One of the most notorious open problems in functional analysis is the invariant subspace problem for Hilbert spaces, which I will state here as a conjecture:

Conjecture 1 (Invariant Subspace Problem, ISP0)Let be an infinite dimensional complex Hilbert space, and let be a bounded linear operator. Then contains a proper closed invariant subspace (thus ).

As stated this conjecture is quite infinitary in nature. Just for fun, I set myself the task of trying to find an equivalent reformulation of this conjecture that only involved finite-dimensional spaces and operators. This turned out to be somewhat difficult, but not entirely impossible, if one adopts a sufficiently generous version of “finitary” (cf. my discussion of how to finitise the infinitary pigeonhole principle). Unfortunately, the finitary formulation that I arrived at ended up being rather complicated (in particular, involving the concept of a “barrier”), and did not obviously suggest a path to resolving the conjecture; but it did at least provide some simpler finitary consequences of the conjecture which might be worth focusing on as subproblems.

I should point out that the arguments here are quite “soft” in nature and are not really addressing the heart of the invariant subspace problem; but I think it is still of interest to observe that this problem is not purely an infinitary problem, and does have some non-trivial finitary consequences.

I am indebted to Henry Towsner for many discussions on this topic.

** — 1. Initial reductions — **

The first reduction is to get rid of the closed invariant subspace , as this will be the most difficult object to finitise. We rephrase ISP0 as

Conjecture 2 (Invariant Subspace Problem, ISP1)Let be an infinite dimensional complex Hilbert space, and let be a bounded linear operator. Then there exist unit vectors such that for all natural numbers .

Indeed, to see that ISP1 implies ISP0, we simply take to be the closed invariant subspace generated by the orbit , which is proper since it is orthogonal to . To see that ISP0 implies ISP1, we let be an arbitrary unit vector in the invariant subspace , and be an arbitrary unit vector in the orthogonal complement .

The claim is obvious if is not separable (just let be arbitrary, and to be a normal vector to the separable space spanned by ), so we may normalise to be . We may also normalise to be a contraction (thus ), and let be the coefficients of .

The next step is to restrict to a compact space of operators. Define a *growth function* to be a monotone increasing function . Given any growth function , we say that a linear contraction with coefficients is *-tight* if one has the bound

For instance, if the matrix is band-limited to the region , it is -tight with . If it is limited to the region , then it is -tight with . So one can view -tightness as a weak version of the band-limited property.

The significance of this concept lies in the following lemma:

Lemma 3 (Sequential compactness)

- (i) Every contraction is -tight with respect to at least one growth function .
- (ii) If is a growth function and is a sequence of -tight contractions, then there exists a subsequence which converges in the strong operator topology to an -tight contraction . Furthermore, the adjoints converge in the strong operator topology to .

*Proof:* To prove (i), observe that if is a contraction and , then

and

and hence by the monotone convergence theorem we can find such that (1), (2). By increasing as necessary one can make monotone.

To prove (ii), we apply the usual Arzelá-Ascoli diagonalisation argument to extract a subsequence that converges componentwise (i.e. in the weak operator topology) to a limit . From Fatou’s lemma we see that is an -tight contraction. From the tightness one can upgrade the weak operator topology convergence to strong operator topology convergence (i.e.

for all ) by standard arguments, and similarly for the adjoints.

We will similarly need a way to compactify the unit vectors . If is a growth function and are natural numbers, we say that a unit vector is *-tight* if one has

for all , with the convention that . Similarly, we say that is *-tight* if one has (3) for all . One has the following variant of Lemma 3:

Lemma 4 (Sequential compactness)Let be a growth function.

- (i) Every unit vector is -tight with respect to at least one increasing sequence . In fact any finite number of unit vectors can be made -tight with the same increasing sequence .
- (ii) If , and for each , is a -tight unit vector, then there exists a subsequence of that converges strongly to an -tight unit vector .

The proof of this lemma is routine and is omitted.

In view of these two lemmas, ISP0 or ISP1 is equivalent to

Conjecture 5 (Invariant Subspace Problem, ISP2)Let be a growth function, and let be an -tight contraction. Then there exist a sequence and a pair of -tight unit vectors such that for all natural numbers .

The compactness given by the -tightness and will be useful for finitising later.

** — 2. Finitising — **

Now we need a more complicated object.

Definition 6Abarrieris a family of finite tuples of increasing natural numbers , such that

- (i) Every infinite sequence of natural numbers has at least one initial segment in ; and
- (ii) If is a sequence in , then no initial segment with lies in .

I learned the terminology “barrier” after asking this question on MathOverflow. Examples of barriers include

- The family of all tuples of increasing natural numbers with ;
- The family of all tuples of increasing natural numbers with ;
- The family of all tuples of increasing natural numbers with .

We now claim that ISP2 is equivalent to the following finitary statement. Let denote the space on .

Conjecture 7 (Finitary invariant subspace problem, FISP0)Let be a growth function, and let be a barrier. Then there exists a natural number such that for every -tight contraction , there exists a tuple in with , and -tight unit vectors , such that for all .

