Quotient Rule

Relevant parts to questions...

• Spotting when to use it: you are given that some contraction $T_{ij}{b}_j$ (or similar) is a tensor for any vector $\mathbf{b}$, and asked to show $T_{ij}$ itself is a tensor.
• The key phrase is “for arbitrary $\mathbf{b}$” - that is what lets you cancel $b_m^{\prime }$ from both sides.
• Reads as a “division” because you are dividing through by an arbitrary tensor to recover $T$.

Quotient Rule (Lemma)

Let $T_{ij}$ be a quantity such that, for every vector $\mathbf{b}$, the contraction $a_i=T_{ij}{b}_j$ is a vector. Then $T_{ij}$ is a rank-2 Tensor Transformation Rule.

The generalisation is immediate: if contracting an unknown quantity against an arbitrary tensor produces a tensor, then the unknown quantity is itself a tensor of the appropriate rank.

Proof

Collect the three assumptions:

• (A1)::$b_j=L_{mj}{b}_m^{\prime }$, since $\mathbf{b}$ is a vector.
• (A2)::$a_k=T_{kj}{b}_j$ holds in every coordinate system.
• (A3)::$a_i^{\prime }=L_{ik}{a}_k$, since $\mathbf{a}$ is a vector.

Writing $a_i^{\prime }$ two ways, first via (A3)→(A2)→(A1):

$$a_i^{\prime }=L_{ik}{a}_k=L_{ik}{T}_{kj}{b}_j=L_{ik}{T}_{kj}{L}_{mj}{b}_m^{\prime }=L_{ik}{L}_{mj}{T}_{kj}{b}_m^{\prime }$$

Second, by applying (A2) in the rotated frame:

$$a_i^{\prime }=T_{im}^{\prime }{b}_m^{\prime }$$

Subtracting:

$$(T_{im}^{\prime }-L_{ik}{L}_{mj}{T}_{kj})b_m^{\prime }=0$$

Because $\mathbf{b}$ is arbitrary, we may choose $b_m^{\prime }\ne 0$, forcing the bracket to vanish:

$$\boxed{T_{im}^{\prime }=L_{ik}{L}_{mj}{T}_{kj}}$$

So $T_{ij}$ obeys the rank-2 tensor transformation rule. ✓

Properties

• Arbitrariness is essential::if $\mathbf{b}$ were fixed, you could only conclude that $T$ agrees with its transform on that one vector, not that it transforms correctly in general.
• Generalises to any rank::if $T_{ij\dots k}{b}^k{c}^j\dots$ is a tensor for arbitrary $\mathbf{b},\mathbf{c},\dots$, then $T_{ij\dots k}$ is.
• The “quotient” analogy::you are recovering $T$ from the product $Tb$ by “dividing” out an arbitrary $b$ - similar in spirit to the calculus quotient rule.

Applications

1. Proving tensor character indirectly, when direct computation of the transformation is hard. Typical examples: the Metric Tensor from $A_i=g_{ik}{A}^k$, or the stress tensor in physics.
2. Showing $\frac{\partial u_i}{\partial x_j}$ is a tensor, via its action on arbitrary $\frac{\partial }{\partial x_j}$-vectors (though lecture 10 prefers the direct chain-rule proof).
3. Inferring rank from contracted rank: if contracting with a rank-$p$ tensor drops you to rank $q$, the original had rank $p+q$.

Needs genuine arbitrariness

The rule fails if $\mathbf{b}$ is restricted (e.g., to a symmetric or traceless class). The whole argument hinges on being able to pick any $\mathbf{b}$ in the rotated frame.

Contract with anything, get a tensor → you started with a tensor.