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for question 1) it's not clear to me why $C^2$ approximating $z \mapsto z^k$ by immersions is possible at all. do you have an example of such a sequence?

edit: Ok, I thought some more on the topological issue of whether it's always possible to deform $z \mapsto z^k$ to an immersion in $C^2$ and I can say that it is definitely NOT possible if $k$ is even. In particular, it's impossible when $k=2$. There is an easy necessary condition for an existence of such deformation. An immersion $f$ of a disk gives a parallelization of the tangent bundle along $f$, i.e we get a map $f'\colon D^2\to V_2(\mathbb R^3)$ where $V_2(\mathbb R^3)$ is the Stiefel manifold of orthonormal 2-frames in $\mathbb R^3$ which is of course just $SO(3)$. This map ought to extend the map on the boundary of the disk $S^1$ which being an immersion already won't change much under a small deformation. That map is essentially given by $z\to kz^{k-1}$ which disregarding the conformal factor $k$ can be thought of as a map $S^1\to SO(2)=V_2^{or}(\mathbb R^2)\subset V_2(\mathbb R^2)=O(2)$. The natural map $V_2(\mathbb R^2)\to V_2(\mathbb R^3)$ corresponds to the standard inclusion $SO(2)\to SO(3)$ on identity component. Now, $\pi_1(SO(2))\cong \mathbb Z$ and $\pi_1(SO(3))\cong \mathbb Z/2$ and the map $\pi_1(SO(2))\to \pi_1(SO(3))$ is well-known to be onto (as is immediate from the long exact sequence for the fibration $SO(2)\to SO(3)\to S^2$).

That means that the map $z\mapsto z^{k-1}$ gives a generator of $\pi_1(SO(3))$ when $k$ is even and thus can not be extended to $D^2$.

(Note that this takes care of both 1) and 2) when $k$ is even.)

I think the above necessary condition should also be sufficient and therefore when $k$ is odd such a deformation should always be possible (at least in $C^0$) as evidenced by $k=1$ of course. I'm not at all an expert on immersions but the subject is very well understood and I hope someone who knows more about it will chime in.

OK, Robert Bryant answered this (see below). Moreover, a direct calculation shows that his example produces a family of immersions $f_t$ with with the induced metrics $ds_t^2$ having $sec\le 0$ for all $t>0$. This settles the original question in the positive for $k$ odd.

for question 1) it's not clear to me why $C^2$ approximating $z \mapsto z^k$ by immersions is possible at all. do you have an example of such a sequence?

edit: Ok, I thought some more on the topological issue of whether it's always possible to deform $z \mapsto z^k$ to an immersion in $C^2$ and I can say that it is definitely NOT possible if $k$ is even. In particular, it's impossible when $k=2$. There is an easy necessary condition for an existence of such deformation. An immersion $f$ of a disk gives a parallelization of the tangent bundle along $f$, i.e we get a map $f'\colon D^2\to V_2(\mathbb R^3)$ where $V_2(\mathbb R^3)$ is the Stiefel manifold of orthonormal 2-frames in $\mathbb R^3$ which is of course just $SO(3)$. This map ought to extend the map on the boundary of the disk $S^1$ which being an immersion already won't change much under a small deformation. That map is essentially given by $z\to kz^{k-1}$ which disregarding the conformal factor $k$ can be thought of as a map $S^1\to SO(2)=V_2^{or}(\mathbb R^2)\subset V_2(\mathbb R^2)=O(2)$. The natural map $V_2(\mathbb R^2)\to V_2(\mathbb R^3)$ corresponds to the standard inclusion $SO(2)\to SO(3)$ on identity component. Now, $\pi_1(SO(2))\cong \mathbb Z$ and $\pi_1(SO(3))\cong \mathbb Z/2\mathbb Z$ and the map $\pi_1(SO(2))\to \pi_1(SO(3))$ is well-known to be onto (as is immediate from the long exact sequence for the fibration $SO(2)\to SO(3)\to S^2$).

That means that the map $z\mapsto z^{k-1}$ gives a generator of $\pi_1(SO(3))$ when $k$ is even and thus can not be extended to $D^2$.

(Note that this takes care of both 1) and 2) when $k$ is even.)

I think the above necessary condition should also be sufficient and therefore when $k$ is odd such a deformation should always be possible (at least in $C^0$) as evidenced by $k=1$ of course. I'm not at all an expert on immersions but the subject is very well understood and I hope someone who knows more about it will chime in.

