What is the difference between a map being linear in linear algebra and a map being linear representation in...












0












$begingroup$


What is the difference between a map being linear in linear algebra and a map being linear representation in linear representation theory?



I know from the answers at the back of the book that the following map:



$$(S(t)f)(x) = f (tx),$$ where $S: R rightarrow S(V)$ and $V$ is the subspace of all polynomials with real coefficients and $t in mathbb{R}, f in V.$



Is not a linear representation (but I do not know how?). Even though I found it is linear when I put instead of $f$, $alpha f + beta f$. I know that for the given map to be a linear representation it must be a homomorphism (I am not very sure from this information, is it correct?) so what are the operations that I should consider here for studying homomorphism?



Could anyone clarify this discrepancies for me please?






linear-algebra abstract-algebra representation-theory







share|cite|improve this question









$endgroup$












  • $begingroup$
    What is $R$ in this case? What is $S(V)$?
    $endgroup$
    – Omnomnomnom
    Jan 12 at 17:19










  • $begingroup$
    R is the reals .... S(V) the group of something(which I do not know) over the vector space V @Omnomnomnom
    $endgroup$
    – hopefully
    Jan 12 at 17:25








  • 1




    $begingroup$
    Which book are you using?
    $endgroup$
    – user458276
    Jan 12 at 17:36










  • $begingroup$
    @user458276 "Linear Representations of groups" for Ernest B. Vinberg
    $endgroup$
    – hopefully
    Jan 12 at 17:56
















0












$begingroup$


What is the difference between a map being linear in linear algebra and a map being linear representation in linear representation theory?



I know from the answers at the back of the book that the following map:



$$(S(t)f)(x) = f (tx),$$ where $S: R rightarrow S(V)$ and $V$ is the subspace of all polynomials with real coefficients and $t in mathbb{R}, f in V.$



Is not a linear representation (but I do not know how?). Even though I found it is linear when I put instead of $f$, $alpha f + beta f$. I know that for the given map to be a linear representation it must be a homomorphism (I am not very sure from this information, is it correct?) so what are the operations that I should consider here for studying homomorphism?



Could anyone clarify this discrepancies for me please?






linear-algebra abstract-algebra representation-theory







share|cite|improve this question









$endgroup$












  • $begingroup$
    What is $R$ in this case? What is $S(V)$?
    $endgroup$
    – Omnomnomnom
    Jan 12 at 17:19










  • $begingroup$
    R is the reals .... S(V) the group of something(which I do not know) over the vector space V @Omnomnomnom
    $endgroup$
    – hopefully
    Jan 12 at 17:25








  • 1




    $begingroup$
    Which book are you using?
    $endgroup$
    – user458276
    Jan 12 at 17:36










  • $begingroup$
    @user458276 "Linear Representations of groups" for Ernest B. Vinberg
    $endgroup$
    – hopefully
    Jan 12 at 17:56














0












0








0





$begingroup$


What is the difference between a map being linear in linear algebra and a map being linear representation in linear representation theory?



I know from the answers at the back of the book that the following map:



$$(S(t)f)(x) = f (tx),$$ where $S: R rightarrow S(V)$ and $V$ is the subspace of all polynomials with real coefficients and $t in mathbb{R}, f in V.$



Is not a linear representation (but I do not know how?). Even though I found it is linear when I put instead of $f$, $alpha f + beta f$. I know that for the given map to be a linear representation it must be a homomorphism (I am not very sure from this information, is it correct?) so what are the operations that I should consider here for studying homomorphism?



Could anyone clarify this discrepancies for me please?






linear-algebra abstract-algebra representation-theory







share|cite|improve this question









$endgroup$




What is the difference between a map being linear in linear algebra and a map being linear representation in linear representation theory?



I know from the answers at the back of the book that the following map:



$$(S(t)f)(x) = f (tx),$$ where $S: R rightarrow S(V)$ and $V$ is the subspace of all polynomials with real coefficients and $t in mathbb{R}, f in V.$



Is not a linear representation (but I do not know how?). Even though I found it is linear when I put instead of $f$, $alpha f + beta f$. I know that for the given map to be a linear representation it must be a homomorphism (I am not very sure from this information, is it correct?) so what are the operations that I should consider here for studying homomorphism?



