It has been discussed several times on this site the phrase "a tensor is something that transforms like a tensor". I'm comfortable with both the mathematical formalism and the physical applications. Even though the physical formalism can be made precise using the mathematical definition of a tensor using multilinearity, the situation is unsatisfactory for me.

In particular, "a tensor is something that transforms like a tensor" very strongly implies the following:

  • There is a big space of "something".
  • This space is equipped with an action of coordinate changes.
  • The linear space of tensors is a subspace of this, characterized by the action behaving a specific way.

The mathematical definition directly constructs the space of tensors in an intrinsic way, without reference to the "something" space at all. This is a discrepancy I'd like to look into. So the question is: Is there a mathematical definition that follows the physical definition more closely?

Another illustrative example: We say that "the difference of two connections is a tensor". What does that mean? Note that connections don't form a linear space, so this has to occur in some larger linear space, such that the affine subspace of connections is parallel to the vector subspace of tensors. Can we define this in a non-adhoc way?

Some extensions of the question: can we do this without mentioning coordinate charts, and can this be extended to spinors?


1 Answer 1


I don't think your implications are the right way to think about this kind of operational definition. Rather, we should consider this as the way physicists talk about the local data attached to various geometric objects over spacetime.

In differential geometry, we have the general notion of a fiber bundle $\pi : B\to M$ over a manifold $M$. Almost all objects we usually talk about live "in" such bundles - they're sections of them or structures on them.

For such bundles, we have that $\pi^{-1}(x)\cong F$ for whatever the fiber $F$ is - a vector space, a space of tensors, etc. While globally $B$ carries generally more structure than just attaching a copy of $F$ to every point in $M$, locally there exist trivializations of these bundles: A cover of $M$ by open sets $U_i\subset M$ for which $\pi^{-1}(U_i) \cong U_i\times F$, i.e. if you only look at one of the $U_i$, the bundle is just attaching a copy of $F$ to every point in $U_i$ in the straightforward way.

Those trivializations come with transition functions $t_{ij} : U_i\cap U_j \to \mathrm{GL}(F)$, where by $\mathrm{GL}$ I mean the appropriate set of invertible structure-preserving transformations of the fiber, e.g. the linear invertible transformations of a vector space when $F$ is a vector space. These functions carry the information about the global structure of $M$ and $B$ - they tell us how to glue together $B$ again from this local data: For any $x\in U_i\cap U_j$, we should consider $(x,f)\in U_i\times F$ and $(x,t_{ij}(x)f)\in U_j\times F$ to be the same thing.

This procedure can be reversed (and is sometimes known as the "cocycle construction" of bundles): We can start with a bunch of $U_i$, attach our desired fiber $F$ to them, specify a bunch of $t_{ij}$, then define $B$ to be the disjoint union $\bigsqcup_i U_i\times F$ quotiented by the relation $$(x_i,f_i)\sim (x_j, f_j) \iff x_i = x_j \quad \land \quad f_i = t_{ij}(x_j)f_j.$$

Now I claim that the specification of the $t_{ij}$ is equivalent to what the physicists are doing when they define objects by their transformation behaviour: When we say that "a vector" $v^\mu$ transforms like $v^\mu \mapsto J^\mu_\nu v^\nu$ under "coordinate transformations", where $J$ is the Jacobian of the transformation, in this language what we mean is that we're building a vector bundle with fiber $\mathbb{R}^n$ - the tuples of numbers $v^\mu$ - by specifying that when $U_i$ and $U_j$ are two different coordinate charts, and $\phi_{ij} : U_i\cap U_j \to U_i \cap U_j$ is the coordinate transformation between them, then the transition function between $U_i\times \mathbb{R}^n$ and $U_j\times \mathbb{R}^n$ is given by the Jacobian of $\phi_{ij}$, $J(\phi_{ij}) : U_i\cap U_j \mapsto \mathrm{GL}(\mathbb{R}^n)$.

This generalizes straightforwardly to tensors (building bundles with fibers $\mathbb{R}^n\otimes\dots\otimes \mathbb{R}^n\cong \mathbb{R}^{n^m}$), gauge fields/connections (fiber is the Lie algebra of the gauge group, the transition functions are the behaviour under local gauge transformations), etc.

Statements like "the difference of two connections is a tensor" just means examining the way the difference of two connections is acted on by the transition functions - it turns out the "non-tensorial" part of the gauge transformation drops out of the difference, since it is independent of the specific value of the connections, and so this lives naturally in the bundle we defined with the tensorial transition functions.

  • $\begingroup$ This is very close to what I wanted! I think to make this independent of coordinate choices we can take a "maximal coordinate atlas" sort of trick, and use the set of all trivializations instead of a chosen few? $\endgroup$
    – Trebor
    Dec 30, 2023 at 15:58
  • $\begingroup$ @Trebor It suffices to choose any cover of $M$, but no one's stopping you from choosing a maximal atlas as your cover, yes. $\endgroup$
    – ACuriousMind
    Dec 30, 2023 at 16:01
  • $\begingroup$ Nice answer. D'you have any examples of geometric objects that don't live in such spaces (or where it simply doesn't make sense to define them in terms of fibre bundles)? $\endgroup$
    – Eletie
    Dec 30, 2023 at 23:24
  • $\begingroup$ Does this construction generalize straightforwardly to spinors? $\endgroup$
    – Jagerber48
    Dec 31, 2023 at 16:52
  • $\begingroup$ @Jagerber48 Sure, the fiber of the spinor bundle is still a vector space, but we have a $\mathrm{SO}(p,q)$-structure now - the transition functions of the tangent bundle are restricted to be valued in $\mathrm{SO}$ instead of $\mathrm{GL}$, this is reduction of the structure group. Their action on the fiber of the spinor bundle is just via the spinor representation map from $\mathrm{SO}(p,q)$ to the linear operators on spinors. $\endgroup$
    – ACuriousMind
    Dec 31, 2023 at 17:04

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