Mixing
The intersection of a maximal toral subalgebra with a simple ideal of a Lie algebra is a maximal toral...
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I'm reading Humphreys' Introduction to Lie Algebras and Representation Theory and I have a question about Corollary 14.1, which reads:
Humphreys Corollary 14.1. Let $L$ be a semisimple Lie algebra, with maximal toral subalgebra $H$ and root system $Phi$. If $L = L_1opluscdotsoplus L_t$ is the decomposition of $L$ into simple ideals, then $H_i = Hcap L_i$ is a maximal toral subalgebra of $L_i$, and the corresponding (irreducible) root system $Phi_i$ may be regarded canonically as a subsystem of $Phi$ in such a way that $Phi=Phi_1cupcdotscupPhi_t$ is the decomposition of $Phi$ into its irreducible components.
I understand most of this. What I'm struggling to understand is why the $Phi_i$ must be pairwise orthogonal. That is, if $alphainPhi_i$, $betainPhi_j$, $ineq j$, why is $(alpha, beta)=0$?
By definition, $(alpha, beta)=kappa(t_{alpha}, t_{beta})$, where $kappa$ denotes the Killing form and, for $gammain H^*$, $t_{gamma}$ is the unique element of $H$ such that $gamma(h)=kappa(t_{gamma}, h)$ for all $hin H$. So, we need to show that $alpha(t_{beta})=0$. Since $alpha(H_k)=0$ for $kneq i$, we really only need to show that $alpha$ annihilates $h_i$, where $t_{beta}=h_1+cdots +h_t$, $h_kin H_k$. From here, I'm not sure how to proceed.
representation-theory lie-algebras root-systems nonassociative-algebras
asked Jan 3 at 7:23
ZilliputZilliput
1,028720
$endgroup$
add a comment |
$begingroup$
I'm reading Humphreys' Introduction to Lie Algebras and Representation Theory and I have a question about Corollary 14.1, which reads:
Humphreys Corollary 14.1. Let $L$ be a semisimple Lie algebra, with maximal toral subalgebra $H$ and root system $Phi$. If $L = L_1opluscdotsoplus L_t$ is the decomposition of $L$ into simple ideals, then $H_i = Hcap L_i$ is a maximal toral subalgebra of $L_i$, and the corresponding (irreducible) root system $Phi_i$ may be regarded canonically as a subsystem of $Phi$ in such a way that $Phi=Phi_1cupcdotscupPhi_t$ is the decomposition of $Phi$ into its irreducible components.
I understand most of this. What I'm struggling to understand is why the $Phi_i$ must be pairwise orthogonal. That is, if $alphainPhi_i$, $betainPhi_j$, $ineq j$, why is $(alpha, beta)=0$?
By definition, $(alpha, beta)=kappa(t_{alpha}, t_{beta})$, where $kappa$ denotes the Killing form and, for $gammain H^*$, $t_{gamma}$ is the unique element of $H$ such that $gamma(h)=kappa(t_{gamma}, h)$ for all $hin H$. So, we need to show that $alpha(t_{beta})=0$. Since $alpha(H_k)=0$ for $kneq i$, we really only need to show that $alpha$ annihilates $h_i$, where $t_{beta}=h_1+cdots +h_t$, $h_kin H_k$. From here, I'm not sure how to proceed.
representation-theory lie-algebras root-systems nonassociative-algebras
asked Jan 3 at 7:23
ZilliputZilliput
1,028720
$endgroup$
$begingroup$
$t_alphain L_i$, $t_betain L_j$. Doesn't that already imply that the composition of $operatorname{ad}(t_alpha)$ and $operatorname{ad}(t_beta)$ is the zero map?
$endgroup$
– Jyrki Lahtonen
Jan 3 at 7:50
$begingroup$
@JyrkiLahtonen I'm probably missing something obvious, but why must we have $t_{alpha}in L_i$ and $t_{beta}in L_j$?
$endgroup$
– Zilliput
Jan 3 at 7:59
$begingroup$
Is your original question, or the one in the comment, answered by one of these: math.stackexchange.com/q/1357074/96384 and math.stackexchange.com/q/2982139/96384 ?
