$\DeclareMathOperator\Tr{Tr}$I have a problem with understanding the proof of Proposition 6.8 in the book ,,Elements of Noncommutative Geometry''. One can find the formulation of this proposition [here](https://ibb.co/V2BHGkY) and the proof [here](https://ibb.co/dDhDHxS). The context is the following: $V$ is a *real* infinite dimensional vector space equipped with a *bilinear* form $g$: let us assume that the complex orthogonal structure $J$ is chosen, namely $J$ is an operator on $V$ such that $J^2=-I$ and $g(Jx,Jy)=g(x,y)$ for $x,y \in V$. This allows us to view $V$ as a *complex* vector space with the action $ix:=Jx$. Once $J$ is chosen, there is a scalar product $\langle x,y \rangle:=g(x,y)+ig(Jx,y)$. The space $V$ with this scalar product will be denoted by $V\_J$. On the other hand, one can consider the complexification $V^{\mathbb{C}}$ of $V$ and equip it with the inner product $\langle \langle x,y \rangle \rangle:=2g^{\mathbb{C}}(\overline{x},y)$ where $g^{\mathbb{C}}$ is the extension of $g$ to the *bilinear* form on $V^{\mathbb{C}}$. One has then the projection $P\_J:=\frac{1}{2}(I-iJ)$ in $V^{\mathbb{C}}$. If we put $W\_J$ to be the range of this projection it turns out that viewing $P\_J$ as an operator $V\_J \to W\_J$ yields a unitary $P\_J$ (this is the reason to choose this constant $2$ in the definition of the inner product on $V^{\mathbb{C}}$). The Schiwger term is defined as $\alpha(A,B):=-\frac{1}{2} \Tr[A\_-,B\_-]$ where $A\_-:=\frac{1}{2}(A+JAJ)$ is the *antilinear* part of $A$ (similarly for $B$) where $A$ is such that $[J,A]$ is Hilbert-Schimdt. Note that we are computing the trace of the commutator so the first impression is that it should be trivially zero-however our operators are antilinear-therefore it is nontrivial.


After explaining the general context let me explain where exactly my problems pop out: I don't see the last two equalities.



> 
> The first one is as follows:
> $$\Tr(iP\_J[A\_-,B\_-]P\_J)+\Tr(-iP\_{-J}[A\_-,B\_-]P\_{-J})=2i\Tr(P\_JA\_-P\_{-J}B\_-P\_J-P\_JB\_-P\_{-J}A\_-P\_J)$$
> Here I don't see how the projection $P\_{-J}$ appears in the middle of these two terms: $P\_{-J}=I-P\_J$ so one can have $P\_J$ in the middle instead of $P\_{-J}$ but still-there is no appearance of $P\_J$ in the middle on the left hand side. I would like to see why this equality is valid
> 
> 
> 



> 
> For the second one, namely:
> $$2i\Tr(P\_JA\_-P\_{-J}B\_-P\_J-P\_JB\_-P\_{-J}A\_-P\_J)=-4i\alpha(A,B)$$
> authors claim that this follows since $P\_J:V\_J \to W\_J$ is a unitary operator. As far as I understood-according to the formulation of the Proposition-$\alpha(A,B)$ is computed as a trace in $V^{\mathbb{C}}$ while the other trace probably takes place in $V\_J$. But somehow these projections $P\_J$ and $P\_{-J}$ go the other way around so I'm confused what is really going on.
> 
> 
> 


I will be very grateful for the explanation!



Another formula for the Schwinger term — problems with a calculation
I combine the two questions in the OP in a single question: find a proof of proposition 6.8,
$$
\alpha(A,B)=\tfrac{1}{8}i\,{\rm Tr}\,J[J,A][J,B].\qquad\qquad(6.8)
$$




---


I note that the Schwinger term $\alpha(A,B)$ is defined as the trace of antilinear operators on a complex vector space, while the trace in (6.8) involves linear operators on a real vector space. The first step is then to rewrite the Schwinger term as a trace of linear operators on a real vector space,
\begin{equation}
\alpha(A,B)=\tfrac{1}{2}i\,{\rm Tr}\,J[A\_-,B\_-]=\tfrac{1}{8}i\,{\rm Tr}\,J[(A+JAJ),(B+JBJ)].\qquad\qquad(1)
\end{equation}
I have substituted the definitions
\begin{equation}
A\_-=\tfrac{1}{2}(A+JAJ),\;\;B\_-=\tfrac{1}{2}(B+JBJ),\qquad\qquad(2)
\end{equation}
where $J^2=-1$.


Since these are traces of linear operators, we can permute them cyclically, ${\rm Tr}\,XY={\rm Tr}\,YX$. I work out the trace in equation 1,
\begin{equation}
{\rm Tr}\,J[(A+JAJ),(B+JBJ)]=4\,{\rm Tr}\,J[A,B],\qquad\qquad(3)
\end{equation}
and compare with
\begin{equation}
{\rm Tr}\,J[J,A][J,B]=2\,{\rm Tr}\,J[A,B].\qquad\qquad(4)
\end{equation}
I thus arrive at
\begin{equation}
\alpha(A,B)=\tfrac{1}{4}i\,{\rm Tr}\,J[J,A][J,B].\qquad\qquad(5)
\end{equation}
This is almost [proposition 6.8](https://ibb.co/V2BHGkY) from the book cited by the OP, except that there the coefficient is $1/8$ instead of $1/4$. I have searched for an [alternative source](https://arxiv.org/abs/hep-th/9402098) by the same authors, where the coefficient (on page 15) is given as $1/4$, like I found, so this is likely a typo in their book.


