According to Dictionary on Lie Superalgebras (page 82), the compact form of $OSP(m,n|\mathbb C)$ Lie superalgebra must satisfy $M^{\text st}H\,M=1$ and $M^{\ddagger}M=1$ (is this the unitarity condition?), this means that the conditions for the corresponding superalgebra are $E^{\text st}H+HE=0$ and $E^{\ddagger}+E=0$ (right? it could be $E^{\ddagger}-E=0$?) due to the exponential map. So $E$ must be anti-Hermitian (right?, see Reduced matrix elements of the orthosymplectic Lie superalgebra (page 32) in which it looks that this condition needs to be applied "by blocks"). Here, $$H=\begin{pmatrix} \mathbb{I}_m & 0\\ 0 & \mathbb{J}_{2p} \end{pmatrix},\quad \mathbb{J}_{2p}=\begin{pmatrix} 0 & \mathbb{I}_p\\ -\mathbb{I}_p & 0 \end{pmatrix}\tag{1}$$

and $E=\begin{pmatrix} A & B\\ C & D \end{pmatrix}$ is an even supermatrix, so $A$ and $D$ are bosonic, and $B$ and $C$ are fermionic. The supertranspose operation is

$$E^{\text st}=\begin{pmatrix} A^{\text t} & C^{\text t}\\ -B^{\text t} & D^{\text t} \end{pmatrix}\tag{2}$$

as defined in page 84 of (1). By imposing

$$E^{\text st}H+HE=0 \tag{3}$$

I get the usual conditions for the $SO(m)\times Sp(n)$ bosonic subalgebra, and others for the fermionic part.

My problem is with the operation $\ddagger$. According to (1) (page 84) (see also Graded Lie algebras: Generalization of Hermitian representations ),

$$E^{\ddagger}=(E^{\text st})^\#$$

The $\#$ operation is a "superconjugation" or superstar. It is not clearly expressed in (1) so I went to CURRENT SUPERALGEBRAS AND UNITARY REPRESENTATIONS (page 18) in which $\#$ is defined as

$$E^{\#}=\begin{pmatrix} A^{*} & -iC^{*}\\ -iB^{*} & D^{*} \end{pmatrix}\tag{4}$$

involving usual conjugation and a transpose!(does this agree with the definitions in (1) and (5)). So it looks like $\#$ is already something like $\ddagger$. With this, a unitary representation of $\mathscr{gl}(m,n|\mathbb C)$, according to (4) must satisfy $E^{\#}+E=0$ from which $C=iB^{*}$. this operation is given also in Cornwell's Group theory in physics, vol. 3 (page 11) but it is not so clear for me.

At the end, $-E=E^{\ddagger}:=(E^{\text st})^\#=-E^{\text st}\Rightarrow E=E^{\text st}?$ and from the condition of $OSP$, $E=-HEH^{-1}$? So this condition with $E^{\text st}H+HE=0$ allows to get the unitary $osp$ superalgebra?

Notice also Superstrings on AdS4 × CP3 as a Coset Sigma-model (page 5) in which eq. (2.4) corresponds to the hermiticity (unitarity) condition.

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    $\begingroup$ Minor comment to the post (v1): In the future please link to abstract pages rather than pdf files. $\endgroup$
    – Qmechanic
    Commented Nov 21, 2019 at 5:44

1 Answer 1


Apparently $E^{\#}\equiv E^{\ddagger}$ and $*\equiv\dagger$, so the condition $$E^{\#}=-E\tag{1}$$ for unitarity, is the same as $$E^{\ddagger}=-E \tag{2}$$ That's why you have a transpose in the definition of $E^{\#}$! SO, actually $C=-iB^{\dagger}$. And, with $E^{\text{st}}H+HE=0$, you get the generators of $uosp(m,n|\mathbb C)$.

In order to "unify" definitions, this is what happens

$$E^{\ddagger}:=(E^{\text{st}})^{\#}:=\begin{pmatrix} A^t & C^t\\ -B^t & D^t \end{pmatrix}^{\#}=\begin{pmatrix} A^{t*} & -iC^{t*}\\ -iB^{t*} & D^{t*} \end{pmatrix}=-E \tag{3}$$


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