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Topic: Invariant subspace problem solved?
Replies: 9   Last Post: Mar 6, 2013 7:34 AM

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David Bernier

Posts: 3,892
Registered: 12/13/04
Re: Invariant subspace problem solved?
Posted: Feb 28, 2013 1:24 AM
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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?

> 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and now they have withdrawn their submission.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:


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

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

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) 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

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.:


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 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).

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, self-adjoint, 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."


The key predicate is this:
"commutes with a nonzero compact operator K".

For Hilbert spaces, at least, the norm-closure in B(H)
of the set of all finite-rank operators _is_ the
set of compact operators.

On an infinite-simensional separable Hilbert space H,
the unitary operators on H will not be compact, for

I'm not up-to-speed 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):

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.

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