$endgroup$
– Torsten Schoeneberg
Jan 3 at 18:37
$begingroup$
@TorstenSchoeneberg As far as I can tell, no.
$endgroup$
– Zilliput
Jan 3 at 18:58
add a comment |
$begingroup$
I'm reading Humphreys' Introduction to Lie Algebras and Representation Theory and I have a question about Corollary 14.1, which reads:
Humphreys Corollary 14.1. Let $L$ be a semisimple Lie algebra, with maximal toral subalgebra $H$ and root system $Phi$. If $L = L_1opluscdotsoplus L_t$ is the decomposition of $L$ into simple ideals, then $H_i = Hcap L_i$ is a maximal toral subalgebra of $L_i$, and the corresponding (irreducible) root system $Phi_i$ may be regarded canonically as a subsystem of $Phi$ in such a way that $Phi=Phi_1cupcdotscupPhi_t$ is the decomposition of $Phi$ into its irreducible components.
I understand most of this. What I'm struggling to understand is why the $Phi_i$ must be pairwise orthogonal. That is, if $alphainPhi_i$, $betainPhi_j$, $ineq j$, why is $(alpha, beta)=0$?
By definition, $(alpha, beta)=kappa(t_{alpha}, t_{beta})$, where $kappa$ denotes the Killing form and, for $gammain H^*$, $t_{gamma}$ is the unique element of $H$ such that $gamma(h)=kappa(t_{gamma}, h)$ for all $hin H$. So, we need to show that $alpha(t_{beta})=0$. Since $alpha(H_k)=0$ for $kneq i$, we really only need to show that $alpha$ annihilates $h_i$, where $t_{beta}=h_1+cdots +h_t$, $h_kin H_k$. From here, I'm not sure how to proceed.
representation-theory lie-algebras root-systems nonassociative-algebras
asked Jan 3 at 7:23
ZilliputZilliput
1,028720
$endgroup$
I'm reading Humphreys' Introduction to Lie Algebras and Representation Theory and I have a question about Corollary 14.1, which reads:
Humphreys Corollary 14.1. Let $L$ be a semisimple Lie algebra, with maximal toral subalgebra $H$ and root system $Phi$. If $L = L_1opluscdotsoplus L_t$ is the decomposition of $L$ into simple ideals, then $H_i = Hcap L_i$ is a maximal toral subalgebra of $L_i$, and the corresponding (irreducible) root system $Phi_i$ may be regarded canonically as a subsystem of $Phi$ in such a way that $Phi=Phi_1cupcdotscupPhi_t$ is the decomposition of $Phi$ into its irreducible components.
I understand most of this. What I'm struggling to understand is why the $Phi_i$ must be pairwise orthogonal. That is, if $alphainPhi_i$, $betainPhi_j$, $ineq j$, why is $(alpha, beta)=0$?
By definition, $(alpha, beta)=kappa(t_{alpha}, t_{beta})$, where $kappa$ denotes the Killing form and, for $gammain H^*$, $t_{gamma}$ is the unique element of $H$ such that $gamma(h)=kappa(t_{gamma}, h)$ for all $hin H$. So, we need to show that $alpha(t_{beta})=0$. Since $alpha(H_k)=0$ for $kneq i$, we really only need to show that $alpha$ annihilates $h_i$, where $t_{beta}=h_1+cdots +h_t$, $h_kin H_k$. From here, I'm not sure how to proceed.
representation-theory lie-algebras root-systems nonassociative-algebras
representation-theory lie-algebras root-systems nonassociative-algebras
asked Jan 3 at 7:23
ZilliputZilliput
1,028720
asked Jan 3 at 7:23
ZilliputZilliput
1,028720
edited Jan 3 at 14:29
user593746
asked Jan 3 at 7:23
ZilliputZilliput
1,028720
asked Jan 3 at 7:23
ZilliputZilliput
1,028720
asked Jan 3 at 7:23
ZilliputZilliput
1,028720
1,028720
$begingroup$
$t_alphain L_i$, $t_betain L_j$. Doesn't that already imply that the composition of $operatorname{ad}(t_alpha)$ and $operatorname{ad}(t_beta)$ is the zero map?
