Dtjytyk

What exactly is the purpose of connection links straped between the rocket and the launch pad

Question about partial fractions with irreducible quadratic factors

As a monk, can you make a melee attack roll using your Strength modifier, but roll damage with your Dexterity modifier?

How does Dispel Magic work against Stoneskin?

"However" used in a conditional clause?

Decoding assembly instructions in a Game Boy disassembler

Touchscreen-controlled dentist office snowman collector game

Is going from continuous data to categorical always wrong?

Does anyone draw a parallel between Haman selling himself to Mordechai and Esav selling the birthright to Yaakov?

My adviser wants to be the first author

What is the definition of "Natural Selection"?

Examples of odd-dimensional manifolds that do not admit contact structure

Does splitting a potentially monolithic application into several smaller ones help prevent bugs?

Coworker uses her breast-pump everywhere in the office

Best mythical creature to use as livestock?

What injury would be of little consequence to a biped but terrible for a quadruped?

How to make readers know that my work has used a hidden constraint?

Is it ok to include an epilogue dedicated to colleagues who passed away in the end of the manuscript?

Make a transparent 448*448 image

Time dilation for a moving electronic clock

Does Linux have system calls to access all the features of the file systems it supports?

Confusion with the nameplate of an induction motor

When two POV characters meet

What is the dot in “1.2.4."

Conjugacy of Borel Subalgebras: Proof in Humphreys' Introduction to Lie Algebras and Representation Theory

Ties between Lie algebras and ring theoryStandard parabolic Lie subalgebras and conjugacyHumphreys Introduction to Lie Algebras - Conjugate Borel subalgebras sl(2,F)S-subalgebra and R-subalgebra.Most general definition of Borel and parabolic Lie algebras?proof of “conjugacy theorem of BSA” following HumphreysIs the theorem of complete reducibility or the abstract Jordan decomposition needed for structure theory of semisimple Lie algebras?Why is $textrm {ad}_{L}t_{alpha}$ nilpotent?Eigenspace for toral in a Borel subalgebraThe codimension of a parabolic subalgebra of a semisimple Lie algebra

0

$begingroup$

In the title referenced above a proof of the conjugacy of Borel subalgebras is given on page 84: We assume $L$ semisimple and let $B$ be a standard Borel subalgebra and $B'$ any other Borel subalgebra. We set $N'$ equal to the set of all nilpotent elements of $B cap B'$ and assume $N'neq 0$ for case 1. (Note that N' is an ideal of $B cap B'$). If $x in N'$ then $ad x$ acts nilpotently on the vector space $B/B cap B'$ whence there exists nonzero $y$ in this vector space that is annihilated by all of $N'$ (by a linear algebra thm). In other words, there is a $y$ in $B$ outside of $B'$ that is sent into $B cap B'$ by any $ad x$ for $x in N'$. But since $[x,y] in [B,B]$ then $ad [x,y]$ is nilpotent on $L$ (since $B$ is standard) and hence in $N'$. Therefore there is a $y$ in $B$ outside of $B'$ that is in the normalizer, $K$, of $N'$. Humphreys then wants to make a symmetric conclusion: That there is a $y'$ in $B'$ outside of $B$ that is in the normalizer, $K$, of $N'$. But how is this possible? Since $[x,y'] in [B',B']$ we may only conclude $ad [x,y']$ is nilpotent on $B'$ and not on all of $L$ since we do not know that $B'$ is standard. Thus we can't know for sure if $[x,y] in N'$ and can't conclude $y'in K$. Thoughts?

linear-algebra representation-theory lie-algebras

share|cite|improve this question

edited Mar 10 at 4:24

J. W. Tanner

3,2301320

asked Mar 10 at 3:59

grahawgrahaw

82

$endgroup$

add a comment | 

0

$begingroup$

In the title referenced above a proof of the conjugacy of Borel subalgebras is given on page 84: We assume $L$ semisimple and let $B$ be a standard Borel subalgebra and $B'$ any other Borel subalgebra. We set $N'$ equal to the set of all nilpotent elements of $B cap B'$ and assume $N'neq 0$ for case 1. (Note that N' is an ideal of $B cap B'$). If $x in N'$ then $ad x$ acts nilpotently on the vector space $B/B cap B'$ whence there exists nonzero $y$ in this vector space that is annihilated by all of $N'$ (by a linear algebra thm). In other words, there is a $y$ in $B$ outside of $B'$ that is sent into $B cap B'$ by any $ad x$ for $x in N'$. But since $[x,y] in [B,B]$ then $ad [x,y]$ is nilpotent on $L$ (since $B$ is standard) and hence in $N'$. Therefore there is a $y$ in $B$ outside of $B'$ that is in the normalizer, $K$, of $N'$. Humphreys then wants to make a symmetric conclusion: That there is a $y'$ in $B'$ outside of $B$ that is in the normalizer, $K$, of $N'$. But how is this possible? Since $[x,y'] in [B',B']$ we may only conclude $ad [x,y']$ is nilpotent on $B'$ and not on all of $L$ since we do not know that $B'$ is standard. Thus we can't know for sure if $[x,y] in N'$ and can't conclude $y'in K$. Thoughts?

linear-algebra representation-theory lie-algebras

share|cite|improve this question

edited Mar 10 at 4:24

J. W. Tanner

3,2301320

asked Mar 10 at 3:59

grahawgrahaw

82

$endgroup$

add a comment | 

$begingroup$

In the title referenced above a proof of the conjugacy of Borel subalgebras is given on page 84: We assume $L$ semisimple and let $B$ be a standard Borel subalgebra and $B'$ any other Borel subalgebra. We set $N'$ equal to the set of all nilpotent elements of $B cap B'$ and assume $N'neq 0$ for case 1. (Note that N' is an ideal of $B cap B'$). If $x in N'$ then $ad x$ acts nilpotently on the vector space $B/B cap B'$ whence there exists nonzero $y$ in this vector space that is annihilated by all of $N'$ (by a linear algebra thm). In other words, there is a $y$ in $B$ outside of $B'$ that is sent into $B cap B'$ by any $ad x$ for $x in N'$. But since $[x,y] in [B,B]$ then $ad [x,y]$ is nilpotent on $L$ (since $B$ is standard) and hence in $N'$. Therefore there is a $y$ in $B$ outside of $B'$ that is in the normalizer, $K$, of $N'$. Humphreys then wants to make a symmetric conclusion: That there is a $y'$ in $B'$ outside of $B$ that is in the normalizer, $K$, of $N'$. But how is this possible? Since $[x,y'] in [B',B']$ we may only conclude $ad [x,y']$ is nilpotent on $B'$ and not on all of $L$ since we do not know that $B'$ is standard. Thus we can't know for sure if $[x,y] in N'$ and can't conclude $y'in K$. Thoughts?

linear-algebra representation-theory lie-algebras

share|cite|improve this question

edited Mar 10 at 4:24

J. W. Tanner

3,2301320

asked Mar 10 at 3:59

grahawgrahaw

82

$endgroup$

share|cite|improve this question

edited Mar 10 at 4:24

J. W. Tanner

3,2301320

asked Mar 10 at 3:59

grahawgrahaw

82

share|cite|improve this question

edited Mar 10 at 4:24

J. W. Tanner

3,2301320

asked Mar 10 at 3:59

grahawgrahaw

82

edited Mar 10 at 4:24

J. W. Tanner

3,2301320

edited Mar 10 at 4:24

J. W. Tanner

3,2301320

asked Mar 10 at 3:59

grahawgrahaw

82

asked Mar 10 at 3:59

grahawgrahaw

82