Normal Subgroups of $SU(n)$ Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Normal subgroups of $U(n)$Are all the unitary irreducible representations of $SU(2)$ have unit determinant?How to show that $pm 1$ is a normal subgroup of $mathbbS^3$ and is the nontrivial normal subgroup?Why is studying maximal subgroups useful?Find all normal subgroups of the followingIntuition behind normal subgroupsHomomorphism with intersection of all Sylow p-subgroups as kernel?what is the smallest non-abelian finite group which has normal, non-abelian subgroups (plural)Counter examples for normal subgroupsExamples of groups with only *non-abelian* normal subgroupsIsomorphism between normal subgroupsSubgroups of semidirect productsImportance of Normal subgroups
Why are there no cargo aircraft with "flying wing" design?
What does '1 unit of lemon juice' mean in a grandma's drink recipe?
Should I discuss the type of campaign with my players?
Can a non-EU citizen traveling with me come with me through the EU passport line?
How to deal with a team lead who never gives me credit?
Models of set theory where not every set can be linearly ordered
What is the longest distance a 13th-level monk can jump while attacking on the same turn?
Why does Python start at index -1 when indexing a list from the end?
What does the "x" in "x86" represent?
How can players work together to take actions that are otherwise impossible?
List *all* the tuples!
Were Kohanim forbidden from serving in King David's army?
Why is black pepper both grey and black?
When to stop saving and start investing?
Disable hyphenation for an entire paragraph
What's the difference between `auto x = vector<int>()` and `vector<int> x`?
Storing hydrofluoric acid before the invention of plastics
Do I really need recursive chmod to restrict access to a folder?
How do I stop a creek from eroding my steep embankment?
What is the musical term for a note that continously plays through a melody?
What causes the vertical darker bands in my photo?
When -s is used with third person singular. What's its use in this context?
Right-skewed distribution with mean equals to mode?
The logistics of corpse disposal
Normal Subgroups of $SU(n)$
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Normal subgroups of $U(n)$Are all the unitary irreducible representations of $SU(2)$ have unit determinant?How to show that $pm 1$ is a normal subgroup of $mathbbS^3$ and is the nontrivial normal subgroup?Why is studying maximal subgroups useful?Find all normal subgroups of the followingIntuition behind normal subgroupsHomomorphism with intersection of all Sylow p-subgroups as kernel?what is the smallest non-abelian finite group which has normal, non-abelian subgroups (plural)Counter examples for normal subgroupsExamples of groups with only *non-abelian* normal subgroupsIsomorphism between normal subgroupsSubgroups of semidirect productsImportance of Normal subgroups
$begingroup$
I was wondering if there is any classification for normal subgroups of $SU(n)$? In particular, I think that the answer is no for $n = 2$ by looking at the covering map onto $SO(3)$, but I was curious if there were any results for arbitrary $n$?
I'm actually looking for finite normal subgroups of the unitary group (a lot to ask for, I know).
group-theory lie-groups lie-algebras
$endgroup$
add a comment |
$begingroup$
I was wondering if there is any classification for normal subgroups of $SU(n)$? In particular, I think that the answer is no for $n = 2$ by looking at the covering map onto $SO(3)$, but I was curious if there were any results for arbitrary $n$?
I'm actually looking for finite normal subgroups of the unitary group (a lot to ask for, I know).
group-theory lie-groups lie-algebras
$endgroup$
add a comment |
$begingroup$
I was wondering if there is any classification for normal subgroups of $SU(n)$? In particular, I think that the answer is no for $n = 2$ by looking at the covering map onto $SO(3)$, but I was curious if there were any results for arbitrary $n$?
I'm actually looking for finite normal subgroups of the unitary group (a lot to ask for, I know).
group-theory lie-groups lie-algebras
$endgroup$
I was wondering if there is any classification for normal subgroups of $SU(n)$? In particular, I think that the answer is no for $n = 2$ by looking at the covering map onto $SO(3)$, but I was curious if there were any results for arbitrary $n$?
I'm actually looking for finite normal subgroups of the unitary group (a lot to ask for, I know).
group-theory lie-groups lie-algebras
group-theory lie-groups lie-algebras
edited Mar 22 '15 at 3:29
MGN
asked Mar 22 '15 at 3:20
MGNMGN
897921
897921
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Here are two general facts.
