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Once I know the proper question, I will be able to solve the problem. (A. Einstein)

Updated question

I would like to prove that the multilinearity condition for a tensor field is equivalent to the transformation rule for its components in a coordinate basis. One way is straightforward, but I haven't been able to find a proof of the converse. For the sake of simplicity, I consider the case of a (1,1) tensor.

Let $\mathfrak{X}(M)$ be the set of tangent vector fields on $M$ and $\mathfrak{X}(M)^*$ the set of dual vector fields. Let $\mathbf{T}$ be a $\mathbb{R}$-bilinear map from $\mathfrak{X}(M)^* \times \mathfrak{X}(M)$ to $\mathbb{R}^M$. When is $\mathbf{T}$ a tensor field? In other words, when is it $\mathbb{R}^M$-bilinear?

For any two maps $(U,x)$ and $(V,y)$ on $M$ such that $U\cap V\neq \emptyset$, we define $J_{x\rightarrow y}{}^\mu{}_\nu |_P:=\frac{\partial {y^\mu}}{\partial {x^\nu}}|_P \in M_2(\mathbb{R})$, or $J^\mu{}_\nu$ for short. Hence, we have $\partial_{y^\mu}=(J^{-1})^\rho {}_\mu \partial_{x^\rho}$ and $dy^\nu=J^\nu{}_\sigma dx^\sigma$. The components of $\mathbf{T}$ in the coordinate map $(U,x)$ are defined by $T_{(x)\mu}{}^\nu:=\mathbf{T}(\partial_{x^\mu}{},dx^\nu)$.

I would like to show that $\mathbb{R}^M$-bilinearity for $\mathbf{T}$ is $equivalent$ to the transformation rules for $T_\mu{}^\nu$:

$``$Lower indices transform with $J^{-1}$ and upper indices transform with $J"$.

  1. Proof of $\implies$:

If $\mathbf{T}$ is $\mathbb{R}^M$-bilinear, since $(J^{-1})^\rho {}_\mu$ and $J^\nu{}_\sigma$ are scalar fields, we have:

$T_{(y)\mu}{}^\nu=\mathbf{T}(\partial_{y^\mu}{},dy^\nu)=\mathbf{T}((J^{-1})^\rho {}_\mu \partial_{x^\rho},J^\nu{}_\sigma dx^\sigma)=(J^{-1})^\rho {}_\mu J^\nu{}_\sigma \mathbf{T}(\partial_{x^\rho},dx^\sigma)=(J^{-1})^\rho {}_\mu J^\nu{}_\sigma T_{(x)\rho}{}^\sigma$

Hence, $T_\mu{}^\nu$ satisfies the transformation rules. $\square$

  1. Proof of $\impliedby$?

Conversely, given $\mathbb{R}$-bilinear $\mathbf{T}$, if $T_\mu{}^\nu$ satisfies the transformation rules for any two maps $(U,x)$ and $(V,y)$, how does one show that $\mathbf{T}$ is $\mathbb{R}^M$-bilinear?

That is, how does one show $\mathbf{T}(a.X,\omega)=\mathbf{T}(X,a.\omega)=a.\mathbf{T}(X,\omega)$ for an arbitrary scalar field $a$, vector field $X$ and dual field $\omega$?

Original question

In general relativity, when asked to prove that an object is (resp. is not) a tensor, we are often expected to show that its components follow (resp. don’t follow) the so-called tensor transformation rules. I’m having difficulties proving the equivalence between the definition of a tensor as a multilinear map and these transformation rules.

Specifically, one way is rather straightforward. Consider a (1,1) tensor for the sake of simplicity. We can show that its components satisfy the transformation rule according to which "in a change of coordinates, lower/covariant indices transform with $J=\frac{\partial x'}{\partial x}$ and upper/contravariant indices transform with $J^{-1}$":

In a local coordinate map $P \mapsto x(P)\in \mathbb{R}^n$, the tangent spaces $T_PM$ and $T_pM^*$ are endowed with the coordinate bases $\partial_\mu$ and $dx^\nu$, respectively. Expressing a form $\omega$ and a vector $X$ in theses bases as $\omega=\omega_\nu dx^\nu$ and $X=X^\mu \partial_\mu$, we can write $T(\omega,X)=T^\mu{}_\nu \omega_\mu X^\nu$, with $T^\mu{}_\nu := T(dx^\mu,\partial_\nu)$.

