Derivations on semisimple Lie algebra

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First recall some definitions :

Let $B$ be a Killing form on Lie algebra $mathfrakg$ over $bf R$ such that
$B(X,Y)doteq Tr(ad_Xad_Y)$.

$mathfrakg$ is semisimple if $B$ is non-degenerate.

Define $partial mathfrakg doteq D[X,Y]=[DX,Y] + [X,DY] $ Clearly it contains $ad(mathfrakg)doteq Xin mathfrakg$.

My question is about proof of $partial mathfrakg=ad(mathfrakg)$ :

I am reading Helgason's book. We can show that $ ad(mathfrakg) = mathfrakg$ easily.

I cannot proceed the proof since $B$ is defined on $mathfrakg$ only.

How can I complete the proof ? Thank you in advance.

differential-geometry lie-algebras

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asked May 20 '13 at 7:03

HK Lee

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up vote
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First recall some definitions :

Let $B$ be a Killing form on Lie algebra $mathfrakg$ over $bf R$ such that
$B(X,Y)doteq Tr(ad_Xad_Y)$.

$mathfrakg$ is semisimple if $B$ is non-degenerate.

Define $partial mathfrakg doteq D[X,Y]=[DX,Y] + [X,DY] $ Clearly it contains $ad(mathfrakg)doteq Xin mathfrakg$.

My question is about proof of $partial mathfrakg=ad(mathfrakg)$ :

I am reading Helgason's book. We can show that $ ad(mathfrakg) = mathfrakg$ easily.

I cannot proceed the proof since $B$ is defined on $mathfrakg$ only.

How can I complete the proof ? Thank you in advance.

differential-geometry lie-algebras

share|cite|improve this question

asked May 20 '13 at 7:03

HK Lee

13.5k31855

add a comment | 

up vote
0
down vote

favorite

up vote
0
down vote

favorite

First recall some definitions :

Let $B$ be a Killing form on Lie algebra $mathfrakg$ over $bf R$ such that
$B(X,Y)doteq Tr(ad_Xad_Y)$.

$mathfrakg$ is semisimple if $B$ is non-degenerate.

Define $partial mathfrakg doteq D[X,Y]=[DX,Y] + [X,DY] $ Clearly it contains $ad(mathfrakg)doteq Xin mathfrakg$.

My question is about proof of $partial mathfrakg=ad(mathfrakg)$ :

I am reading Helgason's book. We can show that $ ad(mathfrakg) = mathfrakg$ easily.

I cannot proceed the proof since $B$ is defined on $mathfrakg$ only.

How can I complete the proof ? Thank you in advance.

differential-geometry lie-algebras

share|cite|improve this question

asked May 20 '13 at 7:03

HK Lee

13.5k31855

First recall some definitions :

Let $B$ be a Killing form on Lie algebra $mathfrakg$ over $bf R$ such that
$B(X,Y)doteq Tr(ad_Xad_Y)$.

$mathfrakg$ is semisimple if $B$ is non-degenerate.

Define $partial mathfrakg doteq D[X,Y]=[DX,Y] + [X,DY] $ Clearly it contains $ad(mathfrakg)doteq Xin mathfrakg$.

My question is about proof of $partial mathfrakg=ad(mathfrakg)$ :

I am reading Helgason's book. We can show that $ ad(mathfrakg) = mathfrakg$ easily.

I cannot proceed the proof since $B$ is defined on $mathfrakg$ only.

How can I complete the proof ? Thank you in advance.

differential-geometry lie-algebras

share|cite|improve this question

asked May 20 '13 at 7:03

HK Lee

13.5k31855

share|cite|improve this question

share|cite|improve this question

asked May 20 '13 at 7:03

HK Lee

13.5k31855

asked May 20 '13 at 7:03

HK Lee

13.5k31855

asked May 20 '13 at 7:03

HK Lee

13.5k31855

13.5k31855

add a comment | 

add a comment | 

1 Answer
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Here's a proof from Humphrey's excellent book:

First of all, show that $I=ad(mathfrak g)subsetpartialmathfrak g$ is an ideal of the Lie algebra $partialmathfrak g$: if $xinmathfrak g$ and $deltainpartialmathfrak g$, then $$[delta,ad(x)]=cdotsinpartialmathfrak g$$ This implies that the Killing form on $mathfrak g$ coincides with that induced by the Killing form on $partialmathfrak g$. Since $mathfrak g$'s Killing form is non degenerate by semisimplicity, we have
$$Ioplus I^perp=partialmathfrak g$$
with respect to the Killing form on the derivations. Now consider $deltain I^perp$. For any $xinmathfrak g$, since both $I$ and $I^perp$ are ideals in $partialmathfrak g$, $$[delta,ad(x)]in Icap I^perp=lbrace 0rbrace$$ equals $ad(delta(x))$ by the calculation above, so for all $xinmathfrak g,~delta(x)inker(ad)=lbrace 0rbrace$. That is, $delta=0$ and $I^perp=lbrace 0rbrace$, so that $ad(mathfrak g)=partialmathfrak g$.