- If $G$ is a connected topological group, then any discrete normal subgroup $N$ of $G$ is in fact contained in its center. This is a standard exercise and the proof is straightforward: by hypothesis, $N$ is setwise fixed by conjugation. But since $N$ is discrete, a connected topological group can't act nontrivially on it, and so $N$ is in fact pointwise fixed by conjugation. This already implies that any finite normal subgroup of $SU(n)$ is contained in its center.
- If $G$ is a connected Lie group and $N$ is a closed normal subgroup, then $N$ is a Lie subgroup, and so its Lie algebra $mathfrakn$ is an ideal of the Lie algebra $mathfrakg$. If $mathfrakg$ is simple (which is the case when $G = SU(n)$), then $mathfrakn = 0$, and so $N$ must in fact be discrete, and then by the previous point we know that $N$ must in fact be central. This implies that any nontrivial closed normal subgroup of $SU(n)$ is contained in its center.
$endgroup$
1
$begingroup$
Thanks! These types of standard results in Lie groups/algebras always seem to answer my questions. Do you have a recommendation for a good textbook to learn Lie groups and Lie algebras side by side? Let's say at an early graduate level.
$endgroup$
– MGN
Mar 22 '15 at 16:02
add a comment |
$begingroup$
Qiaochi's answer deals with closed normal subgroups. However, a priori, there could have been non-closed proper normal subgroups in $SU(n)$ (not contained in its center). One can show that this cannot happen "with bare hands" (this is done for instance in Berger's "Geometry" book in the case of $SO(3)$ merely using elementary geometry), or use the following general theorem:
Theorem. Suppose that $G$ is a compact Hausdorff topological group, which is topologically simple, i.e. contains no proper closed normal subgroups. Then $G$ is simple as an abstract group.
See Theorem 9.90 of
Karl H. Hofmann, Sidney A. Morris, The Structure of Compact Groups. A Primer for the Student – A Handbook for the Expert, de Gruyter Studies in Mathematics, Volume 25, 1998.
Addendum. There exist Hausdorff topologically simple groups which are not simple as abstract groups, see section 3 in
George A. Willis, Compact open subgroups in simple totally disconnected groups, Journal of Algebra, Volume 312, Issue 1 (2007) pages 405-417.
$endgroup$
add a comment |
$begingroup$
I doubt there are any finite normal subgroups besides subgroups contained in the center $lbraceexp(frac2ikpin)mathrmid_Vmid k=1,dots,nrbrace$, as any element $uin SU(n)$ that is not of this kind should have a non discrete conjugacy class. Indeed, its eigenspaces form an orthogonal decomposition of the underlying vector space into at least two subspaces,
$$V=bigoplus_i=1,dots, r^perp E_lambda_i(u)$$ and by conjugation by an element in $SU(n)$ amounts to rotating the eigenspaces, which can be done in a continuous fashion, and hence the conjugacy classes are nondenumerable.
$endgroup$
$begingroup$
Good point! In fact, now that I think about it, $SU(n)$ likely has no normal subgroups at all! Other than those in the center of course.
$endgroup$
– MGN
Mar 22 '15 at 5:04
add a comment |
Your Answer
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%2f1200499%2fnormal-subgroups-of-sun%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Here are two general facts.
- If $G$ is a connected topological group, then any discrete normal subgroup $N$ of $G$ is in fact contained in its center. This is a standard exercise and the proof is straightforward: by hypothesis, $N$ is setwise fixed by conjugation. But since $N$ is discrete, a connected topological group can't act nontrivially on it, and so $N$ is in fact pointwise fixed by conjugation. This already implies that any finite normal subgroup of $SU(n)$ is contained in its center.
- If $G$ is a connected Lie group and $N$ is a closed normal subgroup, then $N$ is a Lie subgroup, and so its Lie algebra $mathfrakn$ is an ideal of the Lie algebra $mathfrakg$. If $mathfrakg$ is simple (which is the case when $G = SU(n)$), then $mathfrakn = 0$, and so $N$ must in fact be discrete, and then by the previous point we know that $N$ must in fact be central. This implies that any nontrivial closed normal subgroup of $SU(n)$ is contained in its center.
$endgroup$
1
$begingroup$
Thanks! These types of standard results in Lie groups/algebras always seem to answer my questions. Do you have a recommendation for a good textbook to learn Lie groups and Lie algebras side by side? Let's say at an early graduate level.
