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This is an extremely technical question about handle attachments.  I am asking why two definitions are equivalent.  My question appears in the second to last paragraph after I've described the two procedures whose equivalence I want to prove.

In the book "Differential Manifolds" by Antoni Kosinski, the following procedure is given for attaching two smooth n-manifolds $N_1, N_2$ along submanifolds of their boundaries:

Let $E := E' \oplus \mathbb{R}^{\geq 0}$ be the "upper half" of a rank n-m riemannian vector bundle over a smooth m- manifold $M$ and let $\phi_i : E \rightarrow N_i$   be imbeddings extending imbeddings $E' \rightarrow \partial N_i$. Now if $\alpha$ is any orientation reversing diffeomorphism $(0,\infty) \rightarrow (0, \infty )$, we make the following identifications:  $x \rightarrow \phi_2\{ \alpha (|\phi_1^{-1}(x)|) \frac{ \phi_1^{-1}(x)}{|\phi_1^{-1}(x)|}\}$ when $x \in \phi_1 (E- M)$ ( I.e. $x$ is an element of the first tubular neighborhood minus the zero section) The quotient space of $N_1 - \phi_1 (M) \sqcup N_2 -\phi_2 (M)$ under this identification is then the desired manifold.

Kosinski then says that "attaching a handle" to a manifold N is just a special case of the above when $N_1 = N$ and $N_2 = \mathbb{D}^n$ but his definition is a bit different:  Let $n= \lambda + \mu$ and write $x \in \mathbb{D}^n$ as $x = (x_\lambda, x_\mu)$. For $0\leq \epsilon < 1$,  Let $T(\epsilon)  := \{ x \in \mathbb{D}^n : |x_\lambda|^2 > \epsilon \}$ .  Now the analogue of the involution map $\alpha$ in the above paragraph is the map $\beta : T(\epsilon)-S^{\lambda-1} \rightarrow T(\epsilon)-S^{\lambda-1}$ given by

$(x_\lambda, x_\mu) \mapsto (\frac{x_\lambda}{|x_\lambda|} (1- |x_\lambda|^2 + \epsilon)^{1/2}, x_\mu \frac{(|x_\lambda|^2-\epsilon)^{1/2} }{(1-|x_\lambda|^2)^{1/2}})$.

Which is not as mysterious as it might appear because it is just the composition of the obvious diffeomorphism $\psi: \mathbb{D}^n - S^{\lambda -1} \rightarrow (\mathbb{D}^{\lambda}-S^{\lambda-1}) \times \mathbb{D}^{\mu}$ followed by an involution in the $\lambda$ coordinate followed by the inverse of $\psi$. 

Now to attach a $\lambda$ handle to a manifold $N$, we choose an imbedding $h: T(\epsilon), \partial T(\epsilon) \rightarrow N, \partial N$ and then identify $x \in T(\epsilon)-S^{\lambda-1}$ with $h\beta (x)$. And finally, the quotient of $N-h(S^{\lambda -1}) \sqcup \mathbb{D}^n - S^{\lambda -1}$ under this identification is the desired handle attachment. 


Ok here's my question: What is the vector bundle structure $\phi: E' \oplus \mathbb{R}^{\geq 0} \rightarrow T(\epsilon)$ which makes the second definition equivalent to the first?  More to the point, why should one exist at all? I have been thinking about this for a long time and I have no idea what it should be. I think I know what $\phi |_{E'}$ should be but I have no idea how to extend it in such a way that the involutions $\alpha$ and $\beta$ will "line up." Thank you for reading such an appallingly technical question.

