Operator representation of the Lorentz generators $M^{\mu\nu}$ To right the J and K generators of the Lorentz group in a compact way, one can write 
$(M^{\mu\nu})^j{}_k=i (g^
{\mu j}g^\nu{}_k - g^{\nu j}g^\mu{}_k)$ 
where on the left hand side, it is helpful to think of l and m as indices/labels, and the j and k as rows/columns for the whole matrix. 
(Apparently) One can write this in an operator representation as 
$M^{\mu\nu} = i(x^\mu \partial^\nu-x^\nu\partial^\mu).$
I am perfectly comfortable with the former representation — I can write down the matrix $M^{\mu\nu}$ matrix for specific $\mu,\nu$, derive an expression for the exponential of that, and explain which Lorentz transformation it corresponds to.
I'm quite confused with the latter representation. Could someone give me some pointers on how to understand it? It is an "operator" representation, what is it operating on, $x^\mu?$
Moreover, maybe this is an index confusion. How does $\partial^\nu x^\sigma= \frac{\partial}{\partial x_\nu} x^\sigma$ equal $g^{\nu\sigma}$?
 A: These operators operate on functions defined on space-time.
To see how isometries of space can correspond to operators on functions, let's first consider the simpler case of one-dimensional translations. The generator of these translations is $\partial/\partial x$:
$$
\exp\left(a\frac\partial{\partial x}\right)f(x)=\sum_{n=0}^\infty\frac{a^n}{n!}\frac{\partial^n f}{\partial x^n}=f(x+a)\;,
$$
as the sum is just the Taylor series for $f$.
To use a slightly more complex example that's closer to your Lorentz transformations, consider rotations in two dimensions. In the representation as matrices of coordinate transformations, the generator of rotations is $\pmatrix{0&-1\\1&0}$, and the rotation matrices are
\begin{eqnarray*}
\exp\left(\alpha\pmatrix{0&1\\-1&0}\right)
&=&
\sum_{n=0}^\infty\frac{\alpha^n}{n!}\pmatrix{0&1\\-1&0}^n
\\
&=&
\sum_{n=0}^\infty(-1)^n\frac{\alpha^{2n}}{(2n)!}\pmatrix{1&0\\0&1}+\sum_{n=0}^\infty(-1)^n\frac{\alpha^{2n+1}}{(2n+1)!}\pmatrix{0&-1\\1&0}
\\
&=&
\cos\alpha\pmatrix{1&0\\0&1}+\sin\alpha\pmatrix{0&-1\\1&0}
\\
&=&
\pmatrix{\cos\alpha&\sin\alpha\\-\sin\alpha&\cos\alpha}\;.
\end{eqnarray*}
In the representation as operators on functions, the generator of rotations is 
$$
\frac{\partial}{\partial\phi}=\frac{\partial x}{\partial\phi}\frac{\partial}{\partial x}+\frac{\partial y}{\partial\phi}\frac{\partial}{\partial y}=y\frac{\partial}{\partial x}-x\frac{\partial}{\partial y}\;.
$$
Its effect is, as above,
$$
\exp\left(\alpha\frac{\partial}{\partial \phi}\right)f(r,\phi)=f(r,\phi+\alpha)\;.
$$
If $f(r,\theta)$ is expressed as $f(x,y)$, the effect is to apply the above spatial rotation matrix to the coordinates.
Your Lorentz transformations are just four-dimensional (and in some cases hyperbolic instead of trigonometric) analogues of this rotation operator.
