Consider the following fragment from Takesaki's book "Theory of operator algebra II":
Questions: (1) How does the proof show that every element of $\mathfrak{m}$ is a linear combination of four elements of $\mathfrak{p}?$
I tried to write $x \in \mathfrak{m}$ as $x = (q_1-q_2) + i(q_3-q_4)$ where $q_1, q_2, q_3, q_4$ are the canonical positive decomposition of $x$ in $\mathscr{M}$. Somehow we have to argue that $q_i \in \mathfrak{m}$.
(2) Is the factor $1/2$ in the proof correct? I think it should be $1/4$ instead.
(3) Later, Takesaki applies this lemma to a weight $\varphi: \mathscr{M}_+ \to [0, \infty]$ and $$\mathfrak{p}_\varphi = \{x \in \mathscr{M}: \varphi(x^*x) < \infty\}.$$
Takesaki then mentions that $\varphi$ extends to a linear functional on $\mathfrak{m}_\varphi$ (where this is defined as in the lemma, with respect to our special cone). How exactly is this extension defined?
You have \begin{align} \sum_{j=1}^n y_j^*x_j=\tfrac14\,\sum_{j=1}^n\sum_{k=0}^3i^k(x_j+i^ky_j)^*(x_j+i^ky_j) =\sum_{k=0}^3\tfrac{i^k}4\,\bigg[\sum_{j=1}^n(x_j+i^ky_j)^*(x_j+i^ky_j)\bigg]. \end{align} The expression inside the square brackets is in $\mathfrak p$, since $\mathfrak p$ is a subcone.
You are right, it should be $1/4$: \begin{align} \sum_{k=1}^n(x_k+y_k)^*(x_k+y_k)-(x_k-y_k)^*(x_k-y_k) &=2\sum_{k=1}^ny_k^*x_k+2\sum_{j=1}^nx_k^*y_k=4a. \end{align} And I don't really see why this would be implied by the Polarization Identity.
When you do the above to the sets coming from a weight $\varphi$, you extend to $\mathfrak m$ by using $$ \varphi\Big(\sum_{j=1}^n y_j^*x_j\Big) =\sum_{k=0}^3\tfrac{i^k}4\,\varphi\bigg[\sum_{j=1}^n[(x_j+i^ky_j)^*(x_j+i^ky_j)\bigg]. $$ The question is whether the above is well-defined. Suppose we have $$ a_1-a_2+i(a_3-a_4)=b_1-b_2+i(b_3-b_4) $$ with $a_j,b_j\in\mathfrak p_\varphi$, $j=1,\ldots,4$. By taking real and imaginary parts, $$ a_1-a_2=b_1-b_2,\qquad a_3-a_4=b_3-b_4. $$ These we can write as $$ a_1+b_2=b_1+a_2,\qquad a_3+b_4=b_3+a_4. $$ As $\varphi$ is defined and additive on $p_\varphi$, $$ \varphi(a_1)+\varphi(b_2)=\varphi(b_1)+\varphi(a_2),\qquad \varphi(a_3)+\varphi(b_4)=\varphi(b_3)+\varphi(a_4). $$ Thus $$ \varphi(a_1)-\varphi(a_2)=\varphi(b_1)-\varphi(b_2),\qquad \varphi(a_3)-\varphi(a_4)=\varphi(b_3)-\varphi(b_4). $$ So $\varphi$ has a unique linear extension to $\mathfrak m$.