Similarly, in a new coordinate map $x'(P)$, we have $T(\omega,X)=T^{\mu'}{}_{\nu'} \omega_{\mu'} X^{\nu'}$. The transformation matrix $J$ gives us the new basis vectors in terms of the old: $dx^{\mu'}=J^{\mu'}{}_\rho dx^\rho$ and $\partial_{\nu'}=(J^{-1})^\sigma {}_{\nu'} \partial_\sigma$. And since $T$ is linear, we obtain the transformation rules:

$T^{\mu'}{}_{\nu'}=T(dx^{\mu'},\partial_{\nu'})=T(J^{\mu'}{}_\rho dx^\rho,(J^{-1})^\sigma {}_{\nu'} \partial_\sigma)=J^{\mu'}{}_\rho (J^{-1})^\sigma {}_{\nu'} T(dx^\rho,\partial_\sigma)=J^{\mu'}{}_\rho (J^{-1})^\sigma {}_{\nu'} T^\rho{}_\sigma \hspace{2mm} \square$

Hence, we see that linearity implies ($\implies$) the transformation rules for tensor components. This is used to prove that a given array of components is not a tensor field by showing that they do not follow the transformation rules (e.g. connection coefficients $\Gamma^\rho{}_{\mu\nu}$).

The converse ($\impliedby$) is assumed to be true, and can be used to prove that a given array of components defines a tensor field by showing that its components follow the transformation rules. However, I have not been able to find a proof…

If an array of components transforms according to the transformation rules for any change of coordinates $x \rightarrow x'$, how do I prove linearity for multiplication by any scalar field? For example, how does one show that "$T$ satisfies the transformation rules for any map $x'(P)$"$\implies T(a.X,\omega)=a.T(X,\omega)$ for an arbitrary scalar field $a$?

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2 Answers 2

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By writing out the coordinate form, it is immediately obvious that any such object with indices you write is multilinear in the components $X^a$ etc. For that you just have to use the fact that multiplication is commutative.

So why is it not necessarily a tensor? The sticking point is that they are, in general, dependent on the coordinate system you pick. The reason why you need the transformation property is to show that you can write it as a multilinear function of abstract coordinate-free vectors rather than just as a multilinear function of coordinates.

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Updated answer:

I think you could state your question more generally: Given a scalar field with indices, e.g., $T^{ij}_k$, does it represent the components of a tensor (in the coordinate basis $x$)? If the answer is yes --- a $(2,1)$-tensor field, $T: T^*M \times T^*M \times TM \rightarrow \mathbb{R}$ does exist, whose components in a coordinate chart $(U,x)$ are $T^{\, ij}_{(x)\,k} = T(dx^i, dx^j, \partial_k)$ --- then one should be able to write down an equation containing the components, $T^{\, ij}_{(x)\,k}$ and $T^{\, lm}_{(y)\,n}$, and show that it reduces to $T = T$. To obtain this equation, affect a change of coordinates to find, e.g., $$ T^{\, lm}_{(y)\,n} = A_i B_j C^k \,\, T^{\, ij}_{(x)\,k} \,\, D^l E^m F_n+ G^{lm}_{(xy),n} \,, $$ where $G^{lm}_{(xy),n}$ depends on both coordinate choices. Now, if the indexed scalar field represents the components of a tensor, then we have $$ T\left(dy^l, dy^m, \frac{\partial}{\partial y^n}\right) = A_i B_j C^k D^l E^m F_n\,\, T\left(dx^i, dx^j, \frac{\partial}{\partial x^k}\right) + G^{lm}_{(xy),n}\,. $$ We can affect a change of basis on the right-hand side version to obtain \begin{align} T\left(dy^l, dy^m, \frac{\partial}{\partial y^n}\right) &= A_i B_j C^k D^l E^m F_n\,\, T\left(\frac{\partial x^i}{\partial y^r} dy^r, \frac{\partial x^j}{\partial y^s} dy^s, \frac{\partial y^t}{\partial x^k} \frac{\partial}{\partial y^t} \right) + G^{lm}_{(xy),n} \\ &= A_i B_j C^k D^l E^m F_n \frac{\partial x^i}{\partial y^r} \frac{\partial x^j}{\partial y^s} \frac{\partial y^t}{\partial x^k}\,\, T\left( dy^r, dy^s, \frac{\partial}{\partial y^t} \right) + G^{lm}_{(xy),n} \end{align} To obtain an equality, we must have: $$ G^{lm}_{(xy),n} = 0 \quad \text{and} \quad \left(A_i D^l \frac{\partial x^i}{\partial y^r} \right) \left(B_j E^m \frac{\partial x^j}{\partial y^s}\right) \left( C^k F_n \frac{\partial y^t}{\partial x^k}\right) = \delta^l_r \, \delta^m_s \, \delta^t_n\,, $$ with the second requirement meaning that, e.g., $A_iD^l = \frac{\partial y^l} {\partial x^i}$.