$endgroup$
– Jyrki Lahtonen
Jan 3 at 7:50
$begingroup$
@JyrkiLahtonen I'm probably missing something obvious, but why must we have $t_{alpha}in L_i$ and $t_{beta}in L_j$?
$endgroup$
– Zilliput
Jan 3 at 7:59
$begingroup$
Is your original question, or the one in the comment, answered by one of these: math.stackexchange.com/q/1357074/96384 and math.stackexchange.com/q/2982139/96384 ?
$endgroup$
– Torsten Schoeneberg
Jan 3 at 18:37
$begingroup$
@TorstenSchoeneberg As far as I can tell, no.
$endgroup$
– Zilliput
Jan 3 at 18:58
add a comment |
$begingroup$
$t_alphain L_i$, $t_betain L_j$. Doesn't that already imply that the composition of $operatorname{ad}(t_alpha)$ and $operatorname{ad}(t_beta)$ is the zero map?
$endgroup$
– Jyrki Lahtonen
Jan 3 at 7:50
$begingroup$
@JyrkiLahtonen I'm probably missing something obvious, but why must we have $t_{alpha}in L_i$ and $t_{beta}in L_j$?
$endgroup$
– Zilliput
Jan 3 at 7:59
$begingroup$
Is your original question, or the one in the comment, answered by one of these: math.stackexchange.com/q/1357074/96384 and math.stackexchange.com/q/2982139/96384 ?
$endgroup$
– Torsten Schoeneberg
Jan 3 at 18:37
$begingroup$
@TorstenSchoeneberg As far as I can tell, no.
$endgroup$
– Zilliput
Jan 3 at 18:58
$begingroup$
$t_alphain L_i$, $t_betain L_j$. Doesn't that already imply that the composition of $operatorname{ad}(t_alpha)$ and $operatorname{ad}(t_beta)$ is the zero map?
$endgroup$
– Jyrki Lahtonen
Jan 3 at 7:50
$begingroup$
$t_alphain L_i$, $t_betain L_j$. Doesn't that already imply that the composition of $operatorname{ad}(t_alpha)$ and $operatorname{ad}(t_beta)$ is the zero map?
$endgroup$
– Jyrki Lahtonen
Jan 3 at 7:50
$begingroup$
@JyrkiLahtonen I'm probably missing something obvious, but why must we have $t_{alpha}in L_i$ and $t_{beta}in L_j$?
$endgroup$
– Zilliput
Jan 3 at 7:59
$begingroup$
@JyrkiLahtonen I'm probably missing something obvious, but why must we have $t_{alpha}in L_i$ and $t_{beta}in L_j$?
$endgroup$
– Zilliput
Jan 3 at 7:59
$begingroup$
Is your original question, or the one in the comment, answered by one of these: math.stackexchange.com/q/1357074/96384 and math.stackexchange.com/q/2982139/96384 ?
$endgroup$
– Torsten Schoeneberg
Jan 3 at 18:37
$begingroup$
Is your original question, or the one in the comment, answered by one of these: math.stackexchange.com/q/1357074/96384 and math.stackexchange.com/q/2982139/96384 ?
$endgroup$
– Torsten Schoeneberg
Jan 3 at 18:37
$begingroup$
@TorstenSchoeneberg As far as I can tell, no.
$endgroup$
– Zilliput
Jan 3 at 18:58
$begingroup$
@TorstenSchoeneberg As far as I can tell, no.
