I am recording here some notes on a nice problem that Sorin Popa shared with me recently. To motivate the question, we begin with the basic observation that the differentiation operator and the position operator in one dimension formally obey the commutator equation

where is the identity operator and is the commutator. Among other things, this equation is fundamental in quantum mechanics, leading for instance to the Heisenberg uncertainty principle.

The operators are unbounded on spaces such as . One can ask whether the commutator equation (1) can be solved using bounded operators on a Hilbert space rather than unbounded ones. In the finite dimensional case when are just matrices for some , the answer is clearly negative, since the left-hand side of (1) has trace zero and the right-hand side does not. What about in infinite dimensions, when the trace is not available? As it turns out, the answer is still negative, as was first worked out by Wintner and Wielandt. A short proof can be given as follows. Suppose for contradiction that we can find bounded operators obeying (1). From (1) and an easy induction argument, we obtain the identity

for all natural numbers . From the triangle inequality, this implies that

Iterating this, we conclude that

for any . Bounding and then sending , we conclude that , which clearly contradicts (1). (Note the argument can be generalised without difficulty to the case when lie in a Banach algebra, rather than be bounded operators on a Hilbert space.)

It was observed by Popa that there is a quantitative version of this result:

Theorem 1Let such that

*Proof:* By multiplying by a suitable constant and dividing by the same constant, we may normalise . Write with . Then the same induction that established (2) now shows that

and hence by the triangle inequality

We divide by and sum to conclude that

giving the claim.

Again, the argument generalises easily to any Banach algebra. Popa then posed the question of whether the quantity can be replaced by any substantially larger function of , such as a polynomial in . As far as I know, the above simple bound has not been substantially improved.

In the opposite direction, one can ask for constructions of operators that are not too large in operator norm, such that is close to the identity. Again, one cannot do this in finite dimensions: has trace zero, so at least one of its eigenvalues must outside the disk , and therefore for any finite-dimensional matrices .

However, it was shown in 1965 by Brown and Pearcy that in infinite dimensions, one can construct operators with arbitrarily close to in operator norm (in fact one can prescribe any operator for as long as it is not equal to a non-zero multiple of the identity plus a compact operator). In the above paper of Popa, a quantitative version of the argument (based in part on some earlier work of Apostol and Zsido) was given as follows. The first step is to observe the following Hilbert space version of Hilbert’s hotel: in an infinite dimensional Hilbert space , one can locate isometries obeying the equation

where denotes the adjoint of . For instance, if has a countable orthonormal basis , one could set

and

where denotes the linear functional on . Observe that (4) is again impossible to satisfy in finite dimension , as the left-hand side must have trace while the right-hand side has trace .

Multiplying (4) on the left by and right by , or on the left by and right by , then gives

From (4), (5) we see in particular that, while we cannot express as a commutator of bounded operators, we can at least express it as the sum of two commutators:

We can rewrite this somewhat strangely as

and hence there exists a bounded operator such that

Moving now to the Banach algebra of matrices with entries in (which can be equivalently viewed as ), a short computation then gives the identity

for some bounded operator whose exact form will not be relevant for the argument. Now, by Neumann series (and the fact that have unit operator norm), we can find another bounded operator such that

and then another brief computation shows that

Thus we can express the operator as the commutator of two operators of norm . Conjugating by for any , we may then express as the commutator of two operators of norm . This shows that the right-hand side of (3) cannot be replaced with anything that blows up faster than as . Can one improve this bound further?

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11 April, 2018 at 8:41 am

AnonymousSorry this is a little off-topic but I was wondering if have you ever considered the hat problem (https://epubs.siam.org/doi/pdf/10.1137/060652774)

and 100-prisoners problem (https://en.wikipedia.org/wiki/100_prisoners_problem).

What is the bigger picture idea of these decentralized control combinatorial problems?

11 April, 2018 at 10:34 am

Mir CeaMaybe the following is too naive and goes nowhere, but anyway, I ask it because the answer is not clear to me. Maybe it has some clarificatory power, regarding whether in the hilbert hotel analogue you presented, it is better or worse, to e.g. move 2/3 of the people three times, rather than 1/2 of the people twice… So the question is: in what sense is it true that applying the same thing you did for two isometries u, v, for say 3 isometries u,v,w, is not going to improve the outcome bound?

11 April, 2018 at 11:46 am

Terence TaoThis could well be worth trying. One model problem would be to try to optimise the numerical constant is in the example. If it turns out that using a different version of the “Hilbert’s hotel” trick gives better numerical constants, it could suggest a direction in which to improve the construction to eventually get an bound.