OK, Robert Bryant answered this (see below). Moreover, a direct calculation shows that his example produces a family of immersions $f_t$ with with the induced metrics $ds_t^2$ having $sec\le 0$ for all $t>0$. This settles the original question in the positive for $k$ odd.

for question 1) it's not clear to me why $C^2$ approximating $z \mapsto z^k$ by immersions is possible at all. do you have an example of such a sequence?

edit: Ok, I thought some more on the topological issue of whether it's always possible to deform $z \mapsto z^k$ to an immersion in $C^2$ and I can say that it is definitely NOT possible if $k$ is even. In particular, it's impossible when $k=2$. There is an easy necessary condition for an existence of such deformation. An immersion $f$ of a disk gives a parallelization of the tangent bundle along $f$, i.e we get a map $f'\colon D^2\to V_2(\mathbb R^3)$ where $V_2(\mathbb R^3)$ is the Stiefel manifold of orthonormal 2-frames in $\mathbb R^3$ which is of course just $SO(3)$. This map ought to extend the map on the boundary of the disk $S^1$ which being an immersion already won't change much under a small deformation. That map is essentially given by $z\to kz^{k-1}$ which disregarding the conformal factor $k$ can be thought of as a map $S^1\to SO(2)=V_2^{or}(\mathbb R^2)\subset V_2(\mathbb R^2)=O(2)$. The natural map $V_2(\mathbb R^2)\to V_2(\mathbb R^3)$ corresponds to the standard inclusion $SO(2)\to SO(3)$ on identity component. Now, $\pi_1(SO(2))\cong \mathbb Z$ and $\pi_1(SO(3))\cong \mathbb Z/2$ and the map $\pi_1(SO(2))\to \pi_1(SO(3))$ is well-known to be onto (as is immediate from the long exact sequence for the fibration $SO(2)\to SO(3)\to S^2$).

That means that the map $z\mapsto z^{k-1}$ gives a generator of $\pi_1(SO(3))$ when $k$ is even and thus can not be extended to $D^2$.

(Note that this takes care of both 1) and 2) when $k$ is even.)

I think the above necessary condition should also be sufficient and therefore when $k$ is odd such a deformation should always be possible (at least in $C^0$) as evidenced by $k=1$ of course. I'm not at all an expert on immersions but the subject is very well understood and I hope someone who knows more about it will chime in.

OK, Robert Bryant answered this (see below). Moreover, a direct calculation shows that his example produces a family of immersions $f_t$ with with the induced metrics $ds_t^2$ having $sec\le 0$ for all $t>0$. This settles the original question in the positive for $k$ odd.

For question 2) you can not expect any upper curvature bounds. Given an immersion you can locally perturb it on an arbitrary small neighborhood of 0 by adding a small thin "finger" to your surface. this will introduce some arbitrary positive (and negative) curvature. Doing this along a given sequence converging in $C^2_{loc}(\mathbb D\backslash \{0\})\cap C^0(\mathbb D)$ will keep such convergence.

for question 1) it's not clear to me why $C^2$ approximating $z \mapsto z^k$ by immersions is possible at all. do you have an example of such a sequence?

edit: Ok, I thought some more on the topological issue of whether it's always possible to deform $z \mapsto z^k$ to an immersion in $C^2$ and I can say that it is definitely NOT possible if $k$ is even. In particular, it's impossible when $k=2$. There is an easy necessary condition for an existence of such deformation. An immersion $f$ of a disk gives a parallelization of the tangent bundle along $f$, i.e we get a map $f'\colon D^2\to V_2(\mathbb R^3)$ where $V_2(\mathbb R^3)$ is the Stiefel manifold of orthonormal 2-frames in $\mathbb R^3$ which is of course just $SO(3)$. This map ought to extend the map on the boundary of the disk $S^1$ which being an immersion already won't change much under a small deformation. That map is essentially given by $z\to kz^{k-1}$ which disregarding the conformal factor $k$ can be thought of as a map $S^1\to SO(2)=V_2^{or}(\mathbb R^2)\subset V_2(\mathbb R^2)=O(2)$. The natural map $V_2(\mathbb R^2)\to V_2(\mathbb R^3)$ corresponds to the standard inclusion $SO(2)\to SO(3)$ on identity component. Now, $\pi_1(SO(2))\cong \mathbb Z$ and $\pi_1(SO(3))\cong \mathbb Z/2$ and the map $\pi_1(SO(2))\to \pi_1(SO(3))$ is well-known to be onto (as is immediate from the long exact sequence for the fibration $SO(2)\to SO(3)\to S^2$).

That means that the map $z\mapsto z^{k-1}$ gives a generator of $\pi_1(SO(3))$ when $k$ is even and thus can not be extended to $D^2$.

(Note that this takes care of both 1) and 2) when $k$ is even.)

I think the above necessary condition should also be sufficient and therefore when $k$ is odd such a deformation should always be possible (at least in $C^0$) as evidenced by $k=1$ of course. I'm not at all an expert on immersions but the subject is very well understood and I hope someone who knows more about it will chime in.

