I am confused about defining the connected sum of two $\color{red}{oriented}$ $n$-dimensional connected smooth manifolds without boundary, where $n\geq 2$. Could anyone help me to clear my confusion? I have written some guesses; maybe you can comment on these also. Thanks in advance.
So, let $M,M'$ be two connected $\color{blue}{orientable}$ $n$-dimensional connected smooth manifolds. Let $\varphi\colon \Bbb B^n\hookrightarrow M$ and $\varphi'\colon \Bbb B^n\hookrightarrow M'$ be two smooth embeddings of closed unit balls. Define $$\sharp(M,M'):=\frac{\big(M\backslash \varphi(\text{int }\Bbb B^n)\big)\sqcup \big(M'\backslash \varphi'(\text{int }\Bbb B^n)\big)}{\varphi(p)\sim \varphi'(p),\ p\in \Bbb S^{n-1}}.$$
Question $(1)$ Is $\sharp(M,M')$ always $\color{blue}{orientable}$?
Question $(2)$ Is it also true that if we $\color{red}{pick\
orientations}$ for $M,M'$, then $\color{red}{one\ can\ orient}$
$\sharp(M,M')$ so that orientation of $\sharp(M,M')$ coincides with
the orientation of $M\backslash \varphi(\text{int }\Bbb B^n)$?
I guess the answers to both questions are positive.
I know the following: If we pick orientations for $M, M'$ and choose $\varphi$ as orientation-preserving and $\varphi'$ as orientation-reversing then $\sharp(M, M')$ has a unique orientation that coincides with the orientations of $M\backslash \varphi(\text{int }\Bbb B^n)$ as well as $M'\backslash \varphi'(\text{int }\Bbb B^n)$. And the proof is based on the following theorem:
Richard Palais Theorem: Let $X$ be a $n$-dimensional smooth manifold without boundary. (If $X$ is orientable, pick an orientation for $X$) Let $\varphi_1,…,\varphi_m\colon \Bbb B^n\hookrightarrow X$ and $\psi_1,…,\psi_m\colon \Bbb B^n\hookrightarrow X$ be two sets of smooth embeddings with pairwise disjoint images (If $X$ is orientable, then we demand that all $\varphi_i,\psi_j$ are orientation-preserving or that all are orientation-reversing). Then there exists an isotopy $F\colon X\times [0,1]\to X$ with $F(-,0)=\text{id}_X$ such that the diffeomorphism $F(-,1)\colon X\to X$ sends $\varphi_i$ to $\psi_i$, i.e., $F\circ \varphi_i=\psi_i$ for all $i=1,…,m$.
For the following two questions, pick orientations for $M, M'$ and choose both $\varphi,\varphi'$ as orientation-preserving smooth embedding.
Question $(3)$ Is it true that if $M$ has an orientation-reversing
diffeomorphism, then $\sharp(M,M')$ is again
$\color{blue}{orientable}$, and $\color{red}{there\ exists\ unique\ orientation}$ for
$\sharp(M, M')$ that coincides with the orientations of $M\backslash
\varphi(\text{int }\Bbb B^n)$ as well as $M'\backslash
\varphi'(\text{int }\Bbb B^n)$?
My guess is that it is again true; probably one needs to pick an orientation for $\sharp(M, M')$ that coincides with the orientation of $M'\backslash \varphi'(\text{int }\Bbb B^n)$. Now, composing an orientation-reversing diffeomorphism $M\to M$ with $F(-,1)\colon M\to M$ (see Richard Palais Theorem), one can produce a diffeomorphic copy of $\sharp(M, M')$ having the same orientation of $\sharp(M, M')$.
Question $(4)$ Is it true that if neither of $M, M'$ has no
orientation-reversing diffeomorphism, then though $\sharp(M, M)$ is
$\color{blue}{orientable}$, it is impossible to $\color{blue}{pick\ an\
orientation}$ for $\sharp(M,M')$ that coincides with the orientations
of $M\backslash \varphi(\text{int }\Bbb B^n)$ as well as $M'\backslash
\varphi'(\text{int }\Bbb B^n)$?
