While talking mathematics with a postdoc here at UCLA (March Boedihardjo) we came across the following matrix problem which we managed to solve, but the proof was cute and the process of discovering it was fun, so I thought I would present the problem here as a puzzle without revealing the solution for now.
The problem involves word maps on a matrix group, which for sake of discussion we will take to be the special orthogonal group of real
matrices (one of the smallest matrix groups that contains a copy of the free group, which incidentally is the key observation powering the Banach-Tarski paradox). Given any abstract word
of two generators
and their inverses (i.e., an element of the free group
), one can define the word map
simply by substituting a pair of matrices in
into these generators. For instance, if one has the word
, then the corresponding word map
is given by
for . Because
contains a copy of the free group, we see the word map is non-trivial (not equal to the identity) if and only if the word itself is nontrivial.
Anyway, here is the problem:
Problem. Does there exist a sequence
of non-trivial word maps
that converge uniformly to the identity map?
To put it another way, given any , does there exist a non-trivial word
such that
for all
, where
denotes (say) the operator norm, and
denotes the identity matrix in
?
As I said, I don’t want to spoil the fun of working out this problem, so I will leave it as a challenge. Readers are welcome to share their thoughts, partial solutions, or full solutions in the comments below.
14 comments
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5 February, 2019 at 11:31 am
Anonymous
Professor Tao, Happy Lunar New Year!
5 February, 2019 at 11:32 am
Jie
Prof Tao, happy lunar new year!
5 February, 2019 at 11:47 am
Andreas Thom
I proved this and more general results in “Convergent sequences in discrete groups”, Canad. Math. Bull. 56 (2013), no. 2, 424-433. The preprint is available at https://arxiv.org/abs/1003.4093.
5 February, 2019 at 11:57 am
Andreas Thom
See also https://mathoverflow.net/questions/229384/solving-equations-in-so3-an-open-problem-by-jan-mycielski/229762
5 February, 2019 at 12:03 pm
Andreas Thom
It is an interesting open problem (as far as I can tell) to determine how fast (in terms of the word length) the image of the word map can converge to the identity matrix. This is related to the shortest length of iterated commutators, a problem that was discussed in
Abdelrhman Elkasapy and Andreas Thom, On the length of the shortest non-trivial element in the derived and the lower central series. J. Group Theory 18 (2015), no. 5, 793–804.
Theorem 5.2. discusses the best asymptotics that we could find. See https://arxiv.org/abs/1311.0138 for the preprint.
5 February, 2019 at 3:44 pm
duetosymmetry
Suppose we consider w_i(e,B) where e is the identity element in SO(3). Then w_i(e,B) can only be of the form B^k for some integer k.
Is it then a necessary condition that there is some sequence of k_i’s such that B^{k_i} converges to the identity for all B in SO(3)?
5 February, 2019 at 11:37 pm
Sean Eberhard
@duetosymmetry No, because $k$ might be (and will be) zero.
8 February, 2019 at 10:40 pm
Anonymous
https://mathoverflow.net/questions/45483/word-maps-on-compact-lie-groups
11 February, 2019 at 5:55 am
Dmitriy Zakharov
For
denote
. It is clear that if
is small enough then
is close to the identity.
Lemma. For any
we have
.
Proof:
Now notice that one of the operators
has rot at most
for any
. So, there is
so that the
-fold commutator
has rot at most
. Therefore, the word
has rot at most
and it is obviously non-trivial.
11 February, 2019 at 9:49 am
Jean Francois LEON
I am sorry but I don’t understand you lemma.
You establish a bound on the norm of the commuting not its rot.
Do I miss something obvious?
[Fixed some typos I spotted in the proof of that lemma. -T]
17 February, 2019 at 12:09 pm
stunient
How does one see that the word is nontrivial?
13 February, 2019 at 7:51 am
jair2018
Reblogged this on Crap.
13 February, 2019 at 8:29 pm
Anonymous
Sorry for the rude blog post,I was experimenting with something on my blog,I called it crap to just label it as a pastebin
18 February, 2019 at 10:53 am
Henry
I believe we can find a sequence of rotations of $S^1$ that converges uniformly to the identity map. Then, using the fact that the group of sphere rotation is isomorphic to $SO(3)$, we can define the elements of our sequence by the maps $(A,B) \mapsto T_n B^{-1} B$, where $T_n$ is the nth element of the uniformly convergent sequence from the circle case.