maximal linear subspaces contained in the cone over the Clifford torus.

Question

Forgot: this is about Find a subspace of $\mathbb{R}^4$ for which $x^T*A*x$ = 0

I was a little surprised to find that, in the cone $x^2 + y^2 = z^2 + w^2$ in $\mathbb R^4,$ there are infinitely many 2-planes passing through the origin and completely contained in the cone. Indeed, take any real vector $(A,B,C,D)$ with $A^2 + B^2 = C^2 + D^2,$ we can make a 2-plne in the cone from the linear span of $$ (A,B,C,D) \; \; \; \rm{and} \; \; (-B,A,-D,C). $$ We get a different 2-plane (I think) from the sapn of $$ (A,B,C,D) \; \; \; \rm{and} \; \; (-B,A,D,-C). $$ I also think that is it, for each nonzero vector in the cone, two 2-planes containing it. So far, I do not see anything in my many quadratic forms books that predicts this, but I could be looking in the wrong places.

I would like confirmation of all that. The cone over the Clifford torus is an important object in the differential geometry of minimal and constant mean curvature submanifolds. An early reference is Blaine Lawson (1970) in the Annals.

Meanwhile, given positive integers $(p,q)$ and the cone in $\mathbb R^{p+q}$ given by $$ x_1^2 + \cdots + x_p^2 \; = \; x_{p+1}^2 + \cdots + x_{p+q}^2, $$ what is the highest dimension of a linear subspace (through the origin) that is entirely contained in the cone? Finally, is there some finiteness result such as I got above, given this many independent vectors in the cone, these complete to a maximal linear subspace in exactly two(?) ways?

A few hours later: managed to relate this to something familiar. In $\mathbb R^3,$ the hyperboloid of one sheet $x^2 + y^2 = z^2 + 1$ is doubly ruled, two families of straight lines. If we intersect the cone $x^2 + y^2 = z^2 + w^2$ with the 3-plane $w=1,$ we get that doubly ruled hyperboloid. Furthermore, the original cone contains the cone over each of those straight lines, giving two $2$-planes for each point in the hyperboloid. http://en.wikipedia.org/wiki/Hyperboloid and http://en.wikipedia.org/wiki/Ruled_surface

Answer 1

A (coordinate) pseudo-Euclidean space $\mathbb{R}^{p,q}$ is just $\mathbb{R}^{p+q}$ with the "dot product"

$$ x\cdot y=x_1y_1+\cdots+x_py_p-x_{p+1}y_{p+1}-\cdots-x_{p+q}y_{p+q} .$$

The null quadric $Q=\{x\mid x\cdot x=0\}\subset\mathbb{R}^{p,q}$ is topologically the cone $C(S^{p-1}\times S^{q-1})$, which means it is $[0,\infty)\times S^{p-1}\times S^{q-1}$ modulo every point in $\{0\}\times S^{p-1}\times S^{q-1}$ identified to a single point. The map $[0,\infty)\times S^{p-1}\times S^{q-1}\to Q$ is given by $(r,u,v)\mapsto (ru,rv)$ where the latter vectors are interpreted as elements of $\mathbb{R}^p\times\mathbb{R}^q$ (a way of thinking about $\mathbb{R}^{p,q}$).

For instance in $\mathbb{R}^{2,1}$ this is $C(S^1\times S^0)$, the double cone sitting between the hyperboloids of one and two sheets (which are the solution sets to the two equations $x\cdot x=\pm1$). The two elements of the $S^0$ represent the two halves of the cone.

As a subset, $Q$ is not closed under addition. Given two vectors $x,y\in Q$, the dot product of their sum will be $(x+y)\cdot(x+y)=2(x\cdot y)$, so they span a linear subspace contained in $Q$ if and only if they are orthogonal with respect to this dot product.

Given a null subspace $U\subset\mathbb{R}^{p,q}$, the projection maps $U\to\mathbb{R}^p$ and $U\to\mathbb{R}^q$ must be one-to-one (for instance a vector in the kernel of $U\to\mathbb{R}^p$ would have to be both null and in $\mathbb{R}^q$, which means it must be zero). Therefore, $\dim U\le\min\{p,q\}$. We can see this upper bound is realizable by simply taking the basis $\{e_i+e_{p+i}:i\le\min\{p,q\}\}$.

Without loss of generality, suppose $p\le q$. Then the projection map $U\to\mathbb{R}^p$ is an isomorphism, which can be inverted and composed with $U\to\mathbb{R}^q$ to obtain an isometry $\beta:\mathbb{R}^p\to\mathbb{R}^q$, and conversely the subspace $U$ may be recovered from $\beta$ as the image $(I+\beta)(\mathbb{R}^p)$.

Therefore, the "moduli space" of maximal null subspaces of $\mathbb{R}^{p,q}$ can be identified with the space of linear isometries $\mathbb{R}^p\to\mathbb{R}^q$. Alternatively, one could use orbit-stabilizer to identify it with the quotient $O(q)/O(q-p)$, i.e. the Stiefel manifold $V_p(\mathbb{R}^q)$ which is a generalized flag variety.

