
Re: Invariant subspace problem solved?
Posted:
Feb 28, 2013 7:56 PM


On 02/28/2013 01:24 AM, David Bernier wrote: > 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 and now they have withdrawn their submission.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 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 nontrivial invariant subspace?
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.

