Mixing
What is the difference between a map being linear in linear algebra and a map being linear representation in...
$begingroup$
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
asked Jan 12 at 17:11
hopefullyhopefully
240114
$endgroup$
add a comment |
$begingroup$
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
asked Jan 12 at 17:11
hopefullyhopefully
240114
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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
asked Jan 12 at 17:11
hopefullyhopefully
240114
$endgroup$
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
asked Jan 12 at 17:11
hopefullyhopefully
240114
asked Jan 12 at 17:11
hopefullyhopefully
240114
asked Jan 12 at 17:11
hopefullyhopefully
240114
asked Jan 12 at 17:11
hopefullyhopefully
240114
asked Jan 12 at 17:11
hopefullyhopefully
240114
240114
$begingroup$
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
add a comment |
$begingroup$
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
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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 $(mathbb{R},+)$
answered Jan 12 at 17:53
MaxMax
14.4k11142
$endgroup$
$begingroup$
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
|
show 4 more comments
Your Answer
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1 Answer
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1 Answer
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$begingroup$
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 $(mathbb{R},+)$
answered Jan 12 at 17:53
MaxMax
14.4k11142
$endgroup$
$begingroup$
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
|
show 4 more comments
$begingroup$
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 $(mathbb{R},+)$
answered Jan 12 at 17:53
MaxMax
14.4k11142
$endgroup$
$begingroup$
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
|
show 4 more comments
$begingroup$
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 $(mathbb{R},+)$
answered Jan 12 at 17:53
MaxMax
14.4k11142
$endgroup$
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 $(mathbb{R},+)$
answered Jan 12 at 17:53
MaxMax
14.4k11142
answered Jan 12 at 17:53
MaxMax
14.4k11142
answered Jan 12 at 17:53
MaxMax
14.4k11142
answered Jan 12 at 17:53
MaxMax
14.4k11142
14.4k11142
$begingroup$
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
|
show 4 more comments
$begingroup$
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
|
show 4 more comments
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$begingroup$
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