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I was recently thinking about what it means to put structure on a set. It seems to me that, in my area (representation theory), the two main ways of imposing structure on a set $X$ are:

  • distinguishing certain permutations of $X$ as structure preserving;

and

  • distinguishing certain test functions (here I think of $X \to \mathbb C$, but that's just because it fits my example) are structure respecting.

For example, given a manifold, we can look at the diffeomorphisms among its permutations, or the smooth functions among its test functions; and, given a vector space, we can look at the linear automorphisms among its permutations, or the functionals among its test functions. Under reasonable hypotheses, the test-functions perspective (more 'analysis'-flavoured, maybe) determines the structure; but does the group perspective (more 'geometry' / Erlangen-flavoured)?

A colleague pointed out that, for manifolds, the diffeomorphism group does determine the manifold, in a very strong sense: http://www.ams.org/mathscinet-getitem?mr=693972. Given that success, I was moved to ask: does the linear automorphism group determine the vector space?

Here's one way of making that informal question precise. Suppose that $V_1$ and $V_2$ are vector spaces (over $\mathbb C$, say) with the same underlying set $X$, and that the set of permutations of $X$ that are linear automorphisms for $V_1$ is the same as the analogous set for $V_2$. Then are $V_1$ and $V_2$ isomorphic?

Another colleague, and The Masked Avenger, both thought of the axiom of choice when I asked this question; but I'm not sure I see it. It's just a curiosity, so I have no particular investment in whether answers assume, negate, or avoid choice.

EDIT: Since I think it may look like I am making some implicit assumptions, I clarify that I do not mean to assume that the vector spaces are finite dimensional, or that the putative isomorphism from $V_1$ to $V_2$ must be the identity as a set map of $X$. Thus, for example, Theo Johnson-Freyd's example (https://mathoverflow.net/a/186494/2383) of letting $X$ be $\mathbb C$, and equipping it with both its usual and its 'conjugate' $\mathbb C$-vector space structure, is perfectly OK.

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    $\begingroup$ Great question! Easy observations: the origin is detectable this way, and the one-dimensional and also the finite dimensional subspaces are also revealed by the collection of linear subspaces of the vector space. And the spaces will have the same rational scalar multiplication. $\endgroup$ – Joel David Hamkins Nov 7 '14 at 3:01
  • $\begingroup$ More fun: what finitely generated modules over a PID? $\endgroup$ – Amritanshu Prasad Nov 7 '14 at 3:35
  • $\begingroup$ Structure means you decide what the category of structure-having sets and structure-preserving maps is. "Generalized test functions" determining the structure is precisely the Yoneda lemma. $\endgroup$ – Qiaochu Yuan Nov 7 '14 at 4:13
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    $\begingroup$ One subtlety here is that as phrased your question requires that $GL(V_1)$ and $GL(V_2)$ be in bijection. One may expect that the groups determine the linear structure (i.e. that bijection being an isomorphism of groups) $\endgroup$ – Reimundo Heluani Nov 7 '14 at 10:10
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    $\begingroup$ @ReimundoHeluani, I am imposing the stronger requirement that $\operatorname{GL}(V_1)$ and $\operatorname{GL}(V_2)$ be equal. $\endgroup$ – LSpice Nov 7 '14 at 18:02
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The dimension of $V$ is the least non-negative integer $n$ such that there exist $v_1,\dotsc, v_n$ in $V$ such that there exists a unique $g\in G:=GL(V)$ that fixes each of $v_1,\dotsc,v_n$. So the isomorphism class of $V$ is determined by the group action of $G$ on $V$.

