Commutativity of induction and inflation of representations
$begingroup$
Let $G$ be a finite group, $H$ a subgroup and $N$ a normal subgroup. Let $chi$ be a representation of $HN/N$. Does $$Ind_{HN}^G Inf_{HN/N}^{HN} chi = Inf_{G/N}^G Ind_{HN/N}^{G/N} chi$$ always hold? Here $Ind$ and $Inf$ are the induction and inflation functors.
If this isn't true, are there any simple sufficient conditions for it to hold (possibly with a sketch proof/reference)?
reference-request finite-groups representation-theory
asked Feb 9 '17 at 14:33
Matt BMatt B
2,8281616
$endgroup$
add a comment |
$begingroup$
Let $G$ be a finite group, $H$ a subgroup and $N$ a normal subgroup. Let $chi$ be a representation of $HN/N$. Does $$Ind_{HN}^G Inf_{HN/N}^{HN} chi = Inf_{G/N}^G Ind_{HN/N}^{G/N} chi$$ always hold? Here $Ind$ and $Inf$ are the induction and inflation functors.
If this isn't true, are there any simple sufficient conditions for it to hold (possibly with a sketch proof/reference)?
reference-request finite-groups representation-theory
asked Feb 9 '17 at 14:33
Matt BMatt B
2,8281616
$endgroup$
add a comment |
$begingroup$
Let $G$ be a finite group, $H$ a subgroup and $N$ a normal subgroup. Let $chi$ be a representation of $HN/N$. Does $$Ind_{HN}^G Inf_{HN/N}^{HN} chi = Inf_{G/N}^G Ind_{HN/N}^{G/N} chi$$ always hold? Here $Ind$ and $Inf$ are the induction and inflation functors.
If this isn't true, are there any simple sufficient conditions for it to hold (possibly with a sketch proof/reference)?
reference-request finite-groups representation-theory
asked Feb 9 '17 at 14:33
Matt BMatt B
2,8281616
$endgroup$
Let $G$ be a finite group, $H$ a subgroup and $N$ a normal subgroup. Let $chi$ be a representation of $HN/N$. Does $$Ind_{HN}^G Inf_{HN/N}^{HN} chi = Inf_{G/N}^G Ind_{HN/N}^{G/N} chi$$ always hold? Here $Ind$ and $Inf$ are the induction and inflation functors.
If this isn't true, are there any simple sufficient conditions for it to hold (possibly with a sketch proof/reference)?
reference-request finite-groups representation-theory
reference-request finite-groups representation-theory
asked Feb 9 '17 at 14:33
Matt BMatt B
2,8281616
asked Feb 9 '17 at 14:33
Matt BMatt B
2,8281616
asked Feb 9 '17 at 14:33
Matt BMatt B
2,8281616
asked Feb 9 '17 at 14:33
Matt BMatt B
2,8281616
asked Feb 9 '17 at 14:33
Matt BMatt B
2,8281616
2,8281616
add a comment |
add a comment |
2 Answers
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$begingroup$
Let $tau$ be an irreducible subrepresentation of $Ind_{HN}^GInf_{HN/N}^{HN}chi$, so by Frobenius reciprocity there's a non-zero map $tau|_{HN}rightarrow Inf_{HN/N}^{HN}chi$. So $N$ acts trivially through $tau$, meaning that we can identify $tau|_{HN}$ with a representation of $HN/N$, and doing so we get a non-zero map $tau|_{HN}rightarrowchi$. Using Frobenius reciprocity again, $tau$ is a subrepresentation of $Ind_{HN/N}^{G/N}chi$, and inflating gives a non-zero map $taurightarrow Inf_{G/N}^GInd_{HN/N}^{G/N}chi$. So every irreducible subrepresentation of the left hand side is a subrepresentation of the right hand side. That means that the only way that the two representations can't be equal is if some subrepresentations appear with different multiplicities, but since $|G/HN|=|G|/|HN|=|G||Hcap N|/|HN|=|G/N|/|H/Hcap N|=|G/N|/|HN/N|$, you can compare dimensions to conclude that the representations genuinely are equal.
answered Feb 9 '17 at 18:23
PL.PL.
1,58188
$endgroup$
$begingroup$
Can you elaborate on what happens with the multiplicities? If $dim tau_1 =dim tau_2$ then the dimension argument doesn't work if say $Ind Inf chi = tau_1 oplus 2 tau_2$ as it allows for $Inf Ind chi = 2tau_1 oplus tau_2$.
