Ted Odell

I was shocked and saddened to hear about a week ago that Ted Odell, a mathematician to whom I owe a lot, died suddenly on January 9th of a heart attack while he was travelling to this year’s joint AMS/MAA meeting in San Diego. He was 65, but seemed a lot younger.

Ted was a world leader in Banach space theory, and in particular in the infinite-dimensional theory. The wry and slightly enigmatic smile you see in the photo was extremely characteristic: if I imagine Ted, I automatically imagine him with exactly that smile. Less clear from the photo, though perhaps it can be guessed from the camera angle, is that he was extremely tall: he belonged to a handful of mathematicians I know who make me feel short (Tom Sanders and Alex Scott being two others).

I first met Ted when I went to my first ever conference, in Strobl am Wolfgangsee in Austria in 1989. I can’t remember how it came about, but I ended up chatting to him, and he started explaining to me in a wonderfully clear way — the kind of explanation you just can’t get from a textbook — how Tsirelson’s space worked. I read in an obituary by András Zsak (which starts on page 30 of this issue of the LMS newsletter) that Ted had a reputation for being kind and encouraging to young mathematicians. He certainly was to me at this conference, taking the time to give this explanation to a graduate student about whom he knew nothing. Most of the next section describes an argument that he sketched out for me on one of those yellow pads of paper that seem to be standard in US maths departments. (I think I’ve still got the yellow sheets that he let me keep, but I’ve no idea where they are.)

Tsirelson’s space

At the time, I was very interested in the so-called distortion problem for Banach spaces. I shall have a lot more to say about that later, but let me briefly formulate the notion of distortion here. Two norms $|.|$ and $\|.\|$ on the same space $X$ are said to be $C$equivalent if there are constants $a$ and $A$ with $A/a\leq C$ such that $a\|x\|\leq |x|\leq A\|x\|$ for every $x\in X$. After rescaling, one can assume that they satisfy the inequalities $\|x\|\leq |x|\leq C\|x\|$, so for convenience I’ll do that. We call the norm $|.|$ a $C$distortion of $\|.\|$ if the above inequality holds, and if it cannot be improved by passing to an infinite-dimensional subspace. That is, $\|x\|\leq |x|\leq C\|x\|$ for every $x\in X$, and for every subspace $Y$ of $X$ and every $\epsilon>0$ you can find $y,z\in Y$ such that $|y|\leq(1+\epsilon)\|y\|$ and $|z|\geq(C-\epsilon)\|z\|$.

It is far from obvious that there exists a constant $C>1$ and a norm that can be $C$-distorted. If you want to get a sense of the difficulty of this problem, then you might like to spend half an hour thinking about it: if you haven’t already seen the answer, then you will get absolutely nowhere.

Of course, I’ve sort of given away that Tsirelson’s space provides an example, but even with that huge clue the problem isn’t trivial, since you have to come up with the $C$-distortion. It was this that Ted explained to me.

I’m not going to give a full explanation with all its details, but I’ll try at least to convey the basic idea, starting with the definition of Tsirelson’s space (or rather, the dual of the space that Tsirelson originally defined, which, as was observed by Figiel and Johnson, is easier to work with). To define the norm, which is on a space of sequences, we need the following notation. If $a=(a_n)$ is a sequence and $E\subset\mathbb{N}$, then $Ea$ is the sequence that takes the value $a_n$ when $n\in E$ and $0$ otherwise. In other words, we write $E$ for the coordinate projection associated with the set $E$. If $E,F\subset\mathbb{N}$, we write $E to mean that $\max E<\min F$: that is, $E$ ends before $F$ starts. Finally, we say that a sequence $(E_1,\dots,E_k)$ of subsets of $\mathbb{N}$ is admissible if $k. (I write $k when formally I should write $\{k\}.) That is, a sequence of sets is admissible if each set in the sequence starts after the previous one has finished, and if the minimum of the first set is bigger than the number of sets in the sequence.

