# Is the Invariant Subspace Problem interesting?

There's an amusing comment in Peter Lax's Functional Analysis book. After a brief description of the Invariant Subspace Problem, he says (paraphrasing) "...this question is still open. It is also an open question whether or not this question is interesting."

To avoid lengthy discussions involving subjective views about what makes math interesting, I'd simply like to know if there are examples of math papers out there that begin with something like, "Suppose the invariant subspace problem has a positive answer..."

Of course, papers that are about the ISP itself don't count!

• Beauty is in the eyes of the beholder. – TCL Dec 10 '10 at 12:12
• Possibly the fact that the problem can be so simply stated and yet seems to need more than just the fundamental tools in the area makes it intrinsically interesting. So even if there are no applications, it may be useful as a yardstick to see the limitations of basic techniques. Of course, saying that the question is interesting in that sense is not the same as saying that it's worth investing tons of energy to solve it. Cf Jacobian Conjecture. – Thierry Zell Dec 10 '10 at 13:40
• For the most part the problem was resolved. The structure of linear operators is important in applications of mathematics to the physical sciences, so it is interesting outside of mathematics. The ideas in General Topology and Finite Group Theory are important too, but like Functional analysis, the circle of ideas inherent in the problems in the field are understood to the extent that they can be, and until some more ideas are added, they will be relatively dormant areas. – Charlie Frohman Dec 10 '10 at 13:50
• @Charlie: when you say "For the most part the problem was resolved" I take it you mean, "whether or not the problem is interesting has been resolved" [and the answer is "yes, it is interesting"], but please correct me if you really meant "for the most part the ISP was resolved" (because that would be surprising). In any case, thanks to all three of you (TCL, TZ, CF) for your thoughful comments. – William DeMeo Dec 10 '10 at 23:53
• Yes, Charlie. Much of the recent work of Pearcy et al has at its roots the wonderful insights of Scott Brown, which I guess are now considered "classical". – Bill Johnson Dec 11 '10 at 16:56

1. The invariant subspace problem for Banach spaces was solved in the negative for Banach spaces by Per Enflo and counterexamples for many classical spaces were constructed by Charles Read. The problem is open for reflexive Banach spaces. On the other hand, S. Argyros and R. Haydon recently constructed a Banach space $X$ s.t. $X^*$ is isomorphic to $\ell_1$ and every bounded linear operator on $X$ is the sum of a scalar times the identity plus a compact operator, hence the invariant subspace problem has a positive solution on $X$.

2. The invariant subspace problem has spurred quite a lot of interesting mathematics. Usually when a positive result is proved, much more comes out, such as a functional calculus for operators. See, e.g., recent papers by my colleague C. Pearcy and his collaborators.

3. In cases where the ISP has a positive solution for a class of operators, there may be a structure theory for the operators. There is, for example, J. Ringrose's classical structure theorem for compact operators on a Banach space. This is a beautiful and useful theorem, which, BTW, I am using currently with T. Figiel and A. Szankowski to relate the Lidskii trace formula to the J. Erdos theorem in Banach spaces.

4. Why is the twin prime conjecture interesting?

• Kevin, the problem with your comment is that BOTH the first, second and third answers you present can be regarded as good answers to the question "Why is the TPC less interesting than the ISP?" and not only, as you argue, to the question "Why is the TPC more interesting than the ISP?" – Gil Kalai Dec 10 '10 at 16:52
• Related to Bill's answer, perhaps one might say that operators on Hilbert spaces have done more than enough to prove their interest, and the invariant subspace problem shows that we still don't understand them. That's a bit vague, but by "understand" I mean something like, "have a good enough description of a general such operator to answer a question as basic as whether it must have an invariant subspace". I don't think that is quite as circular as it may seem. – gowers Dec 10 '10 at 20:53
• @Bill: Thank you for your reply. I was familiar with the Enflo and Read work, but not with Argyros-Haydon, so I appreciate learning about that. Your answers 2 & 3 get more to the point though. I was wondering about the "big" problem (linked to in my question): do all bounded linear operator on an infinite-dimensional separable Hilbert space have non-trivial invariant subspaces. If we know that the answer is yes, does it really change things? Given your answer 3, and gowers comment, I think the answer is "yes" and I (gratefully) accept your answer. (Further answers still welcome though.) – William DeMeo Dec 10 '10 at 23:46
• I'd still like to know of any paper(s) which starts by assuming the (Hilbert space) ISP has a positive/negative answer, and then proves some interesting consequence. – William DeMeo Dec 13 '10 at 0:01
• @William DeMeo: I'd say this is just some kind of different cultural attitude between functional analysts and number theoreticians. Not all branches of mathematics work in the same way. Graph theoreticians love to propose an insanely high number of conjectures, but I would not say proposing many conjectures is either a necessary or sufficient condition for an interesting branch of maths. (and I do like graph theory). – Delio Mugnolo Jan 28 '13 at 21:38

Most of the structure theorems for complex matrices can be expressed solely in terms of invariant subspaces. For example, the statement that every nxn complex matrix is unitarily equivalent to an upper triangular matrix (from which the spectral theorem for normal matrices easily follows) is equivalent to the existence of a chain of invariant subspaces having one of each possible dimension from 0 to n. A matrix is similar to a single Jordan block if and only if its lattice of invariant subspaces is a chain; this allows for the Jordan form to be expressed in terms of invariant subspaces. If you look to infinite-dimensional Hilbert spaces, the sub-Hilbert spaces are closed linear subspaces, and the natural analogue of matrix is a bounded linear operator. If you want to extend the finite-dimensional structure theory to the infinite-dimesnional situation the first natural question to ask is whether every operator has a nontrivial (closed, linear) invariant subspace. This problem was popularized by Paul Halmos in the 1970's and, while the solution may not be important, attempts at solutions have generated a vast amount of important mathematics. For example, the concept of quasidiagonality for C*-algebras, which is very important to that subject, was defined by Halmos as a reducing version of quasitriagularity (a property distilled from several theorems about the invariant subspace problem).

• – user26107 Jan 30 '13 at 20:51
• – marjeta Feb 6 '13 at 10:50

If the invariant subspace problem has a positive answer then every bounded operator $$A \in B(H)$$ can be put in upper triangular form, in the sense that there is a maximal chain $$(E_\lambda)$$ of closed subspaces of $$H$$ such that every $$E_\lambda$$ is invariant for $$A$$.

In $$\mathbb{C}^n$$, a maximal chain of subspaces looks like $$\{0\} = E_0 \subset E_1 \subset \cdots \subset E_n = \mathbb{C}^n,$$ where the dimension of $$E_i$$ is $$i$$, and any operator for which all the $$E_i$$ are invariant is literally upper triangular for an orthonormal basis whose first $$i$$ elements belong to $$E_i$$, for all $$i$$. The infinite dimensional version is a natural generalization and seems to say rather a lot about the structure of $$A$$.

No doubt this result would be considered "known" by experts, but I could not find it explicitly stated anywhere, so I included it near the end of this paper. Also, Matt Kennedy answered this question with a reference to a result that easily implies it.