$endgroup$
– Matt B
Feb 10 '17 at 14:35
add a comment |
$begingroup$
Yes, it holds (as long as your equality sign stands for canonical isomorphism). See Exercise 4.1.11 in Darij Grinberg and Victor Reiner, Hopf algebras and combinatorics, 11 May 2018, arXiv:1409.8356v5 (see the ancillary file for the solution). Note that our $K$, $H$ and $G$ correspond to your $N$, $HN$ and $G$.
answered Feb 9 '17 at 18:58
darij grinbergdarij grinberg
11.2k33167
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add a comment |
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2 Answers
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2 Answers
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$begingroup$
Let $tau$ be an irreducible subrepresentation of $Ind_{HN}^GInf_{HN/N}^{HN}chi$, so by Frobenius reciprocity there's a non-zero map $tau|_{HN}rightarrow Inf_{HN/N}^{HN}chi$. So $N$ acts trivially through $tau$, meaning that we can identify $tau|_{HN}$ with a representation of $HN/N$, and doing so we get a non-zero map $tau|_{HN}rightarrowchi$. Using Frobenius reciprocity again, $tau$ is a subrepresentation of $Ind_{HN/N}^{G/N}chi$, and inflating gives a non-zero map $taurightarrow Inf_{G/N}^GInd_{HN/N}^{G/N}chi$. So every irreducible subrepresentation of the left hand side is a subrepresentation of the right hand side. That means that the only way that the two representations can't be equal is if some subrepresentations appear with different multiplicities, but since $|G/HN|=|G|/|HN|=|G||Hcap N|/|HN|=|G/N|/|H/Hcap N|=|G/N|/|HN/N|$, you can compare dimensions to conclude that the representations genuinely are equal.
answered Feb 9 '17 at 18:23
PL.PL.
1,58188
$endgroup$
$begingroup$
Can you elaborate on what happens with the multiplicities? If $dim tau_1 =dim tau_2$ then the dimension argument doesn't work if say $Ind Inf chi = tau_1 oplus 2 tau_2$ as it allows for $Inf Ind chi = 2tau_1 oplus tau_2$.
$endgroup$
– Matt B
Feb 10 '17 at 14:35
add a comment |
$begingroup$
Let $tau$ be an irreducible subrepresentation of $Ind_{HN}^GInf_{HN/N}^{HN}chi$, so by Frobenius reciprocity there's a non-zero map $tau|_{HN}rightarrow Inf_{HN/N}^{HN}chi$. So $N$ acts trivially through $tau$, meaning that we can identify $tau|_{HN}$ with a representation of $HN/N$, and doing so we get a non-zero map $tau|_{HN}rightarrowchi$. Using Frobenius reciprocity again, $tau$ is a subrepresentation of $Ind_{HN/N}^{G/N}chi$, and inflating gives a non-zero map $taurightarrow Inf_{G/N}^GInd_{HN/N}^{G/N}chi$. So every irreducible subrepresentation of the left hand side is a subrepresentation of the right hand side. That means that the only way that the two representations can't be equal is if some subrepresentations appear with different multiplicities, but since $|G/HN|=|G|/|HN|=|G||Hcap N|/|HN|=|G/N|/|H/Hcap N|=|G/N|/|HN/N|$, you can compare dimensions to conclude that the representations genuinely are equal.
answered Feb 9 '17 at 18:23
PL.PL.
1,58188
$endgroup$
$begingroup$
Can you elaborate on what happens with the multiplicities? If $dim tau_1 =dim tau_2$ then the dimension argument doesn't work if say $Ind Inf chi = tau_1 oplus 2 tau_2$ as it allows for $Inf Ind chi = 2tau_1 oplus tau_2$.
$endgroup$
– Matt B
Feb 10 '17 at 14:35
add a comment |
$begingroup$
Let $tau$ be an irreducible subrepresentation of $Ind_{HN}^GInf_{HN/N}^{HN}chi$, so by Frobenius reciprocity there's a non-zero map $tau|_{HN}rightarrow Inf_{HN/N}^{HN}chi$. So $N$ acts trivially through $tau$, meaning that we can identify $tau|_{HN}$ with a representation of $HN/N$, and doing so we get a non-zero map $tau|_{HN}rightarrowchi$. Using Frobenius reciprocity again, $tau$ is a subrepresentation of $Ind_{HN/N}^{G/N}chi$, and inflating gives a non-zero map $taurightarrow Inf_{G/N}^GInd_{HN/N}^{G/N}chi$. So every irreducible subrepresentation of the left hand side is a subrepresentation of the right hand side. That means that the only way that the two representations can't be equal is if some subrepresentations appear with different multiplicities, but since $|G/HN|=|G|/|HN|=|G||Hcap N|/|HN|=|G/N|/|H/Hcap N|=|G/N|/|HN/N|$, you can compare dimensions to conclude that the representations genuinely are equal.
answered Feb 9 '17 at 18:23
PL.PL.
