# Apparent contradiction with gluing 2-handle to a 4-manifold all being isotopic.

In Hirsch's Differential Topology Chapter 8 Theorem 2.3, it says :

Let $$f, g:\partial Q\approx \partial P$$ be isotopic diffeomorphisms. Then $$P\cup_f Q\approx P\cup_g Q$$. Here $$\approx$$ means diffeomorphic.

Now the question I have is the following. Let's say I have a closed 4-ball and its boundary $$S^3$$. Consider a knot $$k\subseteq S^3$$, and a tubular neighborhood of it, $$N(k)\subseteq S^3$$. Then I can glue a 2-handle $$D^2\times D^2$$ along $$S^1\times D^2$$, if I am given a diffeomorphism $$h:S^1\times D^2\to N(k)$$.

By Hirsch's, isotopic $$h$$'s give diffeomorphic manifolds.

Now Hirsch also says that any two tubular neighborhoods of $$k$$ in $$S^3$$ are isotopic (Chapter 4, Theorem 5.3).

The problem then is if we have two tubular neighborhoods $$f_0:S^1\times D^2\to N(k)$$ and $$f_1:S^1\times D^2\to N(k)$$, this seems to imply that $$f_0$$ and $$f_1$$ are isotopic. This would effectively mean that the choice of framing for the knot doesn't change the diffeomorphism type of the surgered manifold.

How is this argument wrong? I understand this is about discerning between different notions of the statement "tubular neighborhoods are unique up to isotopy", which is what I haven't been able to do.

• An isotopy between two tubular neighborhoods $N(k)$ will indeed give you an isotopy from your $f_0$ to some $f_1$, but not to the $f_1$ that you probably intended --- in particular not to an $f_1$ arising from an inequivalent framing of $k$. Note that your $f$'s implicitly refer to framings, since $S^1\times D^2$ has a canonical framing from its product structure. (You might find it useful to first visualize the case where $k$ is the unknot.) Nov 16, 2021 at 0:41

The definition of an "isotopy of tubular neighborhoods" (bottom of page 111) is not what you might be expecting: yes it's an isotopy, but it only needs to carry one tubular neighborhood to another by a vector bundle isomorphism. This means that if you have a knot $$K\subset S^3$$ with two framings $$h_1,h_2:S^1\times D^2\to N(K)$$, then while there is an isotopy of tubular neighborhoods between both framings' tubular neighborhoods, there's certainly no reason to expect $$h_2^{-1}\circ h_1:S^1\times D^2\to S^1\times D^2$$ to be isotopic to the identity. (This composition is describing the difference between gluing along $$h_1$$ and gluing along $$h_2$$.)
Another way to say it is that the maps $$h_1$$ and $$h_2$$ are not necessarily isotopic, even if the tubular neighborhoods they describe are isotopic as tubular neighborhoods.