Date: Feb 28, 2013 8:12 PM Author: David Bernier Subject: Re: Invariant subspace problem solved? On 02/28/2013 07:56 PM, David Bernier wrote:

> On 02/28/2013 01:24 AM, David Bernier wrote:

>> On 01/28/2013 11:19 AM, José Carlos Santos wrote:

>>> On 28-01-2013 15:35, David Bernier wrote:

>>>

>>>>> I found several texts across the Internet stating that the Invariant

>>>>> Subspace problem for Hilbert spaces was solved by Carl Cowen (USA) and

>>>>> Eva Gallardo (Spain). However, each of these texts is written

>>>>> either in

>>>>> Spanish or in Portuguese. This is rather odd, given the importance of

>>>>> the problem. Did anyone around here read something about this subject?

>>>> [...]

>>>>

>>>> I found something similar to what you found.

>>>>

>>>> From the "Real Sociedad Matemática Española" web-site, a 2013 meeting:

>>>>

>>>> "del 21 al 25 de Enero" 21-25 January?

>>>>

>>>> http://www.rsme.es/content/view/1199/1/

>>>

>>> Interesting. This site is the website of the Royal Spanish Mathematical

>>> Society. It looks serious.

>>>

>>>> I think they mention John von Neumann and the 1930s ...

>> [...]

>>

>> This development seems to be of some interest, and me included,

>> to some degree.

>>

>> The authors announced in early February that a gap in their proof

>> of the invariant subspace problem had been found. They had

>> submitted a paper to a journal containing that claimed proof,

>> and now they have withdrawn their submission.

>>

>> They are working to bridge the gap, and there are results

>> independent of the existence of the gap. Their statement

>> was posted at cafe metematico blog here:

>>

>> http://cafematematico.com/2013/02/05/statement-from-cowen-and-gallardo/

>>

>> ---

>>

>> For complex Hilbert spaces H of dimension n, n finite and n>=2,

>> the related question is if an linear operator T: H-> H

>> that is continuous has a Hilbert sub-space W

>> such that T(W) is contained within W, and

>> W is non-trivial, meaning both

>> dim(W) >= 1 and dim(W) < dim(H) = n.

>>

>> So the sought-for W has 1<= dim(W) <= n-1, if it exists.

>>

>> All n-dimensional Hilbert spaces are isomorphic in the strongest

>> sense, essentially they are all like C^n with

>> (z1, ... z_n).(w1, ... w_n) = sum_{i=1, .. n} z_i w_i^{bar}.

>>

>> Operators on H of finite dimension n>=2 can be thought of

>> as nxn square matrices with entries in C.

>>

>> If A is an nxn matrix, the equation for eigenvalues is:

>> det(A - lambda I) = 0. 'I' is the nxn identity matrix.

>> lambda is the unknown scalar in C.

>>

>> det(A - lambda I) = 0 is equivalent (possibly after mutiplying by +/-1)

>> to lambda^n + a_{n-1} lambda^(n-1) + ... + a_0 = 0,

>> a degree 'n' polynomial equation with indeterminate

>> lambda obtained by fully expanding the determinant:

>> det(A - lambda I)

>> by using the coefficients of A as things that don't

>> depend on lambda (i.e. parameters, coefficients).

>>

>> Then, we apply the fundamental theorem of algebra for

>> C to this degree n polynomial equation.

>>

>> Suppose lambda = r0 is one root of the polynomial.

>> Then r0 is an eigenvalue of the matrix A, and

>> there exists a non-zero vector v in C^n such that

>> Av = r0 v. If k in C is any scalar,

>> A (kv) = k (Av) = k (r0 v) = r0 (kv),

>> so (A - r0 I) (kv) = 0.

>>

>> Or, one could say that for any scalar k in C,

>> A maps kv to (r0 k) v, an element of the 1-dim.

>> space C.v . So W = {kv, k in C the complex numbers},

>> v is non-zero, dim(W) = 1 exactly , W is a non-trivial

>> subspace of C^n, and W is mapped by A to W (although

>> not always onto; this is shown in case A is the zero

>> matrix).

>>

>> In countably infinite dimension, there's essentially one

>> complex Hilbert space, say l^2, the Banach space

>> of square-summable sequences of complex numbers.

