Dimension of the root spaces of a semisimple complex Lie algebra
$begingroup$
I have problems in understanding the proof that the root spaces of a semisimple Lie algebra are all 1-dimensional and that the only multiples of a root $alpha in Phi$ which occur in $Phi$ are $pm alpha$.
The proof which I am referring to is that of page 101 of the book Introduction to Lie algebras. Here is the web page with the book in Google Books: http://books.google.it/books?id=Kbh3-hvDSbIC&pg=PA97&dq=root+space+decomposition&hl=it&sa=X&ei=ZiAbUqCIAunx4QTlooD4Dw&ved=0CFwQ6AEwBQ#v=onepage&q=root%20space%20decomposition&f=false
The problems begin with the discussion of the two cases $s$ even/odd. I agree with the authors that in the even case there must be an $h_alpha$-eigenvector $v in Vsimeq V_s$ of zero eigenvalue (I think this follows from the fact that $h in mathfrak{sl}_2(mathbb{C})$ acts diagonalizably on $V_s$ with eigenvalues $-s,-s+2,dots, s-2,s$ so $0$ is an eigenvalue because $s$ is even by hypothesis). But how do they deduce from this that $alpha(v)=0$? It seems to me - but it is perhaps a wrong conjecture - that they use the fact that the $0$-eigenspace of the action of $h_alpha$ on $L$ is all contained in the direct sum $K oplus mathfrak{sl}(alpha)$, but I don't understand neither why this fact is true nor how the result can be derived from it.
Can anyone help me, please? Thank you!
lie-algebras root-systems
asked Aug 26 '13 at 9:53
DiogenesDiogenes
273110
$endgroup$
add a comment |
$begingroup$
I have problems in understanding the proof that the root spaces of a semisimple Lie algebra are all 1-dimensional and that the only multiples of a root $alpha in Phi$ which occur in $Phi$ are $pm alpha$.
The proof which I am referring to is that of page 101 of the book Introduction to Lie algebras. Here is the web page with the book in Google Books: http://books.google.it/books?id=Kbh3-hvDSbIC&pg=PA97&dq=root+space+decomposition&hl=it&sa=X&ei=ZiAbUqCIAunx4QTlooD4Dw&ved=0CFwQ6AEwBQ#v=onepage&q=root%20space%20decomposition&f=false
The problems begin with the discussion of the two cases $s$ even/odd. I agree with the authors that in the even case there must be an $h_alpha$-eigenvector $v in Vsimeq V_s$ of zero eigenvalue (I think this follows from the fact that $h in mathfrak{sl}_2(mathbb{C})$ acts diagonalizably on $V_s$ with eigenvalues $-s,-s+2,dots, s-2,s$ so $0$ is an eigenvalue because $s$ is even by hypothesis). But how do they deduce from this that $alpha(v)=0$? It seems to me - but it is perhaps a wrong conjecture - that they use the fact that the $0$-eigenspace of the action of $h_alpha$ on $L$ is all contained in the direct sum $K oplus mathfrak{sl}(alpha)$, but I don't understand neither why this fact is true nor how the result can be derived from it.
Can anyone help me, please? Thank you!
lie-algebras root-systems
asked Aug 26 '13 at 9:53
DiogenesDiogenes
273110
$endgroup$
$begingroup$
Unfortunately I cannot see page 101 in the google preview. A good answer for the statement was given here.
$endgroup$
– Torsten Schoeneberg
Jul 5 '17 at 20:11
add a comment |
$begingroup$
I have problems in understanding the proof that the root spaces of a semisimple Lie algebra are all 1-dimensional and that the only multiples of a root $alpha in Phi$ which occur in $Phi$ are $pm alpha$.
