Why must these have integer coefficients?
$begingroup$
We are considering diagonal subgroup of classical groups and their lie algebras. We then consider $l=a_1l_1 + a_2l_2 + ...$ where $l_i(H)$, H in the lie algebra, returns the ith entry of H. We then say that if all $a_1$ are integers, $l$ lifts to $e^l(exp(H))=e^{l(H)}$. Why do they need to be integer coefficients?
lie-groups lie-algebras classical-groups cartan-geometry
asked Dec 2 '18 at 17:01
swedishfishedswedishfished
737
$endgroup$
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$begingroup$
We are considering diagonal subgroup of classical groups and their lie algebras. We then consider $l=a_1l_1 + a_2l_2 + ...$ where $l_i(H)$, H in the lie algebra, returns the ith entry of H. We then say that if all $a_1$ are integers, $l$ lifts to $e^l(exp(H))=e^{l(H)}$. Why do they need to be integer coefficients?
lie-groups lie-algebras classical-groups cartan-geometry
asked Dec 2 '18 at 17:01
swedishfishedswedishfished
737
$endgroup$
add a comment |
$begingroup$
We are considering diagonal subgroup of classical groups and their lie algebras. We then consider $l=a_1l_1 + a_2l_2 + ...$ where $l_i(H)$, H in the lie algebra, returns the ith entry of H. We then say that if all $a_1$ are integers, $l$ lifts to $e^l(exp(H))=e^{l(H)}$. Why do they need to be integer coefficients?
lie-groups lie-algebras classical-groups cartan-geometry
asked Dec 2 '18 at 17:01
swedishfishedswedishfished
737
$endgroup$
We are considering diagonal subgroup of classical groups and their lie algebras. We then consider $l=a_1l_1 + a_2l_2 + ...$ where $l_i(H)$, H in the lie algebra, returns the ith entry of H. We then say that if all $a_1$ are integers, $l$ lifts to $e^l(exp(H))=e^{l(H)}$. Why do they need to be integer coefficients?
lie-groups lie-algebras classical-groups cartan-geometry
lie-groups lie-algebras classical-groups cartan-geometry
asked Dec 2 '18 at 17:01
swedishfishedswedishfished
737
asked Dec 2 '18 at 17:01
swedishfishedswedishfished
737
asked Dec 2 '18 at 17:01
swedishfishedswedishfished
737
asked Dec 2 '18 at 17:01
swedishfishedswedishfished
737
asked Dec 2 '18 at 17:01
swedishfishedswedishfished
737
737
add a comment |
add a comment |
1 Answer
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$begingroup$
Hint: There is an easy way to construct matrices $H$ with just two non-zero entries such that $exp(H)=mathbb I$. For your equation to make sense, you need $e^{l(H)}=1$ for all these matrices.
answered Dec 3 '18 at 12:05
Andreas CapAndreas Cap
11.1k923
$endgroup$
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1 Answer
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$begingroup$
Hint: There is an easy way to construct matrices $H$ with just two non-zero entries such that $exp(H)=mathbb I$. For your equation to make sense, you need $e^{l(H)}=1$ for all these matrices.
answered Dec 3 '18 at 12:05
Andreas CapAndreas Cap
11.1k923
$endgroup$
add a comment |
$begingroup$
Hint: There is an easy way to construct matrices $H$ with just two non-zero entries such that $exp(H)=mathbb I$. For your equation to make sense, you need $e^{l(H)}=1$ for all these matrices.
answered Dec 3 '18 at 12:05
Andreas CapAndreas Cap
11.1k923
$endgroup$
add a comment |
$begingroup$
Hint: There is an easy way to construct matrices $H$ with just two non-zero entries such that $exp(H)=mathbb I$. For your equation to make sense, you need $e^{l(H)}=1$ for all these matrices.
answered Dec 3 '18 at 12:05
Andreas CapAndreas Cap
11.1k923
$endgroup$
Hint: There is an easy way to construct matrices $H$ with just two non-zero entries such that $exp(H)=mathbb I$. For your equation to make sense, you need $e^{l(H)}=1$ for all these matrices.
answered Dec 3 '18 at 12:05
Andreas CapAndreas Cap
11.1k923
answered Dec 3 '18 at 12:05
Andreas CapAndreas Cap
11.1k923
answered Dec 3 '18 at 12:05
Andreas CapAndreas Cap
11.1k923
answered Dec 3 '18 at 12:05
Andreas CapAndreas Cap
11.1k923
11.1k923
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