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Let $(pi , V_pi)$ be an irreducible unitary representation of the (locally compact) group $G$. Let $V_pi^n = V_pi oplus ldots oplus V_pi$ be the $n$-fold direct sum of $V_pi$ on which we have the (unitary) representation $pi^n$ where $$pi^n(g) big(v_1 , ldots , v_n big) := big(pi(g) v_1 , ldots , pi(g) v_n big) .$$
I strongly suspect, that for fixed $v in V_pi^n$ with $v_1 neq 0$ the map $$operatornamePr colon V_pi^n to V_pi colon (v_1 , ldots , v_n ) longmapsto v_1$$is a $G$-intertwining isomorphism bewtween$$left( pi^n , overlinelangle pi(g) v ~ : g in G rangle right) longleftrightarrow big( pi , V_pi big) .$$
The $G$-intertwining property as well as surjectivity are clear but I'm having trouble showing injectivity. I'm grateful for any hints as well as for a counterexample, if there is one. If this makes it any easier, the solution to that problem would be sufficient for the case $G = operatornameSL_2 big(mathbbRbig)$.

representation-theory lie-groups vector-space-isomorphism

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TargonTargon

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$begingroup$

Let $(pi , V_pi)$ be an irreducible unitary representation of the (locally compact) group $G$. Let $V_pi^n = V_pi oplus ldots oplus V_pi$ be the $n$-fold direct sum of $V_pi$ on which we have the (unitary) representation $pi^n$ where $$pi^n(g) big(v_1 , ldots , v_n big) := big(pi(g) v_1 , ldots , pi(g) v_n big) .$$
I strongly suspect, that for fixed $v in V_pi^n$ with $v_1 neq 0$ the map $$operatornamePr colon V_pi^n to V_pi colon (v_1 , ldots , v_n ) longmapsto v_1$$is a $G$-intertwining isomorphism bewtween$$left( pi^n , overlinelangle pi(g) v ~ : g in G rangle right) longleftrightarrow big( pi , V_pi big) .$$
The $G$-intertwining property as well as surjectivity are clear but I'm having trouble showing injectivity. I'm grateful for any hints as well as for a counterexample, if there is one. If this makes it any easier, the solution to that problem would be sufficient for the case $G = operatornameSL_2 big(mathbbRbig)$.

representation-theory lie-groups vector-space-isomorphism

share|cite|improve this question

asked yesterday

TargonTargon

535

$endgroup$

add a comment | 

$begingroup$

Let $(pi , V_pi)$ be an irreducible unitary representation of the (locally compact) group $G$. Let $V_pi^n = V_pi oplus ldots oplus V_pi$ be the $n$-fold direct sum of $V_pi$ on which we have the (unitary) representation $pi^n$ where $$pi^n(g) big(v_1 , ldots , v_n big) := big(pi(g) v_1 , ldots , pi(g) v_n big) .$$
I strongly suspect, that for fixed $v in V_pi^n$ with $v_1 neq 0$ the map $$operatornamePr colon V_pi^n to V_pi colon (v_1 , ldots , v_n ) longmapsto v_1$$is a $G$-intertwining isomorphism bewtween$$left( pi^n , overlinelangle pi(g) v ~ : g in G rangle right) longleftrightarrow big( pi , V_pi big) .$$
The $G$-intertwining property as well as surjectivity are clear but I'm having trouble showing injectivity. I'm grateful for any hints as well as for a counterexample, if there is one. If this makes it any easier, the solution to that problem would be sufficient for the case $G = operatornameSL_2 big(mathbbRbig)$.

representation-theory lie-groups vector-space-isomorphism

share|cite|improve this question

asked yesterday

TargonTargon

535

$endgroup$

share|cite|improve this question

asked yesterday

TargonTargon

535

share|cite|improve this question

asked yesterday

TargonTargon

535

asked yesterday

TargonTargon

535

asked yesterday

TargonTargon

535

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It's not true in general, even for finite dimensional representations of finite groups.

Take for instance $mathfrakS_3$ acting on $mathbbR^3$, take the irreducible subrepresentation $V=mathbfxin mathbbR^3 mid sum x_i = 0$.

Then take $v =(v_1,v_2) in Voplus V$ with $v_1 = (1,1,-2)$, $v_2 = (1,2,-3)$.

