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In the questions Is "semisimple" a dense condition among Lie algebras? and What is the Zariski closure of the space of semisimple Lie algebras?, something equivalent to the following is mentioned: if you have a smoothly varying family of semisimple Lie algebras, all the Lie algebras in the family are isomorphic. e.g. the following quote:

"Because the Cartan classification of isomorphism classes of semisimples is discrete (no continuous families), connected components of the space of semisimples are always contained within isomorphism classes."

I can't see how this follows just from the discreteness of the classification. Can anyone explain why it's true or give a counterexample?

e.g. could you not have a $\mathbb{P}^1$ of semisimple Lie algebras which are generically isomorphic to $\mathfrak{d}_7 \oplus \mathfrak{a}_1$ say, but at one point you get $\mathfrak{e}_8$, or something similar?

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5 Answers 5

up vote 15 down vote accepted

The experts should correct me if there is a fatal mistake in the argument, I am neither an algebraic geometer nor a Lie theorist. I am working over $\mathbb{C}$.

  1. Let $\mathfrak{g}$ be a semisimple Lie algebra, $G$ be the adjoint group, $Aut(\mathfrak{g})$ be the automorphism group. Let $Aut_0 (\mathfrak{g})$ be the subgroup of automorphisms that preserve the decomposition of $\mathfrak{g}$ into simple ideals. It has finite index in the whole automorphism group, since the decomposition into ideals is unique and an automorphism has to map a simple ideal into a simple ideal. The group $Aut_0(\mathfrak{g}$ is the product of the automorphism groups of the simple factors. The automorphism group of any simple Lie algebra has the adjoint group as a finite index subgroup; the quotient is the automorphism group of the Dynkin diagram. The upshot of this discussion is: $Aut(\mathfrak{g})$ has $G$ as a finite index subgroup. In particular, the dimension of $Aut(\mathfrak{g})$ only depends on the dimension of $\mathfrak{g}$!

EDIT: there is a better proof of this step in the literature, e.g. in Procesi's book, page 301.

  1. Let $V$ be the variety of all Lie algebra structures on $C^n$; it has an action of $GL_n$ on it, the stabilizers are the automorphism groups, the orbits are the isomorphism classes. Now let $\mathfrak{g} \in V$ be semisimple and let $O \subset V$ be its $GL_n$-orbit. Now use the orbit closure theorem (Borel, Linear Algebraic Groups, page 53). It says that any $GL_n$-orbit $O$ is open in its closure and that $\bar{O} \setminus O$ consists of orbits of smaller dimension. Hence all Lie algebras in the closure of $O$ which are not isomorphic to $\mathfrak{g}$ must have a larger-dimensional automorphism group. By part 1, they are not semisimple.
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Great, this looks right to me, but I'm also no expert! –  ndkrempel Nov 26 '10 at 21:54
    
Incidentally, it seems to me that this proof shows there are only finitely many semisimple Lie algebras of a given dimension, without having to go through the classification, although perhaps some of the subsidiary results you relied upon need some part of the classification. –  ndkrempel Nov 26 '10 at 22:42
    
This is nice, and I wonder if it really does prove the claim without the classification. –  Theo Johnson-Freyd Nov 27 '10 at 0:08
    
One problem, though, in step 1 is when there are repeated factors. For example, let $g = sl(2) + sl(2) = sl(2) \otimes \mathbb C^2$. Then $Aut(g)$ includes a $GL(2)$ acting on the $\mathbb C^2$ part. Moreover, your $Aut_0(g)$ is the quotient of $Aut(g)$ by this $GL(2)$, I think, and in particular does not have finite index. But maybe I have confused myself. –  Theo Johnson-Freyd Nov 27 '10 at 0:15
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I don't think you get a full $GL(2)$ involved, but only a tranposition swapping the two summands. This is because the summands are actually uniquely determined (unlike the situation with modules). For example the diagonal $sl(2)$ is not an ideal! –  ndkrempel Nov 27 '10 at 0:34
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While I think that what Johannes wrote is right, it misses out on an underlying principle. There's a one sentence explanation for this fact, which is "the tangent space to a Lie algebra in the moduli space of Lie algebras is $H^2(\mathfrak{g},\mathfrak{g})$, the second cohomology valued in the adjoint representation." This cohomology group can also be identified with the space of extensions of Lie algebras of $\mathfrak{g}$ by a copy of the adjoint representation.