11 April, 2018 at 11:11 am

Tobias FritzEquivalently, we can phrase the problem like this: given the constraints and , how small can be made?

Since is equivalent to , this is a problem of noncommutative polynomial optimization, and therefore there is a convergent sequence of (larger and larger) semidefinite programs that gives us the answer for each value of .

At least in principle, these semidefinite programs can not only be solved numerically, but even symbolically. I will try to work out what these semidefinite programs look like, in case that somebody wants to do some computations.

11 April, 2018 at 12:22 pm

Tobias FritzSo here is the proposal in a bit more detail. This is a computer-assisted search, as has also been suggested by Will Sawin, in a systematic form. And Will, you’re right, this approach is for the operators on Hilbert space case.

Suppose we want start with and , and to know whether there are operators and on some nonzero Hilbert space such that and . For the purposes of the following, it is helpful to rewrite these constraints as

and similarly

The left-hand sides of these constraints are noncommutative polynomials . Their particular form is not terribly relevant for the following.

I will roughly follow the exposition of this paper in order to sketch the method. Given a vector in the Hilbert spaces on which these operators act, the moment matrix is an infinite matrix indexed by all monomials , taking the form

for some unit vector in the Hilbert space on which these polynomials act; the particular choice of is not important. As a Gram matrix, this moment matrix is positive semidefinite. Moreover, assuming satisfaction of the above constraints, also the matrix is positive semidefinite for each . By writing as a sum of monomials, the entries of these new matrices are linear combinations of the entries of .

So given operators and satisfying the constraints, we get an infinite positive semidefinite matrix for which a bunch of linearly transformed matrices are also positive semidefinite. The main punchline is that the converse is also true: given positive semidefinite such that the linearly transformed matrices are still positive, then there are operators and satisfying the constraints: simply introduce a scalar product on , introduce the inner product given by the sesquilinear extension of , divide by the null space, and complete. and operate on this Hilbert space in the obvious way by left multiplication. (This is the GNS construction.)

In order to get a sequence of SDPs, one truncates by restricting to a finite subset of monomials.

In this formulation, I can only test whether given and can be realized jointly, and this boils down to a sequence of semidefinite satisfiability problems. I guess that optimizing over one while fixing the other will be a sequence of SDP optimization problems, but I’m not yet sure.

11 April, 2018 at 1:17 pm

Tobias FritzHere’s an attempt at concretely formulating one of these semidefinite relaxations, namely for the truncation corresponding to the set of monomials .

One needs to find a 5×5 positive semidefinite matrix subject to , and subject to positive semidefiniteness of the following two matrices:

Due to the absence of complex numbers in the constraints, it’s enough to assume to be real. Under this assumption, the objective function is to minimize . The resulting optimal value should be a lower bound on in terms of .

However, it’s possible that this truncation is so bad that the resulting lower bound may be negative. Also, it’s very likely that I have a calculation error somewhere, leading to a nonsensical result. It would be much easier if it was possible to set this up directly as a noncommutative polynomial optimization problem in NCSOStools, but I’m not sure whether the problem is of a form that this toolbox can deal with. (It’s more standard to try and

maximizeoperator norms!)14 April, 2018 at 4:19 am

Tobias FritzIt seems that my calculations above were correct, and I have implemented this semidefinite program in the meantime. Unfortunately, my fears turned out to be justified: the resulting lower bounds on are negative.

Nevertheless, using NCSOStools, I am now able to probe whether there exist operators satisfying and for any given pair of values . Thanks to the convergence as the SDP size goes to infinity, the program will exclude (with enough time and memory) every pair of values that cannot be realized.

The computations that I have done so far seem to suggest that is impossible

at all, no matter how large the bounds on and are. Since this contradicts already the Brown-Pearcy result, I have very low confidence in this claim, especially as the SeDuMi solver returns a “Run into numerical problems” message. Could somebody with more experience with SDPs look into this? My Matlab script is here.11 April, 2018 at 11:19 am

Will SawinFor fixed values of , , Popa’s quantitative bound can be truncated to a finite one with only a small loss in the bound on . Moreover I think if any tuple of values is impossible for Banach algebras, there is an argument of this type with finitely many terms that disproves it: Otherwise, take the free algebra on , with the obvious norm, and mod out by the closure of the two-sided ideal generated by , which does not contain any element within of the identity.

So perhaps this problem, like the previous one on word norms, is amenable to computer search to find new such arguments, followed by human interpretation to consolidate gains?

