Every once in a while I have to rederive the formula for the pullback of a Euclidean integral kernel operator by a diffeomorphism. Let me focus on a simple case to make the discussion concrete. Suppose that $u:\mathbb{R} \to \mathbb{R}$ is an orientation-preserving diffeomorphism. Then $u$ determines Hilbert space isometry $U: L^2(\mathbb{R}) \to L^2(\mathbb{R})$ (Lebesgue measure) by pullback. Explicitly: $$ (U\phi)(x) = |u'(x)|^{1/2} \phi(u(x)).$$ Let $k$ be a smooth, compactly-supported function on $\mathbb{R}^2$. Let $K : L^2(\mathbb{R}) \to L^2(\mathbb{R})$ be the corresponding integral kernel operator. That is: $$(K\phi)(x) = \int_\mathbb{R} k(x,y)\phi(y) \ dy.$$ As one would expect, $UKU^{-1}$ is also an integral kernel operator whose kernel $h$ is the "pullback of $k$ along $u$". But what exactly is the formula for this pullback? I find I can never remember and have to work it out every time. The answer is:
$$h(x,y) = |u'(x)|^{1/2} | u'(y)|^{1/2} k(u(x),u(y)).$$
Question: How can one remember that the correct scaling factor is $|u'(x)|^{1/2} | u'(y)|^{1/2}$ instead of some other plausible thing like, say, $|\frac{u'(x)}{u'(y)}|^{1/2}$?
OK I thought about it a bit more later and one way to quickly see what the right formula is would be to look at the case where $K$ is a rank-1 operator $K=\xi \otimes \overline \eta$. That is, the operator defined by $K\phi = \langle \eta,\phi \rangle \xi$, taking the inner product to be conjugate linear in the first slot. The kernel for this operator is then $k(x,y) = \xi(x) \overline{ \eta(y)}$. Anyway, an easy calculation shows
$$UKU^{-1}\phi = U\langle \eta,U^{-1}\phi\rangle \xi = \langle U\eta,\phi \rangle U\xi$$ (since $U$ is an isometry), so $$UKU^{-1} = u^*\xi \otimes \overline{u^*\eta}.$$ This example suggests the correct general formula for the pullback kernel.