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If $\vec{f}\in C^{\infty}(\mathbb{R}^D, \mathbb{R}^K)$, $K=D-d$, such that the Jacobian, $D\vec{f}(\vec{x})$ of $\vec{f}$ has full-rank (i.e. $=K$) on $M=\vec{f}^{-1}(\vec{0})$, then $M$ is a smooth $d$-dimensional submanifold of $\mathbb{R}^D$. This is a consequence of the implicit function theorem.

Question: What sort of smooth manifolds fail to have this representation and are there any classic references for such examples? In other words: are all smooth manifolds the intersection of smooth hypersurfaces?

My intuition tells me that self-intersections are problematic and also possibly orientation-issues. For example, and somewhat informally, consider a figure-eight closed curve in the plane with the point of self-intersection situated at the origin. There is not a unique tangent vector at this point, and thus the only vector that can be orthogonal to all such tangents is the zero vector-so that the gradient of any possible "implicit" function must vanish here--and so $f$ cannot satisfy the hypotheses above. Is this correct?

(I apologize if this is a little elementary--I am self studying differential geometry and it can be a little daunting to navigate the literature, especially with the variety of presentations between intrinsic, extrinsic, local coordinates, etc.)

Edit/Update This nice summary answer on MathOverflow gives a positive answer--locally--which I had expected knowing about the implicit function theorem. But all the classic examples I can think of, e.g. $f(x,y)=x^2+y^2-1$ for the circle, $f(x,y,z) = \left(\sqrt{x^2+y^2}-R)^2\right)+z^2-r^2$, the torus, are all defined globally. (For any of these, if you want to write the manifold in the form, say, $(x,y,F(x,y))$ you have to do this locally, of course.)

Let's consider the circle again. Locally, we know we can define the function $f_\alpha(x,y) = \sqrt{1-x^2}-y$ that has the equation $f_\alpha(x,y)=0$ defining a portion of the manifold. If we were ignorant and started with this, we could easily solve for $y$, and discover the global implicit equation $x^2+y^2-1=0$. I wonder if there are any conditions, and how deep they are, to ensure this is possible in general, if at all...

Nap D. Lover
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  • All smooth manifolds can be put in that form. That's Whitney's embedding theorem. – NDB Jul 12 '23 at 20:33
  • @NDB Sorry that I am being dense but you can you explain a little further. I get that we can map a $d$ dimensional smooth manifold smoothly into $2d$-Euclidean space by Whitney's but how can we construct/obtain $f$ so that $M = {x\in \mathbb{R}^{2d}:f(x)=\vec{0}}$ from this? – Nap D. Lover Jul 13 '23 at 17:22
  • You can take $f(x)=\inf{|x-y|^2\mid y\in M}$. – NDB Jul 13 '23 at 17:27
  • @NDB But that $f$, as defined, is from $\mathbb{R}^{2d}\to \mathbb{R}$ which does not satisfy what I am looking for in the OP. – Nap D. Lover Jul 13 '23 at 19:24
  • Which part it doesn't satisfy? – NDB Jul 13 '23 at 19:26
  • @NDB The codomain should be $\mathbb{R}^K$, $K=D-d$ (or if we use Whitney's, $K=2d-d=d$). – Nap D. Lover Jul 13 '23 at 19:27
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    Ah, I see. You want them to be like complete intersection. – NDB Jul 13 '23 at 19:39
  • @NDB Yes, essentially the question can be rephrased as: can all smooth manifolds be written as the intersection of smooth hypersurfaces? I think I'll edit this into the OP. – Nap D. Lover Jul 13 '23 at 20:10
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    Your manifold has to be stably parallelizable. – Moishe Kohan Jul 14 '23 at 00:18
  • @MoisheKohanonstrike Oh, interesting! How much work is this "stably" qualifier doing?--because on, e.g., Wikipedia, it says that the ordinary $2$-sphere is not parallelizable, thanks to the Hairy ball theorem. But since $f(x,y,z)=x^2+y^2+z^2-1=0$ defines the sphere globally, I am guessing it is stably parallelizable--or did I severely misunderstand something? In any case, this is great motivation to get my hands dirty with tangent bundles etc-so thank you for the comment. – Nap D. Lover Jul 14 '23 at 00:39
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    Google it. For instance, such manifolds are orientable. The question was definitely asked before. – Moishe Kohan Jul 14 '23 at 00:59
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    https://math.stackexchange.com/questions/1810066/is-every-smooth-manifold-the-solution-set-of-finitely-many-equations/1810260#1810260 – Moishe Kohan Jul 14 '23 at 13:48
  • @MoisheKohanonstrike Thanks! I found this and another linked last night--your answer is very instructive. I have marked this question as a duplicate now since I didn't get around to it last night. – Nap D. Lover Jul 14 '23 at 17:58

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