$endgroup$
– MGN
Mar 22 '15 at 16:02
add a comment |
$begingroup$
Here are two general facts.
- If $G$ is a connected topological group, then any discrete normal subgroup $N$ of $G$ is in fact contained in its center. This is a standard exercise and the proof is straightforward: by hypothesis, $N$ is setwise fixed by conjugation. But since $N$ is discrete, a connected topological group can't act nontrivially on it, and so $N$ is in fact pointwise fixed by conjugation. This already implies that any finite normal subgroup of $SU(n)$ is contained in its center.
- If $G$ is a connected Lie group and $N$ is a closed normal subgroup, then $N$ is a Lie subgroup, and so its Lie algebra $mathfrakn$ is an ideal of the Lie algebra $mathfrakg$. If $mathfrakg$ is simple (which is the case when $G = SU(n)$), then $mathfrakn = 0$, and so $N$ must in fact be discrete, and then by the previous point we know that $N$ must in fact be central. This implies that any nontrivial closed normal subgroup of $SU(n)$ is contained in its center.
$endgroup$
1
$begingroup$
Thanks! These types of standard results in Lie groups/algebras always seem to answer my questions. Do you have a recommendation for a good textbook to learn Lie groups and Lie algebras side by side? Let's say at an early graduate level.
$endgroup$
– MGN
Mar 22 '15 at 16:02
add a comment |
$begingroup$
Here are two general facts.
- If $G$ is a connected topological group, then any discrete normal subgroup $N$ of $G$ is in fact contained in its center. This is a standard exercise and the proof is straightforward: by hypothesis, $N$ is setwise fixed by conjugation. But since $N$ is discrete, a connected topological group can't act nontrivially on it, and so $N$ is in fact pointwise fixed by conjugation. This already implies that any finite normal subgroup of $SU(n)$ is contained in its center.
- If $G$ is a connected Lie group and $N$ is a closed normal subgroup, then $N$ is a Lie subgroup, and so its Lie algebra $mathfrakn$ is an ideal of the Lie algebra $mathfrakg$. If $mathfrakg$ is simple (which is the case when $G = SU(n)$), then $mathfrakn = 0$, and so $N$ must in fact be discrete, and then by the previous point we know that $N$ must in fact be central. This implies that any nontrivial closed normal subgroup of $SU(n)$ is contained in its center.
$endgroup$
Here are two general facts.
- If $G$ is a connected topological group, then any discrete normal subgroup $N$ of $G$ is in fact contained in its center. This is a standard exercise and the proof is straightforward: by hypothesis, $N$ is setwise fixed by conjugation. But since $N$ is discrete, a connected topological group can't act nontrivially on it, and so $N$ is in fact pointwise fixed by conjugation. This already implies that any finite normal subgroup of $SU(n)$ is contained in its center.
- If $G$ is a connected Lie group and $N$ is a closed normal subgroup, then $N$ is a Lie subgroup, and so its Lie algebra $mathfrakn$ is an ideal of the Lie algebra $mathfrakg$. If $mathfrakg$ is simple (which is the case when $G = SU(n)$), then $mathfrakn = 0$, and so $N$ must in fact be discrete, and then by the previous point we know that $N$ must in fact be central. This implies that any nontrivial closed normal subgroup of $SU(n)$ is contained in its center.
answered Mar 22 '15 at 8:15
Qiaochu YuanQiaochu Yuan
282k32597945
282k32597945
1
$begingroup$
Thanks! These types of standard results in Lie groups/algebras always seem to answer my questions. Do you have a recommendation for a good textbook to learn Lie groups and Lie algebras side by side? Let's say at an early graduate level.
$endgroup$
– MGN
Mar 22 '15 at 16:02
add a comment |
1
$begingroup$
Thanks! These types of standard results in Lie groups/algebras always seem to answer my questions. Do you have a recommendation for a good textbook to learn Lie groups and Lie algebras side by side? Let's say at an early graduate level.
$endgroup$
– MGN
Mar 22 '15 at 16:02
1
1
$begingroup$
Thanks! These types of standard results in Lie groups/algebras always seem to answer my questions. Do you have a recommendation for a good textbook to learn Lie groups and Lie algebras side by side? Let's say at an early graduate level.