More thoughts:  What really stumps me is the following.  If the two involutions are to "line up,"  each ray emanating from the origin in (a fiber of) $E$ should be mapped to a set of the form $\psi^{-1} \{(t x_{\lambda},x_\mu ) : \epsilon < t^2 < 1 \}$ for fixed $x_\lambda \in S^{\lambda -1}$ and $x_\mu \in \mathbb{D}^{\mu}$ and where $\psi$ is define as above. But these latter sets seem to be elliptical arcs whose closures will not be transversal to the boundary ($S^{n-1}$).  Whereas most rays emanating from the origin in (fibers of) $E$ $do$ meet the boundary transversely!... :(

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If you don't get an answer soon I should be able to write one up sometime on Monday or Thursday. – Ryan Budney Mar 24 '13 at 20:30
@RyanBudney : Wow that is very generous. Thanks, Ryan. – Pliny the ill Mar 24 '13 at 20:32
What kind of illness? In Pliny's day you could have a surfeit of eels. – Will Jagy Mar 25 '13 at 0:05
That and topology-induced anxiety attacks – Pliny the ill Mar 25 '13 at 1:19
up vote 2 down vote accepted

I haven't read the Kosinski text in a while so it's taken me some time to re-acquire my bearings with the notation.

Okay, I see what's going on. My impression is the author started off with a nice geometrically-pleasing trivialization of the tubular neighbourhoods involved, but then once he started writing things down and presenting things in front of a class, the "economics of a time constraint" seems to have caused him to cut out the geometrically-pleasing trivialization and write-out the identifications in a notationally-minimalist way.

Let $M$ be a Riemann manifold with boundary $\partial M$. If $p \in \partial M$ define the inward-pointing half-space $H_p M \subset T_p M$ to be

$$ H_p M = \{ v \in T_p M : \mu(v, \vec n(p)) \leq 0 \}$$

where $\mu$ is the Riemann metric and $\vec n : \partial M \to TM$ is the outward pointing unit normal vector.

So in $D^n$, if $p \in S^{n-1} = \partial D^n$, $H_p D^n = \{ v \in \mathbb R^n : v \cdot p \leq 0 \}$.

Given $p \in \partial D^n$ there is naturally-defined map $G : U \to D^n$ where $U \subset H_p D^n$ is some neighbourhood of the origin in $H_p D^n$. The map is defined by

$$G(q) = q + \left(\sqrt{ 1- |q-(q\cdot p)p|^2}\right) p$$

where $q \cdot p$ is the Euclidean dot product. This all motivates the construction below.

Consider $S^{\lambda-1} \times \{0\} \subset D^{\lambda+\mu}$. Our trivialization of the neighbourhood of $S^{\lambda-1} \times \{0\}$ in $D^{\lambda + \mu}$ is the map

$$H^{\mu+1} \times S^{\lambda-1} \to D^{\mu + \lambda}$$

given by $((q_0, \vec q), p) \longmapsto \left( 1- \sqrt{1-|\vec q|^2} - q_0 \right) p + (0, \vec q)$

where here $H^{\mu+1}$ is the half-space in $\mathbb R^{\mu+1}$, meaning the first coordinate $q_0 \geq 0$, so $\vec q \in \mathbb R^\mu$.

that is the vector space stucture on the tubular neighbourhood. You use the regular euclidean vector space structure on $H^{\mu+1} \subset \mathbb R^{\mu+1}$. Here $(0,\vec q)$ means the vector whose first $\mu$ coordinates are zero, and the remaining coordinates are those of $\vec q$.

Technically, this won't give you exactly the right formula but it will give you the same gluing up to diffeomorphism as your two distance functions (distance to $S^{\lambda -1}$ are isotopic to each other via a radial re-scaling that can be done with a straight-line isotopy.

I hope that makes sense.

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Thanks Ryan. I'm probably just being very dumb but do you maybe want something like $ \left( \sqrt{1-|\vec q|^2} - q_0 \right) p + (0, \vec q)$ instead of $ \left( 1- \sqrt{1-|\vec q|^2} - q_0 \right) p + (0, \vec q)$. Otherwise it seems like it doesn't send the zero section to the circle. – Pliny the ill Mar 27 '13 at 6:20
Also, I think I have a vague idea, but what exactly do you mean by "your two distance functions"? – Pliny the ill Mar 27 '13 at 6:32
Right, I think you're correct I appear to have made some kind of mistake with that formula. I'll think it over again to see what the mistake is. By the "distance function" I mean in all the formulas there's some kind of measure of distance from a point to either the sphere $S^{\lambda-1}$ or a set that corresponds to it in a different model. – Ryan Budney Mar 27 '13 at 17:59

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