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Can an abelian group be a real vector space in more than one way?

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2

$begingroup$

We know that any $mathbb{Z}$ module structure on an abelian group is unique, and furthermore the same is true for $mathbb{Q}$.

Any complex vector space structure is not unique, we can just compose with an automorphism of $mathbb{C}$ (for example the conjugation map).

However, $mathbb{R}$ has trivial automorphism group, see:

Is an automorphism of the field of real numbers the identity map?

So my question is, given an abelian group made into an $mathbb{R}$ vector space, is this the only way we can do this?

We would get another structure if we can embed $mathbb{R}$ into itself, but I'm not sure if this is possible.

If not, then of all the different $mathbb{R}$ structures, must they all have the same dimension?

I know that both of these are not true for general fields.

Thanks in advance.

(Related: number of differents vector space structures over the same field $mathbb{F}$ on an abelian group)

vector-spaces field-theory modules

share|cite|improve this question

asked 8 hours ago

JamesJames

14799 bronze badges

$endgroup$

add a comment | 

2

$begingroup$

We know that any $mathbb{Z}$ module structure on an abelian group is unique, and furthermore the same is true for $mathbb{Q}$.

Any complex vector space structure is not unique, we can just compose with an automorphism of $mathbb{C}$ (for example the conjugation map).

However, $mathbb{R}$ has trivial automorphism group, see:

Is an automorphism of the field of real numbers the identity map?

So my question is, given an abelian group made into an $mathbb{R}$ vector space, is this the only way we can do this?

We would get another structure if we can embed $mathbb{R}$ into itself, but I'm not sure if this is possible.

If not, then of all the different $mathbb{R}$ structures, must they all have the same dimension?

I know that both of these are not true for general fields.

Thanks in advance.

(Related: number of differents vector space structures over the same field $mathbb{F}$ on an abelian group)

vector-spaces field-theory modules

share|cite|improve this question

asked 8 hours ago

JamesJames

14799 bronze badges

$endgroup$

add a comment | 

$begingroup$

We know that any $mathbb{Z}$ module structure on an abelian group is unique, and furthermore the same is true for $mathbb{Q}$.

Any complex vector space structure is not unique, we can just compose with an automorphism of $mathbb{C}$ (for example the conjugation map).

However, $mathbb{R}$ has trivial automorphism group, see:

Is an automorphism of the field of real numbers the identity map?

So my question is, given an abelian group made into an $mathbb{R}$ vector space, is this the only way we can do this?

We would get another structure if we can embed $mathbb{R}$ into itself, but I'm not sure if this is possible.

If not, then of all the different $mathbb{R}$ structures, must they all have the same dimension?

I know that both of these are not true for general fields.

Thanks in advance.

(Related: number of differents vector space structures over the same field $mathbb{F}$ on an abelian group)

vector-spaces field-theory modules

share|cite|improve this question

asked 8 hours ago

JamesJames

14799 bronze badges

$endgroup$

share|cite|improve this question

asked 8 hours ago

JamesJames

14799 bronze badges

share|cite|improve this question

asked 8 hours ago

JamesJames

14799 bronze badges

asked 8 hours ago

JamesJames

14799 bronze badges

asked 8 hours ago

JamesJames

14799 bronze badges

3 Answers
3

active

oldest

votes

1

$begingroup$

Sure. Note that instead of using an automorphism of $mathbb{R}$, you can use an automorphism of the abelian group. Given $mathbb{R}$-vector spaces $V$ and $W$ any abelian group isomorphism $f:Vto W$, you can define a new $mathbb{R}$-vector space structure on $V$ with scalar multiplication $mathbb{R}times Vto V$ given by $(r,v)mapsto f^{-1}(rf(v))$. This will not be the same as the original scalar multiplication on $V$ unless $f$ is $mathbb{R}$-linear.

In particular, as an abelian group, an $mathbb{R}$-vector space $V$ of dimension $n$ is just a $mathbb{Q}$-vector space of dimension $ncdot 2^{aleph_0}$. If $n>0$, this gives loads of automorphisms of $V$ that are not $mathbb{R}$-linear (for instance, for any nonzero $vin V$ and any irrational $alphainmathbb{R}$, $v$ and $alpha v$ are linearly independent over $mathbb{Q}$ so there is an automorphism that exchanges $v$ and $alpha v$, and this cannot be $mathbb{R}$-linear), and thus loads of isomorphic but distinct $mathbb{R}$-vector space structures on $V$. Or, if $0<n,m leq 2^{aleph_0}$, then $ncdot 2^{aleph_0}=mcdot 2^{aleph_0}=2^{aleph_0}$, so $V$ is actually isomorphic as an abelian group to any $mathbb{R}$-vector space $W$ of dimension $m$, and so $V$ can also be given an $mathbb{R}$-vector space structure of dimension $m$.