We now show that ISP2 and FISP0 are equivalent.

**Proof of ISP2 assuming FISP0.** Let be a growth function, and let be an -tight contraction. Let denote the set of all tuples with such that there does not exist -tight unit vectors such that holds for all . Let be those elements of that contain no proper initial segment in .

Suppose first that is not a barrier. Then there exists an infinite sequence such that for all , and thus for all . In other words, for each there exists -tight unit vectors such that for all . By Lemma 4, we can find a subsequence that converge strongly to -tight unit vectors . We conclude that for all , and ISP2 follows.

Now suppose instead that is a barrier. Let be a growth function larger than to be chosen later. Then the -tight contraction is also -tight, as is the restriction of to any finite subspace. Using FISP0, we can thus find with and -tight unit vectors such that

for all , and in particular for all . Note that are almost in , up to an error of . From this and the -tightness of the contraction , we see (if is sufficiently rapid) that and differ by at most for . We conclude that

and so , a contradiction. This yields the proof of ISP2 assuming FISP0.

**Proof of FISP0 assuming ISP2.** Suppose that FISP0 fails. Then there exists a growth function and a barrier such that, for every , there exists an -tight contraction such that there does not exist any tuples in with , and -tight unit vectors , such that for all .

We extend each by zero to an operator on , which is still a -tight contraction. Using Lemma 3, one can find a sequence going to infinity such that converges in the strong (and dual strong) operator topologies to an -tight contraction . Let be a growth function larger than to be chosen later. Applying ISP2, there exists an infinite sequence and -tight unit vectors such that for all .

As is a barrier, there exists a finite initial segment of the above sequence that lies in . For sufficiently large, we have , and also we see from the strong operator norm convergence of to (and thus to for any , as all operators are uniformly bounded) that

for all .

Now we restrict to , and then renormalise to create unit vectors . For large enough, we have

and we deduce (for large enough) that are -tight and for all . But this contradicts the construction of the , and the claim follows.

** — 3. A special case — **

The simplest example of a barrier is the family of -tuples , and one of the simplest examples of an -tight contraction is a contraction that is -band-limited, i.e. the coefficients vanish unless . We thus obtain

Conjecture 8 (Finitary invariant subspace problem, special case, FISP1)Let be a growth function and . Then there exists a natural number such that for every -band-limited contraction , there exists and unit vectors with(i.e. are -close to ) such that for all .

This is perhaps the simplest case of ISP that I do not see how to resolve. (Note that the finite-dimensional operator will have plenty of (generalised) eigenvectors, but there is no particular reason why any of them are “tight” in the sense that they are -close to .) Here is a slightly weaker version that I still cannot resolve:

Conjecture 9 (Finitary invariant subspace problem, special case, FISP2)Let be a growth function, let , and let be a -band-limited contraction. Then there exists and unit vectors such that(i.e. are -close to ) such that for all .

This claim is implied by ISP but is significantly weaker than it. Informally, it is saying that one can find two reasonably localised vectors , such that the orbit of is highly orthogonal to for a very long period of time, much longer than the degree to which are localised.

## 22 comments

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29 June, 2010 at 1:19 pm

Alon“(in particular, involving a strange concept which, for want of a better name, ” – and there it breaks. Intriguing!

[Oops, that was a remnant from an earlier draft of the article. Fixed now. -T]29 June, 2010 at 5:12 pm

MiguelIt is understood that

v,win conjecture 9 are also unit vectors.[Corrected, thanks – T.]30 June, 2010 at 4:29 am

OrrStatement of Lemma 3, “strong operator norm topology”, it seems you mean “strong operator topology”.

[Corrected, thanks – T.]30 June, 2010 at 5:10 pm

Miguel LacruzWhen you say that conjecture 9 is implied by ISP, you mean that this is so in an elementary way or because it is a special case of FISP0?

1 July, 2010 at 8:42 am

Terence TaoFISP2 is a consequence of FISP1, which is in turn a consequence of FISP0.

1 July, 2010 at 2:44 pm

Miguel LacruzHere is an attempt to apply dilation theory in conjecture 9. The idea is that any contraction on the Hilbert space has a unitary power dilation, that is, is a subspace of a larger Hilbert space and there is a unitary operator with the nice property that for all , where represents the orthogonal projection. This suggests the following modification of conjecture 9:

Claim AThere exists and there are unit vectors such that,

for each ,

where denotes the orthogonal projection.

Now, consider the vectors , and notice that

and therefore

for all I wonder if this kind of argument can be turned into a proof that FISP2 is a consequence of claim A. Notice that the spectral theorem can be applied to represent the unitary operator on the Hilbert space as a multiplication operator, that is, where is a measurable function and a.e.

It is clear that claim A has the following equivalent formulation:

Claim BThere exists and there is a function such that,

for each

2 July, 2010 at 6:41 am

Miguel LacruzClaim B should also reflect in the symbol the fact that is a power dilation of a 1-band limited contraction. What can be said about the symbol of the minimal unitary power dilation for a 1-band limited contraction?

