show/hide this revision's text 4 added J-homomorphism explanation

I have two comments which don't fit into the comments field so I'm posting this as an answer.

  1. As Ryan Budney mentioned above the problem becomes trivial if we compose the original embedding $M^d\to \mathbb R^n$ with the canonical inclusion $\mathbb R^n\times{0}\subset \mathbb R^{n+1}$. That means in particular that in the stable range (when $n>2d+1$) there are no obstructions because in that range any two embeddings of $M\to \mathbb R^n$ are ambiently isotopic so if the problem is solvable for one then also for the other. Since $n>2d+1$ we can always isotope our original embedding into $\mathbb R^{n-1}\times {0}\subset \mathbb R^n$ and the claim follows.

  2. When $k=n-d-1$ is odd then $S^k$ is an Eilenberg-Maclane space over $\mathbb Q$ which means that rationally the same exact argument that worked for $k=1$ also works here. To elaborate further, given a trivialization of the normal bundle we have a tubular neighborhood $U\cong D^{k+1}\times M$ and we have an obvious map $f: U\to D^{k+1}$ given by the projection onto the second factor. On the boundary of $U$ the map takes values in $S^k$ and we want to extend it to a map from $W=\mathbb R^n\backslash U$ to $S^k$. Since $S^k$ is rationally equivalent to $K(\mathbb Q,k)$, rationally the homotopy type of $f|_{\partial U}$ is determined by $f^*([S^k]) \in H^*(\partial U, \mathbb Q)$ and the question becomes whether or not this class in the image of $i^*:H^k(W,\mathbb Q)\to H^k(\partial W\cong M\times S^k,\mathbb Q)$. By Alexander duality $H_k(W,\mathbb Q)$ is isomorphic to $H^d(M,\mathbb Q)\cong \mathbb Q)$ with the generator given by $i_*([S^k])$ where $S^k$ is the normal $S^k$ in $\partial W\cong S^k\times M$. Let $\alpha\in H^k(W,\mathbb Q)$ be the dual generator to $i_*([S^k])$. Now, a priori, the original trivialization may have been wrong so that $i^*(\alpha)$ is not equal to $f^*([S^k])$ if $H^k(M)\ne 0$. However, since the evaluation map $SO(k+1) \to S^k$ is a rational isomorphism on $H^k$ we can modify the original trivialization by an appropriate map $M\to SO(k+1)$ which does make $i^*(\alpha)=f^*([S^k])$ meaning that the map extends. What this means is that whatever (if any) obstructions are present in this case they are all torsion.

When $k=1$ then $S^1$ is already a $K(\mathbb Z,1)$ space and the above works on the nose without tensoring with $\mathbb Q$ as Ryan mentioned in a comment above.

I just realized that algori's answer below can not be correct. His(hers?) idea was that if we write $M$ as $(f_1,\ldots, f_{k})=0$ then after a small perturbation we can assume that $0$ is still a regular value $(f_1,\ldots, f_{k-1})=0$ and hence $M$ frame bounds in $N=(f_1,\ldots, f_{k-1})=0$ because it clearly separates $N$. However, this argument does not work because the level set $(f_1,\ldots, f_{k-1})=0$ might not be compact so the fact that $M$ separates it does not imply that it's cobordant to zero. This is not a fake issue since otherwise by Ryan's observation above it would imply that every framed cobordance class is stably trivial which is known not to be the case.

Also, as far as I can tell the effect of the change of the trivialization by a map $M\to SO(n-d)$ changes the framed cobordance class by something in the image of the $J$-homomorphism $J\colon \pi_d(SO(n-d))\to \pi_n(S^{n-d})\cong\Omega^{fr}_d(\mathbb R^n)$.

More explicitly it seems to me that it works as follows. Given any $\alpha\in \pi_d(SO(n-d))$ and $M^d$ as above, take a degree one map $f:M\to S^d$. Then twisting the trivialization by $\alpha\circ f:M\to SO(n-d)$ should give a new framed cobordism class which is different from the original one by $J(\alpha)$. This should be very well-known I'm sure so could somebody in the know please comment on this?