$endgroup$
– MGN
Mar 22 '15 at 16:02
$begingroup$
Thanks! These types of standard results in Lie groups/algebras always seem to answer my questions. Do you have a recommendation for a good textbook to learn Lie groups and Lie algebras side by side? Let's say at an early graduate level.
$endgroup$
– MGN
Mar 22 '15 at 16:02
add a comment |
$begingroup$
Qiaochi's answer deals with closed normal subgroups. However, a priori, there could have been non-closed proper normal subgroups in $SU(n)$ (not contained in its center). One can show that this cannot happen "with bare hands" (this is done for instance in Berger's "Geometry" book in the case of $SO(3)$ merely using elementary geometry), or use the following general theorem:
Theorem. Suppose that $G$ is a compact Hausdorff topological group, which is topologically simple, i.e. contains no proper closed normal subgroups. Then $G$ is simple as an abstract group.
See Theorem 9.90 of
Karl H. Hofmann, Sidney A. Morris, The Structure of Compact Groups. A Primer for the Student – A Handbook for the Expert, de Gruyter Studies in Mathematics, Volume 25, 1998.
Addendum. There exist Hausdorff topologically simple groups which are not simple as abstract groups, see section 3 in
George A. Willis, Compact open subgroups in simple totally disconnected groups, Journal of Algebra, Volume 312, Issue 1 (2007) pages 405-417.
$endgroup$
add a comment |
$begingroup$
Qiaochi's answer deals with closed normal subgroups. However, a priori, there could have been non-closed proper normal subgroups in $SU(n)$ (not contained in its center). One can show that this cannot happen "with bare hands" (this is done for instance in Berger's "Geometry" book in the case of $SO(3)$ merely using elementary geometry), or use the following general theorem:
Theorem. Suppose that $G$ is a compact Hausdorff topological group, which is topologically simple, i.e. contains no proper closed normal subgroups. Then $G$ is simple as an abstract group.
See Theorem 9.90 of
Karl H. Hofmann, Sidney A. Morris, The Structure of Compact Groups. A Primer for the Student – A Handbook for the Expert, de Gruyter Studies in Mathematics, Volume 25, 1998.
Addendum. There exist Hausdorff topologically simple groups which are not simple as abstract groups, see section 3 in
George A. Willis, Compact open subgroups in simple totally disconnected groups, Journal of Algebra, Volume 312, Issue 1 (2007) pages 405-417.
$endgroup$
add a comment |
$begingroup$
Qiaochi's answer deals with closed normal subgroups. However, a priori, there could have been non-closed proper normal subgroups in $SU(n)$ (not contained in its center). One can show that this cannot happen "with bare hands" (this is done for instance in Berger's "Geometry" book in the case of $SO(3)$ merely using elementary geometry), or use the following general theorem:
Theorem. Suppose that $G$ is a compact Hausdorff topological group, which is topologically simple, i.e. contains no proper closed normal subgroups. Then $G$ is simple as an abstract group.
See Theorem 9.90 of
Karl H. Hofmann, Sidney A. Morris, The Structure of Compact Groups. A Primer for the Student – A Handbook for the Expert, de Gruyter Studies in Mathematics, Volume 25, 1998.
Addendum. There exist Hausdorff topologically simple groups which are not simple as abstract groups, see section 3 in
George A. Willis, Compact open subgroups in simple totally disconnected groups, Journal of Algebra, Volume 312, Issue 1 (2007) pages 405-417.
$endgroup$
Qiaochi's answer deals with closed normal subgroups. However, a priori, there could have been non-closed proper normal subgroups in $SU(n)$ (not contained in its center). One can show that this cannot happen "with bare hands" (this is done for instance in Berger's "Geometry" book in the case of $SO(3)$ merely using elementary geometry), or use the following general theorem:
Theorem. Suppose that $G$ is a compact Hausdorff topological group, which is topologically simple, i.e. contains no proper closed normal subgroups. Then $G$ is simple as an abstract group.
See Theorem 9.90 of
Karl H. Hofmann, Sidney A. Morris, The Structure of Compact Groups. A Primer for the Student – A Handbook for the Expert, de Gruyter Studies in Mathematics, Volume 25, 1998.