>>

>> When the topic is Hilbert spaces, by a subspace one

>> always means/assumes a complete (vector) subspace, unless

>> it's stated otherwise. That's always been my working

>> assumption.

>>

>> This also applies, if I remember correctly, to Banach spaces

>> and subspaces, where the assumption would be that a

>> subspace must be norm-complete, a complete metric space.

>>

>> So now if H is l^2 or any other essentially equivalent

>> separable infinite dimensional complex Hilbert space,

>> one supposes given a continuous complex-linear mapping

>> T: H -> H .

>>

>> That's what an operator from H to H is. (However,

>> in quantum mechanics they need mappings that aren't

>> necessarily continuous, so-called unbounded operators).

>>

>> Just like there is a ring of nxn matrices

>> in finite dimension, there is a ring with identity

>> of all the continuous complex-linear mappings

>> T: H -> H that can be. That huge ring is denoted

>> B(H).

>>

>> B(H) is deceptively large.

>> For example, if sigma: N^* -> N^* is a permutation

>> on N^*, and e_i is the sequence

>> (0, 0, ... 1, ... 0, ....)

>> ^ i'th component ,

>>

>> then one can arrange provisionally to map

>> each e_i to e_{sigma(i)} , for i = 1, 2, 3, ad infinitum

>>

>> This can be extended by linearity to all linear combinations

>> of the e_i, i in N^*. Then, by continuity, the mapping

>> can be extended to the closure in l^2 of these linear

>> combinations.

>>

>> This works and will give a bounded operator U_sigma on

>> l^2. This U_sigma has norm 1 in the operator norm.

>> There are 2^\omega permutations on N^*, resulting in

>> 2^\omega = continuum distinct operators U_sigma.

>> If sigma and sigma' are distinct permutations, then

>> for the operators U_sigma and U_sigma' , I get:

>> || U_sigma - U_sigma' || >= 1 (hopefully ... ).

>>

>> { U_sigma, all permutations sigma: N^*->N^*} forms

>> a set of the power of the continuum of points in

>> B(H) any two of which are at an operator-norm distance

>> of at least 1 from each other.

>> It follows that B(H) has no countable dense subset, i.e.

>> B(H) is a non-separable metric space.

>>

>> For any A: H-> H, there's something called the

>> polar decomposition. One lets P = sqrt(A A^*) (I think)

>> and then A = O P , where O is an isometry or a partial

>> isometry and P is a so-called "positive" operator.

>> This might seem like a promising divide-and-conquer

>> approach to the invariant sub-space problem, but

>> it doesn't seem to say much about what happens when,

>> starting with a non-zero x in H, one iterates

>> A, constructing Ax, AAx, A^3 x, etc or using

>> the polar decomposition,

>> (OPx), (OPOP)x, (OPOPOP)x, etc [i.e. which way are things

>> going? we still don't know any better after applying

>> the polar decomposition to A].

>>

>> Per Enflo first constructed an infinite-dim. Banach space

>> and an operator on it that had no non-trivial closed

>> sub-space. See e.g.:

>> archive.numdam.org/article/SAF_1975-1976____A11_0.pdf

>>

>> and/or

>>

>> http://en.wikipedia.org/wiki/Per_Enflo

>>

>>

>> For separable Hilbert spaces, the best known case is when

>> the operator T is self-adjoint, where the so-called

>> spectral theorem in functional analysis applies

>> (Rudin: Functional Analysis).

> [...]

>

> I was reading on problems considered "modulo a compact

> operator K".

>

> If T is a bounded operator on a separable, infinite

> dimensional Hilbert space H,

>

> can one show that for some compact operator K

> on H, T+K has a non-trivial invariant

> subspace?

[...]

Ok. I think we can find a finite-rank K such that

T+K has some non-zero eigenvector, i.e.

one can construct K such that for some v in H,

T+K maps v to a scalar multiple of v.

[Now assuming K may even be compact]

Still, one hasn't found an infinite-dimensional

invariant subspace of T+K, or alternatively

infinitely many "linearly independent"

invariant subspaces

W_1, W_2, W_3, ad infinitum

where dim(W_i) >=1 in each case.

--

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Please specify a *single* volume group to restore.