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edited May 20 '13 at 7:25

answered May 20 '13 at 7:19

Olivier Bégassat

13.1k12171

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1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
3
down vote



accepted

Here's a proof from Humphrey's excellent book:

First of all, show that $I=ad(mathfrak g)subsetpartialmathfrak g$ is an ideal of the Lie algebra $partialmathfrak g$: if $xinmathfrak g$ and $deltainpartialmathfrak g$, then $$[delta,ad(x)]=cdotsinpartialmathfrak g$$ This implies that the Killing form on $mathfrak g$ coincides with that induced by the Killing form on $partialmathfrak g$. Since $mathfrak g$'s Killing form is non degenerate by semisimplicity, we have
$$Ioplus I^perp=partialmathfrak g$$
with respect to the Killing form on the derivations. Now consider $deltain I^perp$. For any $xinmathfrak g$, since both $I$ and $I^perp$ are ideals in $partialmathfrak g$, $$[delta,ad(x)]in Icap I^perp=lbrace 0rbrace$$ equals $ad(delta(x))$ by the calculation above, so for all $xinmathfrak g,~delta(x)inker(ad)=lbrace 0rbrace$. That is, $delta=0$ and $I^perp=lbrace 0rbrace$, so that $ad(mathfrak g)=partialmathfrak g$.

share|cite|improve this answer

edited May 20 '13 at 7:25

answered May 20 '13 at 7:19

Olivier Bégassat

13.1k12171

add a comment | 

up vote
3
down vote



accepted

Here's a proof from Humphrey's excellent book:

First of all, show that $I=ad(mathfrak g)subsetpartialmathfrak g$ is an ideal of the Lie algebra $partialmathfrak g$: if $xinmathfrak g$ and $deltainpartialmathfrak g$, then $$[delta,ad(x)]=cdotsinpartialmathfrak g$$ This implies that the Killing form on $mathfrak g$ coincides with that induced by the Killing form on $partialmathfrak g$. Since $mathfrak g$'s Killing form is non degenerate by semisimplicity, we have
$$Ioplus I^perp=partialmathfrak g$$
with respect to the Killing form on the derivations. Now consider $deltain I^perp$. For any $xinmathfrak g$, since both $I$ and $I^perp$ are ideals in $partialmathfrak g$, $$[delta,ad(x)]in Icap I^perp=lbrace 0rbrace$$ equals $ad(delta(x))$ by the calculation above, so for all $xinmathfrak g,~delta(x)inker(ad)=lbrace 0rbrace$. That is, $delta=0$ and $I^perp=lbrace 0rbrace$, so that $ad(mathfrak g)=partialmathfrak g$.

share|cite|improve this answer

edited May 20 '13 at 7:25

answered May 20 '13 at 7:19

Olivier Bégassat

13.1k12171

add a comment | 

up vote
3
down vote



accepted

up vote
3
down vote



accepted

Here's a proof from Humphrey's excellent book:

First of all, show that $I=ad(mathfrak g)subsetpartialmathfrak g$ is an ideal of the Lie algebra $partialmathfrak g$: if $xinmathfrak g$ and $deltainpartialmathfrak g$, then $$[delta,ad(x)]=cdotsinpartialmathfrak g$$ This implies that the Killing form on $mathfrak g$ coincides with that induced by the Killing form on $partialmathfrak g$. Since $mathfrak g$'s Killing form is non degenerate by semisimplicity, we have
$$Ioplus I^perp=partialmathfrak g$$
with respect to the Killing form on the derivations. Now consider $deltain I^perp$. For any $xinmathfrak g$, since both $I$ and $I^perp$ are ideals in $partialmathfrak g$, $$[delta,ad(x)]in Icap I^perp=lbrace 0rbrace$$ equals $ad(delta(x))$ by the calculation above, so for all $xinmathfrak g,~delta(x)inker(ad)=lbrace 0rbrace$. That is, $delta=0$ and $I^perp=lbrace 0rbrace$, so that $ad(mathfrak g)=partialmathfrak g$.

share|cite|improve this answer

edited May 20 '13 at 7:25

answered May 20 '13 at 7:19

Olivier Bégassat

13.1k12171

Here's a proof from Humphrey's excellent book:

First of all, show that $I=ad(mathfrak g)subsetpartialmathfrak g$ is an ideal of the Lie algebra $partialmathfrak g$: if $xinmathfrak g$ and $deltainpartialmathfrak g$, then $$[delta,ad(x)]=cdotsinpartialmathfrak g$$ This implies that the Killing form on $mathfrak g$ coincides with that induced by the Killing form on $partialmathfrak g$. Since $mathfrak g$'s Killing form is non degenerate by semisimplicity, we have
$$Ioplus I^perp=partialmathfrak g$$
with respect to the Killing form on the derivations. Now consider $deltain I^perp$. For any $xinmathfrak g$, since both $I$ and $I^perp$ are ideals in $partialmathfrak g$, $$[delta,ad(x)]in Icap I^perp=lbrace 0rbrace$$ equals $ad(delta(x))$ by the calculation above, so for all $xinmathfrak g,~delta(x)inker(ad)=lbrace 0rbrace$. That is, $delta=0$ and $I^perp=lbrace 0rbrace$, so that $ad(mathfrak g)=partialmathfrak g$.

share|cite|improve this answer

edited May 20 '13 at 7:25

answered May 20 '13 at 7:19

Olivier Bégassat

13.1k12171

share|cite|improve this answer

share|cite|improve this answer

edited May 20 '13 at 7:25

edited May 20 '13 at 7:25

edited May 20 '13 at 7:25

answered May 20 '13 at 7:19

Olivier Bégassat

13.1k12171

answered May 20 '13 at 7:19

Olivier Bégassat

13.1k12171

answered May 20 '13 at 7:19

Olivier Bégassat

13.1k12171

13.1k12171

add a comment | 

add a comment | 

 

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