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Let $\text{GL}_n^+$ be the group of real invertible $n \times n$ matrices with positive determinant. Set $S=\text{GL}_n^+ \cup \{A \, | \, \text{rank} (A) = n-1 \}$.

Does $S$ form a a submanifold with boundary of $M_n$? ($M_n$ is the space of all square real matrices).

I am not really sure. I know that $S \subseteq \{A \in M_n \, | \, \det A \ge 0 \}$, and that every point of $S$ is a regular point of the determinant function $\det:M_n \to \mathbb{R}$. (The derivative of the determinant at a point $A$, which is the cofactor matrix of $A$, does not vanish when the rank is $n-1$).

However, I am not sure this gives us what we want, since the subset $\{A \in M_n \, | \, \det A \ge 0 \}$ containing points outside of $S$, which are critical points of the determinant function. (when the rank is smaller than $n-1$).

In particular, we can't use directly proposition 5.47 in Lee's book on smooth manifolds (second edition), which says $f^{-1}([b,\infty))$ is a submanifold with boundary (in fact a regular domain) when $b$ is a regular value of $f$.

Specifically, for $n=2$, we have $S=\text{GL}_2^+ \cup \{A \, | \, \text{rank} (A) = 1 \}=M_2 \setminus (\text{GL}_2^- \cup \{0\})$. What can be said in this case?

Asaf Shachar
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  • The presumed boundary, $\text{rank} (A)=n-1$, itself has a boundary, where the rank drops, so it might be a manifold with corners. For $n=2$ you have to remove the origin from $ac-bd=0$ hypersurface, for example. – Conifold Apr 30 '19 at 08:46
  • Well, this is exactly what I am not sure about. I know that the matrices of rank $n-1$ are not closed, because as you say, the rank can drop. But is this a problem? can't the boundary be non-closed inside the ambient manifold? (This is the "differential manifold boundary", which is not always the same as the "topological boundary"). As far as I know, this thing can be only an immersed submanifold, not an embedded one. The purpose of this question is exactly to clarify these points... – Asaf Shachar Apr 30 '19 at 09:18
  • I want to know if this creature falls into the category of smooth submanifolds in some precise way (e.g. with boundary/corners etc). – Asaf Shachar Apr 30 '19 at 09:19
  • I am not sure. For $2\times2$ $\text{rank} (A)=1$ is a submanifold, and so is $\text{rank} (A)=2$ for $3\times3$, see e.g. Stratification of 3×3 Matrices, p.4. It might be true in general. Then you can try to prove that any $\text{rank} (A)=n-1$ matrix has a neighborhood in the ambient diffeomorphic to a half-space, since lower rank matrices are not there. – Conifold Apr 30 '19 at 09:51

1 Answers1

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After some more thought, I think that the answer is positive:

Indeed, set $U=\{A \in M_n \, | \, \text{rank} (A) \ge n-1 \}$. $U$ is an open submanifold of $M_n$. The determinant function $\det:U \to \mathbb{R}$ has only regular points (in $U$). In particular $0$, is a regular value of it.

In particular, by proposition 5.47 in Lee's book on smooth manifolds (second edition), $$U \cap \det \ge 0=\text{GL}_n^+ \cup \{A \in M_n \, | \, \text{rank} (A) = n-1 \}=S$$

is a regular domain in $U$.

Asaf Shachar
  • 25,111