Symmetric Extensions

• Mar 20th 2009, 11:22 AM
Nusc
Symmetric Extensions
Let A be a closed symmetric operator with deficiency indices n_+ and n_-. Then,

(a) A is self-adjoint if and only if n_+ = n_ = 0.

n_+ (A) = dim[ L_+] and n_- (A) = dim [L_-]
What is this saying? From that, how do I approach this problem?

it is invaluable.
• Mar 20th 2009, 02:57 PM
Opalg
Quote:

Originally Posted by Nusc
Let A be a closed symmetric operator with deficiency indices n_+ and n_-. Then,

(a) A is self-adjoint if and only if n_+ = n_ = 0.

n_+ (A) = dim[ K_+] and n_- (A) = dim [K_-]
What is this saying? From that, how do I approach this problem?

If A is a closed symmetric operator with domain D(A) and adjoint A*, then $D(A)\subseteq D(A^*)$. In fact $D(A^*) = D(A)\oplus\mathcal{K}_+\oplus\mathcal{K}_-$. So the condition for A to be selfadjoint (in other words D(A*)=D(A)) is that $\mathcal{K}_+ = \mathcal{K}_- = \{0\}$. The deficiency indices are the dimensions of $\mathcal{K}_\pm$, so the condition for A to be selfadjoint is that they should both be 0.
• Mar 20th 2009, 07:54 PM
Nusc
$D(A^*) = D(A)\oplus\mathcal{K}_+\oplus\mathcal{K}_-$

Sorry, why is this true?

And so show that A has a self-adjoint extension if and only if $n_+ = n_-$ What is meant by self-adjoint extension?

Also,

when they say most operators that occur in real life are unbounded, are they referring to the fact that everything is a signal and can be Fourier transformed (which is unbounded?)

Thanks again,
• Mar 21st 2009, 03:07 PM
Opalg
Quote:

Originally Posted by Nusc
$D(A^*) = D(A)\oplus\mathcal{K}_+\oplus\mathcal{K}_-$

Sorry, why is this true?

Quote:

Originally Posted by Nusc
And so show that A has a self-adjoint extension if and only if $n_+ = n_-$ What is meant by self-adjoint extension?

You really need to go back and check the definitions of these terms in Conway's book. There is a careful explanation of what is meant by a selfadjoint extension in the paragraph following Corollary X.2.10 (page 312).

Quote:

Originally Posted by Nusc
Also,

when they say most operators that occur in real life are unbounded, are they referring to the fact that everything is a signal and can be Fourier transformed (which is unbounded?)

I think that "real life" in that context means "physics". In physics, most operators are differential operators, and differentiation is an unbounded operator. Take another look at Examples 1.11 and 1.12 in Conway's Chapter X. They are the motivating examples for this whole theory.
• Mar 21st 2009, 04:19 PM
Nusc
With regards to the symmetric extension, I should have been more specific.

Why in the literature is symmetric and Hermitian used interchangeably when we're dealing with Complex Hilbert space?

I don't see the connection between the definition of a self-adjoint extension $A \subseteq A^*$ and the dimensions of $\mathcal{L}_+$ and $\mathcal{L}_-$ I said K earlier, that is incorrect. The two textbooks I'm referencing use different notation.

Where L_+ = ker(A*-i)=[ran(A+i)^perp
L_- = ker(A*+1) = [ran(A-i)]^perp

What in your posts need to be changed?
• Mar 22nd 2009, 01:11 PM
Opalg
Quote:

Originally Posted by Nusc
Why in the literature is symmetric and Hermitian used interchangeably when we're dealing with Complex Hilbert space?

No particular reason, it's up to any author to choose their preferred terminology. There's no international regulatory body to enforce standardised terminology.

Quote:

Originally Posted by Nusc
I don't see the connection between the definition of a self-adjoint extension $A \subseteq A^*$ and the dimensions of $\mathcal{L}_+$ and $\mathcal{L}_-$

That is a major theorem due to von Neumann. There is a brief account of it here (see Section 4 in particular), using the notion of the Cayley transform.
• Mar 22nd 2009, 01:30 PM
Nusc
Quote:

Originally Posted by Opalg
No particular reason, it's up to any author to choose their preferred terminology. There's no international regulatory body to enforce standardised terminology.

