Bounding $chi(g)$

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For a character $chi$ of a finite group $G$,



$$|chi(g)|leq dim(chi)$$



Obviously the inequality is strict for $g=1$, and if $dim(chi)=1$ also for $gneq 1$.



Is there a better bound for $chi(g)$ if we assume 1) $gneq 1$, and 2) $dim(chi)>1$?




group-theory finite-groups representation-theory characters




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  • Presumably you mean for $chi$ to be the character of a representation, not the representation itself.
    – Eric Wofsey
    Jul 26 at 0:29














up vote
1
down vote

favorite
2












For a character $chi$ of a finite group $G$,



$$|chi(g)|leq dim(chi)$$



Obviously the inequality is strict for $g=1$, and if $dim(chi)=1$ also for $gneq 1$.



Is there a better bound for $chi(g)$ if we assume 1) $gneq 1$, and 2) $dim(chi)>1$?




group-theory finite-groups representation-theory characters




share|cite|improve this question





















  • Presumably you mean for $chi$ to be the character of a representation, not the representation itself.
    – Eric Wofsey
    Jul 26 at 0:29












up vote
1
down vote

favorite
2









up vote
1
down vote

favorite
2






2





For a character $chi$ of a finite group $G$,



$$|chi(g)|leq dim(chi)$$



Obviously the inequality is strict for $g=1$, and if $dim(chi)=1$ also for $gneq 1$.



Is there a better bound for $chi(g)$ if we assume 1) $gneq 1$, and 2) $dim(chi)>1$?




group-theory finite-groups representation-theory characters




share|cite|improve this question













For a character $chi$ of a finite group $G$,



$$|chi(g)|leq dim(chi)$$



Obviously the inequality is strict for $g=1$, and if $dim(chi)=1$ also for $gneq 1$.



Is there a better bound for $chi(g)$ if we assume 1) $gneq 1$, and 2) $dim(chi)>1$?





group-theory finite-groups representation-theory characters





share|cite|improve this question












share|cite|improve this question




share|cite|improve this question








edited Jul 26 at 2:25
























asked Jul 26 at 0:12









Samuel Plath

673213




673213











  • Presumably you mean for $chi$ to be the character of a representation, not the representation itself.
    – Eric Wofsey
    Jul 26 at 0:29
















  • Presumably you mean for $chi$ to be the character of a representation, not the representation itself.
    – Eric Wofsey
    Jul 26 at 0:29















Presumably you mean for $chi$ to be the character of a representation, not the representation itself.
– Eric Wofsey
Jul 26 at 0:29




Presumably you mean for $chi$ to be the character of a representation, not the representation itself.
– Eric Wofsey
Jul 26 at 0:29










1 Answer
1






active

oldest

votes

















up vote
4
down vote



accepted










No. For instance, taking $chi$ to be a sum of $n$ copies of the trivial character for some $n>1$, then $chi(g)=n$ for all $g$.



It also does not help if you assume additionally that $chi$ is irreducible. For instance, let $G=D_8$ and let $chi$ be the character of its usual two-dimensional representation as symmetries of a square. Then $chi$ is an irreducible character of positive dimension. However, if $gin G$ is the element corresponding to a rotation by $180^circ$, then $gneq 1$ and $chi(g)=-2$.



(In fact, more generally, if $gin Z(G)$ and $chi$ is irreducible over $mathbbC$, then $|chi(g)|$ is always equal to $dim chi$. This is because the representation must send $g$ to a scalar multiple of the identity, by Schur's lemma.)






share|cite|improve this answer























  • Your example is for real irreducible character. I am curious to know about complex irred characters.
    – P Vanchinathan
    Jul 26 at 0:30










  • My example is irreducible over $mathbbC$. If it were reducible, it would be a sum of 1-dimensional representations and thus the image of the representation would be abelian.
    – Eric Wofsey
    Jul 26 at 0:33










  • Yes. I got confused with cyclic groups 2-dim real irred representation. Thanks for being kind to my impatient remark.
    – P Vanchinathan
    Jul 26 at 0:45






  • 1




    Perhaps also worth noting that the set $chi(g)$ is called the center of the character, and the center of the group is the intersections of the centers of the characters.
    – Tobias Kildetoft
    Jul 26 at 5:30










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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
4
down vote



accepted










No. For instance, taking $chi$ to be a sum of $n$ copies of the trivial character for some $n>1$, then $chi(g)=n$ for all $g$.



