Another formula for the Schwinger term — problems with a calculation $\DeclareMathOperator\Tr{Tr}$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 and the proof here. 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!
 A: 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 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 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.