We now define the norm of a sequence $x$ using the following amazing formula. Let $\mathcal{E}$ be the set of admissible sequences of sets.

$\displaystyle \|x\|=\|x\|_\infty\vee\max\{C^{-1}\sum_{i=1}^k\|E_ix\|:k\geq 2, (E_1,\dots,E_k)\in\mathcal{E}\}$

In words, the idea is this. With each $x$ you want to associate a non-negative real number, and you want to make that number as large as possible. You have two options. Either you can settle for the $\ell_\infty$ norm of $x$, or you can chop up $x$ into pieces $E_1x,\dots,E_kx$ (discarding everything up to the $k$th coordinate), play the same game with each piece, add up the results, and divide by $C$.

Note that if $x$ is finitely supported, then there is nothing to be gained by the chopping up if the support of $x$ is entirely contained in one of the pieces, since then you’ll simply have divided by $C$ unnecessarily. So the game terminates, and if you play it optimally, then you get the norm of $x$.

How does one say anything about a norm that is defined by an implicit formula such as the one above? This is what Ted taught me. A first observation is that it is possible to find sequences of unit vectors $x_1<\dots that approximate the unit vector basis of $\ell_1^n$ as closely as you like. (The notation $x means that the support of $x$ ends before the support of $y$ starts.) The proof is simple. If the support of $x_1$ begins after $n$, then the supports of $x_1,\dots,x_n$ themselves form an admissible sequence. It follows from the implicit definition that $\|a_1x_1+\dots+a_nx_n\|\geq C^{-1}\sum_{i=1}^n|a_i|$ for any sequence $a_1,\dots,a_n$ of real numbers.

A nice inductive argument of R. C. James allows one to improve this $C$-equivalence to a $(1+\epsilon)$-equivalence. Let $|.|$ be any norm on $\mathbb{R}^n$ such that $\|x\|_1\leq |x|\leq C\|x\|_1$ for every $x\in\mathbb{R}^n$. Let $m=\sqrt{n}$ (for convenience I’ll assume that $n$ is a perfect square) and divide the standard unit vector basis $e_1,\dots,e_n$ into blocks of size $m$. Then one of two things is guaranteed to happen.

(i) For every $i$, the $i$th block generates a vector $x_i$ such that $|x_i|\leq\sqrt{C}\|x_i\|_1$.

(ii) For some $i$, every vector $x$ generated by the unit vectors in the $i$th block satisfies $\sqrt{C}\|x\|_1\leq |x|\leq C\|x\|_1$.

In the second case, we have obviously improved the equivalence to $\sqrt{C}$ on a subspace of dimension $\sqrt{n}$. But in the first case we have as well, because $|\sum_{i=1}^ma_ix_i|$ is at least $\sum_{i=1}^m|a_i|\|x_i\|_1$ by hypothesis and at most $\sqrt{C}\sum_{i=1}^m|a_i|\|x_i\|_1$ by the triangle inequality.

By iterating this argument a few times, we can make the equivalence as good as we like.

Now comes Ted’s nice idea. If we want a $C$-distortable space, then we can take Tsirelson’s space as defined above (actually, the normal thing to do was to take $C=2$, but Ted wanted to take a general $C$ in order to show that $C$-distortions existed for any $C$) and define an equivalent norm as follows:

$\displaystyle |x|=\max\{\sum_{i=1}^n\|E_ix\|:E_1<\dots

Here, I’m being a bit sloppy about constants: $C$ and $n$ are basically the same, but for the chopping up in the above definition I want an integer. The idea is that you take a vector in Tsirelson’s space and you’re allowed to increase its norm by cutting it up into $n$ pieces and adding the norms of those pieces.

It is easy to show that this is an $n$-equivalent norm on Tsirelson’s space. It is at least as big as the original norm. But also, since $\|E_ix\|\leq \|x\|$ for every $x$, it follows by the triangle inequality that $|x|\leq n\|x\|$. To show that we have a distortion, we need to show that sometimes chopping up a vector and adding the norms of its bits doesn’t help us, and sometimes it does. And we need to show that that is the case inside every subspace of Tsirelson’s space.

