Do “$K/k$ twisted” representations exist?
$begingroup$
Given $k$-representations $V,W$ of a group $G$, where $k$ is a field, $K/k$ a field extension, if we have $Votimes_k Kcong Wotimes_k K$ as $K$-representations, do we have that $Vcong W$?
Being more specific, what about in the case of $V,W$ irreps, $G$ a finite group, with $K/k$ finite and galois?
In characteristic $0$, with character theory, the question can be rephrased as: if the characters of $V$ and $W$ agree, then are $V$ and $W$ isomorphic over their field of definition?
Any reference for these questions would be much appreciated.
abstract-algebra representation-theory extension-field characters
$endgroup$
add a comment |
$begingroup$
Given $k$-representations $V,W$ of a group $G$, where $k$ is a field, $K/k$ a field extension, if we have $Votimes_k Kcong Wotimes_k K$ as $K$-representations, do we have that $Vcong W$?
Being more specific, what about in the case of $V,W$ irreps, $G$ a finite group, with $K/k$ finite and galois?
In characteristic $0$, with character theory, the question can be rephrased as: if the characters of $V$ and $W$ agree, then are $V$ and $W$ isomorphic over their field of definition?
Any reference for these questions would be much appreciated.
abstract-algebra representation-theory extension-field characters
$endgroup$
1
$begingroup$
In general, the functor $$- otimes_k K : k[G]mathrm{-Mod} to K[G]mathrm{-Mod}$$ doesn't reflect isomorphisms (even if this is not explained by the existence of irreps that may not be absolutely irreducible). $$ $$ For algebras instead of modules, see math.stackexchange.com/questions/2578592. Notice that for commutative rings, $- otimes_R S : Rtext{-Mod} to Stext{-Mod}$ reflects isomorphisms if $R to S$ is faithfully flat (see here).
$endgroup$
– Watson
Dec 24 '18 at 12:56
2
$begingroup$
You may want to look at Galois cohomology, namely 1.3.6 here : the twists of $(rho, V)$ should be classified by $$H^1( mathrm{Gal}(K/k) ; mathrm{Aut}_K(rho otimes_k K) ).$$
$endgroup$
– Watson
Dec 24 '18 at 12:56
add a comment |
$begingroup$
Given $k$-representations $V,W$ of a group $G$, where $k$ is a field, $K/k$ a field extension, if we have $Votimes_k Kcong Wotimes_k K$ as $K$-representations, do we have that $Vcong W$?
Being more specific, what about in the case of $V,W$ irreps, $G$ a finite group, with $K/k$ finite and galois?
In characteristic $0$, with character theory, the question can be rephrased as: if the characters of $V$ and $W$ agree, then are $V$ and $W$ isomorphic over their field of definition?
Any reference for these questions would be much appreciated.
abstract-algebra representation-theory extension-field characters
$endgroup$
Given $k$-representations $V,W$ of a group $G$, where $k$ is a field, $K/k$ a field extension, if we have $Votimes_k Kcong Wotimes_k K$ as $K$-representations, do we have that $Vcong W$?
Being more specific, what about in the case of $V,W$ irreps, $G$ a finite group, with $K/k$ finite and galois?
In characteristic $0$, with character theory, the question can be rephrased as: if the characters of $V$ and $W$ agree, then are $V$ and $W$ isomorphic over their field of definition?
Any reference for these questions would be much appreciated.
abstract-algebra representation-theory extension-field characters
abstract-algebra representation-theory extension-field characters
asked Dec 23 '18 at 22:52
user277182user277182
518212
518212
1
$begingroup$
In general, the functor $$- otimes_k K : k[G]mathrm{-Mod} to K[G]mathrm{-Mod}$$ doesn't reflect isomorphisms (even if this is not explained by the existence of irreps that may not be absolutely irreducible). $$ $$ For algebras instead of modules, see math.stackexchange.com/questions/2578592. Notice that for commutative rings, $- otimes_R S : Rtext{-Mod} to Stext{-Mod}$ reflects isomorphisms if $R to S$ is faithfully flat (see here).
