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I asked some time ago:

Let $w(X,Y)$ be a word in $X$ and $Y$ (i.e., an element in the free group on $X$ and $Y$). Let the variables $x$ and $y$ now range among elements of $SL_n(K)$, $K$ an algebraically closed field.

For which $w$ is it the case that, for $x$ generic, the image of the map given by

$y \rightarrow w(x,y)$

is not Zariski-dense?

It seems that the map has Zariski-dense image for most w one can try. At the same time, as I said back when I first asked the question, the image of the map $y \rightarrow w(x,y)$ is not Zariski-dense for $w(x,y) = y x y^{-1}:$ the image of the map is contained in the conjugacy class of $x$.

By the same reasoning, the image of the map $y \rightarrow w(x,y)$ is not Zariski-dense for $x$ generic when $w$ is of the form

$w(x,y) = v(x, u(x,y)^{-1} x u(x,y)),\ (*)$

where $v$ and $u$ are some words. The question can then be made precise: are all examples of this form? That is: is it the case that, for all words $w(x,y)$ not of the special form $(*)$, the image of the map $y\rightarrow w(x,y)$ is Zariski-dense for $x$ generic?

I would be extremely interested in the correct answer, even in the case $n=3$. I suspect that the answer is "yes", at least for $n=2$. At the same time, it would be rather nice if it were "no".

[The most obvious approach may be to take derivatives at the origin. However, while this often proves that a map is surjective, it does not prove that a map is not surjective - and in this problem it leaves too many candidates of possible words w for which the map $y\rightarrow w(x,y)$ might not be surjective but probably really is.

For those who have asked about the characteristic: if the characteristic is finite, you can assume it's much greater than the sum of the absolute values of the exponents of the word we are considering. In other words, let us not consider things like $y\rightarrow y^p, p = char(K)$.

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Sorry, Harald, I'm confused here -- is y -> [x,y] not surjective for a generic x? – JSE Oct 23 '09 at 16:16
No, it is not. The map y -> y x^{-1} y^{-1} is not surjective, since its image is simply the conjugacy class Cl_x of x. Thus, the map y -> [x,y] = x y x^{-1} y^{-1} is not surjective either - its image is x Cl_x. – H A Helfgott Oct 24 '09 at 13:53
One possible way to show that a word w gives non-surjective map is to show that the derivative of the map y -> w(x,yz) at the origin is singular for x and z generic. In fact, I think this is an if-and-only if condition. Now, as to how to show that... – H A Helfgott Oct 25 '09 at 16:03
I'm going to vote this -1 for lack of clarity, and suggest two ways in which it could be improved. Greg Kuperberg mentioned a third. 1: "Word" is used in different ways by different people. Group theorists mean one thing, semigroup theorists another, and universal algebraists another still (generalizing the first two). It's not until later in your question that we see evidence of which meaning you intend. 2: You ask when a map is "not surjective for x generic". That's ambiguous: does it mean "for generic x, it's non-surjective" or "it's false that for generic x, it's surjective"? – Tom Leinster Nov 3 '09 at 23:10
Amendments made. – H A Helfgott Nov 4 '09 at 10:15
up vote 3 down vote accepted

You need some conditions on K to make sense of the question. For instance, y ⟼ y2 is not surjective when K = ℝ and n = 2, because you cannot reach [[-1,0],[0,-2]]. Of course, you might have meant surjectivity in the sense of algebraic groups rather than in the sense of set-theoretic groups. But I think that that just amounts to asking for K to be algebraically closed. There is another problem when K has characteristic p. In this case y ⟼ yp is not surjective; its image is only the diagonalizable matrices.

The conjecture as stated seems plausible when K = ℂ. Or if K has positive characteristic and is algebraically closed, you could perhaps ask for a Zariski-dense image.


Re Tom's comment: I think that Harald's question is clear enough, even though I agree that one has to work a little to see what he means. He must mean that w is an element of the free group in the letters x and y, of course. The second objection is also not essential. The question can be phrased as: For each w, what are the Zariski topology properties of the set of all x for which y ⟼ w(x,y) is surjective? I believe that Chevalley's theorem on constructible images tells you that the set of such x is constructible, so it either contains a Zariski open or it is disjoint from a Zariski open.

Re Harald's comment concerning if-and-only-if conditions. You have probably thought of this, but here goes anyway. If your derivative is non-singular for any z, then the w-map is Zariski dense. Of course, there are interesting maps in algebraic geometry that are Zariski dense but not surjective. You seem to suggest that that cannot happen here, but I do not know how to prove it.

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Thanks for your comment. Yes, as I said in my comment to my own question, I am aware that it is enough for the derivative to be non-singular at any z for the map to be Zariski dense; this would seem to suggest that we have to check that the derivative is singular at every z to be sure that the map is not Zariski dense. (Or is there another way to check this?) I'll be happy to replace "surjective" by "Zariski-dense" - I had either in mind, and both are equally good for my application. I suppose I am thinking of a characteristic considerably greater than the length of the word (or char 0). – H A Helfgott Nov 4 '09 at 10:07
Since all of the sets in sight are constructible, you don't have to check "every" z, just a generic z. This may be useful for a computer calculation, if for instance you are happy with non-rigorous evidence that some w is a counterexample. It is not all that useful theoretically. – Greg Kuperberg Nov 4 '09 at 22:16

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