So, the slick proof is to either prove that this is 0 using the action of the Casimir as in Weibel's book, or to note by the Levi complement theorem that all abelian extensions of a semi-simple Lie algebra split.

After all, if you had a family $\tilde{\mathfrak{g}}$ of such Lie algebras over $\mathbb{A}^1$, then you could look at $\tilde{\mathfrak{g}}/h^2\tilde{\mathfrak{g}}$ (where $h$ is the parameter on $\mathbb{A}^1$), and see that that is an extension of $\mathfrak{g}$ by the adjoint representation.

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This sounds like the theoretically nicer way of showing it, although I don't understand all the details here. For example, how does this proof square with the fact that you can deform a semisimple Lie algebra to the zero algebra, a genuine change in isomorphism type (say by multiplying all the structure constants by an affine parameter $t$). How does the above proof show that there isn't such a jump from one semisimple algebra to another, while not ruling out the jump to the zero algebra? –  ndkrempel Nov 27 '10 at 1:41
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The tangent space measures behavior infinitesimally close to the point. So, while a semi-simple thing can degenerate to a non-semisimple thing, it will never be infinitesimally close to the non-semi-simple thing (in the sense that you can take a neighborhood in the space of Lie algebra structures which only contains isomorphic Lie algebras). On the other hand, there's no open neighborhood around the trivial Lie algebra struture which only contains 0 Lie algebras. –  Ben Webster Nov 27 '10 at 8:41
    
"The tangent space to a Lie algebra in the moduli space of Lie algebras is $H^2(g,g)$, the second cohomology valued in the adjoint representation." I know that formal reasoning and it is not too difficult to see that this group is $0$ for semisimple $g$, but how is it translated into a precise argument? –  Johannes Ebert Nov 27 '10 at 16:28
    
It shows that each orbit corresponding to a semi-simple Lie algebra structure is open in the space of all Lie brackets, and thus no one can be contained in the closure of another. It's not really a different argument from yours, but it gets at it from the general modern machinery of deformation theory. –  Ben Webster Nov 29 '10 at 2:02
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I think that the explanation

"Because the Cartan classification of isomorphism classes of semisimples is discrete (no continuous families), connected components of the space of semisimples are always contained within isomorphism classes"

is a bit simplistic. The real reason, as many people have already mentioned, is Weyl's theorem on complete reducibility, which of course fails in charatcteristic $p$. And it shouldn't come as a surpise that over an algebraically closed field $K$ of characteristic $p>3$ one encounters situations where there are finitely many isoclasses of simple Lie algebras of dimension $N$ and, at the same time, there exist algebraic families of simple $N$-dimensional Lie algebras {$\mathfrak{g}_t|\ t\in K$} over $K$ such that $\mathfrak{g}_t\cong L$ for all $t\ne 0$ and $\mathfrak{g}_0\not\cong L$ for some simple Lie algebra $L$.

Indeed, let $N=p^2-2$. Then it follows from the the classification theory that there are finitely many isoclasses of simple $N$-dimensional Lie algebras over $K$. Now consider the associative $K$-algebra $A$ generated by two elements $x,y$ such that $x^p=y^p=0$ and $[x,y]=1$. This is a fake modular version of the first Weyl algebra, and it is easy to see that it is simple and has dimension $p^2$. It has a finite increasing algebra filtration (with $x,y$ living in degree $1$) such that the corresponding graded algebra $P:={\rm gr}(A)$ is the truncated polinomial ring in $x,y$ with induced Poisson bracket satisfying {$x,y$} $=1.$ Then the Lie algebra $[A,A]/K1$ is isomorphic to $\mathfrak{psl}_p(K)$ whilst the Lie algebra {$P,P$}$/K1$ is nothing but the simple Cartan type Lie algebra $H(2;\underline{1})^{(2)}$. Both Lie algebras are simple of dimension $N$ and the latter is a contraction of the former. Moreover, they are not isomorphic when $p>3$.

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Locally in any continuous family of semisimple Lie algebras, you can choose a (constant) subfamily of Cartan subalgebras etc., and so get a continuous map sending a point in the base to a Cartan matrix, which has integer coefficients.

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If you find non-standard arguments more intuitively appealing you might be interested in a non-standard proof of this result which appears in Springer LNM 881 `Nonstandard analysis' by Lutz and Goze, p.140.

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