Let $M$ be a closed compact manifold.
Is the space of all $C^1$ immersions from $M$ to $\mathbb{R}^m$ ($m> \dim M$) dense in $C^1(M; \mathbb{R}^m)$ (in the $C^1$ topology)?
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4$\begingroup$ No. If you take the codomain to have dimension at least 2dim M then yes. You can find proofs in the second chapter of Hirsch's book on differential topology. $\endgroup$– mmeCommented Sep 21, 2018 at 19:10
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$\begingroup$ I would imagine that this is related to Nash's $C^1$ immersion results... If I throw in a little more regularity by changing $C^1(M,\mathbb{R}^3)$ to $C^1(M,\mathbb{R}^3)\cap W^{2,2}(M,\mathbb{R}^3)$, will the 'no' become a 'yes'? $\endgroup$– TYpCommented Sep 22, 2018 at 16:20
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$\begingroup$ 1) The problem is your topology. Nash's results guarantee a $C^1$ immersion arbitrarily close to your original immersion in the $C^0$ topology. You won't be able to get convergence in $C^1$. 2) Also, you have to assume you have an immersion to begin with. Your set could be empty unless the dimension of the codomain is sufficiently high. (Nonempty when it's at least 2dim M -1 by Whitney's hard immersion theorem, but you get denseness at 2dim M.) $\endgroup$– mmeCommented Sep 22, 2018 at 16:31
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$\begingroup$ Thanks for the remarks. I try to clarify the question: Let $X := C^1(M,\mathbb{R}^3)\cap W^{2,2}(M,\mathbb{R}^3)$, equipped with the norm $\|x\|_X := \|x\|_{C^1} + \|x\|_{W^{2,2}}$. If this space is too troublesome, consider the slightly bigger Banach space $X = W^{2,p}(M,\mathbb{R}^3))$ for a $p>2$. Q: Is $\{x \in X: x \mbox{ an immersion}\}$ dense in $X$ (w.r.t. to the topology of $X$)? $\endgroup$– TYpCommented Sep 22, 2018 at 16:46
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$\begingroup$ No, since your topology is stronger than the $C^1$ topology... I assume here $M$ is supposed to be a surface. $\endgroup$– mmeCommented Sep 22, 2018 at 16:59
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