if $\Pi_1$ and $\Pi_2$ be elliptic planes then $\Pi_1 \oplus \Pi_2$ is still elliptic? - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T20:19:14Z http://mathoverflow.net/feeds/question/112268 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/112268/if-pi-1-and-pi-2-be-elliptic-planes-then-pi-1-oplus-pi-2-is-still if $\Pi_1$ and $\Pi_2$ be elliptic planes then $\Pi_1 \oplus \Pi_2$ is still elliptic? Hassan Jolany 2012-11-13T10:45:39Z 2012-11-14T13:39:43Z <p>Let $\Omega \in\Lambda^{4}\big(V^{ \star }\big)$ be volume form. Define symplectic bilinear form</p> <p>$q: \Pi \oplus \Pi \rightarrow R$ </p> <p>$\big( \alpha ,\beta \big) \longrightarrow \alpha \wedge \beta =q \big( \alpha , \beta \big) \Omega$</p> <p>where $\alpha , \beta \in \Pi$ . We say $\Pi \subset \Lambda^{2}\big(V^{ \star }\big)$ is an elleptic plane if $q\mid_\Pi$ is a nondegenerated determined form. My question is if $\Pi_1$ and $\Pi_2$ be elliptic planes then $\Pi_1 \oplus \Pi_2$ is still elliptic?</p> http://mathoverflow.net/questions/112268/if-pi-1-and-pi-2-be-elliptic-planes-then-pi-1-oplus-pi-2-is-still/112283#112283 Answer by Robert Bryant for if $\Pi_1$ and $\Pi_2$ be elliptic planes then $\Pi_1 \oplus \Pi_2$ is still elliptic? Robert Bryant 2012-11-13T13:49:42Z 2012-11-14T13:39:43Z <p>No. Note that $V$ has dimension $4$. The maximum dimension of an elliptic subspace of $\Lambda^2(V^\ast)$ is $3$, so if $\Pi_1$ and $\Pi_2$ don't intersect, then $\Pi_1\oplus \Pi_2$ is <em>never</em> elliptic.</p> <p>To see why this is so, let $\Omega\in \Lambda^4(V^\ast)$ be a volume form and define a symmetric bilinear form $q$ on $\Lambda^2(V^\ast)$ by the rule $$q(\alpha,\beta)\ \Omega = \alpha\wedge\beta.$$ If $e^i$ ($0\le i\le 3$) is a basis of $V^\ast$ such that $\Omega = e^0\wedge e^1\wedge e^2\wedge e^3$, then setting $$\alpha^1 = e^0\wedge e^1,\quad \alpha^2 = e^0\wedge e^2,\quad \alpha^3 = e^0\wedge e^3$$ and $$\alpha^4 = e^2\wedge e^3,\quad \alpha^5 = e^3\wedge e^1,\quad \alpha^6 = e^1\wedge e^2,$$ one gets that the matrix $Q = (Q^{ij}) = \bigl(Q(\alpha^i,\alpha^j)\bigr)$ of $q$ with respect to this basis has the block form $$Q = \begin{pmatrix} 0_3 &amp; I_3\\ I_3 &amp; 0_3 \end{pmatrix}.$$ Clearly, $Q$ has type $(3,3)$, so $q$ cannot be definite (positive or negative) on any subspace of $\Lambda^2(V^\ast)$ that has dimension greater than $3$.</p> <p>By the way, $\mathrm{SL}(\Omega)\subset \mathrm{GL}(V)$ acts transitively on the set of positive (or negative) definite subspaces of $\Lambda^2(V\ast)$ of a given dimension. In fact, $\mathrm{SL}(\Omega)$ acts transitively on the subspaces of a given $q$-type with one exception: There are two $\mathrm{SL}(\Omega)$ orbits in the set of $q$-null $3$-planes: The first type is the orbit of the span of $\alpha^1,\alpha^2,\alpha^3$ as described in the basis above, i.e., the set of multiples of a single $1$-form. The second type is the orbit of the span of $\alpha^4,\alpha^5,\alpha^6$ as described in the basis above, i.e., spaces of the form $\Lambda^2(W)\subset\Lambda^2(V^\ast)$ where $W$ is a hyperplane in $V^*$.</p> <p>Finally, note that the connected $15$-dimensional group $\mathrm{SL}(\Omega)$ acts on $\Lambda^2(V^\ast)$ preserving $q$ and an orientation of $\Lambda^2(V^\ast)$, which induces a homomorphism $\mathrm{SL}(\Omega)\to\mathrm{SO}(q)$. This homomorphism has $\{\pm I\}\subset \mathrm{SL}(\Omega)$ as kernel and maps onto the identity component of $\mathrm{SO}(q)$ (which, itself, has two components). This is one of the 'exceptional' isomorphisms of the classical groups.</p>