Could anyone clarify this discrepancies for me please?






linear-algebra abstract-algebra representation-theory




linear-algebra abstract-algebra representation-theory






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 12 at 17:11









hopefullyhopefully

183215




183215












  • $begingroup$
    What is $R$ in this case? What is $S(V)$?
    $endgroup$
    – Omnomnomnom
    Jan 12 at 17:19










  • $begingroup$
    R is the reals .... S(V) the group of something(which I do not know) over the vector space V @Omnomnomnom
    $endgroup$
    – hopefully
    Jan 12 at 17:25








  • 1




    $begingroup$
    Which book are you using?
    $endgroup$
    – user458276
    Jan 12 at 17:36










  • $begingroup$
    @user458276 "Linear Representations of groups" for Ernest B. Vinberg
    $endgroup$
    – hopefully
    Jan 12 at 17:56


















  • $begingroup$
    What is $R$ in this case? What is $S(V)$?
    $endgroup$
    – Omnomnomnom
    Jan 12 at 17:19










  • $begingroup$
    R is the reals .... S(V) the group of something(which I do not know) over the vector space V @Omnomnomnom
    $endgroup$
    – hopefully
    Jan 12 at 17:25








  • 1




    $begingroup$
    Which book are you using?
    $endgroup$
    – user458276
    Jan 12 at 17:36










  • $begingroup$
    @user458276 "Linear Representations of groups" for Ernest B. Vinberg
    $endgroup$
    – hopefully
    Jan 12 at 17:56
















$begingroup$
What is $R$ in this case? What is $S(V)$?
$endgroup$
– Omnomnomnom
Jan 12 at 17:19




$begingroup$
What is $R$ in this case? What is $S(V)$?
$endgroup$
– Omnomnomnom
Jan 12 at 17:19












$begingroup$
R is the reals .... S(V) the group of something(which I do not know) over the vector space V @Omnomnomnom
$endgroup$
– hopefully
Jan 12 at 17:25






$begingroup$
R is the reals .... S(V) the group of something(which I do not know) over the vector space V @Omnomnomnom
$endgroup$
– hopefully
Jan 12 at 17:25






1




1




$begingroup$
Which book are you using?
$endgroup$
– user458276
Jan 12 at 17:36




$begingroup$
Which book are you using?
$endgroup$
– user458276
Jan 12 at 17:36












$begingroup$
@user458276 "Linear Representations of groups" for Ernest B. Vinberg
$endgroup$
– hopefully
Jan 12 at 17:56




$begingroup$
@user458276 "Linear Representations of groups" for Ernest B. Vinberg
$endgroup$
– hopefully
Jan 12 at 17:56










1 Answer
1






active

oldest

votes


















1












$begingroup$

For all $t$, $S(t)$ is indeed a linear map; but $tmapsto S(t)$ is not a morphism, because $S(t+t') neq S(t)circ S(t')$ in general.



If you consider $mathbb{R}^*to GL(mathbb{R}[X])$, $tmapsto S(t)$, it will, however be a morphism because $(S(tt')f )(x) = f(tt'x) = f(tcdot (t'x))= S(t)f(t'x) = S(t)(S(t')f)(x) = (S(t)circ S(t') (f))(x)$ so $S(tt') = S(t)circ S(t')$.



A linear representation of a group $G$ is a morphism $Gto GL(V)$, that is, for each $gin G$ you have a linear map $rho_g : Vto V$ (subject to certain conditions). $rho$ is a group morphism, $rho_g$ is linear.



Here $S(t)$ is linear but $tmapsto S(t)$ is not a group morphism from $(mathbb{R},+)$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    What is S(V) here and why $S(t + t^')$$ neq $$S(t) circ S(t^')$, could you give me an example please?
    $endgroup$
    – hopefully
    Jan 12 at 18:01












  • $begingroup$
    Why "$S(t+t') neq S(t)circ S(t')$ in general" , could you give me an example please?
    $endgroup$
    – hopefully
    Jan 12 at 18:08






  • 1




    $begingroup$
    Please think about it and try to see what it would mean for $S(t+t')$ to be $S(t)circ S(t')$. Pick almost any example of polynomial to see
    $endgroup$
    – Max
    Jan 12 at 18:17










  • $begingroup$
    okay thank you so much :)
    $endgroup$
    – hopefully
    Jan 12 at 18:22






  • 1




    $begingroup$
    It's probably the group of invertible linear operatirs (but not $ntimes n$ because $V$ is infinite dimensional)
    $endgroup$
    – Max
    Jan 12 at 18:31












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1 Answer
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active

oldest

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1 Answer
1






active

oldest

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active

oldest

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active

oldest

votes









1












$begingroup$

For all $t$, $S(t)$ is indeed a linear map; but $tmapsto S(t)$ is not a morphism, because $S(t+t') neq S(t)circ S(t')$ in general.