Here’s where I’ll get a bit sketchier. In any subspace we can find unit vectors $x_1,\dots,x_N$ that generate a subspace that’s $(1+\epsilon)$-close to $\ell_1^n$. If we take such vectors and define $x$ to be their average $N^{-1}(x_1+\dots+x_N)$, then we obtain a vector that can’t be improved much by chopping, at least if $N$ is much bigger than $n$. The reason is that, apart from a few edge effects, we’ve basically chopped $x$ up into sums of $n$ consecutive blocks of the $x_i$, and since the $x_i$ generate something very close to a copy of $\ell_1$, $\|x\|$ is very close to the sum of the norms of those sums. So the chopping didn’t achieve anything.

How about a vector where chopping does help? The technique here goes a long way towards explaining the point of the strange definition of Tsirelson’s space. Let’s call a vector of the kind I defined in the previous paragraph an $\ell_1^N$-average. We now inductively construct (inside some arbitrary subspace) a sequence $x_1<\dots of $\ell_1^N$-averages as follows. Once we’ve constructed $x_1,\dots,x_k$, we pick $N_{k+1}$ to be much larger than the maximum of the support of $x_k$ and we let $x_{k+1}$ be an $\ell_1^{N_{k+1}}$-average that starts after $x_k$ finishes.

It turns out that if we want to calculate the norm of the vector $x=x_1+\dots+x_n$ in Tsirelson’s space, we can’t do much better than the obvious decomposition into $x_1,\dots,x_n$, which tells us that the norm is roughly $C^{-1}\sum_{i=1}^n\|x_i\|$.

Why can’t we do better? Well, suppose we chop into $M$ pieces that start after $M$. Then for each $i$ such that $N_i\leq M$ we know that the maximum of the support of $x_{i-1}$ is smaller than $M$, so we’ve completely missed out $x_1,\dots,x_{i-1}$ in the chopping up of $x$. And for each $N_i\geq M$ we know that $N_{i+1}$ is much greater than $M$, so there is nothing to be gained by chopping up $x_{i+1}$ (or any later $x_j$).

But if $\|x\|$ is approximately $C^{-1}\sum_{i=1}^n\|x_i\|$, then $|x|\approx C\|x\|$, since $|x|$ is at least $\sum_{i=1}^n\|x_i\|$.

What we have done with those calculations is show that every subspace contains vectors of two different kinds. One kind is $\ell_1^N$ averages, and the other is sums of $\ell_1^{N_i}$ averages for a sequence $N_1,\dots,N_n$ that grows very rapidly. For the first kind of vector $x$, $\|x\|$ and $|x|$ are roughly equal, whereas for the second kind of vector $x$, $\|x\|\approx C|x|$. So we have shown that Tsirelson’s space, defined with constant $C$, is $C$-distortable (or perhaps it’s safer to say $C/2$-distortable just to provide some elbow room in the calculations).

This shows that for every $C$ there exists a $C$-distortable Banach space. But that raises a very natural question: can we reverse the quantifiers and find a Banach space that is $C$-distortable for every $C$? This question was answered by Thomas Schlumprecht, who had a very close collaboration with Ted over many years, which resulted in a large number of beautiful papers. Let me briefly give the definition of Schlumprecht’s space, the first known arbitrarily distortable space.

The philosophy behind Tsirelson’s space is that since the more pieces you chop a vector into, the bigger you can make the sum of the norms of the pieces, you need some kind of “penalty” associated with the number of pieces you chop into. The penalty is that if you want to chop into $N$ pieces, then you must sacrifice the first $N$ coordinates of your vector. Schlumprecht’s idea was to impose a different kind of penalty: instead of taking a fixed constant $C$, you let $C$ grow with the number of pieces you chop into. Many functions would do for this purpose, but the function $f(x)=\log_2(x+1)$ had some properties that made it convenient in calculations. Accordingly, Schlumprecht’s space is defined by the following formula.

