11

I need to find a representation of $ [M \mathbin\sharp N] \in H_n(M \mathbin\sharp N) $ in terms of the fundamental classes $[M]$ and $[N]$. My idea is that $$ [M \mathbin\sharp N] = [M]+[N]$$

which is based on this MO question, whose answer is a little bit unclear to me because I never made use of triangulation for example.

To prevent some questions, The proof I gave for the fact that $M \mathbin\sharp N$ is orientable iff $M,N$ are, made use of the fact that removing an embedded euclidean nhbd doesn't change orientation behaviour. In fact I think that my question can be rephrased as a request for an homological proof of the orientation behaviour of the connected sum (even though there are lots of different ways to show this).

What I tried is to use the l.e.s of pair $(M \mathbin\sharp N , S^{n-1})$ together with M-V sequence of $M \vee N$ in order to "obtain" the plus, because by M-V we have an iso $$ H_n(M \vee N) \xleftarrow{incl_1 - incl_2} H_n(M) \oplus H_n(N) $$

and by l.e.s. pair we have a s.e.s. (some details can be found here)

$$\require{AMScd} \begin{CD} 0 @>>> H_n(M \mathbin\sharp N) @>>> H_n(M \mathbin\sharp N, S^{n-1}) @>>> H_{n-1}(S^{n-1}) @>>> 0 \\ @. @. @V{\cong}VV \\ @. @. \tilde{H}_n(M \mathbin\sharp N / S^{n-1}) \\ @. @. @V{\cong}VV \\ @. @. \tilde{H}_{n-1}(M \vee N) \end{CD}$$

But I'm unsure how to glue everything together, because the two inclusion with one minus sign would give to me exactly what i want, because I flip one orientation of one of the two manifolds. but these all are intuitive reasonings and I don't know how to formalise them properly.

$$ ++++ $$

following the comments below, it seems that a reasonable way to prove this is to show that $[M]+[N]$ has the characteristic property of the fundamental class. My problem now, (as I double-checked my computations) s that i dont0 have a valid proof for showing that $[M]$ and $[N]$ are elements of $H_n(M\sharp N)$, because trivially $M \not\subset M \sharp N$ obviously, so we need to find something that still represent $[M]$ i the homology group.

An idea was to use Poincaré Duality (see comments) to have a map $H_n(M) \to H_n(M \sharp N)$ but now the problem is that I don't know how to work with such image, because I don't have much control on how does the composition behave. At least, that's my problem right now

Community
  • 1
Riccardo
  • 7,401
  • 1
    What about checking that $[M]+[N]$ gives you the local orientation generator on each point and concluding $[M#N]=[M]+[N]$ from the uniqueness of the orientation class? See Hatcher Lemma 3.27. – archipelago May 26 '15 at 12:08
  • @archipelago I worked your idea a little bit, and it seems leading somewhere, but I cannot prove rigorously the characterising property in the case of points of $S^{n-1}$ (the one identified by the equivalence relation), could you elaborate a little bit this point? – Riccardo May 26 '15 at 15:39
  • In my humble opinion, the cleanest way is to construct the connected sum properly not to have issues on the gluing sphere, e.g. like in http://www.maths.ed.ac.uk/~aar/papers/kervmiln.pdf on the second half of page 505. – archipelago May 26 '15 at 16:14
  • @archipelago ok, I think the only missing proof in my reasoning is how to prove that $[M]+[N]$ is in fact an element of $H_n(M \sharp N)$. It is sufficient to check that $[M]$ and $[N]$ are in it, but I don't know how to show it, at the beginning I though about an "extending by zero" argument, but I feel that it's very sloppy and I cannot convince myself that's correct – Riccardo May 26 '15 at 20:27
  • 5
    You should probably define what $[M]$ and $[N]$ even mean in $H_*(M#N)$. There are no natural inclusions $M$ or $N$ to $M#N$. In fact there are not even say degree 1 maps from say a torus to the connect sum of two tori. There are in fact natural maps from $M#N$ to $M$ or $N$ and if you define $\cap[M]$ and $\cap[N]$ using these maps and Poincare duality you should get the desired result. – PVAL-inactive May 26 '15 at 22:02
  • EDIT : $\cap [M]$ should really be $PD([M])$ – PVAL-inactive May 26 '15 at 22:09
  • @PVAL Could you check my reasoning? because i'm not so confident with PD (seen 3 days ago in the lecture). SO you suggest to consider the "projection" (kind of quotient) from $M \sharp N$ to $M$, on level $0$ of homology, consider the dual of such map, hence a map in cohomology $\pi^* \colon H^0(M) \to H^0(M\sharp N)$ and the PD again, so to obtain a composition $H_n(M) \to H^0(M) \to H^0(M\sharp N) \to H_n(M\sharp N)$. and take the image of $[M]$ through such map. right? – Riccardo May 27 '15 at 09:31
  • @Riccardo An orientation on M, i.e. a choice of generator of its top Z-homology group, determines an orientation on M#N(and also on N). The projection map of M#N onto M has degree 1 so the map in dimension n is an isomorphism of groups. The orientation of N comes from the induced orientation on the connecting sphere which can be seen from the Mayer-Vietoris sequences $0→H_n(N,orM♯N)→H_{n−1}(S_{n−1})→0$. The orientations on M and N can not be arbitrary since an orientation on one determines the other. You can also orient M#N by pulling back the top cohomology of M or N via the projection map. – Joe S May 29 '15 at 11:18
  • @JoeS Intuitively I understand your comment, but I want to prove (or disprove) the equality $[M\sharp N]=[M]+[N]$, and I thought that a way to see it was to study the orientation of the connected sum. So what are you suggesting is to consider the pullback of [M] (homology) via the projection ? because pulling back in cohomology means that I go from the connected sum to M – Riccardo May 29 '15 at 12:57
  • @Riccardo I am a little confused by what you that [M#N] = [M] + [N]. First of all, it is false that the fundamental classes of M and N exist in M#N. So where is this sum taking place? You said that you want to see how orientations of M and N determine orientations of M#N. In homology use the isomorphism of $H_n$(M#N) with $H_n$(M) to find a generator of $H_n$(M#N) when given a generator of $H_n$(M). One this is done, the orientation of N is also determined. In cohomology, just pull back the generator in $H^n$(M) (or N) to $H^n$(M#N). Maybe I don't understand your question. can you explain? – Joe S May 29 '15 at 16:27
  • @JoeS well, my question is enclosed in the first paragraph of the question. And yes, I need to find a way to see [M] and [N] inside $H_n(M\sharp N)$ (even of the form $f_*[M]$ for some $f \colon M \to M\ sharp N$), which is what in other comments people are trying to do but then got no more answers :). MAYBE what I'm trying to do is plain wrong, but what is true (according to my exercise) is that if $x,x'$ homology classes, are representable by manifolds, then their sum is represented by a manifold. so the base case was to prove what I'm asking in the question.I'll add this to the question now – Riccardo May 29 '15 at 17:06

0 Answers0