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A spectrahedron is a convex set defined by a linear matrix inequality (LMI).

Is the boundary of such a set almost always smooth?

By "smooth" I mean that it admits a tangent hyperplane at any point of the surface. I say "almost" because, for the special case of the polytope, the boundary has many "edges". In this special case, the answer would be no.

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    $\begingroup$ Could you be more precise about whay you mean by "almost"? Is there some sense in which the polytope is not typical? $\endgroup$
    – Yemon Choi
    Commented Sep 4, 2016 at 0:17
  • $\begingroup$ I deliberately chose a vague word, the question is open. One possible interpretation is: that happens with probability 1 if the entries of the matrices present in the inequalities defining the set are drawn from a normal distribution. $\endgroup$
    – maroxe
    Commented Sep 4, 2016 at 0:31
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    $\begingroup$ Wikipedia says "Alternatively, the set of n × n positive semidefinite matrices forms a convex cone in $\mathbb{R}^{n × n}$, and a spectrahedron is a shape that can be formed by intersecting this cone with a linear affine subspace." I'm guessing this would just be a straightforward application of transversality/Sard? $\endgroup$
    – Kevin Casto
    Commented Sep 4, 2016 at 0:55
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    $\begingroup$ there are many interesting examples where there are singularities on the border. E.g. spectrahedra describing decompositions of polynomials as sums of squares have singularities corresponding to minimal possible number of squares in such decompositions. $\endgroup$
    – Dima Pasechnik
    Commented Dec 12, 2017 at 14:21

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There are many interesting examples where there are singularities on the border. E.g. spectrahedra describing decompositions of polynomials as sums of squares (s.o.s.) of polynomials have singularities corresponding to minimal possible number of squares in such decompositions.

Specifically, if $f(x_1,...,x_n)=\sum_k q_k(x_1,...,x_n)^2$ then $f=X^\top A X$, where $A$ is a positive semidefinite symmetric matrix and $X$ is the vector of monomials in $x_j$ of degree half the degree of $f$. All such $A$ for a given $f$ form a compact spectrahedron.

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Let $\mathrm{Sym}^2(\mathbb{R}^n)$ be the $\binom{n+1}{2}$ dimensional vector space of $n \times n$ symmetric matrices and let $P \subset \mathrm{Sym}^2(\mathbb{R}^n)$ be the cone of positive semidefinite matrices. A spectrahedron is the intersection of $P$ with an affine linear space $L$. One reasonable interpretation of the "almost always" in your question is "except for a measure $0$ subset of the Grassmannian of $d$-dimensional affine subspaces of $\mathrm{Sym}^2(\mathbb{R}^n)$. With that interpretation, the answer is "yes" for $d \leq 2$, but "no" for $d \geq 3$.

The boundary $\partial (P \cap L)$ is $\partial P \cap L$. The boundary $\partial P$ is singular along the $\binom{n+1}{2}-3$ dimensional space $Q$ of positive semidefinite matrices of rank $\leq n-2$, and smooth at the matrices of rank $n-1$. If $L$ passes through $Q$ then $\partial P \cap L$ is almost surely singular at $Q \cap L$; if $L$ misses $Q$ then $L$ is likely to be singular.

Basic dimension counting (like the proof of Bertini's theorem, though this is in the real setting) shows that, since $Q$ has codimension $3$, a positive volume set of $L$ planes meets $Q$ for $d \geq 3$, but only (using Sard's theorem) a measure $0$ set for $d \leq 2$.

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