Background knowledge: I understand that there are many levels on which to understand tensors. A year or so ago I started trying to learn them with coordinate transformations in mind (just how to work with them generally), but ending up learning a little about lots of other things.

For instance, their definition in abstract algebra, as multilinear maps to a field, universal properties, category theory (about tensors as monoidal categories, and about universal properties), as multidimensional arrays, as elements of tensor product spaces... I have also tried to start getting an idea of what tensors are by looking at examples like the metric tensor and from the view of the usefulness of tensor notation. I have gone into enough depth to do some problems with all of these and understand the idea, but probably won't be familiar and fluent with a ton of examples.

The question is the title. What I want to know is, how do these all tie back into coordinate transforms (or even to each other)?

Monoids seem too detached of an idea, but just going through problems isn't giving me any "aha" moments either. Are abstract definitions of tensors more like useful ways of thinking about tensors for specific things like coordinate transforms or are tensors more of an abstract idea with concrete applications?

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    $\begingroup$ I'm not sure I understand your question, not even whether you are talking about tensors or tensor fields. A tensor is an element of a certain vector space (or of a more general algebraic object), and this is an entirely algebraic construction. A tensor field is an object in differential geometry wich arises from applying the algebraic construction to certain vector bundles. Doing this locally, in a coordinate chart, gives rise to the coordinate transformation approach you seem to be talking about. For a global, coordinate independent approach, the general view is more useful. $\endgroup$
    – Thomas
    Mar 5, 2020 at 19:24
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    $\begingroup$ (continued). The general purpose of tensor fields in differential geometriy is to express geometric facts. A statement about the geometry should not depend on the choice of a coordinate system. This requirement is what leads, in the local coordinate charts, to the transformation rule for tensors. An object which behaves like that does not depend on the choice of a coordinate system. In a coordinate free approach this is automatically true, if you are looking at the 'right' objects. $\endgroup$
    – Thomas
    Mar 5, 2020 at 19:30

1 Answer 1


I am going to give my 2 cents on the matter, please take this with a grain of salt.

The story begins with continuum mechanics, where 18th century scientists realized that numbers were not enough to study three-dimensional objects. (I say "scientists" because I think that there was little difference between mathematicians and physicists at the time). They needed objects "with indices". The position of a point is not a single number, it is three numbers: $(x_1, x_2, x_3)$, let's write it $x_j$, with the convention that $j=1, 2, 3$. Speed and acceleration are similar. But what about stress?

Stress is not the same as position. It is simliar to a vector; at each point it points in the direction in which the structure is subject to strain. But, unlike position, it does not depend only on position in space, it also depends on a further direction that has to be specified at each point. Let's think of a bridge. At any given point, it is subject to a horizontal stress, which is not a big deal, and a vertical one, which is much more relevant. If we compress the bridge, basically nothing happens. But if we apply a vertical strain, so much that we reach a critical level, the bridge collapses downwards.

This may look easy; but what if we have a hump-backed bridge? The stress structure becomes more complicated. We can apply a force normal to the surface, and if we do so strong enough, the bridge will collapse downwards. The collapse is not parallel to the direction of the force, unlike the previous case.

To study these things, they introduced the concept of tensor, and I think that the very name comes from this mechanical context. The stress tensor is a device that, given a point in space and a direction emanating from that point, spits out a vector, telling us the direction and the modulus of the strain we are subjecting the structure to. For the flat bridge, this device is diagonal; to a vertical force there corresponds a vertical strain, to a horizontal force there corresponds a horizontal strain. But it needs not be diagonal in all cases.

Later people discovered that this device can be generalized, and that it is profitable to do so. A Riemannian metric is a device that, given a point and two vectors emanating from that point, spits out a number, the scalar product of the two. This is similar to the above, but not quite the same, and it is a tremendously useful concept in mathematics. This is why all these things have been unified under a single algebraic theory.

  • $\begingroup$ Very interesting. I don’t know if this is a comment or another question, but I’ll say it anyways. Firstly, I’m assume the stress tensor is a bilinear map? This is probably the most common of the more abstract views I’ve come across, thinking of tensors as vectors combined by multilinear maps. $\endgroup$ Mar 7, 2020 at 4:30
  • $\begingroup$ Second, this is what I was asking about, the idea of bilinearity seems like it arises just as a conglomerate of operations and things that have uses for it, but I see 0 significance in it. I know the metric tensor for instance also has to do with the dot product, which is bilinear, but it is also many other things (although they are simpler than bilinearity maybe) like commutative, and it is not clear to me why we emphasize bilinearity. $\endgroup$ Mar 7, 2020 at 4:30
  • $\begingroup$ I mean, I realize that it is a more restrictive condition, but in math that doesn't generally mean it's more useful, and I think that’s what I’m not seeing; what bilinearity has to do with coordinates? I think it must be so obvious that people don’t even say it, but I am probably more of a layman in this area than you think. I’ve kind of been guessing it’s needed in a proof somewhere, but maybe I’m on the entirely wrong track. $\endgroup$ Mar 7, 2020 at 4:30

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