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    $\begingroup$ This is a nice answer, but what if the vector spaces are infinite dimensional? EDIT: no problem, I think; just replace "least non-negative integer" by "least cardinal". $\endgroup$ – LSpice Nov 7 '14 at 4:18
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    $\begingroup$ This does not answer the question posed, which is whether the vector space structure (in the sense that we teach in undergraduate linear algebra) on the underlying set of $V$ can be recovered from how the group $\mathrm{GL}(V)$ sits inside the group of all permutations of the underlying set. $\endgroup$ – Theo Johnson-Freyd Nov 7 '14 at 4:58
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    $\begingroup$ @TheoJohnson-Freyd, actually, it does—the question I asked is I think not the one you have in mind (based on your comment mathoverflow.net/questions/186439/… and answer below mathoverflow.net/a/186494/2383). $\endgroup$ – LSpice Nov 7 '14 at 18:04
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    $\begingroup$ @LSpice: Oh, I'm sorry --- I misunderstood the question. My apologies. $\endgroup$ – Theo Johnson-Freyd Nov 8 '14 at 17:12
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A. Prasad shows that one has access to the dimension $n$ of $V$. Assuming that $n$ is greater than $1$, the addition on $V$ can be recovered from the knowledge of which permutations of $V$ are automorphisms.

First of all, we can recover, for a given subset $F$ of $V$, the span of $F$: it is the set of all vectors $x$ that are fixed by all $g \in GL(V)$ that already fix all the vectors in $F$. In particular, we can use this to determine which subsets are linear subspace; in particular, we know what the zero vector is, which subsets are are $1$-dimensional linear subspaces and which subsets are linear hyperplanes.

Next, we can determine which subsets are affine subspaces and what their translation vector space is: let $H$ be a linear hyperplane of $V$ : we consider the subgroup of $GL(V)$ consisting of the identity and of the automorphisms that fix all the vectors of $H$ but stabilizes no $1$-dimensional linear subspaces that is not included in $H$ (in short: it is the group of transvections with respect to $H$). Then, the affine hyperplanes that are parallel to $H$ are the orbits of $V \setminus H$ under the action of this subgroup. By writing every affine subspace as an intersection of affine hyperplanes, we understand which subsets are affine subspaces and for each such affine subspace we know what its direction is.

From there, we recover addition. First of all, for each vector x, the opposite of $x$ should be $f(x)$ if $f$ is the sole element of the center of $GL(V)$ that satisfies $f^2=id$ and is different from $id_V$, and if no such $f$ exists we have $-x=x$. On the other hand, given non-collinear vectors $x$ and $y$ (non-collinearity is tested by saying that $span(x) \neq span(y)$, $x \neq 0$ and $y \neq 0$), we see that $x+y$ is the sole common point of the affine line that goes through $x$ and is parallel to $span(y)$ and the one that goes through $y$ and is parallel to $span(x)$. Since the addition is associative, we can use this to understand how the addition works on non-zero collinear vectors: given such vectors $x$ and $y$, we choose $z$ that is non-collinear to $x$, and we compute $x+y=((x+z)+y)+(-z)$.

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    $\begingroup$ This is quite interesting—but note that Amri's observation on dimension (mathoverflow.net/a/186442/2383) already answers my question as posed, since the dimension of a vector space (even an infinite-dimensional one, if, as here, we are interested only in abstract vector-space isomorphism) determines it up to isomorphism. $\endgroup$ – LSpice Nov 8 '14 at 4:10
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A simple argument in the finite dimensional case: the commutator subgroup of $GL_n(\mathbb{C})$ is $SL_n(\mathbb{C})$, and the size of the center of $SL_n(\mathbb{C})$ is $n$, as scalar matrices with determinant 1 correspond to roots of unity. (This prompts a question as to whether the general argument can be arranged in a way that uses just the group structure of $GL(V)$, not the action of $GL(V)$ on $V$).

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Vladimir Dotsenko is right:

$GL(V)\cong GL(W)$ iff $V\cong W$.

This is true for any complex (and not only complex) vector spaces. If at least one of the dimensions is finite, then this follows from Dotsenko's argument. If $\dim V$ and $\dim W$ are infinite, this is valid for vector spaces over arbitrary division rings by
Tolstykh's theorem [Theorem 8.3] (the paper contains a complete proof of the theorem and a remark that it "easily follows from general isomorphism theorems proved by O’Meara in [15, Theorem 5.10, Theorem 6.7]").

[15] O. O’Meara, A general isomorphism theory for linear groups, J. Algebra, 44 (1977) 93-142.