It also does not help if you assume additionally that $chi$ is irreducible. For instance, let $G=D_8$ and let $chi$ be the character of its usual two-dimensional representation as symmetries of a square. Then $chi$ is an irreducible character of positive dimension. However, if $gin G$ is the element corresponding to a rotation by $180^circ$, then $gneq 1$ and $chi(g)=-2$.



(In fact, more generally, if $gin Z(G)$ and $chi$ is irreducible over $mathbbC$, then $|chi(g)|$ is always equal to $dim chi$. This is because the representation must send $g$ to a scalar multiple of the identity, by Schur's lemma.)






share|cite|improve this answer























  • Your example is for real irreducible character. I am curious to know about complex irred characters.
    – P Vanchinathan
    Jul 26 at 0:30










  • My example is irreducible over $mathbbC$. If it were reducible, it would be a sum of 1-dimensional representations and thus the image of the representation would be abelian.
    – Eric Wofsey
    Jul 26 at 0:33










  • Yes. I got confused with cyclic groups 2-dim real irred representation. Thanks for being kind to my impatient remark.
    – P Vanchinathan
    Jul 26 at 0:45






  • 1




    Perhaps also worth noting that the set $chi(g)$ is called the center of the character, and the center of the group is the intersections of the centers of the characters.
    – Tobias Kildetoft
    Jul 26 at 5:30














up vote
4
down vote



accepted










No. For instance, taking $chi$ to be a sum of $n$ copies of the trivial character for some $n>1$, then $chi(g)=n$ for all $g$.



It also does not help if you assume additionally that $chi$ is irreducible. For instance, let $G=D_8$ and let $chi$ be the character of its usual two-dimensional representation as symmetries of a square. Then $chi$ is an irreducible character of positive dimension. However, if $gin G$ is the element corresponding to a rotation by $180^circ$, then $gneq 1$ and $chi(g)=-2$.



(In fact, more generally, if $gin Z(G)$ and $chi$ is irreducible over $mathbbC$, then $|chi(g)|$ is always equal to $dim chi$. This is because the representation must send $g$ to a scalar multiple of the identity, by Schur's lemma.)






share|cite|improve this answer























  • Your example is for real irreducible character. I am curious to know about complex irred characters.
    – P Vanchinathan
    Jul 26 at 0:30










  • My example is irreducible over $mathbbC$. If it were reducible, it would be a sum of 1-dimensional representations and thus the image of the representation would be abelian.
    – Eric Wofsey
    Jul 26 at 0:33










  • Yes. I got confused with cyclic groups 2-dim real irred representation. Thanks for being kind to my impatient remark.
    – P Vanchinathan
    Jul 26 at 0:45






  • 1




    Perhaps also worth noting that the set $chi(g)$ is called the center of the character, and the center of the group is the intersections of the centers of the characters.
    – Tobias Kildetoft
    Jul 26 at 5:30












up vote
4
down vote



accepted







up vote
4
down vote



accepted






No. For instance, taking $chi$ to be a sum of $n$ copies of the trivial character for some $n>1$, then $chi(g)=n$ for all $g$.



It also does not help if you assume additionally that $chi$ is irreducible. For instance, let $G=D_8$ and let $chi$ be the character of its usual two-dimensional representation as symmetries of a square. Then $chi$ is an irreducible character of positive dimension. However, if $gin G$ is the element corresponding to a rotation by $180^circ$, then $gneq 1$ and $chi(g)=-2$.



