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

Let $x$ and $y$ now vary in $\operatorname{SL}_n(K)$, 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

$$y \mapsto w(x,y)$$

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 $y \mapsto y x^n y^{-1}$ 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 $x* \mathrm{im}(y\mapsto y x^n y^{-1})$, and, as we just said, $y\mapsto y x^n y^{-1}$ 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.)

*Edit:* The discussion has now moved to this subsequent question:

Surjective maps given by words, redux

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    $\begingroup$ To make an obvious comment, y x^2 y^{-1} is also not surjective. $\endgroup$
    – David E Speyer
    Commented Oct 16, 2009 at 14:33
  • $\begingroup$ I would like to know the answer to this as well! $\endgroup$
    – Andrew Critch
    Commented Oct 16, 2009 at 15:26

4 Answers 4

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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.

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  • $\begingroup$ This is a useful rephrasing - I've been thinking along similar lines. $\endgroup$
    – H A Helfgott
    Commented Oct 16, 2009 at 19:38
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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.

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    $\begingroup$ 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. $\endgroup$
    – H A Helfgott
    Commented Oct 17, 2009 at 19:00
  • $\begingroup$ Right. It follows quite simple from the above examples. $\endgroup$
    – Philipp Lampe
    Commented Oct 17, 2009 at 19:11
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This paper

seems related, although it asks a slightly different question.

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  • $\begingroup$ 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. $\endgroup$
    – H A Helfgott
    Commented Oct 16, 2009 at 16:55
  • $\begingroup$ 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. $\endgroup$
    – Pasha Zusmanovich
    Commented Sep 18, 2010 at 16:29
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The discussion has now moved to

Surjective maps given by words, redux

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