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  • $\begingroup$ Did you mean Theorem 8.3? $\endgroup$ – Todd Trimble Nov 9 '14 at 12:13
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No, of course not, at least not the way the question is posed. For example, consider a one-dimensional complex vector space $V \cong \mathbb C^1$, and the complex conjugate vector space $\bar V$, which as an abelian group is the same as $V$, but has the complex conjugate action. I.e. for an element $v\in V$, the action of $\lambda \in \mathbb C$ on $v$ when thought of as an element of $\bar v$ is the action of $\bar \lambda$ on $v$ when thought of as an element of $V$. These two vector space structures are not literally the same structure, but their automorphism groups are literally the same subgroup of the group of all permutations of the underlying set of $V$.

You can come up with a similar example for any vector space over any field that admits automorphisms. To rule out this problem, we can study the similar question over a field like $\mathbb Q$ or $\mathbb R$ that does not admit any automorphisms, or we can ask whether perhaps this is the only failure to detect the vector space structure (i.e. modify the goal, so that the reconstruction is considered a success if the two vector space structures differ only by some pullback of scalar multiplication along a field automorphism).

Over $\mathbb Q$, the action by $\mathbb Q$ is no data — vector spaces are determined by their structure as additive groups — and Clement de Seguins Pazzis has shown in a previous answer that the additive structure can be determined from the group of automorphisms. At least, he shows this when the dimension is greater than $1$.

Actually, Clement's argument fails in dimension $1$. Consider $V = \mathbb Q^1$. Then $\mathrm{GL}(1,\mathbb Q) = \mathbb Q^\times$ acts simply transitively on $V \smallsetminus \{0\}$. The fundamental theorem of arithmetic then identifies $\mathbb Q^\times$ as the free abelian group with the set of primes as a basis. In particular, $\mathbb Q^\times$ itself has many complicated automorphisms — you could completely rearrange the primes, for example. Via any such automorphism, you can define an addition structure on $\mathbb Q^\times \sqcup \{0\}$ whose automorphism group is the group you started with. Here is an explicit example: write a non-zero rational number as $2^a 3^b \frac p q$ with $a,b\in \mathbb N$ and neither $p$ nor $q$ divisible by either $2$ or $3$. Define $\tau(2^a 3^b \frac p q) = 2^b 3^a \frac p q$, and define $+_\tau$ by $x +_\tau y = \tau(\tau(x)+\tau(y))$. Since $\tau$ is an intertwiner for the $\mathbb Q^\times$-action on $\mathbb Q$, this will define a nontrivial addition on $\mathbb Q$ with the same automorphism group.

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    $\begingroup$ I'm confused—$z \mapsto \overline z$ is a linear isomorphism of 'usual' $\mathbb C$ with 'conjugate' $\mathbb C$, so this doesn't seem like a counterexample. (I didn't require that the isomorphism be the identity map of $X$.) $\endgroup$ – LSpice Nov 7 '14 at 18:00
  • $\begingroup$ In view of your last comment under Prasad's answer, shouldn't the opening sentence be edited a little? $\endgroup$ – Todd Trimble Nov 9 '14 at 14:30
  • $\begingroup$ @LSpice Incidentally, I was led astray by your discussion of the underlying set $X$. Normally a "vector space structure" consists of particular maps --- it makes sense, then, to ask whether two given vector space structures on the same set are equal. The Erlangen program, at best, is about actually giving a manifold some structure in this strict sense. Your question, as I now understand, was whether the isomorphism type of the group $\mathrm{GL}(V)$ determines the isomorphism type of the vector space $V$. This is within a family of common questions, but I would argue is not really ... $\endgroup$ – Theo Johnson-Freyd Nov 10 '14 at 0:56
  • $\begingroup$ ... well formed. Indeed, the problem is that there is no functorial construction of $V$ from $\mathrm{GL}(V)$, so that, in particular, bundles of groups isomorphic to $\mathrm{GL}(V)$ do not lift to bundles of vector spaces isomorphic to $V$. (The first example I know of is a bundle whose base space is the suspension of $\mathbb R \mathbb P^2$.) $\endgroup$ – Theo Johnson-Freyd Nov 10 '14 at 1:00
  • $\begingroup$ @ToddTrimble, I'm not sure that I see an appropriate change, but it's obvious that many people found my wording confusing. Could you suggest a specific, improved wording? $\endgroup$ – LSpice Nov 11 '14 at 21:24
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So I asked a colleague and with his permission I'll post an argument using choice and GCH.