Given a vector $x\in Q$, what is the moduli space of all max null subspaces containing it? Well, by decomposing it $x=(u,v)$ we see the associated isometry $\beta$ is already partially determined; it only needs to be specified on how to map $u^\perp\subset\mathbb{R}^p$ into $v^{\perp}\subset\mathbb{R}^q$, and so this "moduli subspace" is just $V_{p-1}(\mathbb{R}^{p-1})$. Indeed, in the case of $p,q=2$ this is $V_1(\mathbb{R}^1)$, i.e. the $1$-frames in $\mathbb{R}^1$, or equivalently the linear isometries $\mathbb{R}^1\to\mathbb{R}^1$ or equivalently $O(1)/O(0)$ - in all of these cases we get $S^0$, a set of two points, so there are precisely two null 2-planes through any null vector in $\mathbb{R}^{2,2}$.

Answer 2

I don't have anything to add regarding your questions on quadratic forms; anon's answer looks to me to be very illuminating and thorough. But it may be worth mentioning a generalization in a different direction.

Write $x^2+y^2=z^2+w^2$ as $(x-z)(x+z)=(w-y)(w+y)$ and define $s=x-z$, $t=w-y$, $u=w+y$, $v=x+z$ so that the equation of the cone becomes $sv=tu$. Fixing the ratio $s:t$ means $u:v$ has to be fixed to the same ratio. Hence for constants $a$, $b$, not both zero, the two-dimensional plane defined by $at=bs$, $av=bu$ lies in the cone. Fixing the ratio $s:u$ instead, which entails fixing $t:v$ to the same ratio, gives planes $cu=ds$, $cv=dt$ for all pairs $c$, $d$, not both zero.

If $(a_2,b_2)$ is not a scalar multiple of $(a_1,b_1)$ then the planes $a_1t=b_1s$, $a_1v=b_1u$ and $a_2t=b_2s$, $a_2v=b_2u$ intersect only at the origin. The analogous statement holds for the planes $c_1u=d_1s$, $c_1v=d_1t$ and $c_2u=d_2s$, $c_2v=d_2t$ when $(c_2,d_2)$ is not a scalar multiple of $(c_1,d_1)$. On the other hand, the plane $at=bs$, $av=bu$ and the plane $cu=ds$, $cv=dt$ intersect in a line. The two planes together span the three-dimensional linear space tangent to the cone at all points on their common line.

Note that $sv=tu$ is the projectivization of both the hyperbolic paraboloid and the one-sheeted hyperboloid, which are the two doubly-ruled surfaces, apart from the plane. Setting any of $s$, $t$, $u$, or $v$ to a non-zero constant gives the equation of a hyperbolic paraboloid. As you mention in the question, setting one of the original variables to a nonzero constant gives the equation of a one-sheeted hyperboloid.

Arranging the coordinates $s$, $t$, $u$, $v$ in the form of a $2\times2$ matrix, we see that the cone is the determinantal variety, $Y_1(2,2)$, defined by $$ \begin{vmatrix}s & t\\ u & v\end{vmatrix}=0, $$ which is the condition that the rank of the matrix be at most $1$.

Generalizing, let $Y_k(m,n)$ be the set of $m\times n$ real matrices of rank at most $k$, which is characterized by the vanishing of all $(k+1)\times(k+1)$ minors of the matrix, that is, by a set of homogeneous degree-$(k+1)$ equations. The set $Y_k(m,n)$ is a cone since if $M\in Y_k(m,n)$ then $aM\in Y_k(m,n)$ for any scalar $a$. It is the set of matrices that can be written $$ v_1w_1^\mathrm{T}+v_2w_2^\mathrm{T}+\ldots+v_kw_k^\mathrm{T}, $$ where $v_j\in\mathbf{R}^m$, $w_j\in\mathbf{R}^n$ for all $1\le j\le k$. For any $k$-dimensional linear subspace $V$ of $\mathbf{R}^m$ and basis $b_1,\ldots,b_k$ of $V$, the set $$ \left\{b_1w_1^\mathrm{T}+b_2w_2^\mathrm{T}+\ldots+b_kw_k^\mathrm{T}\mid w_1,\ldots,w_k\in\mathbf{R}^n\right\}, $$ which is independent of the choice of basis, is a $kn$-dimensional linear subspace of $\mathbf{R}^{mn}$ that is contained within the cone. Similarly, for any $k$-dimensional subspace $W$ of $\mathbf{R}^n$ and basis $c_1,\ldots,c_k$ of $W$, the set $$ \left\{v_1c_1^\mathrm{T}+v_2c_2^\mathrm{T}+\ldots+v_kc_k^\mathrm{T}\mid v_1,\ldots,v_k\in\mathbf{R}^m\right\} $$ is again independent of the choice of basis and is a $km$-dimensional linear subspace of $\mathbf{R}^{mn}$ that is contained within the cone.

