How can Minkowski spacetime be expressed as a U(2) manifold?

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Minkowski spacetime can be expressed as a U(2) manifold through a unique decomposition of 2x2 Hermitian matrices. The discussion outlines how any unitary matrix can be represented by specific constraints, leading to a general form of the unitary matrix that satisfies certain conditions. The transformation parameters are defined, allowing for a connection between the unitary representation and the spacetime metric. A participant initially sought guidance on expressing the metric in terms of the parameters of the unitary matrix but later resolved the issue by utilizing the determinant of the relevant relations. This highlights the intricate relationship between unitary transformations and the geometry of spacetime.
etotheipi
Homework Statement
It's question 6 of Example Sheet 1, very near the bottom of this document:
https://arxiv.org/pdf/gr-qc/9707012.pdf
An image is attached below
Relevant Equations
N/A
1618677370034.png


Firstly, since ##\{ \mathbb{I}, \sigma_x, \sigma_y, \sigma_z \}## is a basis of the space of ##2 \times 2## Hermitian matrices, and because ##X = t \mathbb{I} + x\sigma_x - y \sigma_y + z \sigma_z##, the map is one-to-one (because each matrix has unique decomposition). It's also easily checked the determinant of ##dX## is ##-ds^2##.

Next, need to show that any unitary ##U## can be expressed as asked. Consider an arbitrary ##U = \begin{pmatrix} a & b \\ c & d \end{pmatrix}##, then by constraining ##U U^{\dagger} = U^{\dagger} U = \mathbb{I}## we may write ##aa^* + bb^* = aa^* + cc^* + dd^* + cc^* = bb^* + dd^* = 1## and ##ac^* + bd^* = ab^* + cd^* = 0##.

It follows from the first four that ##|a| = |d| \implies d = a^* e^{i \varphi_1}## for some ##\varphi_1## and likewise ##|b| = |c| \implies c = b^* e^{i \varphi_2}## for some ##\varphi_2##. Thus, substituting for ##c## and ##d^*## in the equation ##ab^* + cd^*=0## gives\begin{align*}
ab^*(1 + e^{i (\varphi_2 - \varphi_1)}) = 0 &\implies \varphi_2 = \varphi_1 + (2n+1)\pi \\

&\implies e^{i \varphi_2} = e^{i \varphi_1} e^{2n\pi i} e^{i \pi} = - e^{i \varphi_1}
\end{align*}Therefore, defining a new parameter ##\tau## by ##\tau := \varphi_1 / 2##, and similarly defining ##\alpha := a \text{exp}(-i \varphi_1 / 2)## and ##\beta := b \text{exp}(-i \varphi_1 / 2)##, the most general form is indeed ##U = e^{i \tau} \begin{pmatrix} \alpha & \beta \\ -\beta^* & \alpha^* \end{pmatrix}## which satisfies ##|\alpha|^2 + |\beta|^2 = |a|^2 + |b|^2 = 1##.

I am confused how to do the last part, i.e. to express the metric in terms of the parameters of ##U##. From the definition of the Cayley map we can write down ##(1-iX)dU = i(1+U)dX = 2dU(1+U)^{-1}##, but I don't see how that helps. I'd be grateful for a hint about how to get started; thanks :smile:
 
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Actually, I figured out how to do it. You can just take the determinant of the relation they asked you to use, and substitute for ##\mathrm{det}(dU)##. Welp, sorry for posting 🙃
 
I want to find the solution to the integral ##\theta = \int_0^{\theta}\frac{du}{\sqrt{(c-u^2 +2u^3)}}## I can see that ##\frac{d^2u}{d\theta^2} = A +Bu+Cu^2## is a Weierstrass elliptic function, which can be generated from ##\Large(\normalsize\frac{du}{d\theta}\Large)\normalsize^2 = c-u^2 +2u^3## (A = 0, B=-1, C=3) So does this make my integral an elliptic integral? I haven't been able to find a table of integrals anywhere which contains an integral of this form so I'm a bit stuck. TerryW

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