Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Is it true that for every reductive algebraic $G$ over ${\mathbb C}$ with a Lie algebra $\mathfrak g$ there is an open neighborhood $U$ of the identity in $G$ and an algebraic function (in a sense of algebraic geometry) $L: U\to \mathfrak g$ which satisfies the following properties of logarithm:

(1) $L$ is $G$-equivariant with respect to the $G$-action on $G$ by conjugation and the Adjoint $G$-action on $\mathfrak g,$
(2) $L(e)=0,$
(3) $dL$ is an isomorphism at $e$,
(4) For some maximal torus $T$ in $G$, $L(T\cap U)$ lies in the Lie algebra of $T.$

For $G=GL(n,\mathbb C)$, the embedding $L:GL(n,\mathbb C)\to gl(n,\mathbb C)$ works.
For $G=SO(n,\mathbb C)$, the Cayley Transform works: $L(A)= (I-A)(I+A)^{-1}$.
Cayley transform has a version for symplectic matrices as well.

Is there a construction which works for all $G$? If not, are there known ad hoc constructions for exceptional groups?

share|improve this question

2 Answers 2

up vote 3 down vote accepted

Bardsley and Richardson (Etale slices for algebraic transformation groups in characteristic p. Proc. London Math. Soc. (3) 51 (1985), no. 2, 295–317) give a construction which seems to do what you want. It gives less than a "Cayley map" as in Borovoi's answer.

Let $G$ be connected and semisimple over a field of char. 0. Bardsley and Richardson construct a mapping $G \to \operatorname{Lie}(G)$ with nice properties (which I'll indicate below).

Note that we may as well suppose that $G$ is of adjoint type -- indeed, if the problem is solved already for the adjoint group $G_{\operatorname{ad}}$, just take the composite $$G \to G_{\operatorname{ad}} \to \operatorname{Lie}(G).$$

Now, since the characteristic of $k$ is zero and $G$ is of adjoint type, the adjoint representation $V = \operatorname{Lie}(G)$ is a faithful representation of $G$ for which the trace form defined by $\kappa(X,Y) = \operatorname{tr}(X \circ Y)$ -- a non-degenerate form on $\mathfrak{gl}(V)$-- remains non-degenerate on the image $\operatorname{ad}(\operatorname{Lie}(G)) \simeq \operatorname{Lie}(G) \subset \mathfrak{gl}(V)$.

Writing $M$ for the orthogonal complement $M=\operatorname{ad}(\operatorname{Lie}(G))^\perp$ with respect to the form $\kappa$, we have $$\mathfrak{gl}(V) = M \oplus \operatorname{ad}(\operatorname{Lie}(G))$$ as $G$-representations. Write $\pi:\mathfrak{gl}(V) \to \operatorname{ad}(\operatorname{Lie}(G))$ for the projection on the second factor.

Since $G$ is semisimple, $\operatorname{ad}(\operatorname{Lie}(G)) \subset \mathfrak{sl}(V)$ so that the identity mapping $I$ satisfies $\kappa(I,\operatorname{ad}X) = \operatorname{tr}(\operatorname{ad}X) = 0$ for each $X \in \operatorname{Lie}(G)$. Thus $I \in M$.

Write $\lambda$ for the composite mapping $$G \to \operatorname{GL}(V) \subset \mathfrak{gl}(V) \xrightarrow{\pi} \operatorname{ad}( \operatorname{Lie}(G))$$.

Since $I \in M$, evidentally $\lambda(1) = 0$. Since $\pi$ is a $G$-module homomorphism, $\lambda$ is $G$-equivariant. Moreover, by construction $d\lambda_1$ is the identity mapping. Finally, by $G$-equivariance the image under $\lambda$ of a maximal torus $T$ is contained in the $T$-fixed points of $\operatorname{Lie}(G)$, i.e. in $\operatorname{Lie}(T)$.

This verifies the stipulated conditions (1),(2),(3) and (4).

Note that Barsdsley and Richardson go on to show that the restriction of $\lambda$ to the unipotent variety $\mathcal{U} \subset G$ defines a $G$-equivariant isomorphism $\mathcal{U} \xrightarrow{\sim} \mathcal{N}$ where $\mathcal{N} \subset \operatorname{Lie}(G)$ is the nilpotent variety.

Moreover, by Luna's theorem (a proof valid in positive characteristic is given in Bardsley and Richardson's paper) there are $G$-invariant open subset $U \subset G$ and $U' \subset \operatorname{Lie}(G)$ with $1 \in U$ and $0 \in U'$ such that $\lambda_{\mid U}$ defines a surjective etale mapping $U \to U'$. So $\lambda$ need not be birational, but it is fairly nice.

Note that Barsdley and Richardson actually formulate the above construction more generally using representations $V$ of $G$ which are "nice enough" (among other things, the restriction of the traceform on $\mathfrak{gl}(V)$ to the image of $\operatorname{Lie}(G)$ must be non-degenerate), and under suitable assumptions (very roughly: the characteristic should be good for $G$) their construction gives "explicit" Springer isomorphisms $\mathcal{U} \xrightarrow{\sim} \mathcal{N}$ in characteristic $p>0$.

share|improve this answer
There's also a paper by Kostant and Michor studying this composite map. –  Allen Knutson Jul 6 '12 at 17:08
Can we get the classical Cayley map for SO_n from a version of this construction? –  Mikhail Borovoi Jul 7 '12 at 18:39
@Mikhail: I don't know the answer to that question. I'm a bit skeptical, though. The "natural candidate" for $G=SO(V)$ is to apply the construction for the natural representation $V$. In that case, unless I'm making a silly back-of-the-envelope goof, the map seems to be given by the rule $X \mapsto \dfrac{X^2-1}{2X}$, which is similar to the Cayley map but not the same. This map seems not to be birational... –  George McNinch Jul 7 '12 at 21:14
@Allen: Thanks -- I hadn't seen that paper of Kostant and Michor. –  George McNinch Jul 7 '12 at 21:14

A Cayley map is a $G$-equivariant birational isomorphism $\lambda: G\to \mathfrak{g}$ (which does not have to exist). A connected linear algebraic group $G$ over $\mathbb{C}$ is called a Cayley group if it admits a Cayley map, and it is called a stably Cayley group if $G\times (\mathbb{G}_m)^n$ is a Cayley group for some $n=0,1,2,\dots$. These notions were introduced in the paper Cayley groups by Lemire, Popov and Reichstein. As usual, the "stable" question is easier than the original one.

The authors classified Cayley and stably Cayley simple groups. They proved the following result:

Theorem. The stably Cayley simple groups over an algebraically closed field $k$ of characteristic 0 are the following: $SL_2$, $SL_3$, $SO_n$, $Sp_{2n}$, $PGL_n$, and $G_2$. All these groups are Cayley, except $G_2$. The group $G_2$ is not Cayley (V. A. Iskovskikh), but $G_2\times (\mathbb{G}_m)^2$ is Cayley.

Note that the question whether $G_2\times \mathbb{G}_m$ is Cayley is open. Note also that all the groups of types $E_6$, $E_7$, $E_8$ and $F_4$ are not stably Cayley, hence they are not Cayley. In addition, the groups $SL_2$ and $SL_3$ are Cayley, while $SL_4$, $SL_5$, $SL_6$ and so on are not stably Cayley, hence they are not Cayley.

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.