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What is a tensor?

  1. Aug 16, 2011 #1
    Hey all,

    Can anyone describe to me the formal mathematical definition of a tensor? I've done some research on tensors and I've been having trouble piecing together a general definition. Or do tensors vary depending on the mathematical or physical paradigm in question? Also, what will I be using tensors for (I'm a pure math major) and how do they relate to "tensor products?"

    I'll be learning more about tensors and tensor products this year in my study of Algebra, but I wanted a bit of a head start. :) Thanks everyone!


  2. jcsd
  3. Aug 16, 2011 #2


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    This is a very gppd question. To my knowledge, there are different flavors of the tensor product. I'll only describe the tensor product here, since you are a pure math major.

    The tensor product is one of the most important things in pure mathematics, so see that you know it well. You'll encounter it everywhere!!

    In general, consider a bilinear map [itex]f:A\times B\rightarrow C[/itex] between vector spaces. The tensor product is some kind of linear approximation of this map. That is, the bilinear map factors as

    [tex]A\times B\rightarrow A\otimes B\rightarrow C[/tex]

    where the middle space is a vector space which is called the tensor product. In general, every bilinear map factors over [itex]A\otimes B[/itex].

    Tensor products can also be defined over general noncommutative rings or algebra's. They can be used to "change the base ring". For example, let's say we have a vector space V over [itex]\mathbb{R}[/itex]. Let's say that we want to extend V to form a vector space over the complex numbers instead? How do we do this? Well, just form the tensor product [itex]V\times_\mathbb{R} \mathbb{C}[/itex]. This is called "extension of scalars".

    Now you may ask, what exactly is the tensor product? How exactly do we define it? Well, this is actually a non-question. The important thing is the properties that the tensor product has. How exactly it is defined is of no real importance. However, it is only important THAT is can be defined in some way.

    Also read http://www.dpmms.cam.ac.uk/~wtg10/tensors3.html for an extended explanation.
  4. Aug 17, 2011 #3
    Thanks micromass, I appreciate it. I'll be taking a look at that website and take to heart the importance of tensor products. Only a few more weeks until the quarter starts and I start playing with this stuff!
  5. Aug 17, 2011 #4
    A good thing to take forward is that here is (another) example of a process that follows the associative and distributive laws

    \left( {R \otimes S} \right) \otimes T = R \otimes \left( {S \otimes T} \right) \\
    R \otimes \left( {S + T} \right) = R \otimes S + R \otimes T \\
    \left( {R + S} \right) \otimes T = R \otimes T + S \otimes T \\

    But note that the product order is important so in general

    [tex]R \otimes S \ne S \otimes R[/tex]

    Where R, S and T are tensors and [tex] \otimes [/tex] denotes their product
  6. Aug 17, 2011 #5


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    Let V be a finite-dimensional vector space, and V* its dual space. A tensor of type (n,m) is a multilinear map [tex]T:\underbrace{V^*\times\cdots\times V^*}_{\text{$n$ factors}}\times\underbrace{V\times\cdots\times V}_{\text{$m$ factors}}\rightarrow\mathbb R.[/tex] The components of the tensor in a basis [itex]\{e_i\}[/itex] for V are the numbers [tex]T^{i_1\dots i_n}{}_{j_1\dots j_m}=T(e^{i_1},\dots,e^{i_n},e_{j_1},\dots,e_{j_m}),[/tex] where the [itex]e^i[/itex] are members of the dual basis of [itex]\{e_i\}[/itex]. The multilinearity of T ensures that the components will change in a certain way when you change the basis. The rule that describes that change is called "the tensor transformation law".

    An alternative, and in my opinion ridiculous, way to define "tensor" is to say that an association of an (n+m)-tuple of numbers witch each basis is a tensor, if the relationship between tuples associated with different bases is described by the tensor transformation law. If you state the alternative definition in this way, it's awkward, but at least it makes sense. Unfortunately a lot of people try to dumb this down, and "define" the term by saying that a tensor is something that transforms as described by the tensor transformation law. This obviously doesn't make any sense.

    The most interesting situation is when V is the tangent space at a point p of a manifold M. Then the tensor would be called a "tensor of type (n,m) at p". The set of all (n,m) tensors at p can be given the structure of a vector space in an obvious way. Let's call that vector space [itex]T^n_m(p)[/itex]. The union [itex]T^n_m=\bigcup_{p\in M}T^m_n(p)[/itex] is a vector bundle, the bundle of (n,m) tensors. A local section of that bundle is called a tensor field. This means that a tensor field is a function [itex]T:U\rightarrow T^n_m[/itex] such that U is an open subset of M, and for all p in M, T(p) is a tensor at p.

    People often call tensor fields "tensors" when they're being a bit lazy.

    (We can of course generalize the above by allowing the factors of V* and V to appear in a different order in the definition of a tensor).
    Last edited: Aug 17, 2011
  7. Aug 18, 2011 #6
    I like to think of the tensor product as the fundamental piece for the universal property. In my linear alg class, it's quite hard to grasp when you first see it and I think I learned the most about it when I learned the fundamental theorems revolving it.
  8. Aug 20, 2011 #7


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    Suppose that X and Y are finite-dimensional vector spaces over [itex]\mathbb R[/itex]. Suppose that [itex]A\in X^*,\ B\in Y^*[/itex]. Define [itex]A\otimes B(x,y)=A(x)B(y)[/itex] for all [itex](x,y)\in X\times Y[/itex]. The map [itex](A,B)\mapsto A\otimes B[/itex] is a tensor product, right? How do we show this?

    The definition I'm familiar with says that a bilinear function [itex]\tau:U\times V\rightarrow Z[/itex] is a tensor product if for each bilinear function [itex]\sigma:U\times V\rightarrow W[/itex], where W is any vector space, there's a unique linear function [itex]\sigma_*:W\rightarrow Z[/itex] such that [itex]\tau=\sigma_*\circ\sigma[/itex].

    If the case I'm asking about, we can define [itex]\tau:X^*\times Y^*\rightarrow Z[/itex], where [itex]Z=\mathbb R^{X\times Y}[/itex], by [itex]\tau(A,B)=A\otimes B[/itex]. For all [itex](x,y)\in X\times Y[/itex], we have [itex]\tau(A,B)(x,y)=A\otimes B(x,y)=A(x)B(y)[/itex].

    I'm probably missing something really simple, but I don't see how to proceed from here. OK, the next step would be to say this: Let [itex]\sigma:X^*\times Y^*\rightarrow W[/itex] be a bilinear but otherwise arbitrary function. But now what?
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