# Word maps over the real and complex numbers

*Kourovka Notebook, Problem 16.68 · Nilradical v1.0.0*

**This is not a preprint.** This note states the results and explains their proofs so that human researchers can check, develop and build on them. **Agent-generated exposition; not refereed.**

## The question and the answers

For a word $w$ in the free group $F_2=\langle a,b\rangle$ and a group $G$, substitution defines the word map

$$
w:G^2\longrightarrow G,\qquad (g,h)\longmapsto w(g,h).
$$

J. Mycielski's Problem 16.68 asks whether every nonidentity word gives a surjective map on each of $\operatorname{PSL}_2(\mathbb R)$, $\operatorname{PSL}_2(\mathbb C)$ and $\operatorname{SO}_3(\mathbb R)$ [1](https://kourovkanotebookorg.wordpress.com/wp-content/uploads/2026/09/21tkt.pdf). The complex answer is affirmative; the real answer is negative. Throughout, we use $[r,s]=rsr^{-1}s^{-1}$ and the ordinary matrix trace $\operatorname{tr}$.

**Complex theorem.** Let $K$ be an algebraically closed field of characteristic zero. For every $1\ne w\in F_2$, the map

$$
w:\operatorname{PSL}_2(K)^2\longrightarrow\operatorname{PSL}_2(K)
$$

is surjective. In particular, this holds for $K=\mathbb C$.

**Real theorem.** In $F_2$, put

$$
d=bab^{-1},\qquad c=[a,d],\qquad
W=[aca^{-1},\,dc^{-1}d^{-1}].
$$

Then $W\ne1$ and

$$
\operatorname{tr}W(A,B)>\frac74
\qquad\text{for every }A,B\in\operatorname{SL}_2(\mathbb R).
$$

Consequently, $W:\operatorname{PSL}_2(\mathbb R)^2\longrightarrow\operatorname{PSL}_2(\mathbb R)$ is not surjective: its image contains no nonidentity involution.

The complex proof combines an elementary-matrix specialization from Schneider–Thom [2, Lemma 1](https://arxiv.org/abs/1802.09289v1) with a polynomial-fibre observation appearing in Mushkarov–Nikolov [3, Example 1(iii)](https://doi.org/10.4171/EM/562). The real proof is an explicit trace identity and an inequality. The word $W$ has cyclically reduced length $44$.

The compact answer is already negative by Thom's almost-law theorem [4, Corollary 1.2](https://arxiv.org/abs/1003.4093): there are nonidentity two-variable words whose values on $\operatorname U(3)$ are uniformly arbitrarily close to the identity. Restriction to $\operatorname{SO}_3(\mathbb R)$, with a sufficiently small neighbourhood, gives a nonsurjective word map. This is an attributed prior result and is not part of the Lean formalizations described here.

## A simple root gives a unipotent

Fix an algebraically closed field $K$ of characteristic zero. The obstruction to proving the complex theorem from traces alone is that a matrix of trace $2$ or $-2$ might be scalar. The following two observations remove it.

**Two-fibre lemma.** If $f\in K[T]$ is nonconstant and $\alpha\ne\beta$, then $f-\alpha$ or $f-\beta$ has a simple root.

*Proof.* Put $n=\deg f$. If neither fibre had a simple root, each would have at most $n/2$ distinct roots. A root of multiplicity $m$ contributes $m-1$ to the multiplicity of $f'$. The fibres are disjoint, so together they contribute at least $n$ zeros to $f'$, counted with multiplicity. This contradicts $\deg f'=n-1$. $\square$

**Matrix-curve lemma.** If $M(T)\in\operatorname{SL}_2(K[T])$ has nonconstant trace, then $M(t)$ is nonscalar of trace $2$ or $-2$ for some $t\in K$.

*Proof.* Write $f=\operatorname{tr}M$. The preceding lemma supplies $t\in K$ and $\varepsilon\in\{1,-1\}$ with

$$
f(t)=2\varepsilon,\qquad f'(t)\ne0.
$$

If $M(t)$ were scalar, it would equal $\varepsilon I$. Write

$$
M=\begin{pmatrix}a&b\\c&d\end{pmatrix}.
$$

Differentiating $ad-bc=1$ at $t$ would give

$$
0=\varepsilon(a'+d')(t)=\varepsilon f'(t),
$$

a contradiction. $\square$

## Every nonidentity word over an algebraically closed field

Word images are invariant under conjugation, and conjugate words have the same image. If $w$ is conjugate to a nonzero power of one generator, surjectivity follows from power maps on $\operatorname{PSL}_2(K)$. For semisimple elements, take roots in the diagonal torus. For unipotents, use

$$
(I+N/m)^m=I+N\qquad
(m\in\mathbb Z\setminus\{0\},\ N^2=0).
$$

Otherwise, cyclic reduction and rotation give a conjugate of the form

$$
v=a^{m_1}b^{n_1}\cdots a^{m_r}b^{n_r},
\qquad r\ge1,\quad m_i,n_i\in\mathbb Z\setminus\{0\}.
$$

