a tensor, you must have either a "coordinate free" definition or a way of finding its matrix representation in any coordinate system.

I think in each case the "multidimensional array definition" was mentioned, what was meant was the definition that says in a given coordinate system these arrays are tensors and then specifies how these arrays transform under a change of coordinates. It's more convenient to call this the "multidimensional array definition" than the "in a given coordinate system they are arrays that transform according to the tensor transformation law definition" in my humble opinion, but if you like pedantry in all your naming conventions, I suppose that I can oblige :tongue:

 

[Jhenrique]

Conclusion... it's impossible to give a embracing and concise definition about what is a tensor!

 

[1MileCrash]

Yeah, no one knows what the hell a tensor is.

 

[Jhenrique]

Exist a video that seem interesting

I don't understand it very well, 'cause I don't know speak english. What do you think of it?

 

[1MileCrash]

Exist a video that seem interesting

I don't understand it very well, 'cause I don't know speak english. What do you think of it?

It's hard to tolerate how he says "component."

 

[PSarkar]

I think in each case the "multidimensional array definition" was mentioned, what was meant was the definition that says in a given coordinate system these arrays are tensors and then specifies how these arrays transform under a change of coordinates. It's more convenient to call this the "multidimensional array definition" than the "in a given coordinate system they are arrays that transform according to the tensor transformation law definition" in my humble opinion, but if you like pedantry in all your naming conventions, I suppose that I can oblige :tongue:

But then how is the definition coordinate free? You are forced to choose some coordinates in order to work with tensors.

 

[jgens]

But then how is the definition coordinate free? You are forced to choose some coordinates in order to work with tensors.

You modify the definition slightly. The idea is you collect the various coordinate representations of the tensor together in a suitable manner. The result is that you have an object whose existence is independent of a particular choice of coordinates.

 

[WannabeNewton]

But then how is the definition coordinate free? You are forced to choose some coordinates in order to work with tensors.

This depends on what you mean by "work" and in what context but the definition is clearly coordinate-free. I would suggest you take a look at Roman's linear algebra text.

 

[Jhenrique]

Will be the Laplacian, $$\bigtriangledown ^2$$, in actually, a Tensor?? Will be a simplification of Hessian tensor?

 

[Jhenrique]

Come think in 2D. Given 2 values, a and b, it's possible to associate a point P with coordinates (a, b) and trace a straight between the origem and the point P. Ready, we have a vector! So, analogously, given 4 values (a, b, c, d), should be possible to associate these values with straight, and, somehow associate a plane with this straight. This, geometrically, would be a tensor. I think...

 

[PSarkar]

This depends on what you mean by "work" and in what context but the definition is clearly coordinate-free. I would suggest you take a look at Roman's linear algebra text.

Can you explain why the multilinear array definition is coordinate free as this is not obvious to me. If you need to choose coordinates to get the tensor, then the tensor depends on which coordinates you choose. You definitely cannot write one down without choosing coordinates. This definition misses the essence of tensor which is independent of any coordinate and so we should have a definition and representation without coordinates.

I think jgens answer about collecting all the coordinate representation of the tensors together make sense since it will no longer depend on the coordinates. But the down side is you still can't write one down without coordinates.

 

[WannabeNewton]

Can you explain why the multilinear array definition is coordinate free as this is not obvious to me.

Sorry when you said "the definition" I thought you were referring to the usual algebraic definition of a tensor as this is what I take to be "the definition" of a tensor. I find it hard to look at anything relying on coordinate representations as a definition hence I agree that a definition in terms of coordinate components is archaic and inferior for the reasons stated by you and others.