share|cite|improve this answer

edited 7 hours ago

answered 7 hours ago

Eric WofseyEric Wofsey

203k1414 gold badges235235 silver badges368368 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Excellent, thanks! :)
  $endgroup$
  – James
  7 hours ago
  

  
  
  
  

add a comment | 

5

$begingroup$

$Bbb{R}$ and $Bbb{R}^2$ are isomorphic as abelian groups but not as real vector spaces.

share|cite|improve this answer

answered 8 hours ago

Rob ArthanRob Arthan

30.1k44 gold badges2929 silver badges6969 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks, and are of different $mathbb{R}$ dimension.
  $endgroup$
  – James
  8 hours ago
  

  
  
  
  

add a comment | 

-1

$begingroup$

You can non-trivially embed the real field into itself. This is because every field extension has a transcendence basis.

So let $S ⊆ ℝ$ be a transcendence basis for $ℝ/ℚ$. Let $F = ℚ(S)$. Then $ℝ/F$ is an algebraic extension. Now every map $S → S$ naturally extends to a field morphism $F → F$ and then to a field morphism $ℝ → ℝ$.

Since $S$ is uncountable, this yields uncountably many field morphisms $ℝ → ℝ$, all of which give rise to different $ℝ$-linear structures on the abelian group $ℝ$ itself.

share|cite|improve this answer

answered 8 hours ago

k.stmk.stm

11.4k33 gold badges2323 silver badges5050 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks. What about, instead of $mathbb{R}$, we take our field to be F, where F has no transcendence basis? I presume this must happen for some fields like $mathbb{Q}$. Must then the structure be unique or the dimension unique?
  $endgroup$
  – James
  7 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This is wrong, a morphism $Fto F$ does not necessarily extend to a morphism $mathbb{R}tomathbb{R}$.
  $endgroup$
  – Eric Wofsey
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Indeed, any ring homomorphism $mathbb{R}tomathbb{R}$ is the identity, even if you do not require surjectivity. If you look at the answers at math.stackexchange.com/questions/449404/…, none of them actually require the map to be surjective.
  $endgroup$
  – Eric Wofsey
  7 hours ago
  

  
  
  
  

add a comment | 

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1

$begingroup$

Sure. Note that instead of using an automorphism of $mathbb{R}$, you can use an automorphism of the abelian group. Given $mathbb{R}$-vector spaces $V$ and $W$ any abelian group isomorphism $f:Vto W$, you can define a new $mathbb{R}$-vector space structure on $V$ with scalar multiplication $mathbb{R}times Vto V$ given by $(r,v)mapsto f^{-1}(rf(v))$. This will not be the same as the original scalar multiplication on $V$ unless $f$ is $mathbb{R}$-linear.

In particular, as an abelian group, an $mathbb{R}$-vector space $V$ of dimension $n$ is just a $mathbb{Q}$-vector space of dimension $ncdot 2^{aleph_0}$. If $n>0$, this gives loads of automorphisms of $V$ that are not $mathbb{R}$-linear (for instance, for any nonzero $vin V$ and any irrational $alphainmathbb{R}$, $v$ and $alpha v$ are linearly independent over $mathbb{Q}$ so there is an automorphism that exchanges $v$ and $alpha v$, and this cannot be $mathbb{R}$-linear), and thus loads of isomorphic but distinct $mathbb{R}$-vector space structures on $V$. Or, if $0<n,m leq 2^{aleph_0}$, then $ncdot 2^{aleph_0}=mcdot 2^{aleph_0}=2^{aleph_0}$, so $V$ is actually isomorphic as an abelian group to any $mathbb{R}$-vector space $W$ of dimension $m$, and so $V$ can also be given an $mathbb{R}$-vector space structure of dimension $m$.

share|cite|improve this answer

edited 7 hours ago

answered 7 hours ago

Eric WofseyEric Wofsey

203k1414 gold badges235235 silver badges368368 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Excellent, thanks! :)
  $endgroup$
  – James
  7 hours ago
  

  
  
  
  

add a comment | 

1

$begingroup$

Sure. Note that instead of using an automorphism of $mathbb{R}$, you can use an automorphism of the abelian group. Given $mathbb{R}$-vector spaces $V$ and $W$ any abelian group isomorphism $f:Vto W$, you can define a new $mathbb{R}$-vector space structure on $V$ with scalar multiplication $mathbb{R}times Vto V$ given by $(r,v)mapsto f^{-1}(rf(v))$. This will not be the same as the original scalar multiplication on $V$ unless $f$ is $mathbb{R}$-linear.