$endgroup$
– Watson
Dec 24 '18 at 12:56
2
$begingroup$
You may want to look at Galois cohomology, namely 1.3.6 here : the twists of $(rho, V)$ should be classified by $$H^1( mathrm{Gal}(K/k) ; mathrm{Aut}_K(rho otimes_k K) ).$$
$endgroup$
– Watson
Dec 24 '18 at 12:56
add a comment |
1
$begingroup$
In general, the functor $$- otimes_k K : k[G]mathrm{-Mod} to K[G]mathrm{-Mod}$$ doesn't reflect isomorphisms (even if this is not explained by the existence of irreps that may not be absolutely irreducible). $$ $$ For algebras instead of modules, see math.stackexchange.com/questions/2578592. Notice that for commutative rings, $- otimes_R S : Rtext{-Mod} to Stext{-Mod}$ reflects isomorphisms if $R to S$ is faithfully flat (see here).
$endgroup$
– Watson
Dec 24 '18 at 12:56
2
$begingroup$
You may want to look at Galois cohomology, namely 1.3.6 here : the twists of $(rho, V)$ should be classified by $$H^1( mathrm{Gal}(K/k) ; mathrm{Aut}_K(rho otimes_k K) ).$$
$endgroup$
– Watson
Dec 24 '18 at 12:56
1
1
$begingroup$
In general, the functor $$- otimes_k K : k[G]mathrm{-Mod} to K[G]mathrm{-Mod}$$ doesn't reflect isomorphisms (even if this is not explained by the existence of irreps that may not be absolutely irreducible). $$ $$ For algebras instead of modules, see math.stackexchange.com/questions/2578592. Notice that for commutative rings, $- otimes_R S : Rtext{-Mod} to Stext{-Mod}$ reflects isomorphisms if $R to S$ is faithfully flat (see here).
$endgroup$
– Watson
Dec 24 '18 at 12:56
$begingroup$
In general, the functor $$- otimes_k K : k[G]mathrm{-Mod} to K[G]mathrm{-Mod}$$ doesn't reflect isomorphisms (even if this is not explained by the existence of irreps that may not be absolutely irreducible). $$ $$ For algebras instead of modules, see math.stackexchange.com/questions/2578592. Notice that for commutative rings, $- otimes_R S : Rtext{-Mod} to Stext{-Mod}$ reflects isomorphisms if $R to S$ is faithfully flat (see here).
$endgroup$
– Watson
Dec 24 '18 at 12:56
2
2
$begingroup$
You may want to look at Galois cohomology, namely 1.3.6 here : the twists of $(rho, V)$ should be classified by $$H^1( mathrm{Gal}(K/k) ; mathrm{Aut}_K(rho otimes_k K) ).$$
$endgroup$
– Watson
Dec 24 '18 at 12:56
$begingroup$
You may want to look at Galois cohomology, namely 1.3.6 here : the twists of $(rho, V)$ should be classified by $$H^1( mathrm{Gal}(K/k) ; mathrm{Aut}_K(rho otimes_k K) ).$$
$endgroup$
– Watson
Dec 24 '18 at 12:56
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
In fact, for any field extension $K/k$ and any $k$-algebra $A$, if $M$ and $N$ are finite dimensional $A$-modules such that $Motimes_kKcong Notimes_kK$ as $Aotimes_kK$-modules, then $Mcong N$ as $A$-modules. This is the Noether-Deuring Theorem (see, for example, (19.25) in Lam's "First Course in Noncommutative Rings"; he assumes $A$ is finite-dimensional, but that's not essential).
For finite field extensions the proof is very short. If $Motimes_kKcong Notimes_kK$ as $Aotimes_kK$-modules, then, restricting to $A$, $Motimes_kKcong Notimes_kK$ as $A$-modules. But if $[K:k]=n$, then as $A$-modules $Motimes_kKcong M^n$, the direct sum of $n$ copies of $M$, and $Notimes_kKcong N^n$. So $M^ncong N^n$. Now just apply the Krull-Schmidt Theorem to deduce that$Mcong N$.
$endgroup$
add a comment |
$begingroup$
In the case of finite group with semisimple group algebra (i.e. $mathrm{char}, k = 0$ or $mathrm{char}, k$ is coprime to $|G|$), the claim should be true, i.e. $Votimes_k K simeq Wotimes_k K$ implies $V simeq W$.
