The textbook defines a tensor to be an element in $(T^*)^k×T^l→R$. It then expresses tensors as arrays of components with respect to a certain basis, and defines tensor contraction using summation convention. My question is: Can we define tensor contraction without referring to basis, components, and summation, but just the definition of tensor as a multilinear function? If components are inescapable, why bother define tensors as multilinear functions?

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    $\begingroup$ Do you know the so called universality property of the tensor product space and tensor map? It is the theoretical tool exploited for stating abstract (i.e., independent from choices of bases) definitions of this type. I have no spare time now for writing down an understandable answer, but that is the way... $\endgroup$ Aug 26, 2014 at 14:15
  • $\begingroup$ Do you mean the universal mapping property in category theory? I have learnt some category theory, but I haven't seen examples related to tensors yet. $\endgroup$
    – elflyao
    Aug 26, 2014 at 14:19
  • $\begingroup$ Yes, just that kind of property. The contraction is defined suitably using that property applied to the pair tensor product of spaces + tensor product application. $\endgroup$ Aug 26, 2014 at 15:25
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    $\begingroup$ I don't think we need the heavy weaponry here - is not this already basis- and component-free? (It will always be inescapable to talk about the "$x$-th factor" of the tensor product, since we must say which indices are contracted) $\endgroup$
    – ACuriousMind
    Aug 26, 2014 at 18:27

1 Answer 1


I can give you a mathematical reason why you have to use components to define contraction.

The reason is that contraction doesn't work for infinite dimensional vector spaces. You would have to sum over infinitely many components, and this sum might not converge. So there is no possibility of contraction for infinite dimensional tensors.*

But if we had a "basis free" definition of contraction then there would be nothing to stop it working in the infinite dimensional case as well. The definition of finite dimensional is "has a finite basis", and so the definition always has to mention a basis in order to distinguish between the finite and infinite dimensional cases.

So using a basis really is inescapable here.

(In the comments to your question ACuriousMind gives a component free definition of contraction, but this relies on a slightly different definition of tensor than the one you were given. I order to prove that the definitions are the same you again have to assume finite dimensionality and pick a basis, because the definitions aren't the same in the infinite dimensional case.)

If components are inescapable, why bother define tensors as multilinear functions?

In physics there's no "preferred basis". So an answer to a physical question can't be correct if it depends on which basis you pick. After defining contraction, your teacher should have shown that if you were to pick a different basis to write your coordinates in then you would indeed get the same answer as before.

Doing this checking-that-things-don't-depend-on-the-basis is annoying, and so it's often best to give basis free definitions. (Views differ on when one should use a basis. Some people work in a basis all the time; others avoid them like the plague.)

*Example: The tensor $\delta^a_{\;b}$ would have infinitely many ones on its diagonal, so $\delta^a_{\;a}$ wouldn't exist.

  • $\begingroup$ I don't fully understand this (admittedly old) answer. You have two notions of contraction, both of which coincide for finite dimensions, one of which is well-defined in infinite dimensions, with the other definition failing in that domain $-$ I know which one I'd choose. If it bothers the purists, one can simply introduce the qualifier "for finite-dimensional spaces" into the usual componentwise definition of the contraction, with the infinite-dimensional case left open for the cleaner generalization. $\endgroup$ Jul 6, 2017 at 22:18
  • $\begingroup$ @EmilioPisanty It's not quite that there are two notions of contraction but rather two notions of tensor product. The clean definition is via the universal property that maps out of $V\otimes W$ are the same as bilinear maps out of $V\times W$. In this case there is always a contraction $V^*\otimes V \to R$. But some courses take the (easier?) definition that $V\otimes W$ is a bilinear map $V^*\times W^*\to R$. This disagrees with the first definition in the infinite case, and has no notion of contraction in the infinite case. So you're right that it's better to use the clean first definition. $\endgroup$ Jul 7, 2017 at 7:22
  • $\begingroup$ @EmilioPisanty What I was saying in my answer was that if you do use the "bad" second definition then you are forced to pick a basis when you come to define contraction, because contraction only works on finite systems (with this definition of tensor) and finite systems are defined to be those with a finite basis. Any basis-free definition would work in the infinite dimensional case, and therefore can't exist. $\endgroup$ Jul 7, 2017 at 7:25
  • $\begingroup$ Ah, I think I understand the difference now - if it comes all the way from the definition of the tensor product then yes, it needs more care in its handling. $\endgroup$ Jul 7, 2017 at 8:44

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