Best Answer
Palais' disc theorem is not relevant to the question of orientability. It is relevant in arguing that the (oriented) connected sum is independent of the chosen discs up to (oriented) diffeomorphism.
Let the set-up be as in your question. You can find open collar neighborhoods of $\varphi(S^{n-1})$ in $M\setminus\varphi(\mathrm{int} B^n)$ and $\varphi^{\prime}(S^{n-1})$ in $M^{\prime}\setminus\varphi^{\prime}(\mathrm{int} B^n)$, which glue together to give an open bicollar $U$ around the embedded copy of $S^{n-1}$ in $M\# M^{\prime}$. Now $M\#M^{\prime}$ is the union of the three open subsets $M\setminus\varphi(\mathrm{int} B^n)$, $U$ and $M^{\prime}\setminus\varphi^{\prime}(\mathrm{int} B^n)$ , each intersecting only the ones written adjacent to it. Assume you're given an orientation of $M$, hence an orientation of $M\setminus\varphi(\mathrm{int} B^n)$. The intersection $M\setminus\varphi(\mathrm{int} B^n)\cap U$ is connected (here, $n\ge2$ is used) and $U$ is connected (again, $n\ge2$) and orientable, so this determines a unique orientation of $U$ compatible with the orientation on $M\setminus\varphi(\mathrm{int} B^n)$. In the same manner, but reverse, the intersection $U\cap M^{\prime}\setminus\varphi^{\prime}(\mathrm{int} B^n)$ is connected and $M^{\prime}\setminus\varphi^{\prime}(\mathrm{int} B^n)$ is connected (using $n\ge2$ again) and orientable, so this determines a unique orientation of $M^{\prime}\setminus\varphi^{\prime}(\mathrm{int} B^n)$ compatible with the orientation on $U$. These compatible orientations on $M\setminus\varphi(\mathrm{int} B^n)$, $U$ and $M^{\prime}\setminus\varphi^{\prime}(\mathrm{int} B^n)$ glue together to a unique orientation of $M\#M^{\prime}$, which is compatible with the orientation on $M\setminus\varphi(\mathrm{int} B^n)$ coming from the orientation on $M$. This answers questions (1) and (2) affirmatively.
To answer questions (3) and (4), we conduct a more explicit analysis of the above. Note that the uniqueness claim of the previous paragraph means we only have to understand what orientation the unique orientation of $M\#M^{\prime}$ induced by the orientation on $M\setminus\varphi(\mathrm{int} B^n)$ induces on $M^{\prime}\setminus\varphi^{\prime}(\mathrm{int} B^n)$. Choose a collar $c\colon[0,1)\times S^{n-1}\rightarrow M\setminus\varphi(\mathrm{int} B^n)$, which coincides with the given embedding $\varphi\vert_{S^{n-1}}$ at the boundary. The fact that $\varphi\colon B^n\rightarrow M$ is orientation-preserving means that $\varphi\vert_{S^{n-1}}\colon S^{n-1}\rightarrow\partial(M\setminus\varphi(\mathrm{int} B^n))$ is orientation-reversing if the codomain is equipped with the boundary orientation (since $\varphi$ pushes forward inward-pointing tangent vectors along $S^{n-1}$ to outward-pointing tangent vectors along $\partial(M\setminus\varphi(\mathrm{int} B^n))$). This in turn implies that the collar $c$ is orientation-preserving (since the positively oriented tangent vector to $[0,1)$ at $0$ pushes forward to an inward-pointing tangent vector along $\partial(M\setminus\varphi(\mathrm{int} B^n))$).
