2
$\begingroup$

The context of my question is the following.

Up to my knowledge when we are working in a complex Hilbert space $\mathcal{H}$ there exists the notion of adjoint for densily defined operator $A:\mathcal{D}\subseteq \mathcal{H}\to \mathcal{H}$ and it is noted $A^{\ast}$, defined by the property $(Au,v)=(u,A^{\ast}v)$ for all $u\in \mathcal{D}$ and $v$ also in a dense.

But define the traspose $A^{T}$ of an operator $A$ it is not even abailable uless we work in an specific basis of $\mathcal{H}$, suposse $u_{n}$ is such a basis, then the transpose will be definded by $(Au_{i},u_{j})=(A^{T}u_{j},u_{i})$.

But I'm currently working on a specific Hilbert space $\mathcal{H}=\mathcal{F}_{\mathbb{C}}$ which is the complexification a a real Hilbert space $\mathcal{F}$. Clearly the interior product in $\mathcal{F}$ which I call $(,)_{\mathbb{R}}$ is extended to an internal product in $\mathcal{H}$ which I call $(,)_{\mathbb{C}}$, so for operators in $B(\mathcal{H})\simeq B(\mathcal{F})_{\mathbb{C}}$ we have a notion of adjoint, becausse we have an internal product. But now thanks to the structure of complexification we have also a notion of conjuate for vectors and for operators in $B(\mathcal{H})$.

After a lot of computations about this thing I arrive to the idea that in this context one can define the notion of transpose by $A^{T}=\overline{A^{\ast}}$.

Here some of my questions...

  1. ¿Is there analogous of this definition in some textbook? Or ¿do you think it is fine?

  2. I found that, conjugate and adjoit conmute in $B(\mathcal{H})$, i.e. $\overline{(A^{\ast})}=( \overline{A} )^{\ast}$ ¿is it right?

  3. I also found that $\overline{(u,v)_{\mathbb{C}}}=(\overline{u},\overline{v})_{\mathbb{C}}$, for $u,v\in \mathcal{H}$, again the same ¿Is it right?

I hope I were clear. Most of the ideas were partially inspired by this text where you can find the definition of $(,)_{\mathbb{C}}$ in terms of $(,)_{\mathbb{R}}$ and related things.

Many thanks in advance.

$\endgroup$
1
  • $\begingroup$ I believe $\mathcal F$ is called a real form of $\mathcal H$. A real form is a real subspace of $\mathcal H$ so that $(v,w)\in\Bbb R$ for all $v,w\in\mathcal F$ and $\mathcal F+i\mathcal F=\mathcal H$. The data of a real form is equivalent to an anti-unitary involution (the complex conjugation) on $\mathcal H$, which I think is called a real structure. $\endgroup$
    – s.harp
    Jul 16, 2020 at 10:23

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.