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George Melvin
George Melvin
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In Jantzen's AMS text 'Lectures on Quantum Groups' he makes the following remark (p.187, preface to Chapter 9):

"For general (complex semisimple f.d. Lie algebra) $\frak{g}$ we can consider for each simple root $\alpha$... a Lie subalgebra (of $\frak{g}$) isomorphic to $\frak{sl}_{2}$ (ie, the Lie subalgebra $\frak{s}_{\alpha}$ generated by suitable $X_{\alpha}\in \frak{g}_{\alpha}$, $Y_{\alpha}\in\frak{g}_{-\alpha}$). So, if $M$ is a f.d. $\frak{g}$-module, then one can find (for fixed $\alpha$) a basis $v_{1},\ldots,v_{n}$ such that $Y_{\alpha}v_{i}$ is either $0$ or a nonzero multiple of another $v_{j}$ and such that also each $X_{\alpha}v_{h}$ is either $0$ or a nonzero multiple of another $v_{l}$. However, in general, there does not exist a basis that works simultaneously for all simple $\alpha$. (There are exceptions, such as the adjoint representations...)..."

The first part of this statement is standard (we are applying complete reducibility of $M$ as an $\frak{s}_{\alpha}$-module). However, I hope that someone can illuminate the last sentence on 'exceptions' - is this a typo/mis-statement? Or am I missing something here?

It is not possible to obtain a simultaneous basis for the adjoint representation of $\frak{sl}_{3}$: indeed, if $\alpha_{1},\alpha_{2}$ are the simple roots and $\frak{s}_{1},\frak{s}_{2}$ the corresponding $\frak{sl}_{2}$-triples then we can decompose $\frak{sl}_{3}$ as

$\frak{sl}_{3}$$\cong L(1)\oplus L(2)\oplus L(0)\oplus L(1)$

when we consider $\frak{sl}_{3}$ as either a $\frak{s}_{1}$- or $\frak{s}_{2}$-module. Here, $L(n)$ is the irreducible $\frak{sl}_{2}$-module of dimension $n+1$. Also, in both decompositions we have $L(0)$ appears as a subspace of $\frak{h}$ (the $0$-weight space).

If we were to have a simultaneous basis as described above we would need a basis vector $u\in\frak{h}\subset\frak{sl}_{3}$ corresponding to the copy of $L(0)$ appearing in the $\frak{s}_{1}$- and $\frak{s}_{2}$-decompositions of $M$$\frak{sl}_{3}$. This would imply that $\ker ad \;X_{\alpha_{1}}\cap \ker ad \; X_{\alpha_{2}}\cap \frak{h}$ is nonzero, which is impossible (as can be seen by a basic calculation) since we are in characteristic $0$.

Thanks in advance for your comments.

In Jantzen's AMS text 'Lectures on Quantum Groups' he makes the following remark (p.187, preface to Chapter 9):

"For general (complex semisimple f.d. Lie algebra) $\frak{g}$ we can consider for each simple root $\alpha$... a Lie subalgebra (of $\frak{g}$) isomorphic to $\frak{sl}_{2}$ (ie, the Lie subalgebra $\frak{s}_{\alpha}$ generated by suitable $X_{\alpha}\in \frak{g}_{\alpha}$, $Y_{\alpha}\in\frak{g}_{-\alpha}$). So, if $M$ is a f.d. $\frak{g}$-module, then one can find (for fixed $\alpha$) a basis $v_{1},\ldots,v_{n}$ such that $Y_{\alpha}v_{i}$ is either $0$ or a nonzero multiple of another $v_{j}$ and such that also each $X_{\alpha}v_{h}$ is either $0$ or a nonzero multiple of another $v_{l}$. However, in general, there does not exist a basis that works simultaneously for all simple $\alpha$. (There are exceptions, such as the adjoint representations...)..."

The first part of this statement is standard (we are applying complete reducibility of $M$ as an $\frak{s}_{\alpha}$-module). However, I hope that someone can illuminate the last sentence on 'exceptions' - is this a typo/mis-statement? Or am I missing something here?

It is not possible to obtain a simultaneous basis for the adjoint representation of $\frak{sl}_{3}$: indeed, if $\alpha_{1},\alpha_{2}$ are the simple roots and $\frak{s}_{1},\frak{s}_{2}$ the corresponding $\frak{sl}_{2}$-triples then we can decompose $\frak{sl}_{3}$ as

$\frak{sl}_{3}$$\cong L(1)\oplus L(2)\oplus L(0)\oplus L(1)$

when we consider $\frak{sl}_{3}$ as either a $\frak{s}_{1}$- or $\frak{s}_{2}$-module. Here, $L(n)$ is the irreducible $\frak{sl}_{2}$-module of dimension $n+1$. Also, in both decompositions we have $L(0)$ appears as a subspace of $\frak{h}$ (the $0$-weight space).

If we were to have a simultaneous basis as described above we would need a basis vector $u\in\frak{h}\subset\frak{sl}_{3}$ corresponding to the copy of $L(0)$ appearing in the $\frak{s}_{1}$- and $\frak{s}_{2}$-decompositions of $M$. This would imply that $\ker ad \;X_{\alpha_{1}}\cap \ker ad \; X_{\alpha_{2}}\cap \frak{h}$ is nonzero, which is impossible (as can be seen by a basic calculation) since we are in characteristic $0$.

Thanks in advance for your comments.