The proof which I am referring to is that of page 101 of the book Introduction to Lie algebras. Here is the web page with the book in Google Books: http://books.google.it/books?id=Kbh3-hvDSbIC&pg=PA97&dq=root+space+decomposition&hl=it&sa=X&ei=ZiAbUqCIAunx4QTlooD4Dw&ved=0CFwQ6AEwBQ#v=onepage&q=root%20space%20decomposition&f=false
The problems begin with the discussion of the two cases $s$ even/odd. I agree with the authors that in the even case there must be an $h_alpha$-eigenvector $v in Vsimeq V_s$ of zero eigenvalue (I think this follows from the fact that $h in mathfrak{sl}_2(mathbb{C})$ acts diagonalizably on $V_s$ with eigenvalues $-s,-s+2,dots, s-2,s$ so $0$ is an eigenvalue because $s$ is even by hypothesis). But how do they deduce from this that $alpha(v)=0$? It seems to me - but it is perhaps a wrong conjecture - that they use the fact that the $0$-eigenspace of the action of $h_alpha$ on $L$ is all contained in the direct sum $K oplus mathfrak{sl}(alpha)$, but I don't understand neither why this fact is true nor how the result can be derived from it.
Can anyone help me, please? Thank you!
lie-algebras root-systems
asked Aug 26 '13 at 9:53
DiogenesDiogenes
273110
$endgroup$
I have problems in understanding the proof that the root spaces of a semisimple Lie algebra are all 1-dimensional and that the only multiples of a root $alpha in Phi$ which occur in $Phi$ are $pm alpha$.
The proof which I am referring to is that of page 101 of the book Introduction to Lie algebras. Here is the web page with the book in Google Books: http://books.google.it/books?id=Kbh3-hvDSbIC&pg=PA97&dq=root+space+decomposition&hl=it&sa=X&ei=ZiAbUqCIAunx4QTlooD4Dw&ved=0CFwQ6AEwBQ#v=onepage&q=root%20space%20decomposition&f=false
The problems begin with the discussion of the two cases $s$ even/odd. I agree with the authors that in the even case there must be an $h_alpha$-eigenvector $v in Vsimeq V_s$ of zero eigenvalue (I think this follows from the fact that $h in mathfrak{sl}_2(mathbb{C})$ acts diagonalizably on $V_s$ with eigenvalues $-s,-s+2,dots, s-2,s$ so $0$ is an eigenvalue because $s$ is even by hypothesis). But how do they deduce from this that $alpha(v)=0$? It seems to me - but it is perhaps a wrong conjecture - that they use the fact that the $0$-eigenspace of the action of $h_alpha$ on $L$ is all contained in the direct sum $K oplus mathfrak{sl}(alpha)$, but I don't understand neither why this fact is true nor how the result can be derived from it.
Can anyone help me, please? Thank you!
lie-algebras root-systems
lie-algebras root-systems
asked Aug 26 '13 at 9:53
DiogenesDiogenes
273110
asked Aug 26 '13 at 9:53
DiogenesDiogenes
273110
asked Aug 26 '13 at 9:53
DiogenesDiogenes
273110
asked Aug 26 '13 at 9:53
DiogenesDiogenes
273110
asked Aug 26 '13 at 9:53
DiogenesDiogenes
273110
273110
$begingroup$
Unfortunately I cannot see page 101 in the google preview. A good answer for the statement was given here.
$endgroup$
– Torsten Schoeneberg
Jul 5 '17 at 20:11
add a comment |
$begingroup$
Unfortunately I cannot see page 101 in the google preview. A good answer for the statement was given here.
$endgroup$
– Torsten Schoeneberg
Jul 5 '17 at 20:11
$begingroup$
Unfortunately I cannot see page 101 in the google preview. A good answer for the statement was given here.
$endgroup$
– Torsten Schoeneberg
Jul 5 '17 at 20:11
$begingroup$
Unfortunately I cannot see page 101 in the google preview. A good answer for the statement was given here.
$endgroup$
– Torsten Schoeneberg
Jul 5 '17 at 20:11
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
If $xin M$ is such that $h_{alpha}cdot x = [h_{alpha}, x] = 0$, then $xin Kbigoplusmathfrak{sl}(alpha)$, otherwise $xinmathfrak{g}_{calpha}$ for some $calphainPhi$ and $cneq 0$, but then $[h_{alpha},x]= 2cxneq 0$ giving us the contradiction.
