Let $M$ be a non-compact manifold, equipped with a (closed?) submanifold $N\subset M$. The action of $Diff(M)$ on the set of embeddings $N\hookrightarrow M$ induces a map
$$
Diff(M) \rightarrow Emb(N,M).
$$
Is this map a fibration in the sense of Hurewicz?
I am aware of the results of Palais and lately Goodwillie in the case of compact manifolds, but I have no idea about the noncompact case.
$\begingroup$
$\endgroup$
3
-
1$\begingroup$ I don't understand what it means to ask whether a fibration exists over a map. Are you asking whether the map is a fibration? $\endgroup$– Qiaochu YuanCommented Feb 7, 2016 at 17:15
-
$\begingroup$ @Qiaochu Yuan; Yes, that's what I meant thanks, I re-edited the question, sorry I am not a native English speaker. $\endgroup$– s kCommented Feb 7, 2016 at 18:41
-
$\begingroup$ I don't think you're asking whether a particular map is a fibration $\endgroup$– Fernando MuroCommented Feb 7, 2016 at 23:09
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
5
Consider an embedding of $\mathbb R$ into $\mathbb R^3$ which consits of an infinite connected sum of knots, arranged roughly along the $x$-axis. There is a path in the space of proper embeddings $\mathbb R\hookrightarrow\mathbb R^3$ which goes from that knotted line to the unknotted embedding of $\mathbb R$ into $\mathbb R^3$ (where $\mathbb R$ just maps to the $x$-axis). That's the path that "sends all the knotted junk to infinity".
Now, there is no path in $Diff(\mathbb R^3)$ that lifts that path in $Emb(\mathbb R,\mathbb R^3)$. That's because the isn't any diffeomorphism of $\mathbb R^3$ that maps the the $x$-axis to that knotted curve (the proof uses the fundamental group at infinity of $\mathbb R^3$ minus the image of $\mathbb R$).
This shows that your map is not even a Serre fibration.
answered Feb 8, 2016 at 0:56
-
$\begingroup$ Thanks André, but I have sketched a proof that the induced map is a Kan fibration using Siebenmann's isotopy extension theorem. How far can that be right? $\endgroup$– s kCommented Feb 8, 2016 at 6:46
-
$\begingroup$ I assumed the embeddings are proper and the spaces are smooth. $\endgroup$– s kCommented Feb 8, 2016 at 6:58
-
$\begingroup$ I don't understand this example. What is the path of embeddings (and not just the path of images of embeddings)? $\endgroup$ Commented Feb 8, 2016 at 7:59
-
$\begingroup$ In this case, there's not much difference between an image of embedding and an embedding. After fixing base points, you can always take the constant-speed parametrization of an embedded curve. Does what I say make sense, or did I overlook something basic? $\endgroup$ Commented Feb 8, 2016 at 12:49
-
$\begingroup$ @Oscar: he's showing that the isotopy extension theorem fails in this case. $\endgroup$ Commented Feb 11, 2016 at 11:47