OK, Robert Bryant answered this (see below). Moreover, a direct calculation shows that his example produces a family of immersions $f_t$ with with the induced metrics $ds_t^2$ having $sec\le 0$ for all $t>0$. This settles the original question in the positive for $k$ odd.

for question 1) it's not clear to me why $C^2$ approximating $z \mapsto z^k$ by immersions is possible at all. do you have an example of such a sequence?

edit: Ok, I thought some more on the topological issue of whether it's always possible to deform $z \mapsto z^k$ to an immersion in $C^2$ and I can say that it is definitely NOT possible if $k$ is even. In particular, it's impossible when $k=2$. There is an easy necessary condition for an existence of such deformation. An immersion $f$ of a disk gives a parallelization of the tangent bundle along $f$, i.e we get a map $f'\colon D^2\to V_2(\mathbb R^3)$ where $V_2(\mathbb R^3)$ is the Stiefel manifold of orthonormal 2-frames in $\mathbb R^3$ which is of course just $SO(3)$. This map ought to extend the map on the boundary of the disk $S^1$ which being an immersion already won't change much under a small deformation. That map is essentially given by $z\to kz^{k-1}$ which disregarding the conformal factor $k$ can be thought of as a map $S^1\to SO(2)=V_2^{or}(\mathbb R^2)\subset V_2(\mathbb R^2)=O(2)$. The natural map $V_2(\mathbb R^2)\to V_2(\mathbb R^3)$ corresponds to the standard inclusion $SO(2)\to SO(3)$ on identity component. Now, $\pi_1(SO(2))\cong \mathbb Z$ and $\pi_1(SO(3))\cong \mathbb Z/2$ and the map $\pi_1(SO(2))\to \pi_1(SO(3))$ is well-known to be onto (as is immediate from the long exact sequence for the fibration $SO(2)\to SO(3)\to S^2$).

That means that the map $z\mapsto z^{k-1}$ gives a generator of $\pi_1(SO(3))$ when $k$ is even and thus can not be extended to $D^2$.

(Note that this takes care of both 1) and 2) when $k$ is even.)

I think the above necessary condition should also be sufficient and therefore when $k$ is odd such a deformation should always be possible (at least in $C^0$) as evidenced by $k=1$ of course. I'm not at all an expert on immersions but the subject is very well understood and I hope someone who knows more about it will chime in.

OK, Robert Bryant answered this (see below). Moreover, a direct calculation shows that his example produces a family of immersions $f_t$ with with the induced metrics $ds_t^2$ having $sec\le 0$ for all $t>0$. This settles the original question in the negative for $k$ odd.

For question 2) you can not expect any upper curvature bounds. Given an immersion you can locally perturb it on an arbitrary small neighborhood of 0 by adding a small thin "finger" to your surface. this will introduce some arbitrary positive (and negative) curvature. Doing this along a given sequence converging in $C^2_{loc}(\mathbb D\backslash \{0\})\cap C^0(\mathbb D)$ will keep such convergence.

for question 1) it's not clear to me why $C^2$ approximating $z \mapsto z^k$ by immersions is possible at all. do you have an example of such a sequence?

edit: Ok, I thought some more on the topological issue of whether it's always possible to deform $z \mapsto z^k$ to an immersion in $C^2$ and I can say that it is definitely NOT possible if $k$ is even. In particular, it's impossible when $k=2$. There is an easy necessary condition for an existence of such deformation. An immersion $f$ of a disk gives a parallelization of the tangent bundle along $f$, i.e we get a map $f'\colon D^2\to V_2(\mathbb R^3)$ where $V_2(\mathbb R^3)$ is the Stiefel manifold of orthonormal 2-frames in $\mathbb R^3$ which is of course just $SO(3)$. This map ought to extend the map on the boundary of the disk $S^1$ which being an immersion already won't change much under a small deformation. That map is essentially given by $z\to kz^{k-1}$ which disregarding the conformal factor $k$ can be thought of as a map $S^1\to SO(2)=V_2^{or}(\mathbb R^2)\subset V_2(\mathbb R^2)=O(2)$. The natural map $V_2(\mathbb R^2)\to V_2(\mathbb R^3)$ corresponds to the standard inclusion $SO(2)\to SO(3)$ on identity component. Now, $\pi_1(SO(2))\cong \mathbb Z$ and $\pi_1(SO(3))\cong \mathbb Z/2$ and the map $\pi_1(SO(2))\to \pi_1(SO(3))$ is well-known to be onto (as is immediate from the long exact sequence for the fibration $SO(2)\to SO(3)\to S^2$).

That means that the map $z\mapsto z^{k-1}$ gives a generator of $\pi_1(SO(3))$ when $k$ is even and thus can not be extended to $D^2$.

(Note that this takes care of both 1) and 2) when $k$ is even.)

I think the above necessary condition should also be sufficient and therefore when $k$ is odd such a deformation should always be possible (at least in $C^0$) as evidenced by $k=1$ of course. I'm not at all an expert on immersions but the subject is very well understood and I hope someone who knows more about it will chime in.

OK, Robert Bryant answered this (see below). Moreover, a direct calculation shows that his example produces a family of immersions $f_t$ with with the induced metrics $ds_t^2$ having $sec\le 0$ for all $t>0$. This settles the original question in the positive for $k$ odd.

For question 2) you can not expect any upper curvature bounds. Given an immersion you can locally perturb it on an arbitrary small neighborhood of 0 by adding a small thin "finger" to your surface. this will introduce some arbitrary positive (and negative) curvature. Doing this along a given sequence converging in $C^2_{loc}(\mathbb D\backslash \{0\})\cap C^0(\mathbb D)$ will keep such convergence.