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This is a follow-up to the question Connected vertex-transitive graph with the fixed-point property. In particular, it is based on a comment by user bof.

Let $G = (V,E)$ be a graph with $V$ infinite. Suppose $G$ is vertex-transitive, i.e., for every pair of vertices $u,v \in V$ there is an automorphism $\gamma$ of $G$ for which $\gamma(u) = v$. Is it possible that every automorphism of $G$ has at least one fixed vertex?

The previous question dealt with $V$ finite, for which the answer is no (as long as $\#V \geq 2$) by a simple group theory argument. But here it does not seem possible to use a group theory argument-- at least not so easily-- because there are transitive actions on infinite sets in which every permutation has a fixed point.

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    $\begingroup$ Ivanov (see math.stackexchange.com/a/2147305/35400) constructed f.g. groups $G$ of prime exponent $p$ (for some prime $p$) with a cyclic subgroup $C$ of order $p$ such that every conjugacy class of $G$ meets $G$. Choose a Cayley-Abels graph $X$ for $G/C$. Then every element of $G$ fixes a point in $X$. So this would work, provided the automorphism group of $X$ is reduced to $G$. I'd guess it's possible to ensure this. $\endgroup$
    – YCor
    Commented Apr 19, 2022 at 20:46
  • $\begingroup$ @YCor: presumably that's a typo for "... every conjugacy class of $G$ meets $C$"? $\endgroup$
    – Sam Hopkins
    Commented Apr 19, 2022 at 20:52
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    $\begingroup$ yes, sorry: "every conjugacy class of $G$ meets $C$" (actually in a singleton) $\endgroup$
    – YCor
    Commented Apr 19, 2022 at 22:23
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    $\begingroup$ Comment on connectedness (I first thought you forgot to assume connectedness): if the graph is not connected, then using a fixed-point-free permutation of the set of components, there's a fixed-point-free automorphism. So a graph answering the question has to be connected. $\endgroup$
    – YCor
    Commented Apr 20, 2022 at 8:29

1 Answer 1

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Let $\Gamma$ be the graph whose vertices are the 2-dimensional subspaces of $\mathbb{R}^n$ and two will be adjacent if they intersect in a 1-dimensional subspace. Then $PGL(n, \mathbb{R})$ is in the automorphism group of $\Gamma $. I think it is the full automorphism group but that will need checking.

Now any real matrix is similar to its Jordan normal form, which for real matrices is block diagonal with blocks either lower triangular or having size 2x2. Thus it fixes a vertex of $\Gamma$.

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    $\begingroup$ For finite fields these graphs are called Grassmann graphs: en.m.wikipedia.org/wiki/Grassmann_graph $\endgroup$
    – Sam Hopkins
    Commented Apr 20, 2022 at 13:11
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    $\begingroup$ Provided $PGL$ is really the automorphism group of $\Gamma$ (which seems plausible since it's true in the finite field case; see Wiki page above), I think this example works. Nice! $\endgroup$
    – Sam Hopkins
    Commented Apr 20, 2022 at 13:39
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    $\begingroup$ The Wikipedia entry is wrong as in the finite case the full automorphism group is $P\Gamma L(n,q)$ which is $PGL(n,q)$ extended by field automorphisms. I chose my field to be $\mathbb{R}$ as it doesn’t have any nontrivial field automorphisms. The fundamental theorem of projective geometry says that the group of all collineations of the projective space is $PGL(n, \mathbb{R})$. If you can reconstruct the projective space from the graph then this will imply the result. $\endgroup$
    – Michael Giudici
    Commented Apr 20, 2022 at 13:48

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