# Axiom of choice and automorphisms of vector spaces

A classical exercise in group theory is "Show that if a group has a trivial automorphism group, then it is of order $1$ or $2$." I think that the straightforward solution uses that a exponent two group is a vector space over $\operatorname{GF}(2)$, and therefore has nontrivial automorphisms as soon as its dimension is at least $2$ (simply transposing two basis vectors).

My question is now natural:

Is it possible, without the axiom of choice, to construct a vector space $E$ over $\operatorname{GF}(2)$, different from $\{0\}$ or $\operatorname{GF}(2)$, whose automorphism group $\operatorname{GL}(E)$ is trivial?

• @joriki: thanks for the correction. Mar 20, 2011 at 17:40
• @PseudoNeo: I know that as $\mathbb{F}_2$, but once clarified it is fine. Mar 20, 2011 at 20:55
• @Jim: I'm going to wait until the week is up, and I'm going to check on the Rubins' books on AC when I get a chance. If I can't find an answer, I'll probably ask it there. Mar 24, 2011 at 2:33
• @Arturo: I know, it's fine. I'm giving myself a deadline. I would like to either finish this and perhaps work on a short paper generalizing this sort of result, or go back to my M.Sc. thesis topics (which are quite similar to this, actually, but not exactly the same). Otherwise, this question will haunt me for years to come :-) Mar 28, 2011 at 15:32
• @Asaf: Only goes to show you how little forcing I know, that I can't even tell when it's not there! Mar 28, 2011 at 18:10

Nov. 6th, 2011 After several long months a post on MathOverflow pushed me to reconsider this math, and I have found a mistake. The claim was still true, as shown by Läuchli $\small[1]$, however despite trying to do my best to understand the argument for this specific claim, it eluded me for several days. I then proceeded to construct my own proof, this time errors free - or so I hope. While at it, I am revising the writing style.

Jul. 21st, 2012 While reviewing this proof again it was apparent that its most prominent use in generating such space over the field of two elements fails, as the third lemma implicitly assumed $x+x\neq x$. Now this has been corrected and the proof is truly complete.


I got it! The answer is that you can construct such vector space.

I will assume that you are familiar with ZFA and the construction of permutation models, references can be found in Jech's Set Theory $\small[2, \text{Ch}. 15]$ as well The Axiom of Choice $\small{[3]}$. Any questions are welcomed.

Some notations, if $x\in V$ which is assumed to be a model of ZFC+Atoms then:

• $\sym(x) =\{\pi\in\mathscr{G} \mid \pi x = x\}$, and
• $\fix(x) = \{\pi\in\mathscr{G} \mid \forall y\in x:\ \pi y = y\}$

Definition: Suppose $G$ is a group, $\mathcal{F}\subseteq\mathcal{P}(G)$ is a normal subgroups filter if:

1. $G\in\mathcal{F}$;
2. $H,K$ are subgroups of $G$ such that $H\subseteq K$, then $H\in\mathcal{F}$ implies $K\in\mathcal{F}$;
3. $H,K$ are subgroups of $G$ such that $H,K\in\mathcal{F}$ then $H\cap K\in\mathcal{F}$;
4. ${1}\notin\mathcal{F}$ (non-triviality);
5. For every $H\in\mathcal{F}$ and $g\in G$ then $g^{-1}Hg\in\mathcal{F}$ (normality).

Now consider the normal subgroups-filter $\mathcal{F}$ to be generated by the subgroups $\fix(E)$ for $E\in I$, where $I$ is an ideal of sets of atoms (closed under finite unions, intersections and subsets).

Basics of permutation models:

A permutation model is a transitive subclass of the universe $V$ that for every ordinal $\alpha$, we have $x\in\mathfrak{U}\cap V_{\alpha+1}$ if and only if $x\subseteq\mathfrak{U}\cap V_\alpha$ and $\sym(x)\in\mathcal{F}$.

The latter property is known as being symmetric (with respect to $\mathcal{F}$) and $x$ being in the permutation model means that $x$ is hereditarily symmetric. (Of course at limit stages take limits, and start with the empty set)

If $\mathcal{F}$ was generated by some ideal of sets $I$, then if $x$ is symmetric with respect to $\mathcal{F}$ it means that for some $E\in I$ we have $\fix(E)\subseteq\sym(x)$. In this case we say that $E$ is a support of $x$.

Note that if $E$ is a support of $x$ and $E\subseteq E'$ then $E'$ is also a support of $x$, since $\fix(E')\subseteq\fix(E)$.

