Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

We define the affine transformation on distributions by $$\langle A^{*}u, \phi \rangle=\frac{1}{\det(A)}\langle u,\phi(A^{-1}x)\rangle$$

Assume this we should have $$\langle \partial_{i}(A^{*}u), \phi\rangle=\langle A^{*}u,-\partial_{i}\phi\rangle=\frac{1}{\det(A)}\langle u,-\partial_{i}\phi(A^{-1}x)\rangle$$

To make it equal to $$A^{*}(\sum^{n}_{j=1}a_{ji}\partial_{j} u)$$ we would expect $$\langle A^{*}(\sum^{n}_{j=1}a_{ji}\partial_{j}u),\phi\rangle=\frac{1}{\det(A)}\langle \sum^{n}_{j=1}a_{ji}\partial_{j} u,\phi(A^{-1}x)\rangle=\frac{1}{\det(A)}\langle u,-\sum_{j=1}^{n}a_{ji}\partial_{j}\phi(A^{-1}x)\rangle$$

Therefore we should have $$\partial_{i}\phi(A^{-1}x)=\sum_{j=1}^{n}a_{ji}\partial_{j}\phi(A^{-1}x)$$

Now we have $A^{-1}=\frac{1}{\det(A)}B$, where $B_{ij}=A^{*}_{ji}$(the adjoint matrix). Thus we should have $$(A^{-1}x)_{j}=\frac{1}{\det(A)}\sum A^{*}_{ji}x_{i}$$

And by chain rule we should have $$\partial_{i}\phi(A^{-1}x)=\sum \partial_{j}\phi(A^{-1}x)\partial_{i}((A^{-1}x)_{j})$$

But this does not match our result given since $A^{*}_{ji}\not=a_{ji}$; so I want to ask what is wrong in my derivation.

share|cite|improve this question
Just in case someone is going to point out, I think my use of chain rule is wrong but I do not know what is the correct way of deriving this. – Bombyx mori Jul 16 '12 at 1:30
Just in case someone cannot solve this, I solved it. – Bombyx mori Jul 19 '12 at 0:40

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.