In particular, as an abelian group, an $mathbb{R}$-vector space $V$ of dimension $n$ is just a $mathbb{Q}$-vector space of dimension $ncdot 2^{aleph_0}$. If $n>0$, this gives loads of automorphisms of $V$ that are not $mathbb{R}$-linear (for instance, for any nonzero $vin V$ and any irrational $alphainmathbb{R}$, $v$ and $alpha v$ are linearly independent over $mathbb{Q}$ so there is an automorphism that exchanges $v$ and $alpha v$, and this cannot be $mathbb{R}$-linear), and thus loads of isomorphic but distinct $mathbb{R}$-vector space structures on $V$. Or, if $0<n,m leq 2^{aleph_0}$, then $ncdot 2^{aleph_0}=mcdot 2^{aleph_0}=2^{aleph_0}$, so $V$ is actually isomorphic as an abelian group to any $mathbb{R}$-vector space $W$ of dimension $m$, and so $V$ can also be given an $mathbb{R}$-vector space structure of dimension $m$.

share|cite|improve this answer

edited 7 hours ago

answered 7 hours ago

Eric WofseyEric Wofsey

203k1414 gold badges235235 silver badges368368 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Excellent, thanks! :)
  $endgroup$
  – James
  7 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

Sure. Note that instead of using an automorphism of $mathbb{R}$, you can use an automorphism of the abelian group. Given $mathbb{R}$-vector spaces $V$ and $W$ any abelian group isomorphism $f:Vto W$, you can define a new $mathbb{R}$-vector space structure on $V$ with scalar multiplication $mathbb{R}times Vto V$ given by $(r,v)mapsto f^{-1}(rf(v))$. This will not be the same as the original scalar multiplication on $V$ unless $f$ is $mathbb{R}$-linear.

In particular, as an abelian group, an $mathbb{R}$-vector space $V$ of dimension $n$ is just a $mathbb{Q}$-vector space of dimension $ncdot 2^{aleph_0}$. If $n>0$, this gives loads of automorphisms of $V$ that are not $mathbb{R}$-linear (for instance, for any nonzero $vin V$ and any irrational $alphainmathbb{R}$, $v$ and $alpha v$ are linearly independent over $mathbb{Q}$ so there is an automorphism that exchanges $v$ and $alpha v$, and this cannot be $mathbb{R}$-linear), and thus loads of isomorphic but distinct $mathbb{R}$-vector space structures on $V$. Or, if $0<n,m leq 2^{aleph_0}$, then $ncdot 2^{aleph_0}=mcdot 2^{aleph_0}=2^{aleph_0}$, so $V$ is actually isomorphic as an abelian group to any $mathbb{R}$-vector space $W$ of dimension $m$, and so $V$ can also be given an $mathbb{R}$-vector space structure of dimension $m$.

share|cite|improve this answer

edited 7 hours ago

answered 7 hours ago

Eric WofseyEric Wofsey

203k1414 gold badges235235 silver badges368368 bronze badges

$endgroup$

share|cite|improve this answer

edited 7 hours ago

answered 7 hours ago

Eric WofseyEric Wofsey

203k1414 gold badges235235 silver badges368368 bronze badges

answered 7 hours ago

Eric WofseyEric Wofsey

203k1414 gold badges235235 silver badges368368 bronze badges

answered 7 hours ago

Eric WofseyEric Wofsey

203k1414 gold badges235235 silver badges368368 bronze badges

5

$begingroup$

$Bbb{R}$ and $Bbb{R}^2$ are isomorphic as abelian groups but not as real vector spaces.

share|cite|improve this answer

answered 8 hours ago

Rob ArthanRob Arthan

30.1k44 gold badges2929 silver badges6969 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks, and are of different $mathbb{R}$ dimension.
  $endgroup$
  – James
  8 hours ago
  

  
  
  
  

add a comment | 

5

$begingroup$

$Bbb{R}$ and $Bbb{R}^2$ are isomorphic as abelian groups but not as real vector spaces.

share|cite|improve this answer

answered 8 hours ago

Rob ArthanRob Arthan

30.1k44 gold badges2929 silver badges6969 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks, and are of different $mathbb{R}$ dimension.
  $endgroup$
  – James
  8 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

