# Is self-adjoint the same as hyper-maximal symmetric?

I'm reading von Neumanns 1931 book on Quantum Mechanics. Here he presents spectral theory for unbounded operators and quite a bit of the notation and naming has changed in the mean time.

Specifically for him a symmetric operator (for all $x,y\in D(A)$ $\langle Ax,y\rangle = \langle x,Ay\rangle$) is called hermitian and an symmetric operator is called maximal if it admits no proper symmetric extensions.

He characterises maximality via the Cayley Transform, an operator $A$ is maximal iff the Cayley Transform $U(A)$ has image or domain the entire Hilbert space $H$. An operator is defined to be hyper-maximal if the Cayley Transform is unitary, ie has both image and domain all of $H$. He shows that the hyper-maximal symmetric operators are precisely those that admit a spectral decomposition.

For example it is clear that self-adjoint implies maximal: If $A=A^*$ and $B$ is symmetric so that $B\lvert_{D(A)}=A$, if $y\in D(B)$ then for every $x\in D(A)$ you've got: $$|\langle Ax,y\rangle| =| \langle x,By\rangle| ≤\|x\|\,\|By\|$$ it follows $y\in D(A^*)=D(A)$ and $A^*y = By$.

My questions:

What is the modern terminology for maximal?

And

Is hyper-maximal the same as self-adjoint?

The usual term for "maximal" is now "maximally symmetric," meaning that $A$ is symmetric and there are no proper symmetric extensions. von Neumann's work gives the decomposition $$\mathcal{D}(A^*)=\mathcal{D}(A)\oplus\mathcal{N}(A^*-iI)\oplus\mathcal{N}(A^*+iI),$$ which is orthogonal with respect to the following graph inner product $$\langle x,y\rangle_{A^*}=\langle x,y\rangle_H+\langle A^*x,A^*y\rangle_H.$$ If neither of the deficiency spaces $\mathcal{N}(A^*\pm iI)$ is $\{0\}$, then you can always extend $A$ in such a way that the extension remains symmetric. (You can write down such an extension.) So maximally symmetric operators $A$ have either $\mathcal{N}(A^*-iI)=\{0\}$ or $\mathcal{N}(A^*+iI)=\{0\}$. These are maximal, as you have described them. If both are $\{0\}$, then $A^*=A$ because they have the same domain and $A^*$ extends $A$, and this is what you have referred to as hyper-maximal.
• @Bananach : Based on what you wrote, maximal would be where one, but not both, of $\mathcal{N}(A^*\pm i)$ is $\{0\}$, and hypermaximal would be where both are $\{0\}$, which is the case for a self-adjoint operator. – Disintegrating By Parts Jul 18 '20 at 17:53