# How does projective geometry related with Fourier transform via Lorentz transformation?

On one hand, the Lorentz transformation plays an important role in classical electromagnetism. On the other hand, each Lorentz tranformation is also a Möbius transformation of $\Bbb C$ (Needham, Visual Complex Analysis):

$$M(z)=\frac{az+b}{cz+d}$$

But what is the direct link between projective geometry and Fourier transform in physics? Can changing the perspective give rise to uncertainty principle? My knowledge is not very solid, but I guess we would start from the $U(1)$ group in QED then use the Peter - Weyl theorem, but exactly where Möbius transformation plays a role in there? If possible, I'd like to have an intuitive answer.

Related:
What projective groups whose characters form the basis of their corresponding $L^2$ space? on Math

• I think what you're looking for is there is a translation subgroup in the Mobius group for which you can define a transform acting on $L^2 (\mathbb{CP}^1)$ (plug in $z=-d/c$ into the formula). This subgroup becomes a subgroup for Lorentz transformations and you can break up $L^2 (\mathbb{R}^{3,1})$ with respect to this subgroup. It's not a translation subgroup, so not a Fourier transform, but it is "something". Is that what you are asking? – AHusain Jun 16 '18 at 3:01
• What is $L^2(\Bbb{R}^{3,1})$? The theorem makes sure that with any locally compact group there will be a generalized Fourier transform, so why not in this case? Again, my understanding is very rough, so an intuitive answer is very appreciated – Ooker Jun 16 '18 at 3:30
• Yes, there will be a generalized Fourier transform but it won't match with the regular Fourier transform at all. But if it looks like I have interpreted the question correctly, then tell me and I could put a longer explanation into an answer. – AHusain Jun 16 '18 at 3:40
• yes, I just need to see the link between the (generalized?) Möbius transform connects with the (generalized) Fourier transform. The problem is that I see both connect with the Lorentz transform so I think there is a physical thing in that link. I wonder if there is a physical application for the generalized ones. – Ooker Jun 16 '18 at 5:14

We have the subgroup $PSL(2,\mathbb{C}) \simeq SO^+(1,3)$. So yes we can take the subgroup given by the classes of the following elements

$$P_b = \begin{pmatrix} 1&b\\ 0&1 \end{pmatrix}$$

with $b$ real.

Under the specified isomorphism this becomes

$$Q=\begin{pmatrix} t+z&x-iy\\ x+iy&t-z \end{pmatrix} \rightarrow P_b Q P_b^\dagger\\ = \begin{pmatrix} t+2bx+b^2 t + (1-b^2) z & x+bt-bz-i y\\ x+bt-bz+i y & t-z \\ \end{pmatrix}\\ x \to x+bt-bz\\ y \to y\\ t \to bx+(1+\frac{b^2}{2})t - \frac{b^2}{2} z\\ z \to \frac{b^2}{2} t+(1-\frac{b^2}{2})z+b x$$

This therefore acts on functions on $\mathbb{R}^{1,3}$ by acting on the underlying space. So that function space is a representation of this $P_b$ group. We can try to decompose it by characters.

Recall the usual Fourier transform. You have the translation action $x \to x+a$ and the characters of that are the functions $e^{ipx}$ which the function gets decomposed into. They are eigenvectors with eigenvalue $e^{ipa}$ under this transformation.

So take the formula for $P_b Q P_b^\dagger$ and find the analogs for $e^{i p x}$.

See the part of the wiki on parabolic transformations. In particular the sorts of functions that are invariant under this group are given. Those are the ones with eigenvalue $0$ so replace that procedure with other eigenvalues.

But you can see unlike a simple translation and Fourier transform, this is not as intuitive of a transformation. There the two sides were position translations and momentum easily interpreted. Here on one side is this explicit mix of rotations and boosts. Send $b \to \epsilon$, if you want the infinitesimal transformation to write the analog of $\frac{d}{dx} f = ip f$. The other side is the corresponding conserved charge for that 1-parameter group.

The $U(1)$ QED group is a separate group. That refers to an internal symmetry not a symmetry of spacetime. $\psi (t,x,y,z) \to e^{i \theta} \psi (t,x,y,z)$. $(t,x,y,z)$ don't change. That abelian group does not move points around, this abelian group does.

For an uncertainty principle, you just need two operators that don't commute acting on your Hilbert space. But the Hilbert space is something that should be assigned to a timelike slice. This is mixing up different slices.

• I still not fully understand your answer, but I just find out that there is a representation theory of the Galilean group, which I think it's closer to what I'm looking for (nonrelativistic mechanics), and perhaps simpler and more connected to the regular Fourier transform? Can you have some comments on that? Thanks – Ooker Jun 18 '18 at 16:02