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?


2 Answers 2


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 $p>0$.

  • $\begingroup$ There's also a paper by Kostant and Michor studying this composite map. $\endgroup$ Jul 6, 2012 at 17:08
  • $\begingroup$ Can we get the classical Cayley map for SO_n from a version of this construction? $\endgroup$ Jul 7, 2012 at 18:39
  • $\begingroup$ @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... $\endgroup$ Jul 7, 2012 at 21:14
  • $\begingroup$ @Allen: Thanks -- I hadn't seen that paper of Kostant and Michor. $\endgroup$ Jul 7, 2012 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.

EDIT: After I answered the question in 2012, new papers on Cayley groups appeared: Stably Cayley groups in characteristic zero, where, in particular, stably Cayley simple groups over an arbitrary field of characteristic 0 are classified; Stably Cayley semisimple groups, where stably Cayley semisimple groups over an arbitrary field of characteristic 0 are classified; Real reductive Cayley groups of rank 1 and 2, where for all reductive $\mathbb R$-groups of rank 1 and 2 we determine whether they are Cayley (not just stably Cayley) or not.

Concerning the paper "Stably Cayley semisimple groups", note that a product of stably Cayley simple groups is clearly a stably Cayley semisimple group, but it is not true that any stably Cayley semisimple group is a product of stably Cayley simple groups. Indeed, ${SO}_4$ is a Cayley group (this was proved by Cayley), hence it is a stably Cayley group, but it is not simple and not a product of simple groups, hence it is not a product of simple stably Cayley groups. However, it was proved in this paper that any stably Cayley semisimple group $G$ over an algebraically closed field $k$ of characteristic 0 can be written as $$ G=G_1\times_k\dots\times_k G_s\,,$$ where each $G_i$ is either a stably Cayley simple group (from the above theorem), or the stably Cayley semisimple group ${SO}_4$.

In the paper "Real reductive Cayley groups of rank 1 and 2" we show, in particular, that over $\mathbb R$ the group $SU_3$ is Cayley, while $SL_{3,\mathbb R}$ is not Cayley. Note that both groups are stably Cayley because all reductive groups of absolute rank $\le 2$ over a field of characteristic 0 are stably Cayley.


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