In the case $mathrm{char}, k = 0$, the reason is character theory: Namely, the orthogonality of characters still works (see e.g. here the version for non-alg. closed field). So after decomposing $V$ and $W$ into irreps, comparing the decomposition comes down to computing $langle chi_i, chi rangle$ where $chi$ is the common character of $V, W$ (which is invariant under extension of scalars) and $chi_i$ runs over irreducible chracters, as in the case $k=mathbb{C}$.
Note that this argument does not suffice on its own in positive characteristic coprime to $|G|$, i.e. the other semisimple case, since the character orthogonality formula contains the number $d$ of absolutely irreducible components of an irrep, and it's not obvious that this is necessarily nonzero modulo $mathrm{char}, k$.
However, there is a Brauer-Nesbitt Theorem that can be used: It states that two semisimple representations of $G$ are isomorphic whenever their characteristic polynomials (meaning: functions taking $g$ to the char. polynomial of $rho(g)$ when $rho$ is a representation) agree. Since the char. polynomials are invariant under extension of scalars, the same conclusion as in the previous case follows. (See e.g. these notes of Gabor Wiese, Thm 2.4.6 for a more-general version of the Brauer-Nesbitt theorem.)
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3050775%2fdo-k-k-twisted-representations-exist%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
In fact, for any field extension $K/k$ and any $k$-algebra $A$, if $M$ and $N$ are finite dimensional $A$-modules such that $Motimes_kKcong Notimes_kK$ as $Aotimes_kK$-modules, then $Mcong N$ as $A$-modules. This is the Noether-Deuring Theorem (see, for example, (19.25) in Lam's "First Course in Noncommutative Rings"; he assumes $A$ is finite-dimensional, but that's not essential).
For finite field extensions the proof is very short. If $Motimes_kKcong Notimes_kK$ as $Aotimes_kK$-modules, then, restricting to $A$, $Motimes_kKcong Notimes_kK$ as $A$-modules. But if $[K:k]=n$, then as $A$-modules $Motimes_kKcong M^n$, the direct sum of $n$ copies of $M$, and $Notimes_kKcong N^n$. So $M^ncong N^n$. Now just apply the Krull-Schmidt Theorem to deduce that$Mcong N$.
$endgroup$
add a comment |
$begingroup$
In fact, for any field extension $K/k$ and any $k$-algebra $A$, if $M$ and $N$ are finite dimensional $A$-modules such that $Motimes_kKcong Notimes_kK$ as $Aotimes_kK$-modules, then $Mcong N$ as $A$-modules. This is the Noether-Deuring Theorem (see, for example, (19.25) in Lam's "First Course in Noncommutative Rings"; he assumes $A$ is finite-dimensional, but that's not essential).
For finite field extensions the proof is very short. If $Motimes_kKcong Notimes_kK$ as $Aotimes_kK$-modules, then, restricting to $A$, $Motimes_kKcong Notimes_kK$ as $A$-modules. But if $[K:k]=n$, then as $A$-modules $Motimes_kKcong M^n$, the direct sum of $n$ copies of $M$, and $Notimes_kKcong N^n$. So $M^ncong N^n$. Now just apply the Krull-Schmidt Theorem to deduce that$Mcong N$.
$endgroup$
add a comment |
$begingroup$
In fact, for any field extension $K/k$ and any $k$-algebra $A$, if $M$ and $N$ are finite dimensional $A$-modules such that $Motimes_kKcong Notimes_kK$ as $Aotimes_kK$-modules, then $Mcong N$ as $A$-modules. This is the Noether-Deuring Theorem (see, for example, (19.25) in Lam's "First Course in Noncommutative Rings"; he assumes $A$ is finite-dimensional, but that's not essential).
For finite field extensions the proof is very short. If $Motimes_kKcong Notimes_kK$ as $Aotimes_kK$-modules, then, restricting to $A$, $Motimes_kKcong Notimes_kK$ as $A$-modules. But if $[K:k]=n$, then as $A$-modules $Motimes_kKcong M^n$, the direct sum of $n$ copies of $M$, and $Notimes_kKcong N^n$. So $M^ncong N^n$. Now just apply the Krull-Schmidt Theorem to deduce that$Mcong N$.
