MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $w=w(a,b)$ be a non-trivial word in the free group $F_2 = \langle a,b \rangle$ and $w_G \colon G \times G \to G$ be the induced word map for some compact Lie group $G$.

Murray Gerstenhaber and Oskar Rothaus showed in 1962 (see M. Gerstenhaber and O.S. Rothaus, The solution of sets of equations in groups, Proc. Nat. Acad. Sci. U.S.A. 48 (1962), 1531–1533.) that $w_G$ is surjective in a very strong sense for $G=U(n)$ whenever the exponent-sum of $a$ or $b$ in $w$ is non-zero. Using degree-theory and the calculations of the homology of compact Lie groups due to Heinz Hopf, they showed that if the exponent sum of $a$ is non-zero, then $u \mapsto w_{U(n)}(u,v_0)$ is surjective for any fixed $v_0 \in U(n)$.

Earlier, in 1949, Morikuni Gotô (J. Math. Soc. Japan, 1949 vol. 1 pp. 270-272) already showed that the commutator $w=[a,b]$ induces a surjective map $w \colon G \times G \to G$ for any simple compact Lie group $G$. His proof proceeded as follows: Since the commutators are a conjugation invariant set, one can assume that $u$ is in a fixed maximal torus. We can now take a suitable $a$ in that torus and $b$ an element the Weyl group $W_G$ of this torus, such that $b$ does not have $+1$ as an eigenvalue (when acting on the universal cover of the maximal torus). This implies that the action of $1-b$ on the universal covering is an isomorphism. Hence, every element in the maximal torus can be obtained as $a \cdot ba^{-1}b^{-1}$ for some $a$. This strategy can be used for a few other words but does not lead any further.

Question: Are there any other techniques to show that $w_{G}$ is surjective for some $w$ and $G$?

The easiest word for which I cannot answer this is $w= [[a,b],[a^2,b^2]]$.

Question: Is the word-map $w_{PSU(n)}\colon PSU(n) \times PSU(n) \to PSU(n)$ surjective for the word $$w = [[a,b],[a^2,b^2]]?$$

share|cite|improve this question
Second question yes for $PSU(2)$ because yes for $SU(2)$, but by a very ad hoc method: The image of this map is a connected subset of $SU(2)$ and closed under conjugation, so it must be the whole group if it contains $-I$ as well as $I$. It does, because in the interesting subgroup of order $24$ I can think of a pair $a,b$ of elements of order $6$ such that $[[a,b],[a^2,b^2]]$ is the element of order $2$. – Tom Goodwillie Nov 9 '10 at 22:44
That is a nice argument. – Andreas Thom Nov 10 '10 at 6:43

In this paper of Borel it is shown that any non-trivial word is a dominant map from $G \times G$ to $G$ whenever $G$ is a compact connected semisimple Lie group. So the answer to your second question is "yes". I don't recall offhand exactly what techniques are used, but I vaguely recall that one starts with the $SL_2$ case (which contains free subgroups that one can play with) and builds up from there.

EDIT: Dominance would imply surjectivity (or at least that the image is Zariski dense) in an algebraically closed field, but I didn't realise that the question is over the reals, and so Borel's result does not fully resolve the question.

share|cite|improve this answer
I do not understand, how this answers the second question. For any fixed $G$, there are words $w$ such that the image of $w_G$ in $G$ is small in the euclidean topology. – Andreas Thom Nov 10 '10 at 9:33
@Andreas : what would be an example ? – BS. Nov 10 '10 at 10:17
This is Corollary 3.3 in my paper ( For fixed $n$ and $\varepsilon>0$, there exists $w \in F_2$, such that $\|1 - w(u,v)\|< \varepsilon$ for all $u,v \in U(n)$. The word is some iterated commutator of powers of $a$ and $b$. However, for fixed $w$ and $n$ large enough, $w_{PSU(n)}$ or $w_{SU(n)}$ seems to be surjective in many cases. – Andreas Thom Nov 10 '10 at 14:06
Ah, I'm sorry, you're not working over an algebraically closed field. I guess one has to use either real algebraic geometry or homology then, neither of which I can help you with... – Terry Tao Nov 10 '10 at 17:25

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.