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I am trying to find out how a $(p,n-p)$ surgery affects the Euler Characteristic of an orientable, $n-$ dimensional, compact manifold. Call the initial manifold $M$ and the post-op manifold $M'$. This is defined here Surgery theory (wikipedia)

It seems like Mayer-Vietoris is the way to go here (please let me know if this isn't or there are other ways, though). My question is about this map in the M-V sequence for $M'$, for $0<i<n$:

$H_i(S^p\times S^{n-p-1})\rightarrow H_i(M-\{S^p \times D^{n-p})) \oplus H_i(D^{p+1} \times S^{n-p-1} )$

I'd like to show that it is injective ($H_i(S^p\times S^{n-p-1})=\mathbb{Z}$ for $i=p$ and $n -p-1$, and $0$ for all other $0<i<n$). It looks like a cycle in $S^p\times S^{n-p-1}$ that is a boundary in $D^{p+1} \times S^{n-p-1}$, will not be a boundary in $M-\{S^p \times D^{n-p})$, using a few low-dimensional examples but I'm not sure how to show this more convincingly.

Ashley
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    Using MV here is a bit hard. Assume you have triangulated manifolds (this is not true in general but will give you the right answer). Now, just count the (signed) numbers of various simplices: the ones you remove and the ones you add. – Moishe Kohan Jul 20 '14 at 02:24
  • Oh that's very nice, thanks! You don;t have any ideas about more general case do you? – Ashley Jul 20 '14 at 18:33
  • I guess this way may be fairly general..."In dimension greater than 4, the question of whether all topological manifolds have triangulations is an open problem, though it is known that some do not have piecewise-linear triangulations"-wiki – Ashley Jul 20 '14 at 18:52
  • It is now known that in all dimensions $\ge 5$ there are non-triangulable manifolds. However, CW-complex structure exists on manifolds of all dimensions $\ne 4$ (for 4-manifolds this is an open problem). Without CW structure I guess you have to use MV-sequence but I did not think about it. – Moishe Kohan Jul 20 '14 at 23:59

1 Answers1

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Let $U = M\setminus(S^p\times D^{n-p})$ and $V = S^p\times D^{n-p}$. Then $M = U\cup V$ and $U\cap V = S^p\times S^{n-p-1}$, so

\begin{align*} \chi(M) &= \chi(U) + \chi(V) - \chi(U\cap V)\\ &= \chi(U) + \chi(S^p\times D^{n-p}) - \chi(S^p\times S^{n-p-1})\\ &= \chi(U) + \chi(S^p) - \chi(S^p)\chi(S^{n-p-1}). \end{align*}

Now let $W = D^{p+1}\times S^{n-p-1}$ and note that $M' = U\cup W$ and $U\cap W = S^p\times S^{n-p-1}$, so

\begin{align*} \chi(M') &= \chi(U) + \chi(W) - \chi(U\cap W)\\ &= \chi(U) + \chi(D^{p+1}\times S^{n-p-1}) - \chi(S^p\times S^{n-p-1})\\ &= \chi(U) + \chi(S^{n-p-1}) - \chi(S^p)\chi(S^{n-p-1}). \end{align*}

Using the fact that $\chi(U) = \chi(M) - \chi(S^p) + \chi(S^p)\chi(S^{n-p-1})$ we see that

\begin{align*} \chi(M') &= \chi(M) - \chi(S^p) + \chi(S^p)\chi(S^{n-p-1}) + \chi(S^{n-p-1}) - \chi(S^p)\chi(S^{n-p-1})\\ &= \chi(M) - \chi(S^p) + \chi(S^{n-p-1})\\ &= \chi(M) - [1 + (-1)^p] + [1 + (-1)^{n-p-1}]\\ &= \chi(M) - (-1)^p + (-1)^{n-p-1}\\ &= \chi(M) + (-1)^{p+1}[1 + (-1)^n]. \end{align*}

That is,

$$\chi(M') = \begin{cases} \chi(M) & n\ \text{odd}\\ \chi(M) - 2 & n\ \text{even}, p\ \text{even}\\ \chi(M) + 2 & n\ \text{even}, p\ \text{odd}. \end{cases}$$


In particular, surgeries do not change the parity of the Euler characteristic. Another way to see this is to use the fact that if $M$ and $N$ are related by a sequence of surgeries, then they are cobordant, so they have the same Stiefel-Whitney numbers. In particular, $\underline{w_n}(M) = \underline{w_n}(N)$, but this is nothing but the Euler characteristic mod two.

Moreover, any two cobordant manifolds are related by a sequence of surgeries. The above computation shows that if $M$ and $N$ are cobordant, then $\frac{1}{2}|\chi(M) - \chi(N)|$ gives a lower bound on the number of surgeries to obtain one from the other.

Michael Albanese
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