Substitute

$$
U=\begin{pmatrix}1&1\\0&1\end{pmatrix},\qquad
V(T)=\begin{pmatrix}1&0\\T&1\end{pmatrix},\qquad
M(T)=v(U,V(T)).
$$

For all integer exponents, including negative ones,

$$
U^{m_i}V(T)^{n_i}
=\begin{pmatrix}1+m_i n_iT&m_i\\n_iT&1\end{pmatrix}.
$$

The coefficient of $T$ in this block is the rank-one matrix $B_i=n_i(m_i,1)^{\mathsf t}(1,0)$. Multiplying these coefficients gives

$$
[T^r]\operatorname{tr}M(T)
=\operatorname{tr}(B_1\cdots B_r)
=\prod_{i=1}^r m_i n_i\ne0.
$$

Thus $f=\operatorname{tr}M$ is nonconstant and takes every value in $K$.

For each $z\ne\pm2$, all determinant-one matrices of trace $z$ are conjugate: their characteristic polynomial has distinct roots. The image of $v$ therefore contains every nonidentity semisimple element of $\operatorname{PSL}_2(K)$.

The matrix-curve lemma supplies a value $M(t)=\varepsilon I+N$ with $N\ne0$ and $N^2=0$, the latter equality following from Cayley–Hamilton. Its projective class is $[I+\varepsilon N]$, a nonidentity unipotent. All nonidentity unipotents are conjugate over $K$, so they also belong to the image. The conjugators can be taken in $\operatorname{SL}_2(K)$: multiply a $\operatorname{GL}_2(K)$ conjugator by a scalar with the appropriate square to make its determinant one. Finally, $v(1,1)=1$. These cases exhaust $\operatorname{PSL}_2(K)$ and prove the complex theorem. $\square$

## The real trace identity

The real construction substitutes conjugate inputs into the shorter word

$$
w_0=[a[a,b]a^{-1},\,b[b,a]b^{-1}],\qquad
W=w_0(a,bab^{-1}).
$$

This forces the two inputs of $w_0$ to have the same trace, and real conjugacy imposes an additional restriction that makes the bound possible.

**Trace identity.** Let $A,C\in\operatorname{SL}_2(\mathbb R)$ have common trace $x$. Put $U=x^2$ and $p=\operatorname{tr}(AC)$. Then

$$
\begin{aligned}
\operatorname{tr}w_0(A,C)&=P(U,p),\\
P(U,p)&=2+U(p-2)^2(p-1)^2(U-p-2)^3H(U,p),\\
H(U,p)&=U^2(p-1)^2-Up(p^2-2)+p^2.
\end{aligned}
$$

Here is a derivation, so the large factored polynomial can be checked from smaller identities. Write

$$
c=[A,C],\qquad D=A^{-1}C,\qquad
u=AcA^{-1},\qquad v=Cc^{-1}C^{-1},
$$

and set

$$
h=\operatorname{tr}c,\quad d=\operatorname{tr}D,\quad
j=\operatorname{tr}(cD),\quad s=\operatorname{tr}(uv).
$$

The skein and Fricke identities are

$$
\begin{aligned}
\operatorname{tr}(XY)+\operatorname{tr}(XY^{-1})
&=\operatorname{tr}X\operatorname{tr}Y,\\
\operatorname{tr}[X,Y]
&=\alpha^2+\beta^2+\gamma^2-\alpha\beta\gamma-2,
\end{aligned}
$$

where $\alpha=\operatorname{tr}X$, $\beta=\operatorname{tr}Y$ and $\gamma=\operatorname{tr}(XY)$. They give

$$
\begin{aligned}
h&=p^2-Up+2U-2,\\
d&=U-p,\\
j&=U-p(h-1),\\
s&=h^2+d^2+j^2-hdj-2,\\
\operatorname{tr}[u,v]-2&=(s-2)(s+2-h^2).
\end{aligned}
$$

For the formula for $j$, the identity $c^{-1}AC=CA$ first gives $\operatorname{tr}(cAC)=p(h-1)$. Apply the skein identity to $ACA^{-1}$ and $C^{-1}A^{-1}C$: their product is $cD$, and the product of the first with the inverse of the second is $cAC$. Also $uv$ is conjugate to $[c,D]$, which gives the formula for $s$. Expansion now yields

$$
\begin{aligned}
s-2&=U(p-2)(p-1)^2(U-p-2)^2,\\
s+2-h^2&=(p-2)(U-p-2)H(U,p).
\end{aligned}
$$

Multiplying proves the trace identity.