$endgroup$
In fact, for any field extension $K/k$ and any $k$-algebra $A$, if $M$ and $N$ are finite dimensional $A$-modules such that $Motimes_kKcong Notimes_kK$ as $Aotimes_kK$-modules, then $Mcong N$ as $A$-modules. This is the Noether-Deuring Theorem (see, for example, (19.25) in Lam's "First Course in Noncommutative Rings"; he assumes $A$ is finite-dimensional, but that's not essential).
For finite field extensions the proof is very short. If $Motimes_kKcong Notimes_kK$ as $Aotimes_kK$-modules, then, restricting to $A$, $Motimes_kKcong Notimes_kK$ as $A$-modules. But if $[K:k]=n$, then as $A$-modules $Motimes_kKcong M^n$, the direct sum of $n$ copies of $M$, and $Notimes_kKcong N^n$. So $M^ncong N^n$. Now just apply the Krull-Schmidt Theorem to deduce that$Mcong N$.
edited Feb 8 at 13:48
answered Feb 8 at 12:00
Jeremy RickardJeremy Rickard
16.9k11746
16.9k11746
add a comment |
add a comment |
$begingroup$
In the case of finite group with semisimple group algebra (i.e. $mathrm{char}, k = 0$ or $mathrm{char}, k$ is coprime to $|G|$), the claim should be true, i.e. $Votimes_k K simeq Wotimes_k K$ implies $V simeq W$.
In the case $mathrm{char}, k = 0$, the reason is character theory: Namely, the orthogonality of characters still works (see e.g. here the version for non-alg. closed field). So after decomposing $V$ and $W$ into irreps, comparing the decomposition comes down to computing $langle chi_i, chi rangle$ where $chi$ is the common character of $V, W$ (which is invariant under extension of scalars) and $chi_i$ runs over irreducible chracters, as in the case $k=mathbb{C}$.
Note that this argument does not suffice on its own in positive characteristic coprime to $|G|$, i.e. the other semisimple case, since the character orthogonality formula contains the number $d$ of absolutely irreducible components of an irrep, and it's not obvious that this is necessarily nonzero modulo $mathrm{char}, k$.
However, there is a Brauer-Nesbitt Theorem that can be used: It states that two semisimple representations of $G$ are isomorphic whenever their characteristic polynomials (meaning: functions taking $g$ to the char. polynomial of $rho(g)$ when $rho$ is a representation) agree. Since the char. polynomials are invariant under extension of scalars, the same conclusion as in the previous case follows. (See e.g. these notes of Gabor Wiese, Thm 2.4.6 for a more-general version of the Brauer-Nesbitt theorem.)
$endgroup$
add a comment |
$begingroup$
In the case of finite group with semisimple group algebra (i.e. $mathrm{char}, k = 0$ or $mathrm{char}, k$ is coprime to $|G|$), the claim should be true, i.e. $Votimes_k K simeq Wotimes_k K$ implies $V simeq W$.
In the case $mathrm{char}, k = 0$, the reason is character theory: Namely, the orthogonality of characters still works (see e.g. here the version for non-alg. closed field). So after decomposing $V$ and $W$ into irreps, comparing the decomposition comes down to computing $langle chi_i, chi rangle$ where $chi$ is the common character of $V, W$ (which is invariant under extension of scalars) and $chi_i$ runs over irreducible chracters, as in the case $k=mathbb{C}$.
Note that this argument does not suffice on its own in positive characteristic coprime to $|G|$, i.e. the other semisimple case, since the character orthogonality formula contains the number $d$ of absolutely irreducible components of an irrep, and it's not obvious that this is necessarily nonzero modulo $mathrm{char}, k$.
However, there is a Brauer-Nesbitt Theorem that can be used: It states that two semisimple representations of $G$ are isomorphic whenever their characteristic polynomials (meaning: functions taking $g$ to the char. polynomial of $rho(g)$ when $rho$ is a representation) agree. Since the char. polynomials are invariant under extension of scalars, the same conclusion as in the previous case follows. (See e.g. these notes of Gabor Wiese, Thm 2.4.6 for a more-general version of the Brauer-Nesbitt theorem.)
