Cayley Transform for all reductive groups a.k.a an algebraic logarithm - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T06:25:17Z http://mathoverflow.net/feeds/question/101322 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/101322/cayley-transform-for-all-reductive-groups-a-k-a-an-algebraic-logarithm Cayley Transform for all reductive groups a.k.a an algebraic logarithm Adam S Sikora 2012-07-04T17:17:28Z 2012-07-05T19:15:06Z <p>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:</p> <p>(1) $L$ is $G$-equivariant with respect to the $G$-action on $G$ by conjugation and the Adjoint $G$-action on $\mathfrak g,$<br> (2) $L(e)=0,$<br> (3) $dL$ is an isomorphism at $e$,<br> (4) For some maximal torus $T$ in $G$, $L(T\cap U)$ lies in the Lie algebra of $T.$</p> <p>For $G=GL(n,\mathbb C)$, the embedding $L:GL(n,\mathbb C)\to gl(n,\mathbb C)$ works.<br> For $G=SO(n,\mathbb C)$, the Cayley Transform works: $L(A)= (I-A)(I+A)^{-1}$.<br> Cayley transform has a version for symplectic matrices as well.</p> <p>Is there a construction which works for all $G$? If not, are there known ad hoc constructions for exceptional groups?</p> http://mathoverflow.net/questions/101322/cayley-transform-for-all-reductive-groups-a-k-a-an-algebraic-logarithm/101343#101343 Answer by Mikhail Borovoi for Cayley Transform for all reductive groups a.k.a an algebraic logarithm Mikhail Borovoi 2012-07-04T21:14:15Z 2012-07-05T06:15:08Z <p>A <em>Cayley map</em> 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 <em>Cayley group</em> if it admits a Cayley map, and it is called a <em>stably Cayley group</em> 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 <a href="http://arxiv.org/abs/math/0409004" rel="nofollow">Cayley groups</a> by Lemire, Popov and Reichstein. As usual, the "stable" question is easier than the original one.</p> <p>The authors classified Cayley and stably Cayley simple groups. They proved the following result:</p> <blockquote> <p><strong>Theorem.</strong> 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.</p> </blockquote> <p>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.</p> http://mathoverflow.net/questions/101322/cayley-transform-for-all-reductive-groups-a-k-a-an-algebraic-logarithm/101427#101427 Answer by George McNinch for Cayley Transform for all reductive groups a.k.a an algebraic logarithm George McNinch 2012-07-05T19:15:06Z 2012-07-05T19:15:06Z <p>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.</p> <p>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).</p> <p>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 $$G \to G_{\operatorname{ad}} \to \operatorname{Lie}(G).$$</p> <p>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 $\operatorname{ad}(\operatorname{Lie}(G)) \simeq \operatorname{Lie}(G) \subset \mathfrak{gl}(V)$.</p> <p>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.</p> <p>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 $I \in M$.</p> <p>Write $\lambda$ for the composite mapping $$G \to \operatorname{GL}(V) \subset \mathfrak{gl}(V) \xrightarrow{\pi} \operatorname{ad}( \operatorname{Lie}(G))$$.</p> <p>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 $\operatorname{Lie}(T)$.</p> <p>This verifies the stipulated conditions (1),(2),(3) and (4).</p> <p>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.</p> <p>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.</p> <p>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 $p>0$.</p>