$G$-structures of finite type. - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T09:05:50Z http://mathoverflow.net/feeds/question/64064 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/64064/g-structures-of-finite-type $G$-structures of finite type. Leandro 2011-05-05T23:25:23Z 2011-05-09T12:20:30Z <p>A $G$-structure $\pi : B_G \rightarrow M$ is said to be of $finite$ $type$ if $\mathfrak{g}^{(k)} = 0$ for some $k \in \mathbb{N}$, where $\mathfrak{g}^{(k)}$ denotes the $k$th prolongation of the Lie algebra $\mathfrak{g} = T_{e}G$. For finite type $G$-structures, let us call the first $k$ for which $\mathfrak{g}^{(k)} = 0$ the $order$ of the $G$-structure. </p> <p>For example:<br> &bull; $O(n)$-structures (Riemannian metrics) are of finite type and order $1$, because $\mathfrak{o} (n)^{(1)} = 0$. <br> &bull; But $Sp(n)$-structures (symplectic structures) are not of finite type because the group of symplectomorphisms is infinite dimensional.<br> &bull; It can also be shown that $CO(n)$-structures (conformal structures) are of finite type and order $2$ (except if the dimension is $2$, in which case it is not of finite type).</p> <p><b>Are there any finite type $G$-structures of order greater than $2$?</b></p> <p>More generally, are there $G$-structures of any order?</p> <p>Thanks.</p> http://mathoverflow.net/questions/64064/g-structures-of-finite-type/64067#64067 Answer by Robert Bryant for $G$-structures of finite type. Robert Bryant 2011-05-05T23:56:15Z 2011-05-09T12:20:30Z <p>Yes, $G$-structures exist of each finite order. In other words, for every $k\ge1$, there is an $n\ge1$ and a subgroup $G\subset GL(n,\mathbb{R})$ such that its Lie algebra $\frak{g}$ satisfies ${\frak{g}}^{(k-1)}\not=0$ while ${\frak{g}}^{(k)}=0$. </p> <p>There is no known classification of such algebras, but here is a simple example of an algebra ${\frak{g}}_k\subset {\frak{gl}}(k{+}3,\mathbb{R})$ such that ${\frak{g}}_k$ has order $k$: Let $e_1,\ldots, e_{k+3}$ be the standard basis of $\mathbb{R}^{k+3}$, with dual basis $x^1,\ldots, x^{k+3}$. Let ${\frak{g}}_k$ be the (abelian, nilpotent) subalgebra of ${\frak{gl}}(k{+}3,\mathbb{R})$ with basis $l_1,\ldots,l_k$, where $$l_i = e_{i+3}\otimes x^1 + e_{i+2}\otimes x^2.$$ One computes that ${\frak{g}}_1^{(1)}=0$ and that, for $k>1$, the space ${\frak{g}}_k^{(1)}$ has dimension $k{-}1$, with basis $q_2,\ldots,q_k$, where $$q_i = e_{i+3}\otimes (x^1)^2 + 2e_{i+2}\otimes x^1x^2 + e_{i+1}\otimes (x^2)^2.$$ Continuing on in this way, one finds that the dimension of ${\frak{g}}_k^{(j)}$ is $k{-}j$ for $0\le j\le k$.</p> <p>For each $n$, there is an upper bound on the order of the subalgebras of ${\frak{gl}}(n,\mathbb{R})$ of finite type, but I do not know what that is. There are estimates for this upper bound, but I don't think they are very tight.</p> <p>Meanwhile, a theorem of Cartan (originally proved over $\mathbb{C}$ by a classification (but with some omissions), and later completed by others and worked out over $\mathbb{R}$ as well) says that, if $G\subset GL(n,\mathbb{R})$ acts irreducibly on $\mathbb{R}^n$, then $\frak{g}$ has order $1$, $2$, or $\infty$. The list of the irreducibly acting $G\subset GL(n,\mathbb{R})$ that have order $2$ or $\infty$ is known and can be found in my 1996 survey paper, Classical, exceptional, and exotic holonomies: a status report, in Actes de la Table Ronde de Géométrie Différentielle (Luminy, 1992), Sémin. Congr., vol. 1 (1996), pp. 93–165. See the tables in Appendix A.</p>