$endgroup$
add a comment |
$begingroup$
In the case of finite group with semisimple group algebra (i.e. $mathrm{char}, k = 0$ or $mathrm{char}, k$ is coprime to $|G|$), the claim should be true, i.e. $Votimes_k K simeq Wotimes_k K$ implies $V simeq W$.
In the case $mathrm{char}, k = 0$, the reason is character theory: Namely, the orthogonality of characters still works (see e.g. here the version for non-alg. closed field). So after decomposing $V$ and $W$ into irreps, comparing the decomposition comes down to computing $langle chi_i, chi rangle$ where $chi$ is the common character of $V, W$ (which is invariant under extension of scalars) and $chi_i$ runs over irreducible chracters, as in the case $k=mathbb{C}$.
Note that this argument does not suffice on its own in positive characteristic coprime to $|G|$, i.e. the other semisimple case, since the character orthogonality formula contains the number $d$ of absolutely irreducible components of an irrep, and it's not obvious that this is necessarily nonzero modulo $mathrm{char}, k$.
However, there is a Brauer-Nesbitt Theorem that can be used: It states that two semisimple representations of $G$ are isomorphic whenever their characteristic polynomials (meaning: functions taking $g$ to the char. polynomial of $rho(g)$ when $rho$ is a representation) agree. Since the char. polynomials are invariant under extension of scalars, the same conclusion as in the previous case follows. (See e.g. these notes of Gabor Wiese, Thm 2.4.6 for a more-general version of the Brauer-Nesbitt theorem.)
$endgroup$
In the case of finite group with semisimple group algebra (i.e. $mathrm{char}, k = 0$ or $mathrm{char}, k$ is coprime to $|G|$), the claim should be true, i.e. $Votimes_k K simeq Wotimes_k K$ implies $V simeq W$.
In the case $mathrm{char}, k = 0$, the reason is character theory: Namely, the orthogonality of characters still works (see e.g. here the version for non-alg. closed field). So after decomposing $V$ and $W$ into irreps, comparing the decomposition comes down to computing $langle chi_i, chi rangle$ where $chi$ is the common character of $V, W$ (which is invariant under extension of scalars) and $chi_i$ runs over irreducible chracters, as in the case $k=mathbb{C}$.
Note that this argument does not suffice on its own in positive characteristic coprime to $|G|$, i.e. the other semisimple case, since the character orthogonality formula contains the number $d$ of absolutely irreducible components of an irrep, and it's not obvious that this is necessarily nonzero modulo $mathrm{char}, k$.
However, there is a Brauer-Nesbitt Theorem that can be used: It states that two semisimple representations of $G$ are isomorphic whenever their characteristic polynomials (meaning: functions taking $g$ to the char. polynomial of $rho(g)$ when $rho$ is a representation) agree. Since the char. polynomials are invariant under extension of scalars, the same conclusion as in the previous case follows. (See e.g. these notes of Gabor Wiese, Thm 2.4.6 for a more-general version of the Brauer-Nesbitt theorem.)
answered Feb 8 at 5:51
Pavel ČoupekPavel Čoupek
4,57611126
4,57611126
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3050775%2fdo-k-k-twisted-representations-exist%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
1
$begingroup$
In general, the functor $$- otimes_k K : k[G]mathrm{-Mod} to K[G]mathrm{-Mod}$$ doesn't reflect isomorphisms (even if this is not explained by the existence of irreps that may not be absolutely irreducible). $$ $$ For algebras instead of modules, see math.stackexchange.com/questions/2578592. Notice that for commutative rings, $- otimes_R S : Rtext{-Mod} to Stext{-Mod}$ reflects isomorphisms if $R to S$ is faithfully flat (see here).
$endgroup$
– Watson
Dec 24 '18 at 12:56
2
$begingroup$
You may want to look at Galois cohomology, namely 1.3.6 here : the twists of $(rho, V)$ should be classified by $$H^1( mathrm{Gal}(K/k) ; mathrm{Aut}_K(rho otimes_k K) ).$$
$endgroup$
– Watson
Dec 24 '18 at 12:56