In Jantzen's AMS text 'Lectures on Quantum Groups' he makes the following remark (p.187, preface to Chapter 9):

"For general (complex semisimple f.d. Lie algebra) $\frak{g}$ we can consider for each simple root $\alpha$... a Lie subalgebra (of $\frak{g}$) isomorphic to $\frak{sl}_{2}$ (ie, the Lie subalgebra $\frak{s}_{\alpha}$ generated by suitable $X_{\alpha}\in \frak{g}_{\alpha}$, $Y_{\alpha}\in\frak{g}_{-\alpha}$). So, if $M$ is a f.d. $\frak{g}$-module, then one can find (for fixed $\alpha$) a basis $v_{1},\ldots,v_{n}$ such that $Y_{\alpha}v_{i}$ is either $0$ or a nonzero multiple of another $v_{j}$ and such that also each $X_{\alpha}v_{h}$ is either $0$ or a nonzero multiple of another $v_{l}$. However, in general, there does not exist a basis that works simultaneously for all simple $\alpha$. (There are exceptions, such as the adjoint representations...)..."

The first part of this statement is standard (we are applying complete reducibility of $M$ as an $\frak{s}_{\alpha}$-module). However, I hope that someone can illuminate the last sentence on 'exceptions' - is this a typo/mis-statement? Or am I missing something here?

It is not possible to obtain a simultaneous basis for the adjoint representation of $\frak{sl}_{3}$: indeed, if $\alpha_{1},\alpha_{2}$ are the simple roots and $\frak{s}_{1},\frak{s}_{2}$ the corresponding $\frak{sl}_{2}$-triples then we can decompose $\frak{sl}_{3}$ as

$\frak{sl}_{3}$$\cong L(1)\oplus L(2)\oplus L(0)\oplus L(1)$

when we consider $\frak{sl}_{3}$ as either a $\frak{s}_{1}$- or $\frak{s}_{2}$-module. Here, $L(n)$ is the irreducible $\frak{sl}_{2}$-module of dimension $n+1$. Also, in both decompositions we have $L(0)$ appears as a subspace of $\frak{h}$ (the $0$-weight space).

If we were to have a simultaneous basis as described above we would need a basis vector $u\in\frak{h}\subset\frak{sl}_{3}$ corresponding to the copy of $L(0)$ appearing in the $\frak{s}_{1}$- and $\frak{s}_{2}$-decompositions of $\frak{sl}_{3}$. This would imply that $\ker ad \;X_{\alpha_{1}}\cap \ker ad \; X_{\alpha_{2}}\cap \frak{h}$ is nonzero, which is impossible (as can be seen by a basic calculation) since we are in characteristic $0$.

Thanks in advance for your comments.

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George Melvin
George Melvin
  • 1.2k
  • 8
  • 15

A remark in Jantzen's 'Lectures on Quantum Groups'

In Jantzen's AMS text 'Lectures on Quantum Groups' he makes the following remark (p.187, preface to Chapter 9):

"For general (complex semisimple f.d. Lie algebra) $\frak{g}$ we can consider for each simple root $\alpha$... a Lie subalgebra (of $\frak{g}$) isomorphic to $\frak{sl}_{2}$ (ie, the Lie subalgebra $\frak{s}_{\alpha}$ generated by suitable $X_{\alpha}\in \frak{g}_{\alpha}$, $Y_{\alpha}\in\frak{g}_{-\alpha}$). So, if $M$ is a f.d. $\frak{g}$-module, then one can find (for fixed $\alpha$) a basis $v_{1},\ldots,v_{n}$ such that $Y_{\alpha}v_{i}$ is either $0$ or a nonzero multiple of another $v_{j}$ and such that also each $X_{\alpha}v_{h}$ is either $0$ or a nonzero multiple of another $v_{l}$. However, in general, there does not exist a basis that works simultaneously for all simple $\alpha$. (There are exceptions, such as the adjoint representations...)..."

The first part of this statement is standard (we are applying complete reducibility of $M$ as an $\frak{s}_{\alpha}$-module). However, I hope that someone can illuminate the last sentence on 'exceptions' - is this a typo/mis-statement? Or am I missing something here?

It is not possible to obtain a simultaneous basis for the adjoint representation of $\frak{sl}_{3}$: indeed, if $\alpha_{1},\alpha_{2}$ are the simple roots and $\frak{s}_{1},\frak{s}_{2}$ the corresponding $\frak{sl}_{2}$-triples then we can decompose $\frak{sl}_{3}$ as

$\frak{sl}_{3}$$\cong L(1)\oplus L(2)\oplus L(0)\oplus L(1)$

when we consider $\frak{sl}_{3}$ as either a $\frak{s}_{1}$- or $\frak{s}_{2}$-module. Here, $L(n)$ is the irreducible $\frak{sl}_{2}$-module of dimension $n+1$. Also, in both decompositions we have $L(0)$ appears as a subspace of $\frak{h}$ (the $0$-weight space).

If we were to have a simultaneous basis as described above we would need a basis vector $u\in\frak{h}\subset\frak{sl}_{3}$ corresponding to the copy of $L(0)$ appearing in the $\frak{s}_{1}$- and $\frak{s}_{2}$-decompositions of $M$. This would imply that $\ker ad \;X_{\alpha_{1}}\cap \ker ad \; X_{\alpha_{2}}\cap \frak{h}$ is nonzero, which is impossible (as can be seen by a basic calculation) since we are in characteristic $0$.

Thanks in advance for your comments.