I'm having a hard time seeing how the quotient law for tensors is used it the following example of Riley, Hobson and Bence (3rd).

Preceding the example, it reads

Use of the quotient law to test whether a given set of quantities is a tensor is generally much more convenient than making a direct substitution. A particular way in which it is applied is by contracting the given set of quantities, having N subscripts, with an arbitrary $N$th order tensor (i.e. having independently variable components) and determining whether the result is a scalar.

If one contracts a given set of quantities, having $N$ subscripts, with an arbitrary $N$th order tensor, isn't the result then by definition not always a scalar? How is the quotient law used in this statement?

Use the quotient law to show that the elements of $T= \begin{pmatrix} x_2^2 & -x_1x_2 \\ -x_1x_2 & x_1^2\end{pmatrix}$ are the components of a second-order tensor.

The outer product $x_ix_j$ is a second-order tensor. Contracting this with $T_{ij}$ we obtain $$ T_{ij}x_ix_j = x_2^2x_1^2-x_1x_2x_1x_2-x_1x_2x_2x_1+x_1^2x_2^2 $$ which is clearly invariant (a zeroth-order tensor). Hence, by the quotient theorem $T_{ij}$ must also be a tensor.

The quotient law, as mentioned, is introduced as follows. If we know that $\mathbf B$ and $\mathbf C$ are tensors and also $$ A_{pq\ldots k\ldots m} B_{ij \ldots k \ldots n} = C_{pq \ldots mij \ldots n}, $$ does this imply that the $A_{pq\ldots k\ldots m}$ also form the components of a tensor? The quotient law says this is true if the relation holds in all rotated coordinate frames. It then proves it for $N=M=2$. The subscript $k$ that has been contracted may be any of the subscripts in $\mathbf A$ and $\mathbf B$ independently.

Never does the text mention anything about contracting in using the quotient law.

How is the quotient law used in the example?

  • $\begingroup$ Do you know the Einstein summation convention? That is used to denote summing over certain indicies. It states that if an index is repeated, like the $k$ is in your statement, that indicates that you are to sum over $k$. $\endgroup$
    – Teddy38
    Apr 9, 2021 at 21:32
  • $\begingroup$ @Teddy38 Yes, I'm familiar with the Einstein summing convention. I, however, don't understand how the quotient law is applied in the example. $\endgroup$ Apr 9, 2021 at 21:45
  • $\begingroup$ Ah I see. The Quotient Law involves contracting with one index at a time ($k$, say) and not two simultaneously (both $i$ and $j$, in your example). That's interesting, I'll have a think. $\endgroup$
    – Teddy38
    Apr 9, 2021 at 22:44

3 Answers 3


I agree with EE18: The book only shows that the trace of the product of $\mathbf T$ and $\mathbf x\otimes \mathbf x$ is a tensor. So there are several problems:

  1. It is not true that the trace of a second-order quantity is a tensor if and only if the quantitiy is a tensor, so I don't think that we can deduce that the product is a tensor.

  2. Even if the above was true, we could not apply the quotient rule: The tensor $\mathbf x\otimes \mathbf x$ is not invertible, but I think that the tensor $\mathbf B$ in the proof of the quotient rule is assumed to be invertible (see below).


Let $\mathbf A$ and $\mathbf B$ be two second-order quantities. Furthermore, let $\mathbf C$ be two product of $\mathbf A$ and $\mathbf B$. We want to show that if $\mathbf B$ and $\mathbf C$ are tensors, then $\mathbf A$ is a tensor as well. Let $A$ and $A'$ be the representations of the quantity $\mathbf A$ w.r.t. the two orthonormal bases $\mathbf e$ and $\mathbf{e}'$ (in particular, $A$ and $A'$ are 2 times 2 matrices), then the proof in the book can be summarized as follows: $$A'B'=C'=LCL{}^{-1}=LABL{}^{-1}=LAL{}^{-1}LBL{}^{-1}=LAL{}^{-1}B'$$ But we clearly need $\mathbf B$ to be invertible$^1$ to deduce that $LAL{}^{-1}=A'$.

$^1$ Since $\mathbf B$ is a tensor, $B$ is invertible if and only if $B'$ is invertible and in this case we say that $\mathbf B$ is invertible).

