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How do basic results in representation theory change going to positive characteristic fields and/or not algebraically closed?

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6

$begingroup$

I'm working on a project linking graph theory, model theory and representation theory, and am interested in how some results change if we work over fields of positive characteristic or fields that are not algebraically closed.

Let us assume that the characteristic of the field does not divide the order of the group (so by Maschke's Thm the group algebra is still semi-simple).

Consider the following results:

-The sum of squares of dimensions of each irreducible component is the order of the group.

-All irreducible representations of an abelian group are 1-dimensional.

These clearly do not hold over $mathbbR$ - take $mathbbZ_3$ for instance: it has a 1 dimensional representation and a 2 dimensional one in this case. I'm trying to see in the proof of these results where does it use the fact that the underlying field has to be algebraically closed... any ideas?

Thanks!

finite-groups representation-theory abelian-groups

share|cite|improve this question

asked Apr 23 at 8:12

Emil SinclairEmil Sinclair

792

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  By the way, Fulton-Harris has a separate section on real representations of finite groups.
  $endgroup$
  – lisyarus
  Apr 23 at 15:36
  

  
  
  
  

add a comment | 

6

$begingroup$

I'm working on a project linking graph theory, model theory and representation theory, and am interested in how some results change if we work over fields of positive characteristic or fields that are not algebraically closed.

Let us assume that the characteristic of the field does not divide the order of the group (so by Maschke's Thm the group algebra is still semi-simple).

Consider the following results:

-The sum of squares of dimensions of each irreducible component is the order of the group.

-All irreducible representations of an abelian group are 1-dimensional.

These clearly do not hold over $mathbbR$ - take $mathbbZ_3$ for instance: it has a 1 dimensional representation and a 2 dimensional one in this case. I'm trying to see in the proof of these results where does it use the fact that the underlying field has to be algebraically closed... any ideas?

Thanks!

finite-groups representation-theory abelian-groups

share|cite|improve this question

asked Apr 23 at 8:12

Emil SinclairEmil Sinclair

792

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  By the way, Fulton-Harris has a separate section on real representations of finite groups.
  $endgroup$
  – lisyarus
  Apr 23 at 15:36
  

  
  
  
  

add a comment | 

$begingroup$

I'm working on a project linking graph theory, model theory and representation theory, and am interested in how some results change if we work over fields of positive characteristic or fields that are not algebraically closed.

Let us assume that the characteristic of the field does not divide the order of the group (so by Maschke's Thm the group algebra is still semi-simple).

Consider the following results:

-The sum of squares of dimensions of each irreducible component is the order of the group.

-All irreducible representations of an abelian group are 1-dimensional.

These clearly do not hold over $mathbbR$ - take $mathbbZ_3$ for instance: it has a 1 dimensional representation and a 2 dimensional one in this case. I'm trying to see in the proof of these results where does it use the fact that the underlying field has to be algebraically closed... any ideas?

Thanks!

finite-groups representation-theory abelian-groups

share|cite|improve this question

asked Apr 23 at 8:12

Emil SinclairEmil Sinclair

792

$endgroup$

share|cite|improve this question

asked Apr 23 at 8:12

Emil SinclairEmil Sinclair

792

share|cite|improve this question

asked Apr 23 at 8:12

Emil SinclairEmil Sinclair

792

asked Apr 23 at 8:12

Emil SinclairEmil Sinclair

792

asked Apr 23 at 8:12

Emil SinclairEmil Sinclair

792

1 Answer
1

active

oldest

votes

4

$begingroup$

For the second one it's really easy to see where algebraic closure intervenes : you pick an eigenvalue for one of the elements of the group : it might not exist if the field is not algebraically closed !

For the first one you have to recall where this result on the sums of squares comes from : it comes from the fact that the group algebra $k[G]$ decomposes (as an algebra) as a product of $M_n_i(V_i)$ where the $V_i$'s are the irreducible representations, therefore by an analysis of dimensions we get the equality in question.

However if the field isn't algebraically closed, the Artin-Wedderburn theorem only implies that in this decomposition you get matrix algebras over larger division algebras, not over the field itself.

To be more precise on why this first fact relies on the algebraic closedness of $k$, you need to be more specific about which proof you know in the algebraically closed case.

Note, however, that things get much wilder if you remove the hypothesis about the characteristic.

share|cite|improve this answer

edited Apr 23 at 9:54

answered Apr 23 at 8:37

MaxMax

17.2k11144

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  "you get matrix algebras over larger fields" is not true: in general you get matrix algebras over division rings, aka skew fields (e.g. $mathbb H$ may appear if one works over $mathbb R$).
  $endgroup$
  – lisyarus
  Apr 23 at 9:51
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @lisyarus it depends on what you mean by "fields", but sure I will correct to remove any ambiguity
  $endgroup$
  – Max
  Apr 23 at 9:53
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  As far as my knowledge goes, a "field" has a precise ubiquitously-accepted definition...
  $endgroup$
  – lisyarus
  Apr 23 at 9:57
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @lisyarus : some authors disagree that a field has to be commutative (that's where the name "skew field" comes from) - though I agree it is generally understood that they are commutative, hence my modification (and in this context, "skew field" is just a red herring)
  $endgroup$
  – Max
  Apr 23 at 10:10
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @lisyarus : yes I'm sorry I had misread and hadn't seen that "algebraically closed" was there ! Of course if we're away from the characteristic and if we're algebraically closed everything behaves nicely
  $endgroup$
  – Max
  Apr 23 at 12:25
  

  
  
  
  

 | 
show 3 more comments

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4

$begingroup$

For the second one it's really easy to see where algebraic closure intervenes : you pick an eigenvalue for one of the elements of the group : it might not exist if the field is not algebraically closed !