$\displaystyle \|x\|=\|x\|_\infty\vee\max\{\frac 1{f(n)}\sum_{i=1}^n\|E_ix\|:n\geq 2, E_1<\dots

Because $f$ grows slowly, every subspace of Schlumprecht’s space contains $\ell_1^N$ averages, by the James argument. But now if you take sums of $\ell_1^{N_i}$ averages with rapidly increasing $i$, you can get the gain from chopping up into $M$ pieces to grow without limit as $M$ grows.

Incidentally, Tsirelson first constructed his space as an example of an infinite-dimensional space that does not have a subspace isomorphic to $c_0$ or $\ell_p$ for any $1\leq p<\infty$. The fact that it contains copies of $\ell_1^n$ in every subspace and that these copies are generated by disjointly supported vectors rules out everything except $\ell_1$. And $\ell_1$ is ruled out because if it contained a copy of $\ell_1$ then that subspace wouldn’t be distortable.

The distortion problem.

I now want to describe a problem that was for several years my favourite problem in mathematics. The problem is very simple indeed to state, given the definitions in the previous section.

Problem. If $1, is $\ell_p$ $C$-distortable for some $C>1$?

We have seen above that there are distortable spaces. But what about the spaces we know best, the $\ell_p$-spaces? An infinitary analogue of the argument of R. C. James that was used above to obtain $\ell_1^N$-averages can be used to prove that $\ell_1$ is not distortable, and an argument along similar lines shows that $c_0$ is not distortable. But what about a Hilbert space, say?

The fact that Tsirelson’s space can be distorted offers little encouragement, because it relied on the general idea that there are two different kinds of vectors (in the strong sense that they can be found inside every subspace). A Hilbert space is so homogeneous that it looks hard, and perhaps impossible, to identify two different types of vectors.

I came to believe strongly that Hilbert spaces were not distortable. My reason was that there is a simple reformulation of the distortion problem that turns it into a problem in Ramsey theory that is very similar to known Ramsey theorems that have positive answers. Here is the reformulation.

Problem. Let $\epsilon>0$ and let the unit sphere of $\ell_2$ be coloured with finitely many colours. Must there exist an infinite-dimensional subspace $X$ and one of the colours such that every point in the unit sphere of $X$ is within $\epsilon$ of a point with that colour?

This is slightly different from a normal Ramsey theorem in that we don’t find an infinite-dimensional subspace that’s entirely of one colour. Instead, we expand the colours by $\epsilon$ and then try to find a subspace that lives inside one of the (expanded) colours. This is necessary, since otherwise one could colour a vector $x$ red if its first non-zero coordinate is positive and blue if it is negative. But the problem above is a natural “analytic” Ramsey problem.

To see the connection, observe that if you have a $C$-distorted norm on the Hilbert space, then you can have one colour for vectors where the distorted norm is less than $\sqrt{C}$ times the usual norm, and one for the rest. Since every subspace contains vectors that have norms roughly equal to the usual norm and other vectors that have norms roughly equal to $C$ times the usual norm, this gives us a counterexample to the Ramsey-type problem. In the other direction, if you have a two-colouring and no subspace is contained in the $\epsilon$-expansion of either colour, then you can construct a distorted norm. I leave this as an exercise.

Why did I believe so strongly that this Ramsey problem had a positive solution? It is because of a beautiful theorem in Ramsey theory due to Neil Hindman. In its original formulation, it states the following.

Theorem. Let $\mathbb{N}$ be coloured with finitely many colours. Then there exists an infinite sequence $n_1 such that $\sum_{i\in A}n_i$ has the same colour for every non-empty finite subset $A$ of $\mathbb{N}$.

There is also a “finite-unions” version, which says the following.