Lastly if $f$ is a function in $\mathfrak{U}$ and $\pi$ is a permutation in $G$ then $\pi(f(x)) = (\pi f)(\pi x)$.

Start with $V$ a model of ZFC+Atoms, assuming there are infinitely (countably should be enough) many atoms. $A$ is the set of atoms, endow it with operations that make it a vector space over a field $\mathbb{F}$ (If we only assume countably many atoms, we should assume the field is countable too. Since we are interested in $\mathbb F_2$ this assertion is not a big hassle). Now consider $\mathscr{G}$ the group of all linear automorphisms of $A$, each can be extended uniquely to an automorphism of $V$.

Now consider the normal subgroups-filter $\mathcal{F}$ to be generated by the subgroups $\fix(E)$ for $E\in I$, where $E$ a finite set of atoms. Note that since all the permutations are linear they extend unique to $\span(E)$. In the case where $\mathbb F$, our field, is finite then so is this span.

Let $\mathfrak{U}$ be the permutation model generated by $\mathscr{G}$ and $\mathcal{F}$.

Lemma I: Suppose $E$ is a finite set, and $u,v$ are two vectors such that $v\notin\span(E\cup\{u\})$ and $u\notin\span(E\cup\{v\})$ (in which case we say that $u$ and $v$ are linearly independent over $E$), then there is a permutation which fixes $E$ and permutes $u$ with $v$.

Proof: Without loss of generality we can assume that $E$ is linearly independent, otherwise take a subset of $E$ which is. Since $E\cup\{u,v\}$ is linearly independent we can (in $V$) extend it to a base of $A$, and define a permutation of this base which fixes $E$, permutes $u$ and $v$. This extends uniquely to a linear permutation $\pi\in\fix(E)$ as needed. $\square$

Lemma II: In $\mathfrak{U}$, $A$ is a vector space over $\mathbb F$, and if $W\in\mathfrak{U}$ is a linear proper subspace then $W$ has a finite dimension.

Proof: Suppose $W$ is as above, let $E$ be a support of $W$. If $W\subseteq\span(E)$ then we are done. Otherwise take $u\notin W\cup \span(E)$ and $v\in W\setminus \span(E)$ and permute $u$ and $v$ while fixing $E$, denote the linear permutation with $\pi$. It is clear that $\pi\in\fix(E)$ but $\pi(W)\neq W$, in contradiction. $\square$

Lemma III: If $T\in\mathfrak{U}$ is a linear endomorphism of $A$, and $E$ is a support of $T$ then $x\in\span(E)\Leftrightarrow Tx\in\span(E)$, or $Tx=0$.

Proof: First for $x\in \span(E)$, if $Tx\notin\span(E)$ for some $Tx\neq u\notin\span(E)$ let $\pi$ be a linear automorphism of $A$ which fixes $E$ and $\pi(Tx)=u$. We have, if so:

$$u=\pi(Tx)=(\pi T)(\pi x) = Tx\neq u$$

On the other hand, if $x\notin\span(E)$ and $Tx\in\span(E)$ and if $Tx=Tu$ for some $x\neq u$ for $u\notin\span(E)$, in which case we have that $x+u\neq x$ set $\pi$ an automorphism which fixes $E$ and $\pi(x)=x+u$, now we have: $$Tx = \pi(Tx) = (\pi T)(\pi x) = T(x+u) = Tx+Tu$$ Therefore $Tx=0$.

Otherwise for all $u\neq x$ we have $Tu\neq Tx$. Let $\pi$ be an automorphism fixing $E$ such that $\pi(x)=u$ for some $u\notin\span(E)$, and we have: $$Tx=\pi(Tx)=(\pi T)(\pi x) = Tu$$ this is a contradiction, so this case is impossible. $\square$

Theorem: if $T\in\mathfrak{U}$ is an endomorphism of $A$ then for some $\lambda\in\mathbb F$ we have $Tx=\lambda x$ for all $x\in A$.

Proof:

Assume that $T\neq 0$, so it has a nontrivial image. Let $E$ be a support of $T$. If $\ker(T)$ is nontrivial then it is a proper subspace, thus for a finite set of atoms $B$ we have $\span(B)=\ker(T)$. Without loss of generality, $B\subseteq E$, otherwise $E\cup B$ is also a support of $T$.

For every $v\notin\span(E)$ we have $Tv\notin\span(E)$. However, $E_v = E\cup\{v\}$ is also a support of $T$. Therefore restricting $T$ to $E_v$ yields that $Tv=\lambda v$ for some $\lambda\in\mathbb F$.

