Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $w(x,y)$ be a word in $x$ and $y$.

Let $x$ and $y$ now vary in $SL_n(K)$, where $K$ is a field. (Assume, if you wish, that $K$ is an algebraically complete field of characteristic bigger than a constant.)

I would like to know for which words $w$ the map

$y \rightarrow w(x,y)$

isn't surjective (or even dominant - that is, "almost surjective") for $x$ generic.

It is clear, for example, that the map is surjective for $w(x,y)=xy$, and that it isn't surjective for $w(x,y)= y x y^{-1}$, or for $w(x,y) = y x^n y^{-1}$, $n$ an integer: all elements of the image of $y \rightarrow y x^n y^{-1}$ lie in the same conjugacy class. A moment's thought (thanks, Philipp!) shows that $w(x,y) = x y x^n y^{-1}$ isn't surjective either: its image is just $x* im(y\rightarrow y x^n y^{-1})$, and, as we just said, $y\rightarrow y x^n y^{-1}$ isn't surjective.

I would like to know if the only words $w$ for which the map isn't surjective for $x$ generic are the $w$'s of the form $w(x,y) = x^a v(x,y) x^b (v(x,y))^{-1} x^c$, where $v$ is some word and $a,b,c$ are some integers. (This seems to me a sensible guess, though I would actually be quite glad if it weren't true.)

share|improve this question
1  
To make an obvious comment, y x^2 y^{-1} is also not surjective. –  David Speyer Oct 16 '09 at 14:33
    
I would like to know the answer to this as well! –  Andrew Critch Oct 16 '09 at 15:26
    
Thanks, David - I've changed the question above. –  H A Helfgott Oct 16 '09 at 16:54
add comment

4 Answers

I'm going to say some things which might be either (a) obvious, (b) wrong, or (c) useless. (Or some combination!)

You could rephrase the question by asking that the map from SL_k x SL_k to SL_k x SL_k given by (x,y) -> (x,w(x,y)) is dominant. This seems like an improvement, because now you're talking about a map between two spaces of equal dimension.

If the corresponding map on the tangent spaces of Id x Id is an isomorphism, then certainly the map is dominant. (And this map is easy to compute, given w -- we just replace the product in w by the corresponding sum of tangent vectors.)

The map is dominant iff the map on tangent spaces is generically an isomorphism. I don't know how to check this, but my feeling is that it will be easier for this to fail than in the given conjecture.

share|improve this answer
    
This is a useful rephrasing - I've been thinking along similar lines. –  H A Helfgott Oct 16 '09 at 19:38
add comment

At least for n=2, the map y -> xyx-1y-1 is not surjective for generic x.

Let us prove that the map is not surjective for diagonalizable x. Thus, it is not surjective for generic x.

Suppose that x is diagonalizable and let a=(aij) be matrix in SL2(K). We want to solve the equation xyx-1y-1=a. By conjugating with an appropriate matrix, we may wlog assume that x=diag(b,b-1) is a diagonal matrix. Let y=(yij). A short calculation shows that the diagonal entries in the matrix xyx-1y-1 are

a11=y11y22-b2y12y21=1+(1-b2)y12y21, a22=y11y22-b-2y12y21=1+(1-b-2)y12y21.

This means that in the image there are only matrices a with (a11-1)/(a22-1)=-b2.

But xyx-1y-1 has not the form vxkv-1.

share|improve this answer
1  
Philipp, this is very good - however, doesn't this follow directly from the examples we already know? Since f:y-> y x^{-1} y^{-1} isn't surjective, g:y -> x y x^{-1} y^{-1} isn't surjective either: im(g) = x im(f). So, it's essentially the same example. I'll change the phrasing of the question. Thanks. –  H A Helfgott Oct 17 '09 at 19:00
    
Right. It follows quite simple from the above examples. –  Philipp Lampe Oct 17 '09 at 19:11
add comment

This paper http://front.math.ucdavis.edu/math.GR/0211302 seems related, although it asks a slightly different question.

share|improve this answer
    
Thanks for the link, but the question seems pretty different. There it's clear what the answer ought to be - here I am not so positive of my guess. –  H A Helfgott Oct 16 '09 at 16:55
    
Another paper treating a similarly-looking question is: N. Gordeev and U. Rehmann, On multicommutators for simple algebraic groups, J. Algebra 245 (2001), 275-296 mathematik.uni-bielefeld.de/LAG/man/056.html dx.doi.org/10.1006/jabr.2001.8929 . It is proved there that for a simple algebraic group $G$ of sufficiently large Coxeter number, the map $G^{n+1} \to G^n, (g,g_1, ..., g_n) \mapsto ([g,g_1], ..., [g,g_n])$ is dominant. But, again, probably the similarity is superficial. –  Pasha Zusmanovich Sep 18 '10 at 16:29
add comment
up vote 0 down vote accepted

The discussion has now moved to

Surjective maps given by words, redux

share|improve this answer
add comment

Your Answer

 
discard

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.