So it would seem that the group $\pi_n(S^{n-d})/\mathrm{Im }(J)$ is relevant here but I'm having trouble phrasing our obstruction problem in cobordism terms. In particular, is it clear that if we have two frame cobordant manifolds in $\mathbb R^n$ and one can be given by a single equation then so is the other?
show/hide this revision's text 3 added 1360 characters in body

I just realized that algori's answer below can not be correct. His(hers?) idea was that if we write $M$ as $(f_1,\ldots, f_{k})=0$ then after a small perturbation we can assume that $0$ is still a regular value $(f_1,\ldots, f_{k-1})=0$ and hence $M$ frame bounds in $N=(f_1,\ldots, f_{k-1})=0$ because it clearly separates $N$.However, this argument does not work because the level set $(f_1,\ldots, f_{k-1})=0$ might not be compact so the fact that $M$ separates it does not imply that it's cobordant to zero. This is not a fake issue since otherwise by Ryan's observation above it would imply that every framed cobordance class is stably trivial which is known not to be the case.

Also, as far as I can tell the effect of the change of the trivialization by a map $M\to SO(n-d)$ changes the framed cobordance class by something in the image of the $J$-homomorphism $J\colon \pi_d(SO(n-d))\to \pi_n(S^{n-d})\cong\Omega^{fr}_d(\mathbb R^n)$. This should be well-known I'm sure so could somebody in the know comment on this?

So it would seem that the group $\pi_n(S^{n-d})/\mathrm{Im }(J)$ is relevant here but I'm having trouble phrasing our obstruction problem in cobordism terms. In particular, is it clear that if we have two frame cobordant manifolds in $\mathbb R^n$ and one can be given by a single equation then so is the other?
show/hide this revision's text 2 added 14 characters in body

I have two comments which don't fit into the comments field so I'm posting this as an answer.

  1. As Ryan Budney mentioned above the problem becomes trivial if we compose the original embedding $M^d\to \mathbb R^n$ with the canonical inclusion $\mathbb R^n\times{0}\subset \mathbb R^{n+1}$. That means in particular that in the stable range (when $n>2d+1$) there are no obstructions because in that range any two embeddings of $M\to \mathbb R^n$ are ambiently isotopic so if the problem is solvable for one then also for the other. Since $n>2d+1$ we can always isotope our original embedding into $\mathbb R^{n-1}\times {0}\subset \mathbb R^n$ and the claim follows.

  2. When $k=n-d-1$ is odd then $S^k$ is an Eilenberg-Maclane space over $\mathbb Q$ which means that rationally the same exact argument that worked for $k=1$ also works here. To elaborate further, given a trivialization of the normal bundle we have a tubular neighborhood $U\cong D^{k+1}\times M$ and we have an obvious map $f: U\to D^{k+1}$ given by the projection onto the second factor. On the boundary of $U$ the map takes values in $S^k$ and we want to extend it to a map from $W=\mathbb R^n\backslash U$ to $S^k$. Since $S^k$ is rationally equivalent to $K(\mathbb Q,k)$, rationally the homotopy type of $f$ f|_{\partial U}$ is determined by $f^*([S^k]) \in H^*(\partial U, \mathbb Q)$ and the question becomes whether or not this class in the image of $i^*:H^k(W,\mathbb Q)\to H^k(\partial W\cong M\times S^k,\mathbb Q)$. By Alexander duality $H_k(W,\mathbb Q)$ is isomorphic to $H^d(M,\mathbb Q)\cong \mathbb Q)$ with the generator given by $i_*([S^k])$ where $S^k$ is the normal $S^k$ in $\partial W\cong S^k\times M$. Let $\alpha\in H^k(W,\mathbb Q)$ be the dual generator to $i_*([S^k])$. Now, a priori, the original trivialization may have been wrong so that $i^*(\alpha)$ is not equal to $f^*([S^k])$ if $H^k(M)\ne 0$. However, since the evaluation map $SO(k+1) \to S^k$ is a rational isomorphism on $H^k$ we can modify the original trivialization by an appropriate map $M\to SO(k+1)$ which does make $i^*(\alpha)=f^*([S^k])$ meaning that the map extends. What this means is that whatever (if any) obstructions are present in this case they are all torsion.

When $k=1$ then $S^1$ is already a $K(\mathbb Z,1)$ space and the above works on the nose without tensoring with $\mathbb Q$ as Ryan mentioned in a comment above.
show/hide this revision's text 1