We just need to find a bijection between the bases (which exists by choice).

If $V$ and $W$ are finite dimensional, the size of $GL(V)$ already distinguishes them.

Edit to clarify the finite dimensional situation because of all the comments and this even got downvoted: Let $S$ be a set and $V$ and $W$ be two linear vector space structures on $S$. Let us suppose that the group of linear automorphisms of these structures is the same. We are supposed to prove that $V$ is linearly isomorphic to $W$. Lets suppose that they are both finite dimensional vector spaces, say of dimension $n$ and $m$ respectively. If $n=m$ there is nothing to prove. In the contrary case lets proceed by contradiction. The group of automorphisms of $V$ is isomorphic to $GL(n)$. The group of automorphisms of $W$ is isomorphic to $GL(m)$, as $GL(n)\neq GL(m)$ we are done.

Now to the actual non-trivial part of the problem:

If $V$ and $W$ have infinite bases, then we first note that the set of finite parts of an infinite set $X$ has the same cardinality $|X|$ of $X$. The cardinality $|k|$ of our field $k$ (say we have $k=\mathbb{C}$) raised to a finite power is equal to $|k|$. So if $X$ is a basis of the space $V$ we have that $|V| = max(|k|, |X|)$.

Suppose now that the cardinality of the basis $X$ for $V$ is bigger than $|k|$. We see that $|V|=|W|$ implies that $|X|=|Y|$ where $Y$ is a basis of $W$ and we are done (No group of linear transformations was needed here!)

So now we need to check the situation when $|X|=k$ or $X$ is numerable (by CH). That is we want to show that it is absurd to have $X$ numerable and $|Y|=|k|$. Here we will look at the sizes of the groups of automorphisms. As we have that these groups have the sizes of permutations of bases, or $2^{|X|}$ and $2^{|Y|}$. respectively, we have that (in general by GCH) $|X| < |Y|$ implies $2^{|X|} < 2^{|Y|}$ and we are done.

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    $\begingroup$ What it is the notion of 'size' in "the size of $\operatorname{GL}(V)$" in the finite-dimensional case? It can't just mean cardinality, since all $\operatorname{GL}_n(\mathbb C)$'s have the same cardinality. $\endgroup$ – LSpice Nov 7 '14 at 20:03
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    $\begingroup$ Dimension would do for example as mentioned in my comment. You can cook other things for different fields instead of $\mathbb{C}$, but since I used a field of cardinality of the continuum I decided that we could use dimension anyway. $\endgroup$ – Reimundo Heluani Nov 7 '14 at 20:10
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    $\begingroup$ What do you mean by «size» of $GL(V)$ in the finite dimensional case? The center of the group is isomorphic to the multiplicative group of the field, but this is not enough to tell the fields apart in the infite case. $\endgroup$ – Mariano Suárez-Álvarez Nov 7 '14 at 20:25
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    $\begingroup$ How can you tell the dimension of an abstract set of maps? Also, what do you mean by "the groups are not isomorphic on the nose"? In fact, in the question that I posed, they are not just isomorphic but equal. $\endgroup$ – LSpice Nov 7 '14 at 22:19
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    $\begingroup$ I think that I still do not understand the argument. Certainly $\operatorname{GL}_n$ is not equal to $\operatorname{GL}_m$, since they are sets of different kinds of objects; so the question is whether a group abstractly isomorphic to $\operatorname{GL}_n$ can be equal to a group abstractly isomorphic to $\operatorname{GL}_m$, i.e., whether $\operatorname{GL}_n$ and $\operatorname{GL}_m$ (all over $\mathbb C$) are abstractly isomorphic. Although I can believe that they are not, I don't see why it's obvious that they are not. $\endgroup$ – LSpice Nov 8 '14 at 4:08

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