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Let $P_1,\ldots,P_r$ be polynomials over ${\mathbb R}^N$. I am interested in the homotopy type of the semi-algebraic set defined by $$ P_j(x_1,\ldots,x_N)>0,\qquad j=1,\ldots,r . $$

Is there a general theory about that?

Here are motivating examples:

  1. If $r=1$ and $P$ is a homogeneous polynomial, hyperbolic in the direction of some vector $e\ne0$, the connected component of $e$ in $P>0$ is convex (Garding) and therefore has a trivial topology.

  2. If ${\mathbb R}^N=M_n({\mathbb R})$, $r=1$ and $P(M)=\det M$, then $P>0$ is $GL_n^+({\mathbb R})$. Thanks to the polar decomposition, it is homeomorphic to $SO_n({\mathbb R})\times SDP_n$ and its homotopy type is that of $SO_n({\mathbb R})$. For instance, the fundamental group is ${\mathbb Z}$ if $n=2$ and ${\mathbb Z}/2{\mathbb Z}$ if $n\ge3$.

  3. If ${\mathbb R}^N=M_n({\mathbb R})$, $r=n$ and $P_j$ is the $j$th principal minor, then the set $X$ defined by all $P_j(M)>0$ is homeomorphic to $L_n\times U_n^+$, where $L_n$ is the set of lower triangular matrices with unit diagonal and $U_n$ consists of upper triangular matrices with strictly positive diagonal ($LU$ factorization). The homotopy type of $X$ is thus trivial.

I am interested in the special case of the set $TP_n$ of totally positive $n\times n$ matrices. It is defined by the inequalities involving minors $$ M\begin{pmatrix} i_1 & \ldots & i_s \\ j_1 & \ldots & j_s \end{pmatrix} > 0 $$ for every $s=1,\ldots,n$ and every increasing sequences $k\mapsto i_k$ and $k\mapsto j_k$.

I point out that this latter question can be reduced to that of the homotopy type of appropriate subsets $\ell_n$ and $u_n^+$ of $L_n$ and $U_n^+$, defined by the minor equalities above with either the restriction $i_s\le j_1$ (upper triangular case) or $j_s\le i_1$ (lower triangular case). It is known that $TP_n=\ell_n\cdot u_n^+$.

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Why do you have to assume the set is open? Doesn't this follow from continuity of the $P_j$? – Noah Stein Aug 11 '11 at 9:59
Perhaps you mean nonempty and open? Also, from the examples, it looks like you mean "homotopy type" when you say "topology". Is that correct? – S. Carnahan Aug 11 '11 at 10:45
@Noah. Right. I edit – Denis Serre Aug 11 '11 at 11:00
@Carnahan. Right. This is because I am not a topologist. – Denis Serre Aug 11 '11 at 11:00
If we mod out $TP_n$ by $GL(n)^+$, invertible matrices with positive determinant, we get the totally positive Grassmannian, which Postnikov has shown is homeomorphic to an open ball of appropriate dimension: – Sam Hopkins May 10 '14 at 13:47

2 Answers 2

The homotopy groups of the semigroup of determinant one totally positive matrices are all trivial. A proof of this is in the paper (example 6.2)

A. J. Santana and L.A.B. San Martin: The homotopy type of Lie semigroups in semi-simple Lie groups. Monatshefte fur Mathematik, v. 136, n. 2, p. 151-173, 2002.

More generally in this paper it is proved (Theorem 4.15) that the homotopy type of a Lie semigroup in a semi-simple Lie group is that of a compact subgroup. The theorem is proved under an assumption of existence of a "large" subsemigroup necely generated by exponentials. It applies also e.g. to the semigroup of nxn matrices with positive entries and determinant one. This semigroup has the homotopy type of the compact group SO(n-1).

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I cannot speak for totally positive matrices, but since one of Nash's theorem says that every smooth manifold is homeomorphic to a real algebraic variety (yes, I know you have only inequalities, but you can replace equations by inequalities by thickening a bit), I don't think you can get any sort of restriction in the general setting.

Actually, a quick search find this paper:

Which gives a pretty concrete description to the space of total positive matrices; see also

For an introduction to the above.

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