
Re: Invariant subspace problem solved?
Posted:
Feb 28, 2013 1:24 AM


On 01/28/2013 11:19 AM, José Carlos Santos wrote: > On 28012013 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" website, a 2013 meeting: >> >> "del 21 al 25 de Enero" 2125 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 They had submitted a paper to a journal containing that claimed proof, and now they have withdrawn their submission.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/statementfromcowenandgallardo/

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 subspace W such that T(W) is contained within W, and W is nontrivial, meaning both dim(W) >= 1 and dim(W) < dim(H) = n.
So the soughtfor W has 1<= dim(W) <= n1, if it exists.
All ndimensional 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_{n1} lambda^(n1) + ... + 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 nonzero 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 1dim. space C.v . So W = {kv, k in C the complex numbers}, v is nonzero, dim(W) = 1 exactly , W is a nontrivial 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 squaresummable 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 normcomplete, 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 complexlinear 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, socalled unbounded operators).
Just like there is a ring of nxn matrices in finite dimension, there is a ring with identity of all the continuous complexlinear 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 operatornorm distance of at least 1 from each other. It follows that B(H) has no countable dense subset, i.e. B(H) is a nonseparable 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 socalled "positive" operator. This might seem like a promising divideandconquer approach to the invariant subspace problem, but it doesn't seem to say much about what happens when, starting with a nonzero 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 infinitedim. Banach space and an operator on it that had no nontrivial closed subspace. See e.g.: archive.numdam.org/article/SAF_19751976____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 selfadjoint, where the socalled spectral theorem in functional analysis applies (Rudin: Functional Analysis).
I believe the spectral theorem also works for operators T that commute with their adjoint T*, i.e. those such that T . T^* = T^* . T (they are referred to as "normal operators). http://en.wikipedia.org/wiki/Normal_operator
Maybe because B(H) is nonseparable, it seems hard to me to imagine a random, generic, element of B(H). Perhaps we've only seen a fraction of operator behaviour with the unitary, selfadjoint, normal and compact operators ?
Then's there's a quite celebrated result of the Russian/Soviet mathematician Lomonosov.
As given by WordPress blogger Adam Azzam:
(Lomonosov?s Theorem 1973) "If T is an operator that is not a multiple of the identity, and K is a nonzero compact operator that commutes with T, then T has a nontrivial hyper invariant subspace."
Ref.: http://whateversuitsyourboat.wordpress.com/2012/02/05/theinvariantsubspaceproblemandlomonosovstheorempart3of3/
The key predicate is this: "commutes with a nonzero compact operator K".
For Hilbert spaces, at least, the normclosure in B(H) of the set of all finiterank operators _is_ the set of compact operators.
On an infinitesimensional separable Hilbert space H, the unitary operators on H will not be compact, for example.
 I'm not uptospeed or current with big developments in functional analysis.
Hopefully the above can be of some use to some people. 
I think some might be interested to know that Terry Tao had a blog post about the Invariant Subspace Problem for Hilbert spaces in 2010 (some might also enjoy the comments there):
http://terrytao.wordpress.com/2010/06/29/finitaryconsequencesoftheinvariantsubspaceproblem/#more3900
David Bernier  dracut:/# lvm vgcfgrestore File descriptor 9 (/.console_lock) leaked on lvm invocation. Parent PID 993: sh Please specify a *single* volume group to restore.

