$\mathfrak so(1,3)$ generators hermitian or antihermitian?

Question

In Schwartz "Quantum field theory and the standard model" pag 160, the generators of the rotation are Hermitian, while the generators of boosts are anti-Hermitian, as an example:

$ J_1 = \left( \begin{matrix} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & -i \\ 0 & 0 & i & 0 \\ \end{matrix}\right)\,, K_1 = \left( \begin{matrix} 0 & -i & 0 & 0 \\ -i & 0 & 0 & 0 \\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 \end{matrix}\right) $

So you have $\langle \psi | \psi \rangle $ is rotation invariant since $e^{i\theta J_1}$ is Unitary an then

$\langle \psi | \psi \rangle \to \langle \psi | \left(e^{i\theta J_1}\right)^\dagger e^{i\theta J_1} |\psi \rangle = \langle \psi | e^{-i\theta J_1} e^{i\theta J_1} | \psi \rangle = \langle \psi | \psi \rangle $

But for the boost you have that $\left(e^{i\beta K_1}\right)^\dagger = e^{i\beta K_1} $.

1. So why generator of boosts are chosen to be anti-Hermitian instead of Hermitian?

2. Furthermore, using Hermitian generators for rotation you get the bracket $[J_i, J_j] = i\epsilon_{ijk}J_k$, so in this way the $i$ factor makes the Lie algebra $\mathfrak so(1,3)$ not closed in respect to the brackets, unless you re-define the bracket with a $-i$ factor.

Answer 1

These $+$ and $-$ signs in the generators originate from the signs in the Lorentz transformation matrices $\Lambda_{\mu\nu}$ (used for transforming $4$-vectors), which in turn originate from the signs in the Minkowski metric of spacetime.

Consider two examples:

Rotation

A rotation around the $x$-axis (with angle $\theta$) has the Lorentz transformation matrix $$\Lambda=\begin{pmatrix} 1&0&0&0\\ 0&1&0&0\\ 0&0&\cos\theta&-\sin\theta\\ 0&0&\sin\theta&\cos\theta \end{pmatrix}$$ Especially notice the opposite signs of the two off-diagonal coefficients (which assures $y'^2+z'^2=y^2+z^2$).
This matrix can be generated by $$\Lambda=e^{-iJ_x\theta}$$ with the generator $$J_x=\begin{pmatrix} 0&0&0&0\\ 0&0&0&0\\ 0&0&0&-i\\ 0&0&i&0 \end{pmatrix}$$

Boost

A boost along the $x$-axis (with rapidity $\beta$) has the Lorentz transformation matrix $$\Lambda=\begin{pmatrix} \cosh\beta&\sinh\beta&0&0\\ \sinh\beta&\cosh\beta&0&0\\ 0&0&1&0\\ 0&0&0&1 \end{pmatrix}$$ Especially notice the same signs of the two off-diagonal coefficients (which assures $c^2t'^2-x'^2=c^2t^2-x^2$).
This matrix can be generated by $$\Lambda=e^{iK_x\beta}$$ with the generator $$K_x=\begin{pmatrix} 0&-i&0&0\\ -i&0&0&0\\ 0&0&0&0\\ 0&0&0&0 \end{pmatrix}$$