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Mar 25, 2022 at 18:04 comment added Sam Nead The embedding is not isometric, and no such embedding exists. I've added a remark to this effect.
Mar 25, 2022 at 18:03 history edited Sam Nead CC BY-SA 4.0
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Mar 25, 2022 at 14:22 vote accept Dmitry Vilensky
Mar 25, 2022 at 4:47 comment added Dmitry Vilensky Thanks for the answer! I don't see if this embedding is isometric - only the isometric embeddings that are interesting for me. For neighboring vertices, everything seems to be fine here, but are the distances to non-neighbors preserved? This requirement led me to write about the uniqueness of the embedding.
Mar 24, 2022 at 19:59 comment added Steven Stadnicki (And the proof that they generate the free group is probably the canonical example of the Ping-pong lemma. )
Mar 24, 2022 at 19:36 history answered Sam Nead CC BY-SA 4.0