1 July, 2010 at 4:10 pm

Miguel LacruzSure enough, the function in claim B has norm one.

4 July, 2010 at 8:21 am

RichardTerry,

I only read part of this entry, but here goes nuthin’.

Let T: H => H, and let x be in H.

2 3

Consider x, T(x), T (x) [= T(T(x))], T (x), etc.; let V be the subspace of H they generate.

2

If y is in V, then y = a x + a T(x) + a T (x) + … ; then

0 1 2

2

T(y) = T(a x + a T(x) + a T (x) + …)

0 1 2

2 3

= a T(x) + a T (x) + a T (x) + … , which is still in V; so, if y is in V, so is

0 1 2

T(y).

The desired result follows immediately.

2

Take care, T .

— Rich

P.S. When you speak of this comment in the future — and you will — please be kind. (This is, of course, a paraphrase of a famous line from “T and Sympathy”.)

4 July, 2010 at 10:32 am

Miguel Lacruzoh, come on, rich, you know better than that!

5 July, 2010 at 10:17 am

RichardObviously, the subspace I attempted to describe above may => not <= be proper.

Sorry 'bout that.

5 July, 2010 at 6:19 pm

Miguel LacruzThe operator in the proof of ISP2 assuming FISP0 should be called the compression rather than the restriction of to the subspace I say this because the subspace is not necessarily invariant under

21 July, 2010 at 6:50 am

A New Million Dollar Prize? « Gödel’s Lost Letter and P=NP[…] have a non-trivial closed invariant sub-space?” See this for some history. See Terence Tao for a recent discussion of this famous […]

13 September, 2010 at 3:27 pm

Miguel LacruzDear Terry,

You stated the invariant subspace problem for operators on Hilbert space as a conjecture. My comment goes as follows: why a conjecture, with such a few examples of Hilbert space operators, ? As a comparison, with a whole lot of numerical evidence, the Riemann hypothesis, is still just a hypothesis. Could you please comment on this?

All my best,

Miguel

13 September, 2010 at 4:28 pm

Terence TaoWell, as the name suggests, the invariant subspace problem is usually termed a “problem” rather than a “conjecture”, and I think there are opinions in both directions as to what the answer to the problem is. The only reason I called it a conjecture was to be able to talk about the logical relationships between different formulations of that conjecture; it is clear what a statement such as “Conjecture 1 implies Conjecture 2” means, whereas “Problem 1 implies Problem 2” does not make as much sense, because problems are usually not formulated as true-false statements the way a conjecture is.

There is a limit to how much one should read into a name, but I expect that RH is termed a hypothesis due to the immense number of consequences it has. In contrast, I don’t know of many other statements, in functional analysis or otherwise, that would be directly impacted by the solution to the ISP.

14 September, 2010 at 3:32 pm

Miguel LacruzDear Terry,

Thanks for a lot again for your reply, and thanks a lot as well for your clear explanation. Now my comment is the following.

There are a few, but really beautiful results about the existence of invariant subspaces, and they all make very good use of some fixed point theorem. I am thinking about the work of Victor Lomonosov.

I wonder if these results could be reversed to reduce the ISP to, say, a stronger version of Ky Fan’s fixed point theorem. Would that elevate the ISP to the category of the RH?

I ask you please to comment on this.

All my best,

Miguel

27 October, 2010 at 7:00 am

yujiayangDear Terry,

In your proof of ISP2 assuming FISP0,

you claim:

(if F’ is sufficiently rapid) the two vectors differ by at most 1/m for 0=< n =1 your hints can’t be used to prove the estimation,unlucky we also can’t find a way to prove your claim.Obviously,it’s a key step for your proof,do you notice this problem?

All my best,

Yu

27 October, 2010 at 8:44 am

Terence TaoThe easiest way to see this is by (sequential) compactness. If the claim was not true, then for every there would exist an F-tight operator and unit vectors that are within of , such that and differ by at least some which is independent of . There is enough compactness here to make converge to a limit in the strong operator topology, and converge to a unit vector in , after taking a subsequence. But then also converges to in the strong operator topology, and so and both converge to , a contradiction.

One can also argue directly by carefully choosing one’s epsilons in the right order. The point is that because is strongly localised to (with an accuracy that can be made as strong as one wishes by choosing rapid enough) and is -tight, then is strongly localised to some larger (where does not depend on , but rather on , , and how strong one wishes the localisation to be). One can then obtain similar localisations for , and this is enough to obtain the desired approximation if is large enough.

30 October, 2010 at 5:17 am

yujiayangDear Terry,

I compute your second suggestion and find that:

Let G(n)=max{F((n+1)^12),n},

if F'(n)>max{G^n(n),16(n+1)^6}for all n in N ( G^n means n composition of G)

then your claim do hold.

Your hint:”T is F-tight, then Tv is strongly localised to some larger L^2(N(m,1)) …” play a key role in this construction.

Thanks for your wise suggestion!

Best wishes!

Yu

26 January, 2013 at 9:30 am

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