## The restriction imposed by real conjugacy

For $C=BAB^{-1}$, the parameters satisfy

$$
U\ge0,\qquad p\le U-2\quad\text{or}\quad U>4.
$$

To prove this, put $q=\operatorname{tr}[A,B]-2$. The skein identity gives $p=U-2-q$, so it suffices to show $q\ge0$ when $U\le4$. Write

$$
A=\begin{pmatrix}a&b\\c&d\end{pmatrix},\qquad
AB-BA=\begin{pmatrix}k&m\\l&-k\end{pmatrix},\qquad
\delta=a-d.
$$

The determinant and trace identities give

$$
q=k^2+ml,\qquad \delta k+bl+cm=0,\qquad
\delta^2+4bc=U-4.
$$

Consequently,

$$
4b^2q=(2bk-\delta m)^2-(U-4)m^2\ge0.
$$

If $b\ne0$, this proves the claim. If $b=0$, then $\delta^2=U-4\le0$, so $\delta=0$. The upper-right entry of $AB-BA$ is then zero, giving $q=k^2\ge0$. The boundary and degenerate cases are therefore included.

## A uniform lower bound

Apply the trace identity to $A,C=BAB^{-1}$. First suppose $p\le U-2$. If $p<-2$, each term in the defining expression for $H$ is nonnegative and $p^2>0$. Otherwise, put $r=U-p-2\ge0$. Substitution gives

$$
H=(p-1)^2r^2+(p^3-4p+4)r+4>0.
$$

Indeed, $p^3-4p=p(p-2)(p+2)\ge0$ on $[-2,0]\cup[2,\infty)$, while on $[0,2]$,

$$
p^3-4p+4=(p-1)^2(p+2)+2-p>0.
$$

Thus the factored trace gives $P\ge2$ throughout this region.

The remaining region has $U>4$ and $p>U-2$. Set

$$
t=p-2>0,\qquad e=p-U+2,\qquad
0<e<t,\qquad D=4t^2+7t+4.
$$

Then $U=t+4-e$, and

$$
\begin{aligned}
P&=2-(t+4-e)t^2(t+1)^2e^3H,\\
H&=4-(t^3+6t^2+8t+4)e+(t+1)^2e^2.
\end{aligned}
$$

The identity

$$
(4-De)-H=e(t-e)(t+1)^2>0
$$

gives $H\le4-De$. If $H\le0$, again $P\ge2$. If $H>0$, then $4-De>0$, and the exact inequality

$$
27-D^3e^3(4-De)
=(De-3)^2\bigl((De+1)^2+2\bigr)\ge0
$$

yields

$$
2-P\le\frac{27(t+4)t^2(t+1)^2}{D^3}<\frac14.
$$

For the strict final inequality, expand the difference:

$$
\begin{aligned}
D^3-108(t+4)t^2(t+1)^2
={}&64t^6+228t^5+132t^4+43t^3\\
&+348t^2+336t+64>0.
\end{aligned}
$$

Every coefficient is positive. The two regions cover all real inputs, proving $\operatorname{tr}W(A,B)>7/4$ universally.

## The word is nonidentity and omits involutions

One exact evaluation certifies that $W$ is a nonidentity free-group word. Take

$$
A=\begin{pmatrix}1&1\\0&1\end{pmatrix},\qquad
B=\begin{pmatrix}1&0\\2&1\end{pmatrix}.
$$

Here $U=4$ and $p=-2$, so

$$
\operatorname{tr}W(A,B)=P(4,-2)=6045698\ne2.
$$

This single calculation proves nonidentity; the preceding symbolic inequalities establish the bound for every pair.

Now let $X\in\operatorname{SL}_2(\mathbb R)$ lift a nonidentity involution in the projective group. Then $X^2=\pm I$. The case $X^2=I$ forces $X=\pm I$: its minimal polynomial has distinct roots among $1,-1$, and determinant one excludes a pair of opposite eigenvalues. This would make the projective class the identity. Hence $X^2=-I$, and Cayley–Hamilton gives $\operatorname{tr}X=0$.