$Bbb{R}$ and $Bbb{R}^2$ are isomorphic as abelian groups but not as real vector spaces.

share|cite|improve this answer

answered 8 hours ago

Rob ArthanRob Arthan

30.1k44 gold badges2929 silver badges6969 bronze badges

$endgroup$

share|cite|improve this answer

answered 8 hours ago

Rob ArthanRob Arthan

30.1k44 gold badges2929 silver badges6969 bronze badges

answered 8 hours ago

Rob ArthanRob Arthan

30.1k44 gold badges2929 silver badges6969 bronze badges

answered 8 hours ago

Rob ArthanRob Arthan

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-1

$begingroup$

You can non-trivially embed the real field into itself. This is because every field extension has a transcendence basis.

So let $S ⊆ ℝ$ be a transcendence basis for $ℝ/ℚ$. Let $F = ℚ(S)$. Then $ℝ/F$ is an algebraic extension. Now every map $S → S$ naturally extends to a field morphism $F → F$ and then to a field morphism $ℝ → ℝ$.

Since $S$ is uncountable, this yields uncountably many field morphisms $ℝ → ℝ$, all of which give rise to different $ℝ$-linear structures on the abelian group $ℝ$ itself.

share|cite|improve this answer

answered 8 hours ago

k.stmk.stm

11.4k33 gold badges2323 silver badges5050 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks. What about, instead of $mathbb{R}$, we take our field to be F, where F has no transcendence basis? I presume this must happen for some fields like $mathbb{Q}$. Must then the structure be unique or the dimension unique?
  $endgroup$
  – James
  7 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This is wrong, a morphism $Fto F$ does not necessarily extend to a morphism $mathbb{R}tomathbb{R}$.
  $endgroup$
  – Eric Wofsey
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Indeed, any ring homomorphism $mathbb{R}tomathbb{R}$ is the identity, even if you do not require surjectivity. If you look at the answers at math.stackexchange.com/questions/449404/…, none of them actually require the map to be surjective.
  $endgroup$
  – Eric Wofsey
  7 hours ago
  

  
  
  
  

add a comment | 

-1

$begingroup$

You can non-trivially embed the real field into itself. This is because every field extension has a transcendence basis.

So let $S ⊆ ℝ$ be a transcendence basis for $ℝ/ℚ$. Let $F = ℚ(S)$. Then $ℝ/F$ is an algebraic extension. Now every map $S → S$ naturally extends to a field morphism $F → F$ and then to a field morphism $ℝ → ℝ$.

Since $S$ is uncountable, this yields uncountably many field morphisms $ℝ → ℝ$, all of which give rise to different $ℝ$-linear structures on the abelian group $ℝ$ itself.

share|cite|improve this answer

answered 8 hours ago

k.stmk.stm

11.4k33 gold badges2323 silver badges5050 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks. What about, instead of $mathbb{R}$, we take our field to be F, where F has no transcendence basis? I presume this must happen for some fields like $mathbb{Q}$. Must then the structure be unique or the dimension unique?
  $endgroup$
  – James
  7 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This is wrong, a morphism $Fto F$ does not necessarily extend to a morphism $mathbb{R}tomathbb{R}$.
  $endgroup$
  – Eric Wofsey
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Indeed, any ring homomorphism $mathbb{R}tomathbb{R}$ is the identity, even if you do not require surjectivity. If you look at the answers at math.stackexchange.com/questions/449404/…, none of them actually require the map to be surjective.
  $endgroup$
  – Eric Wofsey
  7 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

You can non-trivially embed the real field into itself. This is because every field extension has a transcendence basis.

So let $S ⊆ ℝ$ be a transcendence basis for $ℝ/ℚ$. Let $F = ℚ(S)$. Then $ℝ/F$ is an algebraic extension. Now every map $S → S$ naturally extends to a field morphism $F → F$ and then to a field morphism $ℝ → ℝ$.

Since $S$ is uncountable, this yields uncountably many field morphisms $ℝ → ℝ$, all of which give rise to different $ℝ$-linear structures on the abelian group $ℝ$ itself.

share|cite|improve this answer

answered 8 hours ago

k.stmk.stm

11.4k33 gold badges2323 silver badges5050 bronze badges

$endgroup$

share|cite|improve this answer

answered 8 hours ago

k.stmk.stm

11.4k33 gold badges2323 silver badges5050 bronze badges

answered 8 hours ago

k.stmk.stm

11.4k33 gold badges2323 silver badges5050 bronze badges

answered 8 hours ago

k.stmk.stm

11.4k33 gold badges2323 silver badges5050 bronze badges