Addendum. There exist Hausdorff topologically simple groups which are not simple as abstract groups, see section 3 in
George A. Willis, Compact open subgroups in simple totally disconnected groups, Journal of Algebra, Volume 312, Issue 1 (2007) pages 405-417.
edited Mar 25 '17 at 1:08
answered Mar 24 '17 at 20:24
Moishe KohanMoishe Kohan
48.9k344111
48.9k344111
add a comment |
add a comment |
$begingroup$
I doubt there are any finite normal subgroups besides subgroups contained in the center $lbraceexp(frac2ikpin)mathrmid_Vmid k=1,dots,nrbrace$, as any element $uin SU(n)$ that is not of this kind should have a non discrete conjugacy class. Indeed, its eigenspaces form an orthogonal decomposition of the underlying vector space into at least two subspaces,
$$V=bigoplus_i=1,dots, r^perp E_lambda_i(u)$$ and by conjugation by an element in $SU(n)$ amounts to rotating the eigenspaces, which can be done in a continuous fashion, and hence the conjugacy classes are nondenumerable.
$endgroup$
$begingroup$
Good point! In fact, now that I think about it, $SU(n)$ likely has no normal subgroups at all! Other than those in the center of course.
$endgroup$
– MGN
Mar 22 '15 at 5:04
add a comment |
$begingroup$
I doubt there are any finite normal subgroups besides subgroups contained in the center $lbraceexp(frac2ikpin)mathrmid_Vmid k=1,dots,nrbrace$, as any element $uin SU(n)$ that is not of this kind should have a non discrete conjugacy class. Indeed, its eigenspaces form an orthogonal decomposition of the underlying vector space into at least two subspaces,
$$V=bigoplus_i=1,dots, r^perp E_lambda_i(u)$$ and by conjugation by an element in $SU(n)$ amounts to rotating the eigenspaces, which can be done in a continuous fashion, and hence the conjugacy classes are nondenumerable.
$endgroup$
$begingroup$
Good point! In fact, now that I think about it, $SU(n)$ likely has no normal subgroups at all! Other than those in the center of course.
$endgroup$
– MGN
Mar 22 '15 at 5:04
add a comment |
$begingroup$
I doubt there are any finite normal subgroups besides subgroups contained in the center $lbraceexp(frac2ikpin)mathrmid_Vmid k=1,dots,nrbrace$, as any element $uin SU(n)$ that is not of this kind should have a non discrete conjugacy class. Indeed, its eigenspaces form an orthogonal decomposition of the underlying vector space into at least two subspaces,
$$V=bigoplus_i=1,dots, r^perp E_lambda_i(u)$$ and by conjugation by an element in $SU(n)$ amounts to rotating the eigenspaces, which can be done in a continuous fashion, and hence the conjugacy classes are nondenumerable.
$endgroup$
I doubt there are any finite normal subgroups besides subgroups contained in the center $lbraceexp(frac2ikpin)mathrmid_Vmid k=1,dots,nrbrace$, as any element $uin SU(n)$ that is not of this kind should have a non discrete conjugacy class. Indeed, its eigenspaces form an orthogonal decomposition of the underlying vector space into at least two subspaces,
$$V=bigoplus_i=1,dots, r^perp E_lambda_i(u)$$ and by conjugation by an element in $SU(n)$ amounts to rotating the eigenspaces, which can be done in a continuous fashion, and hence the conjugacy classes are nondenumerable.
answered Mar 22 '15 at 3:33
Olivier BégassatOlivier Bégassat
13.5k12373
13.5k12373
$begingroup$
Good point! In fact, now that I think about it, $SU(n)$ likely has no normal subgroups at all! Other than those in the center of course.
$endgroup$
– MGN
Mar 22 '15 at 5:04
add a comment |
$begingroup$
Good point! In fact, now that I think about it, $SU(n)$ likely has no normal subgroups at all! Other than those in the center of course.
$endgroup$
– MGN
Mar 22 '15 at 5:04
$begingroup$
Good point! In fact, now that I think about it, $SU(n)$ likely has no normal subgroups at all! Other than those in the center of course.
$endgroup$
– MGN
Mar 22 '15 at 5:04
$begingroup$
Good point! In fact, now that I think about it, $SU(n)$ likely has no normal subgroups at all! Other than those in the center of course.
$endgroup$
– MGN
Mar 22 '15 at 5:04
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%2f1200499%2fnormal-subgroups-of-sun%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