By Determinant: Proper vs. Improper#

Take the determinant of both sides of $\eta = \Lambda^T \eta \Lambda$:

$\det(\eta) = \det(\Lambda^T)\,\det(\eta)\,\det(\Lambda)$

Since $\det(\Lambda^T) = \det(\Lambda)$ and $\det(\eta) \neq 0$, we can divide through:

$1 = (\det \Lambda)^2$

$\boxed{\det \Lambda = \pm 1}$

This gives us two classes:

• $\det \Lambda = +1$: Proper Lorentz transformations. These form the subgroup $SO(3,1)$. The "$S$" stands for special, meaning unit determinant — the same convention as $SO(3)$ for rotations.

• $\det \Lambda = -1$: Improper Lorentz transformations. These include parity, time reversal, and combinations thereof. They cannot be continuously connected to the identity (you can’t smoothly go from $\det = +1$ to $\det = -1$), so they don’t have a Lie algebra description — they are discrete.

By the 00-Component: Orthochronous vs. Non-Orthochronous#

Now consider the $\rho = \sigma = 0$ component of $\eta_{\rho\sigma} = \eta_{\mu\nu}\Lambda^\mu{}_\rho\Lambda^\nu{}_\sigma$:

$1 = (\Lambda^0{}_0)^2 - \sum_{i=1}^{3}(\Lambda^i{}_0)^2$

This gives us:

$(\Lambda^0{}_0)^2 = 1 + \sum_{i=1}^{3}(\Lambda^i{}_0)^2 \geq 1$

Since $(\Lambda^0{}_0)^2 \geq 1$, we must have either $\Lambda^0{}_0 \geq 1$ or $\Lambda^0{}_0 \leq -1$. There is no middle ground — this is a discrete split:

• $\Lambda^0{}_0 \geq 1$: Orthochronous — the transformation preserves the direction of time.
• $\Lambda^0{}_0 \leq -1$: Non-orthochronous — the transformation reverses the direction of time.

The Four Components#

Combining these two binary choices, the Lorentz group $O(3,1)$ splits into four disconnected components:

Only $\mathcal{L}^\uparrow_+$ — the proper orthochronous Lorentz group — is connected to the identity. This is the component that has a Lie algebra, and when physicists say “the Lorentz group” without qualification, they almost always mean this component:

$\boxed{\text{Orthochronous Proper Lorentz Transformations} \equiv SO^+(3,1)}$

The other three components are obtained by applying the discrete transformations $P$, $T$, or $PT$ to elements of $SO^+(3,1)$.

Non-Orthochronous Transformations#

When $\Lambda^0{}_0 \leq -1$, the transformation reverses the direction of time. Any non-orthochronous transformation can be written as an orthochronous transformation composed with a discrete inversion. The relevant discrete operations are:

• Time reversal $T$: $(t, x, y, z) \to (-t, x, y, z)$, which has $\det\Lambda = -1$ and $\Lambda^0{}_0 = -1$
• Combined $PT$: $(t, x, y, z) \to (-t, -x, -y, -z)$, which has $\det\Lambda = +1$ and $\Lambda^0{}_0 = -1$

Improper Lorentz Transformations#

Transformations with $\det\Lambda = -1$ are called improper. Any improper transformation can be written as a proper transformation ($\det = +1$) composed with a discrete transformation. Examples include:

• Parity: $(t,x,y,z) \to (t, -x, -y, -z)$ — flips all spatial coordinates, $\det = -1$, orthochronous
• Single-axis reflection: $(t,x,y,z) \to (t, -x, y, z)$ — flips one spatial axis, $\det = -1$, orthochronous
• Time reversal: $(t,x,y,z) \to (-t, x, y, z)$ — flips time, $\det = -1$, non-orthochronous

Notice that parity is improper but orthochronous ($\Lambda^0{}_0 = +1$), while time reversal is both improper and non-orthochronous. These are genuinely different — they live in different disconnected components of the Lorentz group.

Scalar Representation#

For a scalar $\phi$, the index $i$ takes only one value ($i = 1$), so this is a 1-dimensional representation. The generator $(J^{\mu\nu})^i{}_j$ is a $1\times1$ matrix — a single number for each pair $(\mu,\nu)$.

A scalar field is invariant under Lorentz transformations — it does not change:

$\phi \to \Lambda\phi = 1 \cdot \phi = \phi$

(Invariant means the value doesn’t change. Contrast with covariant, which means it transforms in a well-defined way under certain rules — all representations are covariant, but only the scalar is invariant.)

Since $\Lambda = e^0 = 1$ in this representation:

$\delta\phi = 0, \qquad J^{\mu\nu} = 0$

A representation in which all generators are zero is a valid solution of the Lie algebra $[T^a, T^b] = if^{ab}{}_c T^c$ (both sides are trivially zero), and is called the trivial representation.