  • $\begingroup$ Thank you for the answer, but I'm afraid it doesn't use contraction (indeed the two contractions) that the textbook uses: that was what I was hoping for if possible. $\endgroup$
    – EE18
    Dec 19, 2022 at 16:39
  • $\begingroup$ @EE18 "I'm afraid it doesn't use contraction" - What I call "trace" (1) and "product" (2) are precisely what you call "contraction". (1) Given a second-rank tensor, there is only one possible contraction and you obtain it by associating to each basis the trace of the matrix-representation. The trace is basis-independent and hence I call it the trace of the tensor (this is also explained on page 939). (2) Suppose that $\mathbf A$ and $\mathbf B$ are of second order like in our example, then the $C$ in (26.25) is precisely the product of the matrices $A$ and $B$. $\endgroup$
    – Filippo
    Dec 19, 2022 at 17:00

The problem with the textbook is that they summed over both $i$ and $j$ in one go, but the Quotient Law's statement only involves summing over one index at a time, in this case, $k$.

Here's a worked solution to the problem you quoted. We need to test $T_{ij}$ for Tensor character. The textbook tells us to use $x_mx_n$, which we'll write as $$A_{mn}=\begin{pmatrix} x_1^2 & x_1x_2 \\ x_1x_2 & x_2^2\end{pmatrix}\ .$$ So, we need to calculate the following 4 entities and see if they are tensors:

$$ \begin{array}{rcccl} C_{in}&=&T_{ik}A_{kn}&=&T_{i1}A_{1n}+T_{i2}A_{2n}\\ C_{im}&=&T_{ik}A_{mk}&=&T_{i1}A_{m1}+T_{i2}A_{m2}\\ C_{jn}&=&T_{kj}A_{kn}&=&T_{1j}A_{1n}+T_{2j}A_{2n}\\ C_{jm}&=&T_{kj}A_{mk}&=&T_{1j}A_{m1}+T_{2j}A_{m2}\\ \end{array} $$

If all 4 are tensors, then we may conclude that $T_{ij}$ is also a tensor.

As an example, we find $C_{jn}$: $ C_{jn}=T_{kj}A_{kn}=T_{1j}A_{1n}+T_{2j}A_{2n}= \begin{pmatrix} x_1^2x_2^2 & x_1x_2^3 \\ -x_1^3x_2 & -x_1^2x_2^2\end{pmatrix}+ \begin{pmatrix} -x_1^2x_2^2 & -x_1x_2^3 \\ x_1^3x_2 & x_1^2x_2^2\end{pmatrix}=\begin{pmatrix}0&0\\0&0\end{pmatrix}\ ,$

which is clearly a tensor.

Similarly, all the other $C$'s are also second-order $0$ tensors (easily seen as $A$ and $T$ are symmetric). Now, by the Quotient Law, we may conclude that $T$ is a second-order tensor.

I don't know how the textbook justifies their application of the Quotient Law as they are not using it in the way it is stated. Maybe I'm missing something.

  • $\begingroup$ The quotient law says that given $C_{ij}$ and $A_{ij}$ are components of a tensor, then $T_{ij}$ are also components of a tensor if $T_{pk}A_{ik} =C_{pi}$, holds in all rotated coordinate frames, where we run the dummy index $k$ through all index positions. My problem with your example is that I can't think up a case where $T_{ij}$ wouldn't be the components of a tensor. What's the simplest case where $T_{ij}$ fails to be a tensor? Also, are we checking if the relation $T_{pk}A_{ik} =C_{pi}$ holds in all rotated coordinates by changing the position of the dummy index $k$? $\endgroup$ Apr 10, 2021 at 13:56
  • $\begingroup$ Also, how does your example relate to the one in my post? I still don't understand how the example in my post is using the quotient law. $\endgroup$ Apr 10, 2021 at 13:59

A comment (but not answer) to this question: I believe the proof follows if the following theorem is true.

If the contraction of some quantity is tensorial in nature, then that original quantity is tensorial in nature.

If this is true, then one application of the quotient rule followed by this theorem gives the answer. I'm not able to frame this theorem as a special case of the quotient rule though and cannot think how to prove it.


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