Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Consider the Fourier Transform and Inverse Fourier Transform: $$\mathcal{F} f(-s) = \int_{-\infty}^{\infty} e^{-2 \pi i (-s)t} f(t) \ dt = \int_{-\infty}^{\infty} e^{2 \pi ist} f(t) \ dt = \mathcal{F}^{-1}f(s)$$ and $$\mathcal{F}^{-1} f(-t) = \int_{-\infty}^{\infty} e^{2 \pi i s(-t)} f(s) \ ds = \int_{-\infty}^{\infty} e^{-2 \pi ist)} f(s) \ ds = \mathcal{F} f(t)$$

It seems that the Fourier Transform and the Inverse Fourier Transform are symmetric. The reason is that the group and dual group are both $\mathbb{R}$. If the group and dual group are different would this mean that there wouldn't be any symmetry present?

Also could be defined Fourier Transforms over other objects likes rings?

share|cite|improve this question
Not quite sure what you're asking in the first part. Might help to consider the pair $\mathbb{Z}$ and $\mathbb{T}$. – Michael Jun 26 '13 at 19:04

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.