For the first one you have to recall where this result on the sums of squares comes from : it comes from the fact that the group algebra $k[G]$ decomposes (as an algebra) as a product of $M_n_i(V_i)$ where the $V_i$'s are the irreducible representations, therefore by an analysis of dimensions we get the equality in question.

However if the field isn't algebraically closed, the Artin-Wedderburn theorem only implies that in this decomposition you get matrix algebras over larger division algebras, not over the field itself.

To be more precise on why this first fact relies on the algebraic closedness of $k$, you need to be more specific about which proof you know in the algebraically closed case.

Note, however, that things get much wilder if you remove the hypothesis about the characteristic.

share|cite|improve this answer

edited Apr 23 at 9:54

answered Apr 23 at 8:37

MaxMax

17.2k11144

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  "you get matrix algebras over larger fields" is not true: in general you get matrix algebras over division rings, aka skew fields (e.g. $mathbb H$ may appear if one works over $mathbb R$).
  $endgroup$
  – lisyarus
  Apr 23 at 9:51
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @lisyarus it depends on what you mean by "fields", but sure I will correct to remove any ambiguity
  $endgroup$
  – Max
  Apr 23 at 9:53
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  As far as my knowledge goes, a "field" has a precise ubiquitously-accepted definition...
  $endgroup$
  – lisyarus
  Apr 23 at 9:57
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @lisyarus : some authors disagree that a field has to be commutative (that's where the name "skew field" comes from) - though I agree it is generally understood that they are commutative, hence my modification (and in this context, "skew field" is just a red herring)
  $endgroup$
  – Max
  Apr 23 at 10:10
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @lisyarus : yes I'm sorry I had misread and hadn't seen that "algebraically closed" was there ! Of course if we're away from the characteristic and if we're algebraically closed everything behaves nicely
  $endgroup$
  – Max
  Apr 23 at 12:25
  

  
  
  
  

 | 
show 3 more comments

4

$begingroup$

For the second one it's really easy to see where algebraic closure intervenes : you pick an eigenvalue for one of the elements of the group : it might not exist if the field is not algebraically closed !

For the first one you have to recall where this result on the sums of squares comes from : it comes from the fact that the group algebra $k[G]$ decomposes (as an algebra) as a product of $M_n_i(V_i)$ where the $V_i$'s are the irreducible representations, therefore by an analysis of dimensions we get the equality in question.

However if the field isn't algebraically closed, the Artin-Wedderburn theorem only implies that in this decomposition you get matrix algebras over larger division algebras, not over the field itself.

To be more precise on why this first fact relies on the algebraic closedness of $k$, you need to be more specific about which proof you know in the algebraically closed case.

Note, however, that things get much wilder if you remove the hypothesis about the characteristic.

share|cite|improve this answer

edited Apr 23 at 9:54

answered Apr 23 at 8:37

MaxMax

17.2k11144

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  "you get matrix algebras over larger fields" is not true: in general you get matrix algebras over division rings, aka skew fields (e.g. $mathbb H$ may appear if one works over $mathbb R$).
  $endgroup$
  – lisyarus
  Apr 23 at 9:51
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @lisyarus it depends on what you mean by "fields", but sure I will correct to remove any ambiguity
  $endgroup$
  – Max
  Apr 23 at 9:53
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  As far as my knowledge goes, a "field" has a precise ubiquitously-accepted definition...
  $endgroup$
  – lisyarus
  Apr 23 at 9:57
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @lisyarus : some authors disagree that a field has to be commutative (that's where the name "skew field" comes from) - though I agree it is generally understood that they are commutative, hence my modification (and in this context, "skew field" is just a red herring)
  $endgroup$
  – Max
  Apr 23 at 10:10
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @lisyarus : yes I'm sorry I had misread and hadn't seen that "algebraically closed" was there ! Of course if we're away from the characteristic and if we're algebraically closed everything behaves nicely
  $endgroup$
  – Max
  Apr 23 at 12:25
  

  
  
  
  

 | 
show 3 more comments

$begingroup$

For the second one it's really easy to see where algebraic closure intervenes : you pick an eigenvalue for one of the elements of the group : it might not exist if the field is not algebraically closed !

For the first one you have to recall where this result on the sums of squares comes from : it comes from the fact that the group algebra $k[G]$ decomposes (as an algebra) as a product of $M_n_i(V_i)$ where the $V_i$'s are the irreducible representations, therefore by an analysis of dimensions we get the equality in question.

However if the field isn't algebraically closed, the Artin-Wedderburn theorem only implies that in this decomposition you get matrix algebras over larger division algebras, not over the field itself.

To be more precise on why this first fact relies on the algebraic closedness of $k$, you need to be more specific about which proof you know in the algebraically closed case.

Note, however, that things get much wilder if you remove the hypothesis about the characteristic.

share|cite|improve this answer

edited Apr 23 at 9:54

answered Apr 23 at 8:37

MaxMax

17.2k11144

$endgroup$

share|cite|improve this answer

edited Apr 23 at 9:54

answered Apr 23 at 8:37

MaxMax

17.2k11144

answered Apr 23 at 8:37

MaxMax

17.2k11144

answered Apr 23 at 8:37

MaxMax

17.2k11144