$endgroup$
– Zilliput
Jan 3 at 18:58
add a comment |
1 Answer
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As @JyrkiLahtonen commented, it suffices to show that $t_{alpha}in L_i$ and $t_{beta}in L_j$, since this implies that $operatorname{ad}t_{alpha}operatorname{ad}t_{beta}(x)=[t_{alpha}[t_{beta}x]]in L_icap L_j=0$ for all $xin L$, i.e., $operatorname{ad}t_{alpha}operatorname{ad}t_{beta}=0$, so $$(alpha, beta)=kappa(t_{alpha}, t_{beta})=operatorname{tr}(operatorname{ad}t_{alpha}operatorname{ad}t_{beta})=operatorname{tr}(0)=0.$$
To see that $t_{alpha}in L_i$, let $kappa_i$ denote the Killing form of $L_i$. By Humphreys Lemma 5.1, $kappa_i=kappavert_{L_itimes L_i}$. By Humphreys Corollary 8.2, $kappavert_{Htimes H}$ and $kappa_ivert_{H_itimes H_i}$ are nondegenerate. This allows us to identify $H$ with $H^*$, by associating to $gammain H^*$ the unique element $t_{gamma}in H$ such that $gamma(h)=kappa(t_{gamma}, h)$ for all $hin H$. Similarly, we can identify $H_i$ with $H_i^*$, by associating to $deltain H_i^*$ the unique element $u_{delta}in H_i$ such that $delta(h_i)=kappa_i(u_{delta}, h_i)$ for all $h_iin H_i$. With this in mind, $u_{alpha}in H_i$ and $alpha(h_i)=kappa_i(u_{alpha}, h_i)$ for all $h_iin H_i$. However, by an argument similar to that in the previous paragraph, $kappa(L_i, L_k)=0$ for $ineq k$, so given $h=h_1+cdots+h_tin H$, $h_kin H_k$, we have $$alpha(h)=alpha(h_i)=kappa_i(u_{alpha}, h_i)=kappavert_{L_itimes L_i}(u_{alpha}, h_i)=kappa(u_{alpha}, h_i)=kappa(u_{alpha}, h).$$ Then, by uniqueness, $t_{alpha}=u_{alpha}in H_isubseteq L_i$. Similarly, $t_{beta}in L_j$.
answered Jan 3 at 22:09
ZilliputZilliput
1,028720
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add a comment |
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$begingroup$
As @JyrkiLahtonen commented, it suffices to show that $t_{alpha}in L_i$ and $t_{beta}in L_j$, since this implies that $operatorname{ad}t_{alpha}operatorname{ad}t_{beta}(x)=[t_{alpha}[t_{beta}x]]in L_icap L_j=0$ for all $xin L$, i.e., $operatorname{ad}t_{alpha}operatorname{ad}t_{beta}=0$, so $$(alpha, beta)=kappa(t_{alpha}, t_{beta})=operatorname{tr}(operatorname{ad}t_{alpha}operatorname{ad}t_{beta})=operatorname{tr}(0)=0.$$
To see that $t_{alpha}in L_i$, let $kappa_i$ denote the Killing form of $L_i$. By Humphreys Lemma 5.1, $kappa_i=kappavert_{L_itimes L_i}$. By Humphreys Corollary 8.2, $kappavert_{Htimes H}$ and $kappa_ivert_{H_itimes H_i}$ are nondegenerate. This allows us to identify $H$ with $H^*$, by associating to $gammain H^*$ the unique element $t_{gamma}in H$ such that $gamma(h)=kappa(t_{gamma}, h)$ for all $hin H$. Similarly, we can identify $H_i$ with $H_i^*$, by associating to $deltain H_i^*$ the unique element $u_{delta}in H_i$ such that $delta(h_i)=kappa_i(u_{delta}, h_i)$ for all $h_iin H_i$. With this in mind, $u_{alpha}in H_i$ and $alpha(h_i)=kappa_i(u_{alpha}, h_i)$ for all $h_iin H_i$. However, by an argument similar to that in the previous paragraph, $kappa(L_i, L_k)=0$ for $ineq k$, so given $h=h_1+cdots+h_tin H$, $h_kin H_k$, we have $$alpha(h)=alpha(h_i)=kappa_i(u_{alpha}, h_i)=kappavert_{L_itimes L_i}(u_{alpha}, h_i)=kappa(u_{alpha}, h_i)=kappa(u_{alpha}, h).$$ Then, by uniqueness, $t_{alpha}=u_{alpha}in H_isubseteq L_i$. Similarly, $t_{beta}in L_j$.