Let $v,u\notin\span(E)$ linearly independent over $\span(E)$. We have that: $Tu=\alpha u, Tv=\mu v$, and $v+u\notin\span(E)$ so $T(v+u)=\lambda(v+u)$, for $\lambda\in\mathbb F$. \begin{align} 0&=T(0) \\ &= T(u+v-u-v)\\ &=T(u+v)-Tu-Tv \\ &=\lambda(u+v)-\alpha u-\mu v=(\lambda-\alpha)u+(\lambda-\mu)v \end{align} Since $u,v$ are linearly independent we have $\alpha=\lambda=\mu$. Due to the fact that for every $u,v\notin\span(E)$ we can find $x$ which is linearly independent over $\span(E)$ both with $u$ and $v$ we can conclude that for $x\notin E$ we have $Tx=\lambda x$.

For $v\in\span(E)$ let $x\notin\span(E)$, we have that $v+x\notin\span(E)$ and therefore: \begin{align} Tx &= T(x+u - u)\\ &=T(x+u)-T(u)\\ &=\lambda(x+u)-\lambda u = \lambda x \end{align}

We have concluded, if so that $T=\lambda x$ for some $\lambda\in\mathbb F$. $\square$

Set $\mathbb F=\mathbb F_2$ the field with two elements and we have created ourselves a vector space without any nontrivial automorphisms. However, one last problem remains. This construction was carried out in ZF+Atoms, while we want to have it without atoms. For this simply use the Jech-Sochor embedding theorem $\small[3, \text{Th}. 6.1, \text p. 85]$, and by setting $\alpha>4$ it should be that any endomorphism is transferred to the model of ZF created by this theorem.

(Many thanks to t.b. which helped me translating parts of the original paper of Läuchli.
Additional thanks to Uri Abraham for noting that an operator need not be injective in order to be surjective, resulting a shorter proof.)

Bibliography

1. Läuchli, H. Auswahlaxiom in der Algebra. Commentarii Mathematici Helvetici, vol 37, pp. 1-19.

2. Jech, T. Set Theory, 3rd millennium ed., Springer (2003).

3. Jech, T. The Axiom of Choice. North-Holland (1973).

• Congratulations -- it certainly looks cool :-) I don't know enough about this stuff to understand this proof, but I want to use the opportunity to learn about it, and I know I will have at least two questions when I do: a) how does this relate to Randall Dougherty's answer? b) does it tell us anything about the amount of choice we need to make this impossible? perhaps also for other steps in the chain of implications in my "answer"? Mar 28, 2011 at 8:55
• joriki: The method he specifies is a similar method to this one. The usage of adding Cohen reals and then treating them as atoms is known as symmetric extensions, Jech has an embedding theorem that takes permutation models of ZFA to a symmetric extension of their kernel (not fully, but to a prescribed length in the construction hierarchy). Essentially the answers are very similar in their nature. Mar 28, 2011 at 9:05
• joriki: As for the second question, I'm not sure. I think that assuming $AC_\kappa$ you can ensure that up to cardinality $\kappa$ no counter examples can arise, but from that point onwards it might be open. I'm not sure. Mar 28, 2011 at 9:06
• @Asaf: Do you know whether the Boolean prime ideal theorem holds in your model? Mar 28, 2011 at 14:58
• joriki: I don't know, but I believe that the answer is negative. I'm not sure how to prove that, though. I also think that countable choice fails in this model. Mar 28, 2011 at 15:17

This is not really an answer, more like a set of comments that's too long to put in the comments.

Taking the question literally, whether in the absence of choice such a vector space can be constructed, this thread, in particular the post near the end by Randall Dougherty, answers in the affirmative. (I don't know enough about forcing to check whether the construction works.)

Then the question becomes how much choice is required and whether the non-existence of non-trivial vector spaces over $\mathbb{F}_2$ with trivial automorphism group is equivalent to some more well-known choice-like axiom. In the last post in the thread, Herman Rubin claims that this follows from the Boolean prime ideal theorem and conjectures that it's equivalent to it. I tried for a while to find a proof in either direction but didn't get much further than equipping the set of all ideals of a Boolean algebra with the structure of a vector space over $\mathbb{F}_2$.

