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The variety $X_n$ of singular $n\times n$ real matrices is stratified by smooth strata $X_{n,k}$ where $k$ is the rank. Choose a rank $k$ matrix $A\in X_{n,k}$. Is there a local diffeomorphism sending $X_n$ locally (near $A$) to the Cartesian product of the stratum $X_{n,k}$ by the determinantal variety of complementary rank $X_{n-k}$, with reference?

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    $\begingroup$ It seems to me that you can use the so called "Slice Theorem": en.wikipedia.org/wiki/Slice_theorem_%28differential_geometry%29 I think you can restrict your question to the set of $n \times n$ matrices such that $trace(X.X^t) = 1$. The group $G$ is the product $SO(n) \times SO(n)$ acting by left and right multiplications. Notice that $A$ is a smooth point of the strata $X_{n,k}$ so locally the strata agree with the $G$-orbit. Then it is necessary to check that the isotropy is $SO(n-k) \times SO(n-k)$ acts as you need on the normal space. $\endgroup$ – Holonomia Feb 9 '16 at 23:08
  • $\begingroup$ Well $A$ is a smooth point of the stratum itself but the determinantal variety itself is typically singular at $A$ so this is a bit more complicated than the smooth differential-topological situation. $\endgroup$ – Mikhail Katz Feb 10 '16 at 8:03
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    $\begingroup$ What if you apply the slice theorem at $A$ as a point of the smooth manifold $S := trace(X X^t) =1$ (the sphere) so you get a local $G$-equivariant splitting of $S$ near $A$. Then you restrict the splitting of the slice theorem to the determinantal variety $X_n$. Since the involved diffeomorphisms are $G$-equivariant and since the determinantal structure (included the singularities) is determined by the $G$-orbits it seems to me that you get something similar to what you are looking for. $\endgroup$ – Holonomia Feb 10 '16 at 21:30
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The result you state is more classical than the following reference, but the following reference is quite well-known. This is Example 5.3 of the following.

MR1623714 32J05
MacPherson, R.; Procesi, C.
Making conical compactifications wonderful.
Selecta Math. (N.S.) 4 (1998), no. 1, 125–139.

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    $\begingroup$ What MacPherson and Procesi do there seems to be the same thing as in my comment above i.e. using the slice theorem, etc. $\endgroup$ – Holonomia Feb 11 '16 at 6:52

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