If you consider $mathbb{R}^*to GL(mathbb{R}[X])$, $tmapsto S(t)$, it will, however be a morphism because $(S(tt')f )(x) = f(tt'x) = f(tcdot (t'x))= S(t)f(t'x) = S(t)(S(t')f)(x) = (S(t)circ S(t') (f))(x)$ so $S(tt') = S(t)circ S(t')$.



A linear representation of a group $G$ is a morphism $Gto GL(V)$, that is, for each $gin G$ you have a linear map $rho_g : Vto V$ (subject to certain conditions). $rho$ is a group morphism, $rho_g$ is linear.



Here $S(t)$ is linear but $tmapsto S(t)$ is not a group morphism from $(mathbb{R},+)$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    What is S(V) here and why $S(t + t^')$$ neq $$S(t) circ S(t^')$, could you give me an example please?
    $endgroup$
    – hopefully
    Jan 12 at 18:01












  • $begingroup$
    Why "$S(t+t') neq S(t)circ S(t')$ in general" , could you give me an example please?
    $endgroup$
    – hopefully
    Jan 12 at 18:08






  • 1




    $begingroup$
    Please think about it and try to see what it would mean for $S(t+t')$ to be $S(t)circ S(t')$. Pick almost any example of polynomial to see
    $endgroup$
    – Max
    Jan 12 at 18:17










  • $begingroup$
    okay thank you so much :)
    $endgroup$
    – hopefully
    Jan 12 at 18:22






  • 1




    $begingroup$
    It's probably the group of invertible linear operatirs (but not $ntimes n$ because $V$ is infinite dimensional)
    $endgroup$
    – Max
    Jan 12 at 18:31
















1












$begingroup$

For all $t$, $S(t)$ is indeed a linear map; but $tmapsto S(t)$ is not a morphism, because $S(t+t') neq S(t)circ S(t')$ in general.



If you consider $mathbb{R}^*to GL(mathbb{R}[X])$, $tmapsto S(t)$, it will, however be a morphism because $(S(tt')f )(x) = f(tt'x) = f(tcdot (t'x))= S(t)f(t'x) = S(t)(S(t')f)(x) = (S(t)circ S(t') (f))(x)$ so $S(tt') = S(t)circ S(t')$.



A linear representation of a group $G$ is a morphism $Gto GL(V)$, that is, for each $gin G$ you have a linear map $rho_g : Vto V$ (subject to certain conditions). $rho$ is a group morphism, $rho_g$ is linear.



Here $S(t)$ is linear but $tmapsto S(t)$ is not a group morphism from $(mathbb{R},+)$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    What is S(V) here and why $S(t + t^')$$ neq $$S(t) circ S(t^')$, could you give me an example please?
    $endgroup$
    – hopefully
    Jan 12 at 18:01












  • $begingroup$
    Why "$S(t+t') neq S(t)circ S(t')$ in general" , could you give me an example please?
    $endgroup$
    – hopefully
    Jan 12 at 18:08






  • 1




    $begingroup$
    Please think about it and try to see what it would mean for $S(t+t')$ to be $S(t)circ S(t')$. Pick almost any example of polynomial to see
    $endgroup$
    – Max
    Jan 12 at 18:17










  • $begingroup$
    okay thank you so much :)
    $endgroup$
    – hopefully
    Jan 12 at 18:22






  • 1




    $begingroup$
    It's probably the group of invertible linear operatirs (but not $ntimes n$ because $V$ is infinite dimensional)
    $endgroup$
    – Max
    Jan 12 at 18:31














1












1








1





$begingroup$

For all $t$, $S(t)$ is indeed a linear map; but $tmapsto S(t)$ is not a morphism, because $S(t+t') neq S(t)circ S(t')$ in general.



If you consider $mathbb{R}^*to GL(mathbb{R}[X])$, $tmapsto S(t)$, it will, however be a morphism because $(S(tt')f )(x) = f(tt'x) = f(tcdot (t'x))= S(t)f(t'x) = S(t)(S(t')f)(x) = (S(t)circ S(t') (f))(x)$ so $S(tt') = S(t)circ S(t')$.



A linear representation of a group $G$ is a morphism $Gto GL(V)$, that is, for each $gin G$ you have a linear map $rho_g : Vto V$ (subject to certain conditions). $rho$ is a group morphism, $rho_g$ is linear.



Here $S(t)$ is linear but $tmapsto S(t)$ is not a group morphism from $(mathbb{R},+)$






share|cite|improve this answer









$endgroup$



For all $t$, $S(t)$ is indeed a linear map; but $tmapsto S(t)$ is not a morphism, because $S(t+t') neq S(t)circ S(t')$ in general.