(In fact, more generally, if $gin Z(G)$ and $chi$ is irreducible over $mathbbC$, then $|chi(g)|$ is always equal to $dim chi$. This is because the representation must send $g$ to a scalar multiple of the identity, by Schur's lemma.)






share|cite|improve this answer















No. For instance, taking $chi$ to be a sum of $n$ copies of the trivial character for some $n>1$, then $chi(g)=n$ for all $g$.



It also does not help if you assume additionally that $chi$ is irreducible. For instance, let $G=D_8$ and let $chi$ be the character of its usual two-dimensional representation as symmetries of a square. Then $chi$ is an irreducible character of positive dimension. However, if $gin G$ is the element corresponding to a rotation by $180^circ$, then $gneq 1$ and $chi(g)=-2$.



(In fact, more generally, if $gin Z(G)$ and $chi$ is irreducible over $mathbbC$, then $|chi(g)|$ is always equal to $dim chi$. This is because the representation must send $g$ to a scalar multiple of the identity, by Schur's lemma.)







share|cite|improve this answer















share|cite|improve this answer



share|cite|improve this answer








edited Jul 26 at 1:01


























answered Jul 26 at 0:28









Eric Wofsey

162k12189300




162k12189300











  • Your example is for real irreducible character. I am curious to know about complex irred characters.
    – P Vanchinathan
    Jul 26 at 0:30










  • My example is irreducible over $mathbbC$. If it were reducible, it would be a sum of 1-dimensional representations and thus the image of the representation would be abelian.
    – Eric Wofsey
    Jul 26 at 0:33










  • Yes. I got confused with cyclic groups 2-dim real irred representation. Thanks for being kind to my impatient remark.
    – P Vanchinathan
    Jul 26 at 0:45






  • 1




    Perhaps also worth noting that the set $chi(g)$ is called the center of the character, and the center of the group is the intersections of the centers of the characters.
    – Tobias Kildetoft
    Jul 26 at 5:30
















  • Your example is for real irreducible character. I am curious to know about complex irred characters.
    – P Vanchinathan
    Jul 26 at 0:30










  • My example is irreducible over $mathbbC$. If it were reducible, it would be a sum of 1-dimensional representations and thus the image of the representation would be abelian.
    – Eric Wofsey
    Jul 26 at 0:33










  • Yes. I got confused with cyclic groups 2-dim real irred representation. Thanks for being kind to my impatient remark.
    – P Vanchinathan
    Jul 26 at 0:45






  • 1




    Perhaps also worth noting that the set $chi(g)$ is called the center of the character, and the center of the group is the intersections of the centers of the characters.
    – Tobias Kildetoft
    Jul 26 at 5:30















Your example is for real irreducible character. I am curious to know about complex irred characters.
– P Vanchinathan
Jul 26 at 0:30




Your example is for real irreducible character. I am curious to know about complex irred characters.
– P Vanchinathan
Jul 26 at 0:30












My example is irreducible over $mathbbC$. If it were reducible, it would be a sum of 1-dimensional representations and thus the image of the representation would be abelian.
– Eric Wofsey
Jul 26 at 0:33




My example is irreducible over $mathbbC$. If it were reducible, it would be a sum of 1-dimensional representations and thus the image of the representation would be abelian.
– Eric Wofsey
Jul 26 at 0:33












Yes. I got confused with cyclic groups 2-dim real irred representation. Thanks for being kind to my impatient remark.
– P Vanchinathan
Jul 26 at 0:45




Yes. I got confused with cyclic groups 2-dim real irred representation. Thanks for being kind to my impatient remark.
– P Vanchinathan
Jul 26 at 0:45




1




1




Perhaps also worth noting that the set $chi(g)$ is called the center of the character, and the center of the group is the intersections of the centers of the characters.
– Tobias Kildetoft
Jul 26 at 5:30




Perhaps also worth noting that the set $chi(g)$ is called the center of the character, and the center of the group is the intersections of the centers of the characters.
– Tobias Kildetoft
Jul 26 at 5:30












 

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