answered Jan 3 at 22:09
ZilliputZilliput
1,028720
$endgroup$
add a comment |
$begingroup$
As @JyrkiLahtonen commented, it suffices to show that $t_{alpha}in L_i$ and $t_{beta}in L_j$, since this implies that $operatorname{ad}t_{alpha}operatorname{ad}t_{beta}(x)=[t_{alpha}[t_{beta}x]]in L_icap L_j=0$ for all $xin L$, i.e., $operatorname{ad}t_{alpha}operatorname{ad}t_{beta}=0$, so $$(alpha, beta)=kappa(t_{alpha}, t_{beta})=operatorname{tr}(operatorname{ad}t_{alpha}operatorname{ad}t_{beta})=operatorname{tr}(0)=0.$$
To see that $t_{alpha}in L_i$, let $kappa_i$ denote the Killing form of $L_i$. By Humphreys Lemma 5.1, $kappa_i=kappavert_{L_itimes L_i}$. By Humphreys Corollary 8.2, $kappavert_{Htimes H}$ and $kappa_ivert_{H_itimes H_i}$ are nondegenerate. This allows us to identify $H$ with $H^*$, by associating to $gammain H^*$ the unique element $t_{gamma}in H$ such that $gamma(h)=kappa(t_{gamma}, h)$ for all $hin H$. Similarly, we can identify $H_i$ with $H_i^*$, by associating to $deltain H_i^*$ the unique element $u_{delta}in H_i$ such that $delta(h_i)=kappa_i(u_{delta}, h_i)$ for all $h_iin H_i$. With this in mind, $u_{alpha}in H_i$ and $alpha(h_i)=kappa_i(u_{alpha}, h_i)$ for all $h_iin H_i$. However, by an argument similar to that in the previous paragraph, $kappa(L_i, L_k)=0$ for $ineq k$, so given $h=h_1+cdots+h_tin H$, $h_kin H_k$, we have $$alpha(h)=alpha(h_i)=kappa_i(u_{alpha}, h_i)=kappavert_{L_itimes L_i}(u_{alpha}, h_i)=kappa(u_{alpha}, h_i)=kappa(u_{alpha}, h).$$ Then, by uniqueness, $t_{alpha}=u_{alpha}in H_isubseteq L_i$. Similarly, $t_{beta}in L_j$.
answered Jan 3 at 22:09
ZilliputZilliput
1,028720
$endgroup$
add a comment |
$begingroup$
As @JyrkiLahtonen commented, it suffices to show that $t_{alpha}in L_i$ and $t_{beta}in L_j$, since this implies that $operatorname{ad}t_{alpha}operatorname{ad}t_{beta}(x)=[t_{alpha}[t_{beta}x]]in L_icap L_j=0$ for all $xin L$, i.e., $operatorname{ad}t_{alpha}operatorname{ad}t_{beta}=0$, so $$(alpha, beta)=kappa(t_{alpha}, t_{beta})=operatorname{tr}(operatorname{ad}t_{alpha}operatorname{ad}t_{beta})=operatorname{tr}(0)=0.$$
To see that $t_{alpha}in L_i$, let $kappa_i$ denote the Killing form of $L_i$. By Humphreys Lemma 5.1, $kappa_i=kappavert_{L_itimes L_i}$. By Humphreys Corollary 8.2, $kappavert_{Htimes H}$ and $kappa_ivert_{H_itimes H_i}$ are nondegenerate. This allows us to identify $H$ with $H^*$, by associating to $gammain H^*$ the unique element $t_{gamma}in H$ such that $gamma(h)=kappa(t_{gamma}, h)$ for all $hin H$. Similarly, we can identify $H_i$ with $H_i^*$, by associating to $deltain H_i^*$ the unique element $u_{delta}in H_i$ such that $delta(h_i)=kappa_i(u_{delta}, h_i)$ for all $h_iin H_i$. With this in mind, $u_{alpha}in H_i$ and $alpha(h_i)=kappa_i(u_{alpha}, h_i)$ for all $h_iin H_i$. However, by an argument similar to that in the previous paragraph, $kappa(L_i, L_k)=0$ for $ineq k$, so given $h=h_1+cdots+h_tin H$, $h_kin H_k$, we have $$alpha(h)=alpha(h_i)=kappa_i(u_{alpha}, h_i)=kappavert_{L_itimes L_i}(u_{alpha}, h_i)=kappa(u_{alpha}, h_i)=kappa(u_{alpha}, h).$$ Then, by uniqueness, $t_{alpha}=u_{alpha}in H_isubseteq L_i$. Similarly, $t_{beta}in L_j$.