If you consider $mathbb{R}^*to GL(mathbb{R}[X])$, $tmapsto S(t)$, it will, however be a morphism because $(S(tt')f )(x) = f(tt'x) = f(tcdot (t'x))= S(t)f(t'x) = S(t)(S(t')f)(x) = (S(t)circ S(t') (f))(x)$ so $S(tt') = S(t)circ S(t')$.



A linear representation of a group $G$ is a morphism $Gto GL(V)$, that is, for each $gin G$ you have a linear map $rho_g : Vto V$ (subject to certain conditions). $rho$ is a group morphism, $rho_g$ is linear.



Here $S(t)$ is linear but $tmapsto S(t)$ is not a group morphism from $(mathbb{R},+)$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 12 at 17:53









MaxMax

16.2k11144




16.2k11144












  • $begingroup$
    What is S(V) here and why $S(t + t^')$$ neq $$S(t) circ S(t^')$, could you give me an example please?
    $endgroup$
    – hopefully
    Jan 12 at 18:01












  • $begingroup$
    Why "$S(t+t') neq S(t)circ S(t')$ in general" , could you give me an example please?
    $endgroup$
    – hopefully
    Jan 12 at 18:08






  • 1




    $begingroup$
    Please think about it and try to see what it would mean for $S(t+t')$ to be $S(t)circ S(t')$. Pick almost any example of polynomial to see
    $endgroup$
    – Max
    Jan 12 at 18:17










  • $begingroup$
    okay thank you so much :)
    $endgroup$
    – hopefully
    Jan 12 at 18:22






  • 1




    $begingroup$
    It's probably the group of invertible linear operatirs (but not $ntimes n$ because $V$ is infinite dimensional)
    $endgroup$
    – Max
    Jan 12 at 18:31


















  • $begingroup$
    What is S(V) here and why $S(t + t^')$$ neq $$S(t) circ S(t^')$, could you give me an example please?
    $endgroup$
    – hopefully
    Jan 12 at 18:01












  • $begingroup$
    Why "$S(t+t') neq S(t)circ S(t')$ in general" , could you give me an example please?
    $endgroup$
    – hopefully
    Jan 12 at 18:08






  • 1




    $begingroup$
    Please think about it and try to see what it would mean for $S(t+t')$ to be $S(t)circ S(t')$. Pick almost any example of polynomial to see
    $endgroup$
    – Max
    Jan 12 at 18:17










  • $begingroup$
    okay thank you so much :)
    $endgroup$
    – hopefully
    Jan 12 at 18:22






  • 1




    $begingroup$
    It's probably the group of invertible linear operatirs (but not $ntimes n$ because $V$ is infinite dimensional)
    $endgroup$
    – Max
    Jan 12 at 18:31
















$begingroup$
What is S(V) here and why $S(t + t^')$$ neq $$S(t) circ S(t^')$, could you give me an example please?
$endgroup$
– hopefully
Jan 12 at 18:01






$begingroup$
What is S(V) here and why $S(t + t^')$$ neq $$S(t) circ S(t^')$, could you give me an example please?
$endgroup$
– hopefully
Jan 12 at 18:01














$begingroup$
Why "$S(t+t') neq S(t)circ S(t')$ in general" , could you give me an example please?
$endgroup$
– hopefully
Jan 12 at 18:08




$begingroup$
Why "$S(t+t') neq S(t)circ S(t')$ in general" , could you give me an example please?
$endgroup$
– hopefully
Jan 12 at 18:08




1




1




$begingroup$
Please think about it and try to see what it would mean for $S(t+t')$ to be $S(t)circ S(t')$. Pick almost any example of polynomial to see
$endgroup$
– Max
Jan 12 at 18:17




$begingroup$
Please think about it and try to see what it would mean for $S(t+t')$ to be $S(t)circ S(t')$. Pick almost any example of polynomial to see
$endgroup$
– Max
Jan 12 at 18:17












$begingroup$
okay thank you so much :)
$endgroup$
– hopefully
Jan 12 at 18:22




$begingroup$
okay thank you so much :)
$endgroup$
– hopefully
Jan 12 at 18:22




1




1




$begingroup$
It's probably the group of invertible linear operatirs (but not $ntimes n$ because $V$ is infinite dimensional)
$endgroup$
– Max
Jan 12 at 18:31




$begingroup$
It's probably the group of invertible linear operatirs (but not $ntimes n$ because $V$ is infinite dimensional)
$endgroup$
– Max
Jan 12 at 18:31


















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