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Consider the orthogonal group $G=O(n)$ as a subset of the vector space of $n\times n$ real matrices. Let $C=C(G)$ denote the Euclidean cone over $G$, i.e., the space of matrices of the form $tA, A\in G, t\in {\mathbb R}$.

Question. What are vector subspaces in $C$? For instance, are there 3-dimensional linear subspaces? (One can ask the same question for cones over other linear Lie groups $G$.)

Update: @Dmitri noticed that in dimensions divisible by $4$ and $8$ there are 4-dimensional and 8-dimensional linear subspaces coming from quaternions and octonions. Thus, the question is: What are other linear subspaces, in particular, are there linear subspaces (which are not lines) for odd $n$? Are there linear subspaces which are not contained in cones over subgroups?

Trivial examples of subspaces in $C(G)$ are planes, obtained as cones over $O(2)$. Diagonal embedding of $O(2)$ to $O(2n)$ (and its conjugates) yield examples of planes in $C(O(2n))$. Are there other examples of linear subspaces (of dimension $>1$) in $C(O(n))$? My motivation for the question comes from hyperbolic geometry (where the question is about affine subspaces of $O(n,1)$), but I find the linear algebra question about the orthogonal group intriguing, because it is so basic. The question is somewhat similar to the one in Spaces of matrices with same eigenvalue/Great circles in O(n)-orbits

Alternatively, and algebraic geometers like it better, one can work over ${\mathbb C}$ and projectivize everything. Thus, I am asking about linear subspaces of the variety $X=PO(n)\subset {\mathbb P}^{N}$, $N=n^2-1$. In other words, the question is about Fano varieties $F_k(X)$ for various $k$. Algebraic geometers studied Fano varieties $F_k(X)$ of projective varieties since 19th century, so, maybe somebody looked at the case of homogeneous $X$, in particular, projective varieties $X$ which are algebraic subgroups. Unfortunately, google search did not yield anything useful.

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3 Answers 3

up vote 4 down vote accepted

Suppose that $\mathbb R^n$ is a module of the Clifford algebra $C_m$, with generators $e_1, \dots, e_m$, and relations $e_i^2 = -1$, and $e_ie_j+e_je_i = 0$. After a base change in $\mathbb R^n$, the images of the $e_i$ in $GL_n(\mathbb R)$ can be taken to be orthogonal. Then it is immediate to check that the subspace generated by these images is contained in the cone over $O_n$, and has dimension $m$.

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Thanks, Angelo! –  Misha Mar 19 '12 at 0:19

Angelo's answer essentially gives a description of all such linear spaces and also describes the only way they can happen. To see this, suppose that you have a linear subspace $L\subset C$ that passes through $A\in O(n)$. Then multiplying everything on the left by $A^{-1}$, you can assume that $L$ passes through the identity element $I\in O(n)$. Suppose that $L$ has dimension $1{+}m$. For any nonzero element $a\in L$, the matrix $a^T a$ must be a positive multiple of $I$, so there exists a positive definite quadratic function $f:L\to \mathbb{R}$ such that $a^T a = f(a) I$. Let $a_0 = I, a_1,\ldots,a_m$ be an $f$-orthonormal basis of $L$. Computation shows that $a_i^T\ a_j+a_j^T\ a_i = 2\delta_{ij}\ I$. In particular, it follows that each $a_i$ for $i>0$ is skew-symmetric and hence that $a_ia_j+a_ja_i = -2\delta_{ij}\ I$ for $i,j>0$ and distinct. Thus, the $a_i$ for $i>0$ generate the action of an orthogonal Clifford algebra on $\mathbb{R}^n$.

A similar argument can be made over the complexes, but, there, you have to watch out for degenerate quadratic forms.

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Thanks, Robert! –  Misha Mar 19 '12 at 0:19
    
@Misha: You are welcome. Note, by the way, that this shows that the maximum dimension of such an $L$ is $1{+}m(n)$ where $m(n)$ is the largest integer such that $\mathbb{R}^n$ is a module over $Cl\bigl(m(n)\bigr)$. Thus, $m(n)$ is easy to determine when $n$ is given by consulting the classical theory of Clifford algebras. In particular $m(2k{+}1) = 0$, $m(4k{+}2)=1$, $m(8k{+}4)=3$, $m(16k{+}8)=7$, and $m(16k) = m(k)+8$. –  Robert Bryant Mar 19 '12 at 15:15

Just a remark, $C$ can contain four-planes and even eight-planes, this follows from existence of quaternions and octonions, again you can consider diagonal actions. These planes generalse two-planes that you describe that can be seen as planes coming from complex numbers.

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Good point, Dima! Is there anything else, in particular, not contained in a cone over a subgroup? –  Misha Mar 18 '12 at 15:39
    
This is great, thanks, Angelo and Robert! So, all linear subspaces in $CO(n)$ are "algebraic" in nature. I would have preferred to split my vote between two of your answers, but I cannot... –  Misha Mar 19 '12 at 0:19

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