Theorem. Let the finite subsets of $\mathbb{N}$ be coloured with finitely many colours. Then there exist subsets $E_1 such that $\bigcup_{i\in A}E_i$ has the same colour for every non-empty finite subset $A$ of $\mathbb{N}$.

Here the notation $E means that the maximum element of $E$ is less than the minimum element of $F$.

To see that the sums version implies the unions version, think of each subset of $\mathbb{N}$ as the binary expansion of a positive integer. That is, the subset $A$ corresponds to the positive integer $\sum_{i\in A}2^{i-1}$. Given a colouring of the finite subsets of $\mathbb{N}$, pull it over to the integers using this correspondence, and apply the first version of Hindman’s theorem. This doesn’t quite finish things off, because it gives us some sets that may well not satisfy the condition $E_1. However, given any infinite set $X$ of integers and any $k$, we can always find a sum of distinct elements of $X$ that is divisible by $2^k$. This follows easily from the pigeonhole principle: you just find $2^k$ integers that are the same mod $2^k$ and add them up. Therefore, using the set $X$ that Hindman’s theorem has given us, we can find finite sums that correspond to sets $E$ with arbitrarily large minimum, from which it follows that we can find a sequence of (disjoint) finite sums that correspond to sets $E_1.

The finite-unions version of the theorem can in turn be rephrased in terms of 01-sequences. If $x$ and $y$ are two 01-sequences, write $x to mean that the last non-zero entry of $x$ comes before the first non-zero entry of $y$. Then the finite-unions version of the theorem is trivially equivalent to the following statement.

Theorem. Let the finitely supported 01-sequences be coloured with finitely many colours. Then there exist sequences $x_1 such that $\sum_{i\in A}x_i$ has the same colour for every non-empty finite subset $A$ of $\mathbb{N}$.

This is what got me excited. It seems very much like a combinatorial analogue of a positive solution to the distortion problem: a sequence space is replaced by a set of 01-sequences, but that shouldn’t be a huge deal.

I did in fact manage to pursue these thoughts to prove a theorem in the space that most resembles the space of finitely supported 01-sequences, namely $c_0$ (the space of real sequences that tend to zero, with the $\ell_\infty$ norm). It states the following.

Theorem. Let $\epsilon>0$ and let the unit sphere of $c_0$ be coloured with finitely many colours. Then there exists an infinite-dimensional subspace that lies entirely in the $\epsilon$-expansion of one of the colours.

For what it’s worth, the theorem above is the one result of mine that appears in an Elsevier journal.

There are two things that make $c_0$ particularly combinatorial. One is that it can be discretized very naturally and easily: you just take all points with coordinates that are multiples of $1/n$ for some $n$. The other is that if you add two disjointly supported vectors, then you basically just concatenate them. I won’t say much about the proof here — just that it used ultrafilters (as does one of the nicest proofs of Hindman’s theorem) and it involved a subtlety, in that the obvious discrete problem that you might be tempted to write down has a negative answer because of the sign-of-first-non-zero-coordinate example. So even for the discrete problem one has to prove an approximate result — the $\epsilon$ doesn’t just come from the approximation of the continuous problem by the discrete one.

Try as I might, I just couldn’t find a way of doing without the combinatorial crutch that I had in $c_0$. And yet the homogeneity of $\ell_2$ and the fact that it was a very beautiful and natural sequence space made it impossible for me to give up the dream that there was some vast analytic generalization of Hindman’s theorem waiting to be proved.

What did finally cause me to give up that dream was the announcement in I would guess 1993 (since the paper was published in 1994) that Odell and Schlumprecht had constructed a counterexample to the distortion problem. Very surprisingly (to me at least), they deduced that every $\ell_p$ with $1 is arbitrarily distortable from the fact that Schlumprecht’s space is arbitrarily distortable. To do this, they had to apply a very clever non-linear map to take the colourings that correspond to distortions of Schlumprecht’s space and convert them into colourings that correspond to distortions of $\ell_p$. (I think I may be oversimplifying here — it was a long time ago and I don’t remember much more than I am writing here.)

