# Fourier transform on Minkowski space

Physicists Some people like to define the "Fourier transform" on Minkowski space as $$\hat f(\xi) = \int e^{i \eta(x,\xi)} f(x) dx$$, where $$\eta(x,\xi)$$ is the Minkowski form. I'm used to thinking of the Fourier transform as a canonical isomorphism $$L^2(K) \to L^2(\hat K)$$ where $$K$$ is a locally compact abelian group and $$\hat K$$ is its Pontryagin dual. But this "Minkowski-Fourier transform" doesn't seem to arise in this way.

Questions:

1. Is there an abstract framework in which to understand this "Minkowski-Fourier transform"? For instance, is there a general theory of "Fourier transforms" on spaces equipped with a nondegenerate symmetric bilinear form? Is there a relationship between this "Minkowski-Fourier transform" and the representations of a suitable "Heisenberg group"?

2. Which properties of the usual Fourier transform on Euclidean space are shared by the "Minkowski-Fourier transform"? For instance, what is a precise statement of the Fourier inversion formula in this context?

3. Is there a good reference for the mathematical properties of the "Minkowski-Fourier transform"?

Perhaps it's worth adding that physicists seem to be a bit blase about using this "Minkowski-Fourier transform", and treat it as though it were an ordinary Fourier transform.

• Up to a harmless sign change isn't this an instance of the Fourier transform of locally compact abelian groups $K$ with $K=\mathbb R^n$? Physically this is just the translations on Minkowski space (including translations in time). – AlexArvanitakis Apr 13 '20 at 21:24
• Ah I see what is happening. Implicit in physicist textbooks is the use of the isomorphism between Minkowski space and its dual in every formula that does not contain indices. So for example by default I'd parse the $k\cdot x$ in Peskin and Schroeder as $k_\mu x^\mu$ (as I mentioned above) but you could also write it as $k^\mu x^\nu \eta_{\mu\nu}$ in Einstein summation notation. This simultaneously defines what you'd call a "Euclidean" Fourier transform (as a function of $k_\mu=\eta_{\mu\nu}k^\nu$) or your "Minkowskian" Fourier transform (as a function of $k^\mu=\eta^{\mu\nu}k_\nu$). – AlexArvanitakis Apr 13 '20 at 21:44
• Thanks, that's clarifying! This is strange -- I should have noticed that if you write $x^\mu$ for position and $p_\mu$ for momentum, then you don't need any kind of pairing to write down $p_\mu x^\mu$! It's only if (for some reason) you decide to write things in terms of $p^\mu$ rather than $p_\mu$ that a pairing is needed... I have seen the Fourier transform written out explicitly with the signs from the Minkowski form, but it was in a more math-flavored source, so it seems they were just (somewhat perversely!) working with the dual of momentum space! – Tim Campion Apr 13 '20 at 21:56
• As AlexArvanitakis explained, it is just the canonical Fourier transform up to a non-canonical identification $K \cong \hat{K}$ of a vector space with its dual using the Lorentzian inner product. – Igor Khavkine Apr 13 '20 at 22:13
• This sort of thing comes up even for Fourier series, because we do not work directly with the Pontryagin dual, which is a space of functions, but we choose an isomorphism with a well-known space. For Fourier series we do not use $\hat{\mathbb{Z}}$ directly, but $\mathbb{R}/\mathbb{Z}$ or $\mathbb{R}/2\pi\mathbb{Z}$, depending on what isomorphism we choose. – Robert Furber Apr 14 '20 at 12:23

Well you seem to have worked it out but I wrote most of this before your comment happened: I claim there isn't any material difference between your "Minkowski space Fourier transform" and the usual Fourier transform on $${\mathbb R}^n$$: in fact write $$\hat f(\xi)\equiv \int e^{i\eta(x,\xi)} f(x) dx$$ for any non-degenerate bilinear form $$\eta$$. Then there exists another such form $$\eta^{-1}$$ so $$\eta(x,\eta^{-1}\zeta)= \langle x,\zeta\rangle$$ where $$\zeta \in ({\mathbb R}^n)^\star$$. Clearly $$\hat f(\eta^{-1} \zeta)=({\mathcal F}f)(\zeta)\,,$$ where $$\mathcal F$$ is the usual --- Euclidean'' --- Fourier transform.
In physics texts this $$\zeta$$ variable is the down-index momentum ($$k_\mu$$ in e.g. Peskin and Schroeder) while $$\xi$$ is the up-index momentum ($$k^\mu$$ in e.g. Peskin and Schroeder). Derivatives play perfectly nice with up/down index notation, which allows one to be blase about whether the Fourier transform involves the up- or down-index $$k$$.
• I should mention that it is essential, and also implicit, in physics contexts that any indexful expression has a "fundamental" definition, from which all others arise via index gymnastics. This isn't very important for quantum field theory --- because all metrics are flat --- but it's crucial in general relativity. For momenta in particular one defaults to the index-down $k_\mu$ (as I did above). This can be important if e.g. you have a space of the form ${\mathbb R}^n\times T^m$, in which case you need to be very precise about what lattice the "momenta along the toroidal directions" lie in. – AlexArvanitakis Apr 13 '20 at 22:24