# What is a tensor?

1. Nov 20, 2013

### 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.

2. Nov 20, 2013

### 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.

3. Nov 20, 2013

### 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.

4. Nov 20, 2013

### Staff: Mentor

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.

5. Nov 20, 2013

### D H

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

6. Nov 20, 2013

### D H

Staff Emeritus
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.

7. Nov 20, 2013

### Staff: Mentor

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

8. Nov 20, 2013

### 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.

9. Nov 20, 2013

### PSarkar

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.

10. Nov 20, 2013

### 1MileCrash

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.

11. Nov 20, 2013

### jgens

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:

Last edited: Nov 20, 2013
12. Nov 20, 2013

### PSarkar

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.

13. Nov 20, 2013

### 1MileCrash

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.

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14. Nov 20, 2013

### jgens

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.

Last edited: Nov 20, 2013
15. Nov 20, 2013

### PSarkar

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 any one know which approach is "the right one" in pure maths. I think it is the algebra way?

16. Nov 20, 2013

### jgens

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.

Last edited: Nov 20, 2013
17. Nov 20, 2013

### wotanub

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."

18. Nov 20, 2013

### 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.

19. Nov 20, 2013

### PSarkar

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.

20. Nov 20, 2013

### 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.