The simplest form of the game would be to find identities implied by the equation of the form a term that, when are replaced with and replaced with , and all subtractions signs removed, would produce a number smaller than .

11 April, 2018 at 11:44 am

Terence TaoWill, could you elaborate on what the “obvious norm” is? I tried using the spectral radius but could not verify the additivity. In any event I do agree that the problem has a similar feel to Polymath14 and a computer assisted search for bounds may well be the way forward.

11 April, 2018 at 12:14 pm

Will SawinI was thinking of the least norm on the free algebra where have norm and has norm . The norm of an element is then the minimum over algebraic expressions in for that element of the “obvious” norm of that expression. It’s equivalent to view this as the norm where each monomial has measure the product of the norms of its elements.

My idea is very similar to Tobias Fritz’s. I think the only difference is his works for algebras of operators on Hilbert spaces, while mine should work for arbitrary Banach algebras, and his involves semidefinite programming while mine would involve linear programming.

11 April, 2018 at 2:13 pm

Terence TaoOK, I see what you’re saying now, thanks!

11 April, 2018 at 4:48 pm

Terence TaoHere is a model problem one could try to attack numerically. Let’s forget about for now and try proving a lower bound on assuming , in which one is only “allowed” to use words in of a fixed length (involving copies of and copies of – note from the scaling symmetry that the powers of have to be in balance). For instance, one can rewrite the Wintner-Wielandt argument by observing the identity

assuming , where there are iterated commutators; expanding this out into noncommutative monomials of degree in , we conclude

in any Banach algebra norm. But is best possible here? In other words, for given , what is the minimal norm of the coefficients of a homogeneous noncommutative polynomial of degree that evaluates to when ? Presumably one could actually work this out numerically for small values of through linear algebra (there are noncommutative monomials, which after applying can be expressed in terms of the basis for ).

12 April, 2018 at 6:16 am

Will SawinThis is sharp if and only if for each homogeneous noncommutative monomial of degree , , where is the coefficient of in the canonical form of the monomial. (If direction by linearity, only if by writing as a linear combination of the monomials that appear in the iterated commutators expression ).

I believe I have checked this for but first I disproved it several times so I can’t be completely sure.

Noncommutative polynomials with equal degree in and are actually a commutative ring, generated by $XD$. To understand this function on monomials analytically it would be helpful to understand what it looks like in the basis of powers of $XD$, but I haven’t managed this yet.

12 April, 2018 at 7:53 am

Terence TaoIt looks like of all the noncommutative monomials of degree , when expressed in terms of the basis for , the coefficient of 1 never exceeds (attained for ). This suggests to me that the best inequality one could ever hope for using degree monomials for is , thus to get a bound of the form one would need to use monomials of degree . Given that is not so much smaller than , this also hints that Popa’s lower bound is perhaps sharp except for the constant.

12 April, 2018 at 8:32 am

Will SawinOne thing that remains is to explain why to finish the argument in the setting requires using the bounds for each value of simultaneously, so that a term like appears in the final bound, rather than just an individual term (where we could take much larger than .)

11 April, 2018 at 12:37 pm

Andreas ThomThanks for this interesting post. In the proof of Theorem 1, I think you get

before dividing by and summing from to .

[Corrected, thanks – T.]12 April, 2018 at 5:37 am

AulaLet be a positive integer and a ring of characteristic . The identity matrix of has trace , but in it is equal to zero, so in principle it is possible to find matrices such that .

12 April, 2018 at 7:10 am

AnonymousWhat about operators satisfying ?

12 April, 2018 at 7:24 am

AnonymousMore exactly, my questión is the following, are they necessarily isometries? In the negative case, what can we say about their invariant subspaces?

13 April, 2018 at 7:26 am

Terence TaoOne could conceivably build a sequence of examples showing that Popa’s bound is essentially sharp through a recursive construction. For instance, if whenever one can solve with and , one can somehow create new operators (possibly matrix combinations of the existing ones) with , and , then one can iterate (starting from an example with small enough) to show that Popa’s bound is sharp up to constants. Maybe a variant of the 2 x 2 matrix construction in the blog post would work for this?

13 April, 2018 at 12:56 pm

Tobias FritzThere is something that confuses me in the construction sketched in the second part of the post. I have been able to verify the first commutator equation involving the 2×2-matrices, but I am unable to follow the second one. I’m getting that the relevant equation for is . Is this supposed to be implied by ?

14 April, 2018 at 6:44 am

Terence TaoOops, there was a typo; should have been . Fixed now.