SL(2, ℂ) and the Lorentz Group

A real Minkowski vector can be encoded as a Hermitian two-by-two matrix whose determinant is its spacetime norm. Acting on that matrix with $SL(2,\mathbb C)$ constructs every proper orthochronous Lorentz transformation, with exactly two matrices representing the same transformation. This double cover explains how a two-component spinor can transform consistently while acquiring a minus sign under a $2\pi$ rotation.

Required background. Lorentz Transformations fixes the group components and passive boosts. We use the Pauli identity $(\mathbf a\cdot\boldsymbol\sigma)(\mathbf b\cdot\boldsymbol\sigma) =(\mathbf a\cdot\mathbf b)I+i(\mathbf a\times\mathbf b)\cdot\boldsymbol\sigma$.

Minkowski vectors as Hermitian matrices

For real components $x^\mu$, with $x^0=ct$, define

$$X=x^0I+x^1\sigma^1+x^2\sigma^2+x^3\sigma^3 =\begin{pmatrix} x^0+x^3&x^1-ix^2\\ x^1+ix^2&x^0-x^3 \end{pmatrix}.$$

This is an explicit use of upper vector components, not an implicit Minkowski contraction with a lowered sigma matrix. Every Hermitian two-by-two matrix has a unique expansion of this form, and

$$\det X=(x^0)^2-|\mathbf x|^2.$$

For $A\in SL(2,\mathbb C)$, meaning $\det A=1$, let

$$X'=AXA^\dagger .$$

The result is Hermitian, real-linear in $x^\mu$, and has the same determinant as $X$. It therefore defines a real Lorentz matrix $\Lambda(A)$ by $x'=\Lambda(A)x$. The assignment respects composition:

$$\Lambda(A_2A_1)=\Lambda(A_2)\Lambda(A_1).$$

The Hermitian conjugate is essential. Replacing it by the inverse would describe a different action and would not give the boosts below.

Why the image preserves orientation and the future

A future timelike vector corresponds to a positive-definite $X$: its eigenvalues are $x^0\pm|\mathbf x|>0$. Congruence by an invertible $A$ preserves positive definiteness. Future null vectors similarly correspond to nonzero positive-semidefinite matrices of rank one. Thus the map preserves the future cone.

Moreover $SL(2,\mathbb C)$ is connected. One way to see this is polar decomposition: $A=HU$, where $H$ is positive Hermitian with determinant one and $U\in SU(2)$. The path $H^t$ connects $H$ to $I$, and $SU(2)$ is connected. The determinant of $\Lambda(A)$ cannot change continuously between $+1$ and $-1$, so the image is in $SO^+(1,3)$.

This construction does not produce parity or time reflection. Those disconnected transformations need an extension of the connected-group action, not another choice of $A$ within $SL(2,\mathbb C)$.

Rotation and boost lifts

For a right-handed spatial rotation through $\theta$ about a unit vector $\mathbf n$, take

$$A_R=\exp(-i\theta\,\mathbf n\cdot\boldsymbol\sigma/2) =I\cos(\theta/2)-i\mathbf n\cdot\boldsymbol\sigma\sin(\theta/2).$$

For example, about $z$, conjugation sends $\sigma^1$ to $\cos\theta\,\sigma^1+\sin\theta\,\sigma^2$. It leaves $x^0$ unchanged and gives the ordinary spatial rotation.

For the passive boost used in the toolkit, the primed frame moves with velocity $c\tanh\xi\,\mathbf n$. The lift is

$$A_B=\exp(-\xi\,\mathbf n\cdot\boldsymbol\sigma/2) =I\cosh(\xi/2)-\mathbf n\cdot\boldsymbol\sigma\sinh(\xi/2).$$

Along $z$, $A_B=\operatorname{diag}(e^{-\xi/2},e^{\xi/2})$. The diagonal entries of $X$ then give

$$\begin{aligned} x'^0&=\cosh\xi\,x^0-\sinh\xi\,x^3,\\ x'^3&=-\sinh\xi\,x^0+\cosh\xi\,x^3, \end{aligned}$$

while $x'^1=x^1$ and $x'^2=x^2$. This checks the rapidity sign against a physical frame convention. The active boost that sends a rest momentum toward $+\mathbf n$ uses the inverse lift.