To show that $vin K$, note that $mathfrak{h} = Kbigoplus span({h_{alpha}})$ (exercise), it is also easy to show that $vinmathfrak{h}$. Finally, since $M=Kbigoplus mathfrak{sl}(alpha)bigoplus W$ and $vin Vsubseteq W$, $v$ must lie in $K$.
answered Dec 29 '18 at 16:31
Sid CarolineSid Caroline
1,5982515
$endgroup$
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%2f476423%2fdimension-of-the-root-spaces-of-a-semisimple-complex-lie-algebra%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
If $xin M$ is such that $h_{alpha}cdot x = [h_{alpha}, x] = 0$, then $xin Kbigoplusmathfrak{sl}(alpha)$, otherwise $xinmathfrak{g}_{calpha}$ for some $calphainPhi$ and $cneq 0$, but then $[h_{alpha},x]= 2cxneq 0$ giving us the contradiction.
To show that $vin K$, note that $mathfrak{h} = Kbigoplus span({h_{alpha}})$ (exercise), it is also easy to show that $vinmathfrak{h}$. Finally, since $M=Kbigoplus mathfrak{sl}(alpha)bigoplus W$ and $vin Vsubseteq W$, $v$ must lie in $K$.
answered Dec 29 '18 at 16:31
Sid CarolineSid Caroline
1,5982515
$endgroup$
add a comment |
$begingroup$
If $xin M$ is such that $h_{alpha}cdot x = [h_{alpha}, x] = 0$, then $xin Kbigoplusmathfrak{sl}(alpha)$, otherwise $xinmathfrak{g}_{calpha}$ for some $calphainPhi$ and $cneq 0$, but then $[h_{alpha},x]= 2cxneq 0$ giving us the contradiction.
To show that $vin K$, note that $mathfrak{h} = Kbigoplus span({h_{alpha}})$ (exercise), it is also easy to show that $vinmathfrak{h}$. Finally, since $M=Kbigoplus mathfrak{sl}(alpha)bigoplus W$ and $vin Vsubseteq W$, $v$ must lie in $K$.
answered Dec 29 '18 at 16:31
Sid CarolineSid Caroline
1,5982515
$endgroup$
add a comment |
$begingroup$
If $xin M$ is such that $h_{alpha}cdot x = [h_{alpha}, x] = 0$, then $xin Kbigoplusmathfrak{sl}(alpha)$, otherwise $xinmathfrak{g}_{calpha}$ for some $calphainPhi$ and $cneq 0$, but then $[h_{alpha},x]= 2cxneq 0$ giving us the contradiction.
To show that $vin K$, note that $mathfrak{h} = Kbigoplus span({h_{alpha}})$ (exercise), it is also easy to show that $vinmathfrak{h}$. Finally, since $M=Kbigoplus mathfrak{sl}(alpha)bigoplus W$ and $vin Vsubseteq W$, $v$ must lie in $K$.
answered Dec 29 '18 at 16:31
Sid CarolineSid Caroline
1,5982515
$endgroup$
If $xin M$ is such that $h_{alpha}cdot x = [h_{alpha}, x] = 0$, then $xin Kbigoplusmathfrak{sl}(alpha)$, otherwise $xinmathfrak{g}_{calpha}$ for some $calphainPhi$ and $cneq 0$, but then $[h_{alpha},x]= 2cxneq 0$ giving us the contradiction.
To show that $vin K$, note that $mathfrak{h} = Kbigoplus span({h_{alpha}})$ (exercise), it is also easy to show that $vinmathfrak{h}$. Finally, since $M=Kbigoplus mathfrak{sl}(alpha)bigoplus W$ and $vin Vsubseteq W$, $v$ must lie in $K$.
answered Dec 29 '18 at 16:31
Sid CarolineSid Caroline
1,5982515
edited Dec 29 '18 at 20:57
answered Dec 29 '18 at 16:31
Sid CarolineSid Caroline
1,5982515
answered Dec 29 '18 at 16:31
Sid CarolineSid Caroline
1,5982515
answered Dec 29 '18 at 16:31
Sid CarolineSid Caroline
1,5982515
1,5982515
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%2f476423%2fdimension-of-the-root-spaces-of-a-semisimple-complex-lie-algebra%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
$begingroup$
Unfortunately I cannot see page 101 in the google preview. A good answer for the statement was given here.
$endgroup$
– Torsten Schoeneberg
Jul 5 '17 at 20:11