1,58188
$endgroup$
Let $tau$ be an irreducible subrepresentation of $Ind_{HN}^GInf_{HN/N}^{HN}chi$, so by Frobenius reciprocity there's a non-zero map $tau|_{HN}rightarrow Inf_{HN/N}^{HN}chi$. So $N$ acts trivially through $tau$, meaning that we can identify $tau|_{HN}$ with a representation of $HN/N$, and doing so we get a non-zero map $tau|_{HN}rightarrowchi$. Using Frobenius reciprocity again, $tau$ is a subrepresentation of $Ind_{HN/N}^{G/N}chi$, and inflating gives a non-zero map $taurightarrow Inf_{G/N}^GInd_{HN/N}^{G/N}chi$. So every irreducible subrepresentation of the left hand side is a subrepresentation of the right hand side. That means that the only way that the two representations can't be equal is if some subrepresentations appear with different multiplicities, but since $|G/HN|=|G|/|HN|=|G||Hcap N|/|HN|=|G/N|/|H/Hcap N|=|G/N|/|HN/N|$, you can compare dimensions to conclude that the representations genuinely are equal.
answered Feb 9 '17 at 18:23
PL.PL.
1,58188
answered Feb 9 '17 at 18:23
PL.PL.
1,58188
answered Feb 9 '17 at 18:23
PL.PL.
1,58188
answered Feb 9 '17 at 18:23
PL.PL.
1,58188
1,58188
$begingroup$
Can you elaborate on what happens with the multiplicities? If $dim tau_1 =dim tau_2$ then the dimension argument doesn't work if say $Ind Inf chi = tau_1 oplus 2 tau_2$ as it allows for $Inf Ind chi = 2tau_1 oplus tau_2$.
$endgroup$
– Matt B
Feb 10 '17 at 14:35
add a comment |
$begingroup$
Can you elaborate on what happens with the multiplicities? If $dim tau_1 =dim tau_2$ then the dimension argument doesn't work if say $Ind Inf chi = tau_1 oplus 2 tau_2$ as it allows for $Inf Ind chi = 2tau_1 oplus tau_2$.
$endgroup$
– Matt B
Feb 10 '17 at 14:35
$begingroup$
Can you elaborate on what happens with the multiplicities? If $dim tau_1 =dim tau_2$ then the dimension argument doesn't work if say $Ind Inf chi = tau_1 oplus 2 tau_2$ as it allows for $Inf Ind chi = 2tau_1 oplus tau_2$.
$endgroup$
– Matt B
Feb 10 '17 at 14:35
$begingroup$
Can you elaborate on what happens with the multiplicities? If $dim tau_1 =dim tau_2$ then the dimension argument doesn't work if say $Ind Inf chi = tau_1 oplus 2 tau_2$ as it allows for $Inf Ind chi = 2tau_1 oplus tau_2$.
$endgroup$
– Matt B
Feb 10 '17 at 14:35
add a comment |
$begingroup$
Yes, it holds (as long as your equality sign stands for canonical isomorphism). See Exercise 4.1.11 in Darij Grinberg and Victor Reiner, Hopf algebras and combinatorics, 11 May 2018, arXiv:1409.8356v5 (see the ancillary file for the solution). Note that our $K$, $H$ and $G$ correspond to your $N$, $HN$ and $G$.
answered Feb 9 '17 at 18:58
darij grinbergdarij grinberg
11.2k33167
$endgroup$
add a comment |
$begingroup$
Yes, it holds (as long as your equality sign stands for canonical isomorphism). See Exercise 4.1.11 in Darij Grinberg and Victor Reiner, Hopf algebras and combinatorics, 11 May 2018, arXiv:1409.8356v5 (see the ancillary file for the solution). Note that our $K$, $H$ and $G$ correspond to your $N$, $HN$ and $G$.
answered Feb 9 '17 at 18:58
darij grinbergdarij grinberg
11.2k33167
$endgroup$
add a comment |
$begingroup$
Yes, it holds (as long as your equality sign stands for canonical isomorphism). See Exercise 4.1.11 in Darij Grinberg and Victor Reiner, Hopf algebras and combinatorics, 11 May 2018, arXiv:1409.8356v5 (see the ancillary file for the solution). Note that our $K$, $H$ and $G$ correspond to your $N$, $HN$ and $G$.
answered Feb 9 '17 at 18:58
darij grinbergdarij grinberg
11.2k33167
$endgroup$
Yes, it holds (as long as your equality sign stands for canonical isomorphism). See Exercise 4.1.11 in Darij Grinberg and Victor Reiner, Hopf algebras and combinatorics, 11 May 2018, arXiv:1409.8356v5 (see the ancillary file for the solution). Note that our $K$, $H$ and $G$ correspond to your $N$, $HN$ and $G$.
answered Feb 9 '17 at 18:58
darij grinbergdarij grinberg
11.2k33167
edited Jan 6 at 15:33
answered Feb 9 '17 at 18:58
darij grinbergdarij grinberg
11.2k33167
answered Feb 9 '17 at 18:58
darij grinbergdarij grinberg
11.2k33167
answered Feb 9 '17 at 18:58
darij grinbergdarij grinberg
11.2k33167
11.2k33167
add a comment |
add a comment |
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