So to return to (my restatement) of your original question: an object $T^{ij}_k$ represents the components of a tensor if the form of its transformation equation from one basis to another (the first equation above) has no additive terms and involves the index object being multiplied by exactly those partials given by the rules for "tranformation of a tensor.''

Original answer:

To take your counterexample specifically, the transformation rule for Christoffel symbols (see, e.g., Schuller's Lecture 7 at 1h 9m) is: $$ \Gamma_{(y)\,jk}^{\,i} = \frac{\partial y^{i}}{\partial x^q} \frac{\partial x^{s}}{\partial y^j} \frac{\partial x^{p}}{\partial y^k} \Gamma_{(x)\,sp}^{\,q} + \frac{\partial y^{i}}{\partial x^q} \frac{\partial^2 x^{q}}{\partial y^j \partial y^k} $$ Just by inspection, this cannot be a tensor because, as you showed in your (1,1)-tensor example, linearity in each slot of a tensor would yield a single term with the desired partials multiplying the tensor component.

So if you find that the coordinate transformation of some object with indices looks like the first term on the right-hand side above (a seeming tensor component, multiplied by partials of the two different coordinates), then that is indeed a tensor. To see this, simply assume that the index object is a tensor acting on basis vectors/covectors in the new basis, and then pull in the partial derivative terms. You find it to be precisely the action of a tensor on the basis vectors of the original coordinate basis (i.e., the left-hand side).

But if instead you obtain something that includes extra terms, like for the Christoffel symbol above, then it cannot be a tensor.

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  • $\begingroup$ “ So if you find that the coordinate transformation of some object with indices looks like the first term on the right-hand side above… Simply assume that it is a tensor“ — I think the OP is asking for a proof of this, not an assumption. $\endgroup$
    – Aiden
    Dec 2, 2023 at 16:24
  • $\begingroup$ Maybe my wording is poor, I will edit slightly. I simply mean that an equation of the form of above Christoffel symbol transformation, but without the second term on the right hand side, is exactly what one would find if $\Gamma^{\,q}_{(x)\, sp}$ are the components of a tensor. Which you can see by treating it as such. $\endgroup$
    – Ben H
    Dec 2, 2023 at 16:32
  • $\begingroup$ I agree: if a tensor is multilinear map, then its components obey the transformation rule. However the OP is looking for a proof of the converse: if some array obeys the transformation rule, then it is a multilinear map. $\endgroup$
    – Aiden
    Dec 2, 2023 at 16:39
  • $\begingroup$ That’s right, it is the converse which I can’t figure out. Thanks! $\endgroup$
    – gusifang
    Dec 2, 2023 at 18:13
  • $\begingroup$ I've tried to add something else. Not sure if it's adequate. I think the more general question is: is a given scalar field with indices the components of a tensor. Or, is there a tensor for which this scalar field with indices is its components. The answer should be: yes if it "transforms as a tensor" or no if it does not. $\endgroup$
    – Ben H
    Dec 2, 2023 at 18:28

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