Product Of The Unit Disk

Let $D = \{ (x,y)\in\mathbb{R}^2\ :\ x^2+y^2\leq 1 \}$ be the unit disk, and consider the 4-dimensional convex body $D\times D$. Describe its faces and finde a non-zero polynomial in 4 unknowns that vanishes in this boundary. Is $D\times D$ a basic closed semialgebraic set? Is it a spectrahedron?

First, some interesting/useful stuff.

Claim. The direct product of finitely many non-empty convex bodies is also convex. (Just for fun)
Pf. Let $A_1,\ldots,A_m$ be convex non-empty sets in $\mathbb{R}^n$ and $B:=A_1\times\dots\times A_m$. For all $x,y\in B$ and all $\lambda \in [0,1]$ is $z_i:=\lambda x_i + (1-\lambda) y_i \in A_i$ and, hence, $z=\lambda x + (1-\lambda)y\in B$.
Note that the converse is also true, ie if B is convex then all $A_i$ are convex.

A basic closed semialgebraic set S in $\mathbb{R}^n$ is the solution set of a finite system of non-strict polynomial inequalities, ie

\begin{align} S = \{ x\in\mathbb{R}^n\ :\ f_1(x)\geq 0,\ldots,f_m(x)\geq 0\} \end{align}

for a positive integer m and polynomials $f_1\ldots,f_m\in\mathbb{R}[x_1,\ldots,x_n]$.

The unit disk D has the following representation as a spectrahedron,

\begin{align} D = \left\{ (x,y)\in\mathbb{R}^2\ :\ \begin{bmatrix} 1-x & y \\ y& 1+x \end{bmatrix} \right\}. \end{align}

(Recall this from Pablo's lecture.)
Now, the direct product of D with a copy of itself is the set

\begin{align} D \times D = \{ (x,y,v,w)\in\mathbb{R}^4\ :\ x^2+y^2\leq 1\ \wedge\ v^2+w^2\leq 1 \} = \left\{ (x,y,v,w)\in\mathbb{R}^4\ :\ A_1:=\begin{bmatrix} 1-x & y \\ y & 1+x \end{bmatrix}\succeq 0\ \wedge A_2:=\begin{bmatrix} 1-v & w \\ w & 1+v \end{bmatrix}\succeq 0 \right\} \end{align}

Recall that a block diagonal matrix is positive semidefinite if and only if the blocks are positive semidefinite. Therefore, $D\times D$ is a spectrahedron with representation

\begin{align} D\times D = \left\{ (x,y,v,w)\in\mathbb{R}^4\ :\ A:= \begin{bmatrix} 1-x & y & 0 & 0 \\ y & 1+x & 0 & 0 \\ 0 & 0 & 1-v & w \\ 0 & 0 & w & 1+v \end{bmatrix}\succeq 0 \right\}. \end{align}

Since a matrix is positive semidefinite iff all principal minors are non-negative, and the number of principal minors is finite, every spectrahedron is a basic closed semialgebraic set.

A point lies on the boundary of a spectrahedron iff all principal minors are non-negative and the determinant vanishes at this point or, equivalently, the defining matrix is rank deficient at this point, ie the determinant is a (non-zero) polynomial that vanishes on the boundary. We get that the boundary is a subset of the (complex) variety defined by the determinant of A,

\begin{align} \partial(D\times D) \subset V(\det A) = V(\det (A_1)\cdot\det(A_2)) = V((1-x^2-y^2)(1-v^2-w^2)). \end{align}

Note that the determinant of a block diagonal matrix is the product of the determinants of the blocks.

Now, let's compute the faces. Every face of a spectrahedron is exposed, ie a face F is the intersection of the set S with an affine hyperplane H such that S is contained in one of the defined halfspaces.

A point $(x_0,y_0,v_0,w_0)$ lies on the boundary of $D\times D$ iff the determinant vanishes at this point, ie the rank of $A(x_0,y_0,v_0,w_0)$ is at most 3. Up to symmetry, there are only two possible cases.

  • $rank A = 0$ iff A is the zero matrix iff both diagonal blocks are the zero matrix, but $1-x$ and $1+x$ have no common roots (the same holds for $A_2$).
  • $rank A = 1$ iff one diagonal block has rank 1 and the other rank 0, see above.
  • $rank A = 2$ iff both blocks have rank 1 (by above, no other case is possible), ie both, $(x_0,y_0)$ and $(v_0,w_0)$, are zero-dimensional faces of D.
  • $rank A = 3$ iff one block has rank 1 and the other block has rank 2, ie the direct product of a point on the boundary of D with (the relative interior of) D.

Let $(x_0,y_0)$ any but fixed point on the boundary of D, ie $1-x_0^2-y_0^2=0$, then the unique tangent (ie hyperplane in the plane) at this point is $H = \{(x,y)\in\mathbb{R}^2\ :\ x_0x+y_0y=1 \}$. Obviously, the intersection of H (as a hyplerplane in $\mathbb{R}^4$) with $D\times D$ is a face of the set. More precisely,

\begin{align} H\cap (D\times D) = \{ (x,y,v,w)\in\mathbb{R}^2\ :\ 1-v^2-w^2\geq 0 \} = \{(x_0,y_0)\}\times D. \end{align}

The same holds for any point $(v_0,w_0)$ on the boundary of D.

Summing up, the faces of $D\times D\subset\mathbb{R}^4$ are

  • the two-dimensional, rank 3 faces $\{(x_0,y_0)\}\times D$ for all $(x_0,y_0)\in\partial D$ and $\{(v_0,w_0)\}\times D$ for all $(v_0,w_0)\in\partial D$
  • the zero-dimensional, rank 2 faces $\{(x_0,y_0,v_0,w_0)\}$ for all $(x_0,y_0)\in\partial D$ and $(v_0,w_0)\in\partial D$

Remarks. [as an exercise ;-) ]
Let S be a spectrahedron and F a face of S. If A is in the relative interior of F and B on the relative boundary of F, then every other element in the interior of F has the same rank as A and the rank of A is greater than the rank of B (but, in general, the rank on the relative boundary is not constant!).

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