A typical Lorentz scalar in particle physics is the rest mass of a particle — all observers agree on its value regardless of their reference frame.

Vector Representation#

A contravariant 4-vector $V^\mu$ transforms as:

$V^\mu \to \Lambda^\mu{}_\nu\,V^\nu$

and a covariant 4-vector $V_\mu$ transforms as:

$V_\mu \to \Lambda_\mu{}^\nu\,V_\nu$

with $\Lambda$ satisfying the Lorentz invariance condition $\eta = \Lambda^T\eta\Lambda$. The spacetime coordinates $x^\mu$ and the four-momentum $p^\mu = (E, \mathbf{p})$ are the most important examples of contravariant 4-vectors.

This is a 4-dimensional representation: each generator $J^{\mu\nu}$ is a $4\times4$ matrix, denoted $(J^{\mu\nu})^\rho{}_\sigma$. The explicit form of the generator is:

$\boxed{(J^{\mu\nu})^\rho{}_\sigma = i\left(\eta^{\mu\rho}\delta^\nu_\sigma - \eta^{\nu\rho}\delta^\mu_\sigma\right)}$

This formula is antisymmetric in $\mu\nu$ (as it must be, since $J^{\mu\nu} = -J^{\nu\mu}$), and it’s the unique generator consistent with the infinitesimal form of $V'^\mu = \Lambda^\mu{}_\nu V^\nu$.

To see this, consider an infinitesimal Lorentz transformation $\Lambda^\mu{}_\nu = \delta^\mu_\nu + \omega^\mu{}_\nu$:

$\Lambda^\mu{}_\nu V^\nu = (\delta^\mu_\nu + \omega^\mu{}_\nu)V^\nu = V^\mu + \omega^\mu{}_\nu V^\nu \equiv V^\mu + \delta V^\mu$

Comparing with the general formula $\delta V^\rho = -\frac{i}{2}\omega_{\mu\nu}(J^{\mu\nu})^\rho{}_\sigma V^\sigma$ and substituting the explicit generator, one can verify that the two expressions agree: $\delta V^\rho = \omega^\rho{}_\sigma V^\sigma$. The circle closes.

Boosts#

A boost along the $x$-axis can be written in terms of velocity or rapidity. In terms of velocity, where $\beta^i = v^i$ (in natural units with $c = 1$) and $\gamma^i = (1 - (\beta^i)^2)^{-1/2}$:

$t \to \gamma^x(t + \beta^x x), \qquad x \to \gamma^x(x + \beta^x t)$

Since $-1 < \beta < 1$, we can write $\beta^i = \tanh\eta^i$ where $-\infty < \eta^i < \infty$ is the rapidity. The same transformation becomes:

$t \to (\cosh\eta^x)\,t + (\sinh\eta^x)\,x, \qquad x \to (\sinh\eta^x)\,t + (\cosh\eta^x)\,x$

The rapidity parameterization is nicer for several reasons: rapidities add under composition of collinear boosts (unlike velocities), and the hyperbolic functions make the analogy with rotations transparent — $\cos\theta, \sin\theta$ for rotations become $\cosh\eta, \sinh\eta$ for boosts.

Boosts in the $y$ and $z$ directions follow identically, with the $\cosh$ and $\sinh$ appearing in the appropriate row/column.

Boost Matrices#

The explicit $4\times4$ matrices for boosts along each axis are:

Boost along $x$ ($\beta$-form and $\eta$-form):

$\Lambda(\beta^x) = \begin{pmatrix} \gamma^x & \beta^x\gamma^x & 0 & 0 \\ \beta^x\gamma^x & \gamma^x & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix}, \qquad \Lambda(\eta^x) = \begin{pmatrix} \cosh\eta^x & \sinh\eta^x & 0 & 0 \\ \sinh\eta^x & \cosh\eta^x & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix}$

Boost along $y$:

$\Lambda(\beta^y) = \begin{pmatrix} \gamma^y & 0 & \beta^y\gamma^y & 0 \\ 0 & 1 & 0 & 0 \\ \beta^y\gamma^y & 0 & \gamma^y & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix}, \qquad \Lambda(\eta^y) = \begin{pmatrix} \cosh\eta^y & 0 & \sinh\eta^y & 0 \\ 0 & 1 & 0 & 0 \\ \sinh\eta^y & 0 & \cosh\eta^y & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix}$

Boost along $z$:

$\Lambda(\beta^z) = \begin{pmatrix} \gamma^z & 0 & 0 & \beta^z\gamma^z \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ \beta^z\gamma^z & 0 & 0 & \gamma^z \end{pmatrix}, \qquad \Lambda(\eta^z) = \begin{pmatrix} \cosh\eta^z & 0 & 0 & \sinh\eta^z \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ \sinh\eta^z & 0 & 0 & \cosh\eta^z \end{pmatrix}$

The pattern: the boost mixes the time component (row/column 0) with the spatial component in the boost direction, leaving the other two spatial components untouched. The $\cosh$ sits on the diagonal and the $\sinh$ on the off-diagonal — compare with rotations, where $\cos$ and $\sin$ play the same role.