Apparently interchanging self-adjoint and symmetric there is a distinction in unbounded operators.

So I read somewhere that unbounded operators are just specific cases in Functional analysis, what is really meant by unbounded? I don't think unbounded necessarily mean infinite dimensiona, after all Hilbert spaces in general are infinite dimensional.

Then to prove A is a maximal symmetric operator that is not self-adjoint if and only if either n_+ = 0 and n_->0 or n_+ >0 and n_- = 0. Conway says it follows from the previous theorem:

Let A be a closed symmetric operator. If W is a partial isometry with initial space in L_+ and final subsapce in L_-, let

D_W = {f + g+ Wg: f in dom A, g in intialy W}
and define A_W on D_W by

A_W(f + g + Wg) = Af + ig -Wg.

I don't see how the theorem can be applied to prove this statement.
• Mar 24th 2009, 01:14 PM
Nusc
Now I'm confused, what is meant by a closed extension?
• Mar 24th 2009, 01:27 PM
Opalg
Quote:

Originally Posted by Nusc
Now I'm confused, what is meant by a closed extension?

You'll never understand this stuff unless you go back and absorb the basic definitions.

1. What is meant by a closed operator?

2. What does it mean to say that an operator T is an extension of an operator S?

Make sure that you are confident about what those two definitions mean. Then put them both together and you will know what it means to say that T is a closed extension of S.
• Mar 24th 2009, 01:33 PM
Nusc
An operator T:H->K is closed if its graph is closed in H direct sum K.

Let T_1 and T be operators on H. If gra (T ) subset gra (T_1) , then T_1 is said to be an extension of T.

An operator A is closable if it has a closed extension. Every closable operator has a smallest closed extension, called its closure - A*

If A is symmetric, A* is a closed extension of A, so the smallest closed extension A** of A must be contained in A*. Thus for symmetric operators, we have

A contained A** contained in A*.

For closed symmetric operators,

A=A** contained A*

Hence I've shown that the closre of symmetric operator is symmetric.

Now how do I show that the self-adjoint operator is a maximal symmetric operator?
• Mar 24th 2009, 01:57 PM
Opalg
Quote:

Originally Posted by Nusc
how do I show that the self-adjoint operator is a maximal symmetric operator?

If A and B are closed (densely-defined) symmetric operators, write $A\subseteq B$ to mean that B is an extension of A. There is a result (which you should be able to find in any book on the subject—it's not a difficult result to prove) which says that A* is an extension of B*, so that $A\subseteq B\subseteq B^*\subseteq A^*$. This tells you that every symmetric extension of A must lie "between" A and A*.

But if A is selfadjoint then A*=A, and there is no room between A and A*. So there are no symmetric proper extensions of A. In other words, A is maximal symmetric.
• Mar 24th 2009, 04:17 PM
Nusc
To show that B* $\subseteq$ A*.

If B is a symmetric extension of operator A, then B $\subset$ A* - every symmetric extension of the operator A is the restriction of the adjoint operator A*.

Since B $\subseteq$ B*, we get $A \subseteq B \subseteq B* \subseteq A*$

Quote:

Originally Posted by Opalg

You really need to go back and check the definitions of these terms in Conway's book. There is a careful explanation of what is meant by a selfadjoint extension in the paragraph following Corollary X.2.10 (page 312).

In my textbook, they go over the Cayley transform after the theorem. How else can I do this?

We know that for A to be self-adjoint D(A) = D(A*) when K_+ = K_-=0.

$A \subset A*$

The dimensions of K_+/- must be n_+/-=0 for A to be self-adjoint,

why doesn't the self-adjoint extension have the dim K_+/- = 0?
• Mar 28th 2009, 04:16 PM
Nusc
Quote:

Originally Posted by Nusc
And so show that A has a self-adjoint extension if and only if $n_+ = n_-$ What is meant by self-adjoint extension?

Isn't this just von Neumann's theorem?