# Unitarity/Hermiticity condition for $osp(m,n|\mathbb{C})$ superalgebra

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.

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

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