Is faithful positive sesqulinear form an inner product?

Question

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?

Answer 1

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 $\sigma(w,v) = \overline{\sigma(v,w)}$.

Answer 2

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. $$\begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}$$

Answer 3

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}