Note that the cone itself is an algebraic variety of dimension $k(m+n-k)$. To specify the $k$-dimensional linear subspace $V\subset\mathbf{R}^m$ requires $k(m-k)$ parameters: take as basis vectors the rows of a $k\times m$ matrix $[I\mid M]$, where $I$ is the $k\times k$ identity. (One, of course, cannot always achieve a basis of this form. Instead of columns $1$ through $k$, one may specify some other set of $k$ columns as the pivot columns, with the remaining columns containing the $k(m-k)$ parameters. This amounts to the choice of a particular coordinate chart, and every linear subspace will be covered by at least one such chart.) After specifying $V$, one needs to specify $k$ vectors $w_i\in\mathbf{R}^n$, which requires $kn$ parameters, for a total of $k(m+n-k)$ parameters. Alternatively, specifying the $k$-dimensional subspace $W\in\mathbf{R}^n$ requires $k(n-k)$ parameters; then the $k$ vectors $v_i\in\mathbf{R}^m$ need to be specified, for, again, a total of $k(m+n-k)$ parameters. Observe that $V$ is an element of the Grassmannian $\operatorname{\bf Gr}(k,\mathbf{R}^m)$ and that $W$ is an element of $\operatorname{\bf Gr}(k,\mathbf{R}^n)$.

Define $K=\min(m,n)$. Then $0\le k\le K$. We may understand the determinantal variety $Y_k(m,n)$ as a way of realizing the set of elements of $\mathbf{R}^m\otimes\mathbf{R}^n$ of tensor rank at most $k$. I began thinking about these things while pondering the questions here, here, and here about geometric interpretations of the tensor product, something I still don't feel I fully understand. Nevertheless, thinking about determinantal varieties does give some insight into how $\mathbf{R}^m\otimes\mathbf{R}^n$ is built up from spaces isomorphic to $\mathbf{R}^m$, $(\mathbf{R}^m)^2$, ..., $(\mathbf{R}^m)^K$ and also from spaces isomorphic to $\mathbf{R}^n$, $(\mathbf{R}^n)^2$, ..., $(\mathbf{R}^n)^K$.

The $(kn)$-dimensional and $(km)$-dimensional subspaces of $\mathbf{R}^m\otimes\mathbf{R}^n$ defined above are $V\otimes\mathbf{R}^n$ and $\mathbf{R}^m\otimes W$. Their intersection is $V\otimes W$. Furthermore, \begin{align} (V_1\otimes\mathbf{R}^n)\cap(V_2\otimes\mathbf{R}^n)&=(V_1\cap V_2)\otimes\mathbf{R}^n\\ (\mathbf{R}^m\otimes W_1)\cap(\mathbf{R}^m\otimes W_2)&=\mathbf{R}^m\otimes(W_1\cap W_2). \end{align} As a consequence, if $V_1$ and $V_2$ intersect only at $0$, then so do the linear spaces $V_1\otimes\mathbf{R}^n$ and $V_2\otimes\mathbf{R}^n$. In $\mathbf{R}^2$, any two distinct one-dimensional linear subspaces intersect only at $0$, and therefore so do your $2$-planes, $V_1\otimes\mathbf{R}^2, V_2\otimes\mathbf{R}^2\subset Y_1(2,2)\subset\mathbf{R}^2\otimes\mathbf{R}^2$ as well as your $2$-planes $\mathbf{R}^2\otimes W_1, \mathbf{R}^2\otimes W_2\subset Y_1(2,2)\subset\mathbf{R}^2\otimes\mathbf{R}^2$. On the other hand $(V\otimes\mathbf{R}^2)\cap(\mathbf{R}^2\otimes W)=V\otimes W$, which is one-dimensional. Similarly, if, for example, $\dim(V_1)=\dim(V_2)=2$ and $\dim(V_1\cap V_2)=1$ then, while $V_1\otimes\mathbf{R}^n$ and $V_2\otimes\mathbf{R}^n$ are subsets of $Y_2(m,n)$, the linear space $(V_1\cap V_2)\otimes\mathbf{R}^n$ is a subset also of $Y_1(m,n)$. Observe the inclusion of determinantal varieties $$ \{0\}=Y_0(m,n)\subset Y_1(m,n)\subset\ldots\subset Y_K(m,n)=\mathbf{R}^m\otimes\mathbf{R}^n. $$ If, to take an example, $m=3$ and $n=4$, the dimensions of these varieties are $$ 0,\ 6,\ 10,\ 12; $$ $Y_1(3,4)$ is ruled by linear spaces of dimensions $3$ and $4$; $Y_2(3,4)$ is ruled by linear spaces of dimensions $6$ and $8$; $Y_3(3,4)=\mathbf{R}^3\otimes\mathbf{R}^4$ is ruled by linear spaces of dimensions $9$ and $12$.