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edited May 7 2012 at 19:26 |
Andreas, I believe so, and for free groups with more than 2 generators too. As well as the analogous question for more than 3 elements: for example, if $w=fxf^{-1}gyg^{-1}hzh^{-1}$.
To see if $x=gyg^{-1}hzh^{-1}$, it is enough to study first the cases $x=1$, $y=1$, $z=1$, and if none of these elements is trivial, to see if it is possible to "glue" a sphere out of 3 discs labelled on the boundary with $x$, $y$, and $z$, respecting the labels, and so that the orientations of the discs labelled with $y$ and $z$ were the same, and the orientation of the disc labelled with $x$ was the inverse.
The length of the $1$-skeleton of such a sphere will be $(|x|+|y|+|z|)/2$, so there are finitely many cases to consider.
More details.
If such a "sphere" (more properly it would be called a "combinatorial sphere") glued of 3 faces exists, it is more or less clear that $x$ is the product of a conjugate of $y$ and a conjugate of $z$.
Suppose now $x=gyg^{-1}hzh^{-1}$. Then it is possible to use 3 faces with contour labels $x$, $y$, and $z$, and additional "degenerate" faces with contour labels "$a1a^{-1}1$", "$b1b^{-1}1$", "$a1a^{-1}$", "$b1b^{-1}$", "$a^{-1}1a$", "$b^{-1}1b$", "$111$" to glue a "big" sphere as above, except with additional "degenerate" faces.
Then it is possible to "collapse" all edges labelled with "1". The sphere will "fall apart" into several spheres. Assuming that none of the words $x$, $y$, $z$ is trivial, none of the 3 "essential" faces can be left on its own, so all 3 will be in the same smaller sphere. Other faces in this smaller sphere will be digons with contour labels "$aa^{-1}$" and "$bb^{-1}$". It is possible to "eliminate" this these bigons one by one, leaving a sphere with only 3 faces as desired.
Remark.
Alternatively to combinatorial complexes, you can think in terms of pictures in the sense of Rourke, as in
Colin P. Rourke, Presentations and the trivial group, Topology of low-dimensional manifolds (Berlin), Lecture Notes in Math., vol. 722, Springer, 1979, Proc. Second Sussex Conf., Chelwood Gate, 1977, pp. 134–143. Johannes Huebschmann, Aspherical 2-complexes and an unsettled problem of J. H. C. Whitehead, Math. Ann. 258 (1981), 17–37.
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edited May 7 2012 at 19:09 |
Andreas, I believe so, and for free groups with more than 2 generators too. As well as the analogous question for more than 3 elements: for example, if $w=fxf^{-1}gyg^{-1}hzh^{-1}$.
To see if $x=gyg^{-1}hzh^{-1}$, it is enough to study first the cases $x=1$, $y=1$, $z=1$, and if none of these elements is trivial, to see if it is possible to "glue" a sphere out of 3 discs labelled on the boundary with $x$, $y$, and $z$, respecting the labels, and so that the orientations of the discs labelled with $y$ and $z$ were the same, and the orientation of the disc labelled with $x$ was the inverse.
The length of the $1$-skeleton of such a sphere will be $(|x|+|y|+|z|)/2$, so there are finitely many cases to consider.
More details.
If such a "sphere" (more properly it would be called a "combinatorial sphere") glued of 3 faces exists, it is more or less clear that $x$ is the product of a conjugate of $y$ and a conjugate of $z$.
Suppose now $x=gyg^{-1}hzh^{-1}$. Then it is possible to use 3 faces with contour labels $x$, $y$, and $z$, and additional "degenerate" faces with contour labels "$a1a^{-1}1$", "$b1b^{-1}1$", "$a1a^{-1}$", "$b1b^{-1}$", "$a^{-1}1a$", "$b^{-1}1b$", "$111$" to glue a "big" sphere as above, except with additional "degenerate" faces.
Then it is possible to "collapse" all edges labelled with "1". The sphere will "fall apart" into several spheres. Assuming that none of the words $x$, $y$, $z$ is trivial, none of the 3 "essential" faces can be left on its own, so all 3 will be in the same smaller sphere. Other faces in this smaller sphere will be digons with contour labels "$aa^{-1}$" and "$bb^{-1}$". It is possible to "eliminate" this bigons one by one, leaving a sphere with only 3 faces as desired.
Remark.
Alternatively to combinatorial complexes, you can think in terms of pictures in the sense of Rourke, as in
Colin P. Rourke, Presentations and the trivial group, Topology of low-dimensional manifolds (Berlin), Lecture Notes in Math., vol. 722, Springer, 1979, Proc. Second Sussex Conf., Chelwood Gate, 1977, pp. 134–143. Johannes Huebschmann, Aspherical 2-complexes and an unsettled problem of J. H. C. Whitehead, Math. Ann. 258 (1981), 17–37.