The Consequences of the Axiom of Choice website has a number of related results that don't settle the question but at least give a non-trivial upper bound on the amount of choice required: We don't need full choice, since Form 333, the axiom of odd choice, which is weaker than full choice in ZF, is equivalent to [333 A], "In every vector space $B$ over the two element field, every subspace of $B$ has a complementary subspace", which is at least as strong as the result in question (since we just need a complementary subspace for a one-dimensional subspace). [333 B], "Every quotient group of an Abelian group each of whose non-unit elements has order $2$ has a set of representatives", is another equivalent. Here is the original paper by Keremedis from which these results are taken.

Some other related choice-like principles:

[1 BJ] "In every vector space over the two element field, every generating set contains a basis." : This is equivalent to full choice; the proof is in the Keremedis paper linked to above and is rather neat.

Form 28(p): "Every vector space $V$ over $\mathbb{Z}_p$ has the property that every linearly independent subset can be extended to a basis. ($\mathbb{Z}_p$ is the $p$ element field.)"

Form 66: "Every vector space over a field has a basis." (Apparently it's unknown whether this is equivalent to full choice.)

Form 95(F): "Existence of Complementary Subspaces over a Field $F$: If $F$ is a field, then every vector space $V$ over $F$ has the property that if $S \subseteq V$ is a subspace of $V$, then there is a subspace $S' \subseteq V$ such that $S\cap S' = \{0\}$ and $S \cup S'$ generates $V$." (This implies odd choice.)

Form 109: "Every field $F$ and every vector space $V$ over $F$ has the property that each linearly independent set $A \subseteq V$ can be extended to a basis."

Form 429(p): "Every vector space over $\mathbb{Z}_p$ has a basis. ($\mathbb{Z}_p$ is the $p$ element field.)"

(The full table of implications can be viewed at the website by entering the form numbers in the form.)

[Edit:]

Out of my failed attempts at equipping the (non-distributive, non-complemented) lattice of subspaces of $E$ with some structure that would guarantee the existence of a maximal ideal (which left me doubting whether there really is a connection to the Boolean prime ideal theorem), this consideration emerged: We have the following chain of implications and equivalences:

$E$ has a basis $\Rightarrow$ There is an inner product on $E$ $\Rightarrow$ Every subspace of $E$ has a complement $\Rightarrow$ Every finite-dimensional subspace of $E$ has a complement $\Leftrightarrow$ Every one-dimensional subspace of $E$ has a complement $\Rightarrow$ There is a one-dimensional subspace of $E$ that has a complement $\Leftrightarrow$ There is a non-trivial linear functional on $E$ $\Rightarrow$ There is a non-trivial automorphism of $E$.

(Here $E$ in each case stands for "every vector space over $\mathbb{F}_2$ with more than two elements", not just a particular one; else the last implication wouldn't hold.)

I don't see any obvious implications in the other direction except for the ones indicated. It seems likely that different steps along this chain require different levels of choice; perhaps progress towards the overall solution might be made by proving further implications in the other direction or by proving that one of the steps in the chain is equivalent to or implies or is implied by a well-known choice-like principle. The only such connections I know of so far are that "Every subspace of $E$ has a complement" is equivalent to the axiom of odd choice (see above) and of course that the axiom of choice implies "$E$ has a basis", and hence the entire chain.

• joriki: You gave me some insights and some ideas to work with. I've been busting my brain for a few days and got stuck every single time, now I think I know why. I work in a model without choice but with PIT... I'll see what happens in other models. :-) Mar 24, 2011 at 20:41
• @Asaf Karagila: Glad to be of service. "I've been busting my brain for a few days and got stuck every single time" -- that pretty much sums up my situation, too :-) In case you need a vector space structure on the ideals of a Boolean algebra, let me know, but I'm afraid that will be the easier part of any proof... Mar 24, 2011 at 20:48
• @joriki: We need a complement to a 2-dimensional subspace, surely: then we can do a swap in the two-dimensional subspace, and leave the complement alone to get a nontrivial automorphism. Mar 24, 2011 at 20:54
• @Arturo: The problem is that $\mathbb{F}_2$ has only 0,1 scalars so proving a nontrivial linear automorphism (in the Jech construction, at least) is contradictory is somewhat of a pain. Mar 24, 2011 at 20:57
• @Arturo: I think the mystery is resolved by the suboptimal encoding of the implication table in Consequences. The table differentiates between implications provable in ZF and provable in a weaker system; multiple choice doesn't imply AC in the weaker system, so the status of the implication is unknown in the weaker system; and apparently that leads to the overall status of the implication being classified as "unknown", even though the equivalence is provable in ZF. (BTW, here's a link to the Blass paper: math.lsa.umich.edu/~ablass/bases-AC.pdf) Mar 25, 2011 at 4:39