Every pair of projective inputs has special-linear lifts. If its $W$-value were a nonidentity involution, the matrix word evaluated on those lifts would therefore have trace zero, contradicting the strict bound. For example, the class of

$$
J=\begin{pmatrix}0&-1\\1&0\end{pmatrix}
$$

is omitted. This proves the real theorem. $\square$

## Credit and the formal record

Bandman–Zarhin proved earlier complex surjectivity results [5](https://arxiv.org/abs/1407.3447); the all-word question is stated explicitly by Gordeev–Plotkin [6, Problem 0.1](https://doi.org/10.1142/S0218196726400060). The complex argument above uses the specialization of Schneider–Thom and the scalar polynomial observation of Mushkarov–Nikolov. Differentiating the determinant connects a simple root to a nonidentity projective unipotent.

For the real group, Gordeev–Kunyavskii–Plotkin discuss the surjectivity question and prove that every nonidentity word attains all split semisimple elements. They also show that attaining an involution would imply that every semisimple element is attained [7, Question 2.5 and Proposition 2.6](https://arxiv.org/abs/1808.02303). The explicit word $W$ and its trace bound give the obstruction here. Thom's compact result supplies the third answer to Problem 16.68.

The solution and formalization are by **Nilradical v1.0.0**, using Lean and mathlib. Both the real and complex results have complete proofs with the standard free-group and projective special linear group definitions. The [complex endpoint](../../Kourovka/Problem1668/Complex/WordMaps/Surjectivity.lean) proves `WordMaps.word_surjective` for every algebraically closed field of characteristic zero and specializes it in `WordMaps.complex_word_surjective`. The [real endpoint](../../Kourovka/Problem1668/Real/RealWord/Counterexample.lean) proves `RealWord.word_ne_one`, `RealWord.tr_value_gt_seven_fourths`, `RealWord.word_not_surjective` and `RealWord.exists_nontrivial_nonsurjective_word`.

The real formal proof explicitly omits the displayed class of $J$. Omission of **every** nonidentity involution is the consequence of the trace bound proved above; it is not a separately exported endpoint. The compact case is attributed to Thom and is not formalized in these projects. The [formalization guide](../../Kourovka/Problem1668/README.md) and [verification record](../nilradical-16.68/README.md) identify the sources and their verification evidence. The frozen developments passed independent statement comparison, proof replay and external kernel checking.

**Final-stage approver: Aluna Rizzoli.** Both Lean statements were approved on 21 September 2026 against the frozen source snapshots. That approval concerns their statement scope; it does not constitute refereeing of this exposition or a priority assessment.

## References

1. E. I. Khukhro and V. D. Mazurov (eds.), *The Kourovka Notebook*, 21st ed., September 2026 update, Problem 16.68 (J. Mycielski), p. 102. [Editors' text](https://kourovkanotebookorg.wordpress.com/wp-content/uploads/2026/09/21tkt.pdf).

2. J. Schneider and A. Thom, *Word images in symmetric and unitary groups are dense*, arXiv:1802.09289v1, Lemma 1. Published, with revised title, in *Pacific J. Math.* **311** (2021), 475–504. [Original preprint](https://arxiv.org/abs/1802.09289v1).

3. O. Mushkarov and N. Nikolov, *A Picard little theorem for entire functions of matrices*, *Elem. Math.* **81** (2026), 34–39, Example 1(iii). [DOI](https://doi.org/10.4171/EM/562).

4. A. Thom, *Convergent sequences in discrete groups*, *Canad. Math. Bull.* **56** (2013), 424–433, Corollary 1.2. [Author version](https://arxiv.org/abs/1003.4093).

5. T. Bandman and Yu. G. Zarhin, *Surjectivity of certain word maps on $\operatorname{PSL}(2,\mathbb C)$ and $\operatorname{SL}(2,\mathbb C)$*, *Eur. J. Math.* **2** (2016), 614–643. [Author version](https://arxiv.org/abs/1407.3447).

6. N. Gordeev and E. Plotkin, *The group $\operatorname{SL}_2(\mathbb C)$: Word maps and related topics* (2026), Problem 0.1. [DOI](https://doi.org/10.1142/S0218196726400060).

7. N. Gordeev, B. Kunyavskii and E. Plotkin, *Geometry of word equations in simple algebraic groups over special fields*, *Russian Math. Surveys* **73** (2018), 753–796, Question 2.5 and Proposition 2.6. [Author version](https://arxiv.org/abs/1808.02303).