Another way of stating the remarkable result that Odell and Schlumprecht proved is the following. I’ll state it for $\ell_2$ because there it is nicest.

Theorem. For every $\epsilon>0$ there exist subsets $A$ and $B$ of the unit sphere of $\ell_2$ with the following properties.

(i) Every infinite dimensional subspace of $\ell_2$ contains a point in $A$ and a point in $B$.

(ii) $|\langle x,y\rangle| <\epsilon$ for every $x\in A$ and $y\in B$.

In other words, the two sets $A$ and $B$ are almost orthogonal to each other and yet both intersect every single infinite-dimensional subspace. Even getting the inner product to be at most $1-\epsilon$ is formidably difficult, or at least seems so: it would be fascinating if somebody could come up with a fundamentally different construction.

For me, the whole episode had some important lessons for how mathematical research works. I became obsessed with the distortion problem, and ultimately failed to solve it. And yet, out of that obsession, and also from what I learned during that conversation with Ted Odell, came an understanding of distortability that ended up yielding me a number of other results. On the other hand, I was a little too convinced that the problem had a positive answer, and that meant that there was no chance of my solving it — though the solution of Odell and Schlumprecht was so ingenious that I don’t think I would have solved it even if I had gone all out to find a counterexample.

Conclusion.

I have tried in this post to give some idea of what Ted Odell contributed to mathematics, not by going through his numerous mathematical achievements one by one, but in a more indirect way by conveying the impact he had on me personally. Many other people will have different stories to tell, but the themes that will run through them are that he was a wonderful mathematician, generous with his ideas, and very supportive to younger mathematicians.

I’d like to end by mentioning a beautiful problem that arises very naturally from the above discussion and that is, as far as I know, still unsolved. It is all that remains of my hopes for proving an analytic Ramsey theorem for a non-combinatorial Banach space.

Problem. Does there exist a distortable Banach space that is not arbitrarily distortable? In particular, if Tsirelson’s space is defined with constant $C$, can it be distorted by substantially more than $C$?

9 Responses to “Ted Odell”

1. Ghoussoub Says:

Equally shocked and saddened: Ted Odell: Rest in piece old friend

2. Ghoussoub Says:

Reblogged this on Piece of Mind and commented:
Shocked and saddened. Rest in peace old friend.

3. Monday Highlights | Pseudo-Polymath Says:

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[…] Death, space, and an argument recalled. […]

5. Bill Johnson Says:

Ted was an excellent mathematician and an even better person. He affected many people with his kindness and generosity. I miss him very much.

6. Daniel Freeman Says:

The small acts of kindness that Ted did for many people made a big difference and are greatly appreciated. He always introduced himself to young mathematicians who he did not know at conferences, and he always introduced young mathematicians to other people as well. He was very giving of his time and ideas, and he enjoyed including people in his research and sharing problems that he was interested in. Ted was considerate and respectful to everyone around him. For example, he would take some of the secretaries at UT out to lunch to show them that they were appreciated. Ted was a great friend, mentor, and role model for me and many others. I owe so much to him, and there were many people at his memorial service who were saying the same thing. I miss him very much.

7. Remembering Ted Odell — The Endeavour Says:

[…] Tim Gower’s tribute to Ted […]

8. Bill, Joram, Olek, Ted and Bob | Piece of Mind Says:

[…] I had re-blogged what Tim Gowers wrote about Ted Odell. And when Tim writes, you can rest your pen. I have known Ted since 1982, when I spent a year at the University of Texas at Austin. Ted and his wife Gail were wonderful hosts. I remember him telling me how intrigued he was, when he, who was born in Woodstock, NY (yes the famous one) saw upon his arrival to Texas a cowboy tying his horse before entering a McDonalds. He had never seen another one since. […]

9. Remembering Ted Odell Says:

[…] Tim Gowers’ tribute to Ted […]