Timeline for Extension of an involutive automorphism
Current License: CC BY-SA 3.0
13 events
| when toggle format | what | by | license | comment | |
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| Dec 27, 2014 at 16:16 | vote | accept | Hebe | ||
| Dec 24, 2014 at 15:59 | comment | added | Paul Levy | If $\tau$ is the `standard' such choice of involution then yes, there is such an extension. To see this, let $\alpha_1,\alpha_2,\alpha_3,\alpha_4$ be a basis for the roots in $\mathfrak{so}_9$, so $\alpha_1,\alpha_2,\alpha_3,\alpha_3+2\alpha_4$ is a basis for $\mathfrak{so}_9$. Then the graph automorphism of $D_4$ which swaps $\alpha_3$ and $\alpha_3+2\alpha_4$ is given by the reflection $s_4$ in $W(B_4)$. On the other hand, the involution of $\mathfrak{so}_8$ which swaps $\alpha_1$ and $\alpha_3$ can't be lifted to an automorphism of $\mathfrak{so}_9$. | |
| Dec 24, 2014 at 0:06 | history | edited | Jim Humphreys | CC BY-SA 3.0 |
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| Dec 23, 2014 at 13:17 | comment | added | Hebe | For example, the canonical embedding $so(8,\mathbb{C})$ to $so(9,\mathbb{C})$. Let $\tau$ be an involutive automorphism of $so(8,\mathbb{C})$ which defines the symmetric pair $(so(8,\mathbb{C}),so(7,\mathbb{C}))$. Can $\tau$ be extended to an automorphism of $so(9,\mathbb{C})$? I cannot not find one, but I am not sure whether it exists. | |
| Dec 23, 2014 at 13:07 | comment | added | Hebe | Good example. But I am thinking of the case that $\mathfrak{g}$ is simple and $\mathfrak{g}'$ is reductive with its commutator $[\mathfrak{g}',\mathfrak{g}']$ simple. | |
| Dec 23, 2014 at 13:00 | comment | added | Paul Levy | No: let ${\mathfrak g}$ be of type $G_2$ and ${\mathfrak g}'=\mathfrak{sl}_2\oplus\mathfrak{sl}_2$ the span of the $\alpha_1$- and $\hat\alpha$-copies of $\mathfrak{sl}_2$. Then there is an involution of ${\mathfrak g}'$ which swaps the two copies of $\mathfrak{sl}_2$, but this can't be lifted to an involution of $G_2$ (except in characteristic 3) since they come from roots of different length. On the other hand, if $\mathfrak{g}$ and $\mathfrak{g}'$ are simple then the statement may be true. There is a triality auto of $\mathfrak{so}_8$ which doesn't lift to an auto of $\mathfrak{so}_9$. | |
| Dec 23, 2014 at 12:51 | comment | added | Hebe | Thanks for your example. Then can $\tau$ be always extended to an automorphism (not necessary involutive) of $\mathfrak{g}$? | |
| Dec 22, 2014 at 21:39 | comment | added | Paul Levy | Even for inner automorphisms, and ${\mathfrak g}',{\mathfrak g}$ semisimple of equal rank, the answer is no. Let ${\mathfrak g}=\mathfrak{so}_{2n+1}$, ${\mathfrak g}'=\mathfrak{so}_{2n}$. Let $\{\alpha_1,\ldots,\alpha_n\}$ be a root basis in ${\mathfrak g}$ and $\{\alpha_1,\ldots,\alpha_{n-1},\alpha_{n-1}+2\alpha_n\}$ a root basis in ${\mathfrak g}'$. Then the involution of ${\mathfrak g}'$ which satisfies $e_{\pm\alpha_i}\mapsto e_{\pm\alpha_i}$ for $1\leq i\leq n-1$ and $e_{\pm(\alpha_{n-1}+2\alpha_n)}\mapsto -e_{\pm(\alpha_{n-1}+2\alpha_n)}$ only lifts to an order 4 auto of ${\mathfrak g}$. | |
| Dec 22, 2014 at 21:16 | comment | added | Paul Levy | While I agree that there are lots of pairs $({\mathfrak g},{\mathfrak g}')$ for which the answer to the question is no, I don't think your suggested pair is such an example. In particular, it isn't true that the centralizers of these $\mathfrak{sl}_2$s are of different dimensions. (In both cases we get a copy of $\mathfrak{sl}_6$.) If we take ${\mathfrak g}'$ to be the span of the 3 rank 1 Lie subalgebras given by $\alpha_1,\alpha_6$ and $\alpha_2$, with the involution swapping the $\alpha_2$ and the $\alpha_6$, then this should work. | |
| Dec 22, 2014 at 16:55 | comment | added | Jim Humphreys | In examples where $\mathfrak{g}$ has a pseudo-Levi proper subalgebra of the same semisimple rank, I'd again expect some negative answers, e.g., $A_2 \subset G_2$ and $D_8 \subset E_8$ have graph automorphisms of order 2 which probably couldn't extend to involutive inner automorphisms. This too would need some checking. | |
| Dec 22, 2014 at 10:43 | comment | added | Hebe | Thank you for your comment and answer. Yes, you are right. In general, it is not true. But what if the semisimple ranks of \mathfrak{g} and \mathfrak{g}' are equal? | |
| Dec 21, 2014 at 20:32 | history | edited | Jim Humphreys | CC BY-SA 3.0 |
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| Dec 21, 2014 at 19:06 | history | answered | Jim Humphreys | CC BY-SA 3.0 |