# Why are embedded spheres removed in the connected sum but not in the handle attachment of (smooth) manifolds?

So i am currently studying differential manifolds and morse-theory. When i came across the connected sum, i learned that we glue two manifolds $$M_1$$ and $$M_2$$ along the boundaries of removed disks such that we obtain the quotient manifold $$M_1 \# M_2 = (M_1 \setminus h_1(S_1)) \cup (M_2 \setminus h_2(S_2))$$ where $$h_1,h_2$$ are embeddings $$h_i: \mathbb{R}^n \to M_i, i = 1,2$$.

Now when i learned about handle attachments in the theory of smooth manifolds, there didn't seem to be any consideration about removing anything from the handlebody $$M$$ prior to attaching a $$\lambda$$-handle $$H_\lambda = D^\lambda \times D^{n-\lambda}$$ via the attaching-map $$\varphi_\lambda : \partial D^\lambda \times D^{n-\lambda} \to \partial M.$$

Even though, as it seems, handle attachment is just a special case of the connected sum of two manifolds along submanifolds.

Furthermore, when i read about surgery, we removed (embedded) spheres $$S^{\lambda-1}$$ from two Manifolds $$M_1, M_2$$ and pasted them along their tubular neighborhoods if i'm not mistaken (Kosinski's Differential Manifolds p. 112).

However, according to Kosisnki, attaching a $$\lambda-$$handle along the embedded sphere $$S^{\lambda-1}$$ apparently becomes precisely surgery on $$S^{\lambda-1}$$, when restricted to the boundaries.

My first question(s): Am i correct in the way i phrased the above situations? What am i seem to be missing? Where am i mistaken? Why is there no removal of embedded submanifolds in the context of handle attachments?

My second question: Am i correct, that the difference between attaching a $$\lambda$$-handle and surgery (on $$S^{\lambda-1}$$) is that they are equivalent once we restrict the tubular neighborhoods of $$S^{\lambda-1}$$ and the embedded submanifold $$h(S^{\lambda-1})$$ to the boundaries of both of them respectively?

Highly appreciating any help! Thank you very much.

Starting with $$M_1$$ and $$M_2$$, and forming their disjoint union $$M_1 \coprod M_2$$, the connected sum can be thought of as obtained in two steps from $$M_1 \coprod M_2$$:

• Step 1: Remove an $$n$$-handle $$S^0 \times D^n$$ (whose boundary is homeomorphic to $$S^0 \times \partial D^n = S^0 \times S^{n-1}$$); more specifically, remove one $$D^n$$ from $$M_1$$ and remove another $$D_n$$ from $$M_2$$.
• Step 2: Attach a 0-handle $$D^1 \times S^{n-1}$$ (whose boundary is homeomorphic to $$\partial D^1 \times S^{n-1} = S^0 \times S^{n-1})$$.

This is indeed a special case of surgery. The general case of surgery on an $$n$$-manifold goes like this:

• Step 1: Remove a $$k$$-handle $$S^{n-k} \times D^k$$ (whose boundary is homeomorphic to $$S^{n-k} \times S^{k-1}$$)
• Step 2: Attach an $$n-k+1$$ handle $$D^{n-k+1} \times S^{k-1}$$ (whose boundary is homeomorphic to $$S^{n-k} \times S^{k-1}$$)

So it's not true that handle attachment is a special case of connected sum. Instead, handle attachment is one of two steps of a generalization of connected sum, i.e. it is one of two steps of a surgery.

Your second question is not so coherent. As I've said, attaching a handle is not the same as surgery, it is instead only one half of a full surgery operation.

What you might have read regarding tubular neighborhoods is that identifying the boundary of one manifold to the boundary of another one can also be described topologically as identifying the tubular neighborhood of one boundary with the tubular neighborhood of the other, in a kind of "inside-out" manner. The reason that one uses the method of gluing tubular neighborhoods of boundaries is that it is easier to describe the smooth atlas.

• Hello @Lee, thanks for your consideration. Is there any particular reason you've chosen the $3$-disk or rather a $3$-handle in your given example? The same would apply to lower dimensions, wouldn't it? Edit: ah, you might have chosen the $3$-handle since attaching a $0$-handle results in the disjoint union. that might be the answer, i hope. – Zest Jan 9 at 4:54
• Oops, I betrayed my low-dimensional background by default. I rewrote the answer in any dimension. – Lee Mosher Jan 9 at 13:47