What is the commutator in $gl(m|n)$? For Lie algebra $gl(m)$, the commutator is
\begin{align}
[E_{ij}, E_{kl}] = \delta_{jk}E_{il} - \delta_{li}E_{kj}.
\end{align}
What is the commutator $[E_{ij}, E_{kl}]$ in Lie superalgebra $gl(m|n)$? Thank you very much.
 A: If $A=A_0\oplus A_1$ is a superalgebra over a field k, where $A_0$ denotes the set of even elements and $A_1$ denotes the set of odd elements, then we can always construct an Lie superalgebra which is $A$ as a $k$ super vector space and has bracket defined by the $super$commutator:
$$[a,b]=ab-(-1)^{\overline{a}\overline{b}}ba$$
where if $x\in A$ is homogeneous then $\overline{x}$ is the degree of $x$.  Here we are only defining the bracket on homogeneous elements and extending linearly.  You can check this satisfies the (super) Jacobi identity.
To form $\mathfrak{gl}(m|n)$, take as $A$ the superalgebra of endomorphisms of the supervector space $k^{m|n}$, which I will denote $\text{Mat}_{m|n}(k)$.  Define an element to be even if it is parity preserving, and odd if it is parity reversing.  If we choose a basis $e_1,\dots,e_m$ on $k^m$ for the even part and $e_{m+1},\dots,e_{m+n}$ on $k^n$ the odd part, the even elements of $\text{Mat}_{m|n}(k)$ are matrices of the form $\begin{bmatrix}A & 0\\0 & D\end{bmatrix}$ and odd elements are those of the form $\begin{bmatrix}0 & B\\C & 0\end{bmatrix}$, where we are writing the matrices in block form according the the decomposition $k^{m|n}=k^m\oplus k^n$.  The Lie superalgebra we get from this superalgebra is $\mathfrak{gl}(m|n)$. 
A: If $E_{ab}\ $ (with $ \ a,b=1,2,...,m+n$), are the standard generators of the general Lie superalgebra $\mathcal{gl}(m|n)$, that is $E_{ab}$ is the $(m+n)\times(m+n)$ matrix with $1$ on the $(a,b)$-entry and zero everywhere else, then the supercommutator of $\mathcal{gl}(m|n)$ is given by: 
$$
\mathbf{[}E_{ab},E_{cd}\mathbf{]}=\delta_{bc}E_{ad}-(-1)^{(|a|-|b|)(|c|-|d|)}\delta_{ad}E_{cb}
$$
where
$$
|a|=
\left\{
\begin{array}{l}
0, \ \ a=1,2,...,m \\
1, \ \ a=m+1,m+2,...,m+n
\end{array}
\right.
$$
and the supercommutator $\mathbf{[}.,.\mathbf{]}$ is the usual commutator $[.,.]$ if at least one of $E_{ab},E_{cd}$ is even and the anticommutator $\{.,.\}$ if they are both odd.