Then with $sigma = (1 2)$ you have $sigma v - v$ which is sent to $sigma v_1 - v_1=0$ but $sigma v - v = (sigma v_1 - v_1, sigma v_2 - v_2) = (0, (1,-1,0))neq 0$

(To construct the counterexample I thought of the easiest way for $displaystylesum_g lambda_g gv_1 = 0$ but $displaystylesum_g lambda_g g$ not $0$ on $V$, then it suffices to find $v_2$ not in the kernel; for this I looked for examples of $g-1$ not being $0$ (that is, $g$ is not in the kernel of the representation), but having a zero (that is, $g$ having a fixed point) - this led me to my example)

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MaxMax

15.1k11143

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0

$begingroup$

It's not true in general, even for finite dimensional representations of finite groups.

Take for instance $mathfrakS_3$ acting on $mathbbR^3$, take the irreducible subrepresentation $V=mathbfxin mathbbR^3 mid sum x_i = 0$.

Then take $v =(v_1,v_2) in Voplus V$ with $v_1 = (1,1,-2)$, $v_2 = (1,2,-3)$.

Then with $sigma = (1 2)$ you have $sigma v - v$ which is sent to $sigma v_1 - v_1=0$ but $sigma v - v = (sigma v_1 - v_1, sigma v_2 - v_2) = (0, (1,-1,0))neq 0$

(To construct the counterexample I thought of the easiest way for $displaystylesum_g lambda_g gv_1 = 0$ but $displaystylesum_g lambda_g g$ not $0$ on $V$, then it suffices to find $v_2$ not in the kernel; for this I looked for examples of $g-1$ not being $0$ (that is, $g$ is not in the kernel of the representation), but having a zero (that is, $g$ having a fixed point) - this led me to my example)

share|cite|improve this answer

answered yesterday

MaxMax

15.1k11143

$endgroup$

add a comment | 

0

$begingroup$

It's not true in general, even for finite dimensional representations of finite groups.

Take for instance $mathfrakS_3$ acting on $mathbbR^3$, take the irreducible subrepresentation $V=mathbfxin mathbbR^3 mid sum x_i = 0$.

Then take $v =(v_1,v_2) in Voplus V$ with $v_1 = (1,1,-2)$, $v_2 = (1,2,-3)$.

Then with $sigma = (1 2)$ you have $sigma v - v$ which is sent to $sigma v_1 - v_1=0$ but $sigma v - v = (sigma v_1 - v_1, sigma v_2 - v_2) = (0, (1,-1,0))neq 0$

(To construct the counterexample I thought of the easiest way for $displaystylesum_g lambda_g gv_1 = 0$ but $displaystylesum_g lambda_g g$ not $0$ on $V$, then it suffices to find $v_2$ not in the kernel; for this I looked for examples of $g-1$ not being $0$ (that is, $g$ is not in the kernel of the representation), but having a zero (that is, $g$ having a fixed point) - this led me to my example)

share|cite|improve this answer

answered yesterday

MaxMax

15.1k11143

$endgroup$

add a comment | 

$begingroup$

It's not true in general, even for finite dimensional representations of finite groups.

Take for instance $mathfrakS_3$ acting on $mathbbR^3$, take the irreducible subrepresentation $V=mathbfxin mathbbR^3 mid sum x_i = 0$.

Then take $v =(v_1,v_2) in Voplus V$ with $v_1 = (1,1,-2)$, $v_2 = (1,2,-3)$.

Then with $sigma = (1 2)$ you have $sigma v - v$ which is sent to $sigma v_1 - v_1=0$ but $sigma v - v = (sigma v_1 - v_1, sigma v_2 - v_2) = (0, (1,-1,0))neq 0$

(To construct the counterexample I thought of the easiest way for $displaystylesum_g lambda_g gv_1 = 0$ but $displaystylesum_g lambda_g g$ not $0$ on $V$, then it suffices to find $v_2$ not in the kernel; for this I looked for examples of $g-1$ not being $0$ (that is, $g$ is not in the kernel of the representation), but having a zero (that is, $g$ having a fixed point) - this led me to my example)

share|cite|improve this answer

answered yesterday

MaxMax

15.1k11143

$endgroup$

share|cite|improve this answer

answered yesterday

MaxMax

15.1k11143

answered yesterday

MaxMax

15.1k11143

answered yesterday

MaxMax

15.1k11143