Rotations and boosts generate $SO^+(1,3)$, and each has a lift above, so the map is onto that group. Equivalently, map a future unit timelike vector to rest with a boost; any remaining transformation fixes rest and is a rotation.

The kernel has exactly two elements

Suppose $AXA^\dagger=X$ for every Hermitian $X$. Setting $X=I$ first gives $AA^\dagger=I$. The remaining condition becomes $AX=XA$ for every Hermitian matrix, so $A$ commutes with all Pauli matrices. It must be a scalar matrix $A=aI$. Finally $\det A=a^2=1$, hence

$$\ker\Lambda=\{I,-I\},\qquad SO^+(1,3)\simeq SL(2,\mathbb C)/\{I,-I\}.$$

If $A_1$ and $A_2$ induce the same Lorentz transformation, $A_2^{-1}A_1$ lies in this kernel. There are exactly two lifts, not an arbitrary phase freedom. The six real parameters of $SL(2,\mathbb C)$ encode the three rotations and three boosts.

Under the lift of a continuously performed full rotation, $A_R(2\pi)=-I$ and $A_R(4\pi)=I$. A spinor transforming by $A$ therefore returns to itself only after $4\pi$, although the associated vector matrix has already returned after $2\pi$.

Spinor signs and quantum probabilities

A common overall sign does not change a quantum ray or a bilinear such as $AXA^\dagger$. It can matter as a relative phase if only one coherently interfering branch is rotated. The double cover is therefore compatible with both the invariance of probabilities under a global phase and the possibility of detecting relative spinor phases.

The two-component action $z\mapsto Az$ is not unitary for boosts: for the $z$ boost, the squared component norm $z^\dagger z$ of $z=(1,0)^T$ changes by $e^{-\xi}$. This finite-component norm is not the conserved integrated one-particle norm. The latter also involves the spacetime argument, hypersurface measure, and current, as explained by Relativistic Currents.

There is a second inequivalent complex two-component action $z\mapsto(A^\dagger)^{-1}z$. On rotations the two actions agree, but their boost signs are opposite. They provide the two chiral pieces of a Dirac spinor. A field’s transformation law must also specify its argument: $z'(x')=Az(x)$, with $x'=\Lambda(A)x$, is equivalent to $z'(x)=Az(\Lambda(A)^{-1}x)$. Continue with Weyl Spinors for the two chiral representations and their equations.

Exercises

1. Let $X=zz^\dagger$ for a nonzero two-component column $z$. Show that its associated vector is future null and that this remains true after $z\mapsto Az$.

Solution

$X$ has rank one and is positive semidefinite. Thus $\det X=0$ and $x^0=\operatorname{tr}X/2=z^\dagger z/2>0$. After the transformation, $X'=(Az)(Az)^\dagger$ has the same properties because $A$ is invertible.

2. For a passive boost along $z$ with $\tanh\xi=3/5$, compute $A_B$ and its action on $X=I$.

Solution

$\cosh\xi=5/4$, $\sinh\xi=3/4$, and $e^\xi=2$. Therefore $A_B=\operatorname{diag}(1/\sqrt2,\sqrt2)$ and $X'=\operatorname{diag}(1/2,2)$, corresponding to $x'^0=5/4$, $x'^3=-3/4$. Its determinant remains one.

3. Why do the matrices $e^{i\varphi}A$ not supply an arbitrary continuous family of lifts in $SL(2,\mathbb C)$?

Solution

Although the phase cancels in $AXA^\dagger$, its determinant is $\det(e^{i\varphi}A)=e^{2i\varphi}$. Staying in $SL(2,\mathbb C)$ requires $e^{2i\varphi}=1$, leaving only $A$ and $-A$.

References

• H. K. Dreiner, H. E. Haber, and S. P. Martin, “Two-component spinor techniques and Feynman rules for quantum field theory and supersymmetry,” Physics Reports 494, 1–196, 2010, doi:10.1016/j.physrep.2010.05.002; corrected arXiv:0812.1594v6, 2022 — two-component Lorentz representations.
• S. Weinberg, The Quantum Theory of Fields, Vol. I: Foundations, Cambridge University Press, 1995 — Lorentz transformations, their covering group, and particle states.
• P. Woit, Quantum Field Theory for Mathematicians, Columbia University course notes, 2024, chapter 10 — vectors, spinors, and the Lorentz real form.