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Suppose that $g$ is a complex semi-simple Lie algebra and $g'$ its reductive subalgebra.

If $\tau$ is an involutive automorphism of $g'$, can $\tau$ be extended to an involutive automorphism of $g$ in general?

If not in general, for what kind of pair $(g, g')$ or $\tau$, this is true?

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    $\begingroup$ This may be a reasonable question, but you have to keep in mind that g' can be considerably smaller than g. Moreover, g' might have an involutive outer automorphism while g doesn't; so there might be a complicated mix of inner/outer here. I'm not optimistic about getting a useful criterion for an arbitrary pair (g,g'). $\endgroup$
    – Jim Humphreys
    Commented Dec 21, 2014 at 15:30

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[REVISED]

The answer to the first question is NO. It's difficult as a rule to show that an involutive automorphism of $\mathfrak{g'}$ fails to extend to $\mathfrak{g}$, so it may be better to consider first the possible automorphisms of a simple Lie algebra $\mathfrak{g}$ (up to conjugacy in the adjoint group). This is thoroughly worked out in the context of many other classification problems, in the classic 1978 book Differential Geometry, Lie Groups, and Symmetric Spaces by S. Helgason (reprinted as GSM 34, AMS, 2001). In Chapter X, $\S5$, Helgason studies automorphisms in the style of Victor Kac, which like other topics in this long chapter requires quite a bit of notation. But at the end of the section, Helgason explains in detail the possible automorphisms of $\mathfrak{g}$ having order 2, up to conjugacy. Here the fixed point subalgebra $\mathfrak{g}_0$ for $\mathfrak{g}$ of type $E_8$ can only be of type $D_8$ or else $A_1 \oplus E_7$.

This already shows that the pseudo-Levi subalgebra of type $D_8$ can't have a nontrivial involutive automorphism which extends to one in $E_8$, though $D_8$ does have a nontrivial graph automorphism. (The situation for an example such as $A_2$ embedded in $G_2$ is similar.)

As Paul Levy notes in his comments, there are other configurations (not involving simple subalgebras of simple Lie algebras) which also cause trouble here. I've included a couple of doubtful examples of this type in my earlier answer and comments, though as Paul comments it's not possible to rule out existence of some extensions by looking at centralizer dimensions alone.

All of this cautions against expecting to find many positive answers for pairs $(\mathfrak{g}, \mathfrak{g}')$.

[EARLIER TRY]

Start with a simple Lie algebra $\mathfrak{g}$ over $\mathbb{C}$ of type $E_6$. Using Bourbaki numbering for the vertices of its Dynkin diagram (with 2 assigned to the "extra" vertex), let $\mathfrak{g}'$ be the direct sum of the two rank 1 Lie subalgebras corresponding to the vertices 2 and 6. The subalgebra $\mathfrak{g}'$ has an involutive automorphism which interchanges these two copies of $\mathfrak{sl}_2$. But there seems to exist no extension to such an automorphism of $\mathfrak{g}$, essentially because the rank 1 subalgebras involved do not play interchangeable roles in the overall structure of $\mathfrak{g}$. (For example, their centralizers have different dimensions.) [EDIT: As Paul notes, this parenthetic comment is false.]

Clearly this would require some work to make rigorous, but such a configuration of the pair $(\mathfrak{g}, \mathfrak{g}')$ indicates the kind of challenge you'd face here and also in your follow-up question.

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  • $\begingroup$ 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? $\endgroup$
    – Hebe
    Commented Dec 22, 2014 at 10:43
  • $\begingroup$ 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. $\endgroup$
    – Jim Humphreys
    Commented Dec 22, 2014 at 16:55
  • $\begingroup$ 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. $\endgroup$
    – Paul Levy
    Commented Dec 22, 2014 at 21:16
  • $\begingroup$ 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 ${\mathfrak g}$. $\endgroup$
    – Paul Levy
    Commented Dec 22, 2014 at 21:39
  • $\begingroup$ Thanks for your example. Then can $\tau$ be always extended to an automorphism (not necessary involutive) of $\mathfrak{g}$? $\endgroup$
    – Hebe
    Commented Dec 23, 2014 at 12:51

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