Note: I do not believe this is simply a matter of convention regarding what is considered a positive angle, handedness of the coordinates, nor the ordering of matrix multiplication for vector operations. All of those are standard in other parts of MTW.
My question: is my modified definition a correct. Or is that given by MTW correct?
In exercise 9.13 of Gravitation, by Misner, Thorne and Wheeler, the component definition of the generator matrices of the rotation group is given as $\left(\mathcal{K}_{l}\right)_{mn}=\epsilon_{lmn},$ where $\epsilon_{lmn}$ is the Levi-Civita symbol. This appears to be incorrect. I propose that the definition should be $\left(\mathcal{K}_{l}\right)_{mn}=-\epsilon_{lmn}$
Unless my mind is playing tricks on me, the definition given by MTW results in
$$\mathcal{K}_{1}=\begin{bmatrix}0 & 0 & 0\\ 0 & 0 & 1\\ 0 & -1 & 0 \end{bmatrix};\mathcal{K}_{2}=\begin{bmatrix}0 & 0 & -1\\ 0 & 0 & 0\\ 1 & 0 & 0 \end{bmatrix};\mathcal{K}_{3}=\begin{bmatrix}0 & 1 & 0\\ -1 & 0 & 0\\ 0 & 0 & 0 \end{bmatrix}.$$
Something called the complex structure of $\mathbb{R}^2$ is introduced in Modern Differential Geometry of Curves and Surfaces with Mathematica, Third Edition, by Alfred Gray, Elsa Abbena, Simon Salamon. It is defined as $\mathcal{J}\left(p_1,p_2\right)=\left(-p_2,p_1\right),$ which is a rotation by $\pi/2$. Its matrix $\mathcal{J}$ and whole number powers thereof are
$$\mathcal{J}=\begin{bmatrix}0 & -1\\ 1 & 0 \end{bmatrix};\mathcal{J}^{2}=-\mathcal{I};\mathcal{J}^{3}=-\mathcal{J};\mathcal{J}^{4}=\mathcal{I}=\mathcal{J}^{0}.$$
Raising $e$ to a matrix power is defined to be formally identical to the Taylor series expansion of $e^{x}.$ So if our matrix is $\theta\mathfrak{m},$ where $\theta$ is a scalar, we have
$$e^{\theta\mathfrak{m}}=\theta^{0}\mathfrak{m}^{0}+\theta\mathfrak{m}+\frac{\theta^{2}}{2}\mathfrak{m}^{2}+\ldots=\sum_{n=0}^{\infty}\frac{\theta^{n}}{n!}\mathfrak{m}^{n}.$$
The familiar Taylor Series expansion of the complex exponential function is
$$e^{\theta\text{i}}=\left(1-\frac{\theta^{2}}{2}+\frac{\theta^{4}}{4!}-\dots\right)+\mathrm{i}\left(\theta-\frac{\theta^{3}}{3!}+\frac{\theta^{5}}{5!}-\dots\right) =\cos\theta+\mathrm{i}\sin\theta$$
Matching terms we see that
$$\begin{aligned} e^{\mathcal{J}\theta}&=\mathcal{I}\left(1-\frac{\theta^{2}}{2}+\frac{\theta^{4}}{4!}-\dots\right)+\mathcal{J}\left(\theta-\frac{\theta^{3}}{3!}+\frac{\theta^{5}}{5!}-\dots\right)\\ &=\mathcal{I}\cos\theta+\mathcal{J}\sin\theta\\ &=\begin{bmatrix}1 & 0\\ 0 & 1 \end{bmatrix}\cos\theta+\begin{bmatrix}0 & -1\\ 1 & 0 \end{bmatrix}\sin\theta\\ &=\begin{bmatrix}\cos\theta & -\sin\theta\\ \sin\theta & \cos\theta \end{bmatrix}, \end{aligned}$$
which is a rotation in the Euclidean plane by $\theta$.
As can be seen in the screen capture below, the sub-matrices of the $\mathcal{K}_l$ formed of non-zero rows and columns, and the powers of those sub-matrices are equal to the matrix $\pm\mathcal{J},$ and its powers.
The exercise gives the definition
$$\mathcal{R}_{x}\left(\theta\right)\equiv\exp\left(\mathcal{K}_{1}\theta\right)=\sum_{n=0}^{\infty}\frac{\theta^{n}}{n!}\left(\mathcal{K}_{1}\right)^{n},$$
and asks us to show that this is a rotation matrix which produces a rotation by $\theta$ about the $x$-axis. And similarly for the $y$- and $z$-axes.
To simplify things, we define
$$\mathcal{I}_{1}=\begin{bmatrix}0 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{bmatrix}.$$
Thus we have
$$\left(\mathcal{K}_{1}\right)^{0}=\mathcal{I};\left(\mathcal{K}_{1}\right)^{1}=\mathcal{K}_{1};\left(\mathcal{K}_{1}\right)^{2}=-\mathcal{I}_{1};\left(\mathcal{K}_{1}\right)^{3}=-\mathcal{K}_{1};\left(\mathcal{K}_{1}\right)^{4}=\mathcal{I}_{1}.$$
Using these to expand our exponential gives
$$\begin{aligned} \exp\left(\mathcal{K}_{1}\theta\right)&=\mathcal{I}-\mathcal{I}_{1}+\mathcal{I}_{1}\left(1-\frac{\theta^{2}}{2}+\frac{\theta^{4}}{4!}-\dots\right)+\mathcal{K}_{1}\left(\theta-\frac{\theta^{3}}{3!}+\frac{\theta^{5}}{5!}-\dots\right)\\ &=\mathcal{I}-\mathcal{I}_{1}+\mathcal{I}_{1}\cos\theta+\mathcal{K}_{1}\sin\theta\\ &=\begin{bmatrix}1 & 0 & 0\\ 0 & \cos\theta & 0\\ 0 & 0 & \cos\theta \end{bmatrix}+\begin{bmatrix}0 & 0 & 0\\ 0 & 0 & \sin\theta\\ 0 & -\sin\theta & 0 \end{bmatrix}\\ &=\begin{bmatrix}1 & 0 & 0\\ 0 & \cos\theta & \sin\theta\\ 0 & -\sin\theta & \cos\theta \end{bmatrix} \end{aligned}.$$
But this is a rotation about the $x$-axis by $-\theta.$ The other two matrices also produce rotations by $-\theta.$
$$ \exp\left(\mathcal{K}_{2}\theta\right)=\begin{bmatrix}\cos\theta & 0 & -\sin\theta\\ 0 & 0 & 0\\ \sin\theta & 0 & \cos\theta \end{bmatrix}\\ \exp\left(\mathcal{K}_{3}\theta\right)=\begin{bmatrix}\cos\theta & \sin\theta & 0\\ -\sin\theta & \cos\theta & 0\\ 0 & 0 & 0 \end{bmatrix} .$$
