Is faithful positive sesqulinear form an inner product? As in the title: does a positive sesqulinear form need to be conjugate-symmetric?
Background: The question comes from an attempt to understand the proof of the Stinespring representation theorem. It consists in defining an inner product by constructing a positive definite sesqulinear from, which is later "faithfulized" on a quotient space by a standard argument. However the author of the book I'm reading (Takesaki) doeas not bother to check for symmetry. The symmetry would also follow if every linear completely positive map between C*-algebras preserved *-operation (???).
Can anyone help? 
 A: I'm assuming that the use of "sesquilinear" indicates that we're dealing with a complex vector space. When "sesquilinear" is used to include real bilinear forms, the symmetry need not hold in the real case.
For a sesquilinear map $\sigma$ - linear in the first argument and antilinear in the second - on a complex vector space, we have the polarisation formula
$$4\sigma(v,w) = \sigma(v+w,v+w) - \sigma(v-w,v-w) + i\sigma(v+iw,v+iw) - i\sigma(v-iw,v-iw).\tag{1}$$
If one considers maps that are antilinear in the first argument and linear in the second, change the signs of the last two terms. The verification of $(1)$ is a simple (although a wee bit tedious) expansion using the sesquilinearity of $\sigma$.
Now if $\sigma$ is a sesquilinear form, i.e. a sesquilinear map with codomain $\mathbb{C}$, the hermitian symmetry $\sigma(w,v) = \overline{\sigma(v,w)}$ for all $v,w$ is equivalent to the condition $\sigma(u,u) \in \mathbb{R}$ for all $u$.
Clearly hermitian symmetry implies $\sigma(u,u) = \overline{\sigma(u,u)}$ and hence $\sigma(u,u) \in \mathbb{R}$. Conversely, if $\sigma(u,u) \in \mathbb{R}$ for all $u$, then $(1)$ shows
\begin{align}
4\operatorname{Re} \sigma(v,w) &= \sigma(v+w,v+w) - \sigma(v-w,v-w)\\
&= \sigma(w+v,w+v) - \sigma(w-v,w-v)\\
&= 4\operatorname{Re} \sigma(w,v)
\end{align}
and
\begin{align}
4\operatorname{Im} \sigma(v,w) &= \sigma(v+iw,v+iw) - \sigma(v-iw,v-iw)\\
&= \sigma(-i(v+iw),-i(v+iw)) - \sigma(i(v-iw),i(v-iw))\\
&= \sigma(w-iv,w-iv) - \sigma(w+iv,w+iv)\\
&= -4\operatorname{Im} \sigma(w,v),
\end{align}
so $\sigma(w,v) = \overline{\sigma(v,w)}$.
A: In finite dimensions, positive-definite sesquilinear forms correspond to positive-definite matrices, and there are of course positive-definite matrices which are not conjugate-symmetric, e.g.
$$\begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}$$
A: The polarization identity for sesquilinear forms is compactly written in Arveson form as $\sigma(u,v)=\frac{1}{4}\sum_{n=0}^{3}i^n\sigma(u+i^nv,u+i^nv)$. Using this form and the fact that $\sigma(w,w)$ is real,
\begin{align}
\overline{4\sigma(u,v)}&=\overline{\sum_{n=0}^{3}i^{n}\sigma(u+i^nv,u+i^nv)} \\
 &=\sum_{n=0}^{3}(-i)^{n}\sigma(u+i^nv,u+i^nv) \\
 &=\sum_{n=0}^{3}(-i)^{n}\sigma((-i)^n u+v,(-i)^n u+v) \\
 &=\sum_{n=0}^{3}i^n\sigma(v+i^{n}u,v+i^{n}u)=4\sigma(v,u).
\end{align}