Rotation Matrices#

Rotations don’t touch the time component at all — they act purely in the spatial $3\times3$ block:

Rotation about $x$-axis:

$\Lambda(\theta^x) = \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & \cos\theta^x & \sin\theta^x \\ 0 & 0 & -\sin\theta^x & \cos\theta^x \end{pmatrix}$

Rotation about $y$-axis:

$\Lambda(\theta^y) = \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & \cos\theta^y & 0 & -\sin\theta^y \\ 0 & 0 & 1 & 0 \\ 0 & \sin\theta^y & 0 & \cos\theta^y \end{pmatrix}$

Rotation about $z$-axis:

$\Lambda(\theta^z) = \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & \cos\theta^z & \sin\theta^z & 0 \\ 0 & -\sin\theta^z & \cos\theta^z & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix}$

These are just the familiar $3\times3$ rotation matrices embedded in the lower-right $3\times3$ block of a $4\times4$ matrix, with the time-time component equal to 1 and all time-space components equal to 0.

Boost Generators#

$K^i = -i\frac{\partial\Lambda(\eta^i)}{\partial\eta^i}\bigg|_{\eta^i=0}$

Using $\cosh(0) = 1$ and $\sinh(0) = 0$:

$K^x = -i\begin{pmatrix} 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix}, \quad K^y = -i\begin{pmatrix} 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix}, \quad K^z = -i\begin{pmatrix} 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 \end{pmatrix}$

Notice that the $K^i$ are symmetric matrices (the inner $4\times4$ part, before the $-i$). This reflects the fact that boosts are not unitary transformations — the Lorentz group is non-compact, and boosts push you along a hyperbola rather than around a circle.

Rotation Generators#

$J^i = -i\frac{\partial\Lambda(\theta^i)}{\partial\theta^i}\bigg|_{\theta^i=0}$

Using $\cos(0) = 1$ and $\sin(0) = 0$:

$J^x = i\begin{pmatrix} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & -1 \\ 0 & 0 & 1 & 0 \end{pmatrix}, \quad J^y = i\begin{pmatrix} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \end{pmatrix}, \quad J^z = i\begin{pmatrix} 0 & 0 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix}$

The $J^i$ are antisymmetric (the inner part, before the $i$). This reflects the fact that rotations are unitary — the rotation group $SO(3)$ is compact.

The Lorentz Algebra#

These six generators satisfy the following commutation relations:

$[J^i, J^j] = i\epsilon^{ijk}J^k$

$[J^i, K^j] = i\epsilon^{ijk}K^k$

$[K^i, K^j] = -i\epsilon^{ijk}J^k$

Each of these relations has a clear physical meaning:

• $[J^i, J^j] = i\epsilon^{ijk}J^k$: The rotation generators close among themselves and satisfy the $\mathfrak{su}(2)$ algebra. This is simply the statement that $J^i$ are angular momenta — rotations form a subgroup.

• $[J^i, K^j] = i\epsilon^{ijk}K^k$: The boosts transform as a vector under rotations. If you rotate your coordinate system, the boost generators rotate accordingly. This is expected on physical grounds — a boost “in the $x$-direction” should become a boost “in the $y$-direction” under a $90°$ rotation about $z$.

• $[K^i, K^j] = -i\epsilon^{ijk}J^k$: This is the crucial relation. The commutator of two boosts gives a rotation, not another boost. Boosts do not form a subgroup. The minus sign (compared to $[J^i, J^j]$) is physically significant: it’s the reason the Lorentz group is non-compact, the reason finite-dimensional unitary representations don’t exist, and ultimately the reason we need spinors and the Dirac equation.

If this third relation had a plus sign instead — $[K^i, K^j] = +i\epsilon^{ijk}J^k$ — then $(J^i, K^i)$ would generate $SO(4)$, a compact group with perfectly well-behaved finite-dimensional unitary representations. The minus sign makes all the difference.

Connecting to the Tensor Notation#

The six generators $J^i$ and $K^i$ can be packaged into the antisymmetric tensor $J^{\mu\nu}$ via:

$J^i = \frac{1}{2}\epsilon^{ijk}J^{jk}, \qquad K^i = J^{0i}$

This is useful because the Lorentz transformation takes the compact form $\Lambda = \exp(-\tfrac{i}{2}\omega_{\mu\nu}J^{\mu\nu})$, where the six independent components of $\omega_{\mu\nu}$ are the three rotation angles and three boost rapidities.