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This might be a stupid question. The constant rank level set theorem says that if I have a smooth map $f:M\to N$ and a regular point $p\in N$, then if on $U_p$ the rank of the differential is constant, then the preimage of $p$ is a submanifold. The regular level set theorem only demands that the differential with respect to the preimages of $p$ is surjective. For the constant rank level set theorem to imply regular level set theorem, I need to know the surjectivity on the point $p$ implies the constant rank on some neighborhood of $p$. But I think this is not generally true. I'd appreciate any help!

jk001
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  • "preimage of $p\in M$" doesn't make sense when the map goes $M\to N$. Anyway, the set of surjective linear maps $\mathbb{R}^m\to\mathbb{R}^n$ is open: surjectivity (i.e., rank $n$) means there is an $n\times n$ submatrix which is invertible, and invertibility is open (as it is inverse image of nonzero under the polynomial function "det"). – user10354138 Feb 16 '21 at 14:50
  • Could you explain in what sense the set of maps is open? In what topology? Or do you mean that they are all open maps? I'm not sure if I understand the last part, "invertibility is open." – jk001 Feb 16 '21 at 15:03
  • I mean open in the space of matrices $\mathbb{R}^{m\times n}$ (usual topology). By invertibility is open I mean $GL_n(\mathbb{R})$ is open in $\mathbb{R}^{n\times n}$. If $df_p$ has full rank, so does nearby $df_{p'}$ (use a chart to get locally $df\colon U\to\mathbb{R}^{m\times n}$ and project to $\mathbb{R}^{n\times n}$, etc.). – user10354138 Feb 16 '21 at 15:13
  • Sorry I still don't quite see it...I think what you're saying is that the set of invertible matrices is open. This implies that the set of differential maps (which is a linear map from Euclidean space to Euclidean space, hence invertible matrices) is open. Now, how do I see that your explanation implies the statement in question?.. – jk001 Feb 16 '21 at 15:51

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Suppose $q\in M$ is a preimage of $p$ and thtat the map at $q$ has maximal rank. You can then find charts $V\subset M$ and $U\subset N$ such that $f(U)\subset V$ (because $f$ is smooth).

Consider now $g$ to be $f$ in the coordinates given by these charts, say $g = (g_1..., g_n)$. The rank of $f$ is the same as the rank of $g$ and the latter one can be computed in the Jacobian: $Dg = \left(\dfrac{dg_i}{dx_j}\right)$. That the rank is maximal it means that you can find a subdeterminant of maximal size (the biggest of the matrix dimensions) that does not vanish.

Notice that if you were working with a smaller rank, and not the maximal rank, that you can find a subdeterminant of the appropiate size that does not vanish is still true, but not equivalent to your matrix having that given rank since nothing assures you there is not a bigger subdeterminant which also doesnt vanish. That is, the rank might jump up. This of course doesnt occur with maximal rank since no bigger determinant can be found.

Since the determinant of this subsquare matrix is a polynomial in the $\dfrac{dg_i}{dx_j}$, the condition of it not vanishing is an open condition. That is, there is a neighborgood $U_q$ around $q$ such that the rank is constant since it is the maximal possible rank. Varying $q$ on the fiber of $p$, and making a union over all the previous neighborhoods $U_q$, implies that indeed there is an open neighborhood around it such that the rank is constant there and now you can apply the constant rank theorem to conclude the preimage is actually a regular submanifold.

MEEL
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  • Could you explain how we can decompose $g$ into $(g_1,g_2,\dots,g_n)$? $f$ is a map from $M$ to $N$ and there is a chart from $V$ to $\mathbb{R}^n$. Do you mean each $g_i$ is a map from $U$ to $\mathbb{R}$, as is usually called $f^{i}$? – jk001 Feb 17 '21 at 00:40
  • If $\phi$ and $\psi$ are the charts in $M$ and $N$, then $g = \psi\circ f \circ \phi^{-1}: \phi(V)\rightarrow \psi(U)$, which are open sets of respective euclidean spaces (whose dimensions are those of $M$ and $N$, say $m$ and $n$. Then, it looks like $g(x_1,..., x_m) = (g_1(x_1,..., x_m),...,g_n(x_1,..., x_m))$, and each $g_1,..., g_n$ is a smooth function to $\mathbb{R}$. – MEEL Feb 17 '21 at 02:31
  • Oh, thanks! Wouldn't $Dg$ be exactly the same matrix as $Df$ at $q$, by definition? I mean if we consider the same charts – jk001 Feb 17 '21 at 12:00