answered Jan 3 at 22:09
ZilliputZilliput
1,028720
$endgroup$
As @JyrkiLahtonen commented, it suffices to show that $t_{alpha}in L_i$ and $t_{beta}in L_j$, since this implies that $operatorname{ad}t_{alpha}operatorname{ad}t_{beta}(x)=[t_{alpha}[t_{beta}x]]in L_icap L_j=0$ for all $xin L$, i.e., $operatorname{ad}t_{alpha}operatorname{ad}t_{beta}=0$, so $$(alpha, beta)=kappa(t_{alpha}, t_{beta})=operatorname{tr}(operatorname{ad}t_{alpha}operatorname{ad}t_{beta})=operatorname{tr}(0)=0.$$
To see that $t_{alpha}in L_i$, let $kappa_i$ denote the Killing form of $L_i$. By Humphreys Lemma 5.1, $kappa_i=kappavert_{L_itimes L_i}$. By Humphreys Corollary 8.2, $kappavert_{Htimes H}$ and $kappa_ivert_{H_itimes H_i}$ are nondegenerate. This allows us to identify $H$ with $H^*$, by associating to $gammain H^*$ the unique element $t_{gamma}in H$ such that $gamma(h)=kappa(t_{gamma}, h)$ for all $hin H$. Similarly, we can identify $H_i$ with $H_i^*$, by associating to $deltain H_i^*$ the unique element $u_{delta}in H_i$ such that $delta(h_i)=kappa_i(u_{delta}, h_i)$ for all $h_iin H_i$. With this in mind, $u_{alpha}in H_i$ and $alpha(h_i)=kappa_i(u_{alpha}, h_i)$ for all $h_iin H_i$. However, by an argument similar to that in the previous paragraph, $kappa(L_i, L_k)=0$ for $ineq k$, so given $h=h_1+cdots+h_tin H$, $h_kin H_k$, we have $$alpha(h)=alpha(h_i)=kappa_i(u_{alpha}, h_i)=kappavert_{L_itimes L_i}(u_{alpha}, h_i)=kappa(u_{alpha}, h_i)=kappa(u_{alpha}, h).$$ Then, by uniqueness, $t_{alpha}=u_{alpha}in H_isubseteq L_i$. Similarly, $t_{beta}in L_j$.
answered Jan 3 at 22:09
ZilliputZilliput
1,028720
answered Jan 3 at 22:09
ZilliputZilliput
1,028720
answered Jan 3 at 22:09
ZilliputZilliput
1,028720
answered Jan 3 at 22:09
ZilliputZilliput
1,028720
1,028720
add a comment |
add a comment |
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$begingroup$
$t_alphain L_i$, $t_betain L_j$. Doesn't that already imply that the composition of $operatorname{ad}(t_alpha)$ and $operatorname{ad}(t_beta)$ is the zero map?
$endgroup$
– Jyrki Lahtonen
Jan 3 at 7:50
$begingroup$
@JyrkiLahtonen I'm probably missing something obvious, but why must we have $t_{alpha}in L_i$ and $t_{beta}in L_j$?
$endgroup$
– Zilliput
Jan 3 at 7:59
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Is your original question, or the one in the comment, answered by one of these: math.stackexchange.com/q/1357074/96384 and math.stackexchange.com/q/2982139/96384 ?
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– Torsten Schoeneberg
Jan 3 at 18:37
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@TorstenSchoeneberg As far as I can tell, no.
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– Zilliput
Jan 3 at 18:58