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edited May 7 2012 at 19:02 |
Andreas, I believe so, and for free groups with more than 2 generators too. As well as the analogous question for more than 3 elements: for example, if $w=fxf^{-1}gyg^{-1}hzh^{-1}$.
To see if $x=gyg^{-1}hzh^{-1}$, it is enough to study first the cases $x=1$, $y=1$, $z=1$, and if none of these elements is trivial, to see if it is possible to "glue" a sphere out of 3 discs labelled on the boundary with $x$, $y$, and $z$, respecting the labels, and so that the orientations of the discs labelled with $y$ and $z$ were the same, and the orientation of the disc labelled with $x$ was the inverse.
The length of the $1$-skeleton of such a sphere will be $(|x|+|y|+|z|)/2$, so there are finitely many cases to consider.
UPDATE: more
More details.
If such a "sphere" (more properly it would be called a "combinatorial sphere") glued of 3 faces exists, it is more or less clear that $x$ is the product of a conjugate of $y$ and a conjugate of $z$.
Suppose now $x=gyg^{-1}hzh^{-1}$. Then it is possible to use 3 faces with contour labels $x$, $y$, and $z$, and additional "degenerate" faces with contour labels "$a1a^{-1}1$", "$b1b^{-1}1$", "$a1a^{-1}$", "$b1b^{-1}$", "$a^{-1}1a$", "$b^{-1}1b$", "$111$" to glue a "big" sphere as above, except with additional "degenerate" faces.
Then it is possible to "collapse" all edges labelled with "1". The sphere will "fall apart" into several spheres. Assuming that none of the words $x$, $y$, $z$ is trivial, none of the 3 "essential" faces can be left on its own, so all 3 will be in the same smaller sphere. Other faces in this smaller sphere will be digons with contour labels "$aa^{-1}$" and "$bb^{-1}$".
It is possible to "eliminate" this bigons one by one, leaving a sphere with only 3 faces as desired.
Remark.
Alternatively to combinatorial complexes, you can think in terms of pictures in the sense of Rourke, as in
Colin P. Rourke, Presentations and the trivial group, Topology of low-dimensional manifolds (Berlin), Lecture Notes in Math., vol. 722, Springer, 1979, Proc. Second Sussex Conf., Chelwood Gate, 1977, pp. 134–143. Johannes Huebschmann, Aspherical 2-complexes and an unsettled problem of J. H. C. Whitehead, Math. Ann. 258 (1981), 17–37.
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edited May 7 2012 at 18:55 |
UPDATE: more details. If such a "sphere" (more properly it would be called a "combinatorial sphere") glued of 3 faces exists, it is more or less clear that $x$ is the product of a conjugate of $y$ and a conjugate of $z$. Suppose now $x=gyg^{-1}hzh^{-1}$. Then it is possible to use 3 faces with contour labels $x$, $y$, and $z$, and additional "degenerate" faces with contour labels "$a1a^{-1}1$", "$b1b^{-1}1$", "$a1a^{-1}$", "$b1b^{-1}$", "$a^{-1}1a$", "$b^{-1}1b$", "$111$" to glue a "big" sphere as above, except with additional "degenerate" faces. Then it is possible to "collapse" all edges labelled with "1". The sphere will "fall apart" into several spheres. Assuming that none of the words $x$, $y$, $z$ is trivial, none of the 3 "essential" faces can be left on its own, so all 3 will be in the same smaller sphere. Other faces in this smaller sphere will be digons with contour labels "$aa^{-1}$" and "$bb^{-1}$". It is possible to "eliminate" this bigons one by one, leaving a sphere with only 3 faces as desired.
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answered May 7 2012 at 15:37 |
Andreas, I believe so, and for free groups with more than 2 generators too. As well as the analogous question for more than 3 elements: for example, if $w=fxf^{-1}gyg^{-1}hzh^{-1}$.
To see if $x=gyg^{-1}hzh^{-1}$, it is enough to study first the cases $x=1$, $y=1$, $z=1$, and if none of these elements is trivial, to see if it is possible to "glue" a sphere out of 3 discs labelled on the boundary with $x$, $y$, and $z$, respecting the labels, and so that the orientations of the discs labelled with $y$ and $z$ were the same, and the orientation of the disc labelled with $x$ was the inverse. The length of the $1$-skeleton of such a sphere will be $(|x|+|y|+|z|)/2$, so there are finitely many cases to consider.
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