Next, choose a collar $c^{\prime}\colon(-1,0]\times S^{n-1}\rightarrow M^{\prime}\setminus\varphi^{\prime}(\mathrm{int} B^n)$. The two collars glue together to the bicollar $C\colon(-1,1)\times S^{n-1}\stackrel{\sim}{\rightarrow}U\subseteq M\#M^{\prime}$. If we now follow the construction from earlier explicitly, the orientation on $M$ determines the orientation of $U$ obtained by pushing forward the standard orientation of $(-1,1)\times S^{n-1}$ along $C$, by the previous paragraph. This in turm means the compatible orientation on $M^{\prime}\setminus\varphi^{\prime}(\mathrm{int} B^n)$ is uniquely determined by making $c^{\prime}$ orientation-preserving. Reversing the arguments from the previous paragraph, this is equivalent to $\varphi^{\prime}$ being orientation-reversing (note crucially the difference between one collar neighborhood defined on $[0,1)\times S^{n-1}$ and the other on $(-1,0]\times S^{n-1}$). From this, we obtain that two orientations on $M$ and $M^{\prime}$ induce an orientation on $M\#M^{\prime}$, agreeing with the given ones on $M\setminus\varphi(\mathrm{int} B^n)$ and $M^{\prime}\setminus\varphi^{\prime}(\mathrm{int} B^n)$ respectively, if and only if the embedded discs are oppositely oriented.
I want to stress that this is the expected behavior. To illustrate this, let's think of $\mathbb{R}^n$ as the union of upper half-space $\mathbb{H}_+^n$ and lower half-space $\mathbb{H}_-^n$, which is topologically the local model for gluing manifolds along their boundaries. The intersection $\mathbb{H}_+^n\cap\mathbb{H}_-^n=\partial\mathbb{H}_+^n=\partial\mathbb{H}_-^n$ is a hyperplane and reflection along this hyperplane is a homeomorphism between $\mathbb{H}_+^n$ and $\mathbb{H}_-^n$, which is the identity on the boundary, but it reverses orientations! So if you want to glue upper and lower half-space along their boundary to get $\mathbb{R}^n$ in the oriented category, they need to be oppositely oriented. Equivalently, you glue them along an orientation-reversing diffeomorphism. This manifests in the above proof, which is basically noting that if you identify the boundaries in the given manner, what looks outward-pointing in one gluing component has to look inward-pointing in the other and vice versa.
Edit: Ok, so if you have embeddings $\varphi\colon B^n\rightarrow M$ and $\varphi^{\prime}\colon B^n\rightarrow M^{\prime}$, let $M\#_{(\varphi,\varphi^{\prime})}M^{\prime}$ denote the specific connected sum constructed from this data. Palais' theorem implies that if $\psi\colon B^n\rightarrow M$ and $\psi^{\prime}\colon B^n\rightarrow M^{\prime}$ are two more embeddings, such that $\varphi,\psi$ and $\varphi^{\prime},\psi^{\prime}$ are respectively equioriented, then $M\#_{(\varphi,\varphi^{\prime})}M^{\prime}\cong M\#_{(\psi,\psi^{\prime})}M^{\prime}$. Thus, we can write $M\#_{(\pm,\pm)}M^{\prime}$, where the signs denote if the respective embeddings preserves or reverses orientations, and obtain 4 well-defined diffeomorphism types. Now if $(\varphi,\psi)$ is a pair of embeddings, then $(\varphi r,\psi r)$, where $r\colon B^n\rightarrow B^n$ is a reflection (so orientation-reversing), is a pair of embeddings whose signs are opposite and you can see immediately that $M\#_{(\varphi,\psi)}M^{\prime}$ and $M\#_{(\varphi r,\psi r)}M^{\prime}$ are described by the same equivalence relation on the same space, hence equal. Thus, there really are only two diffeomorphism types worth considering, $M\#_+M^{\prime}$ and $M\#_-M^{\prime}$, where the sign now denotes whether these are realized by a pair of embeddings that have the same or the opposite effect on the respective orientation. Conventionally, one writes $M\#M^{\prime}$ to mean the diffeomorphism type $M\#_-M^{\prime}$, because this is the one that comes with a natural orientation compatible with both $M$ and $M^{\prime}$, as discussed previously.
These two diffeomorphism types can be genuinely different. The standard example is that $\mathbb{CP}^2\#\mathbb{CP}^2$ and $\mathbb{CP}^2\#-\mathbb{CP}^2$ have different signatures, so they aren't even homeomorphic. However, if $M$ has an orientation-reversing diffeomorphism $\alpha\colon M\rightarrow M$ and $(\varphi,\psi)$ realizes $M\#_-M^{\prime}$, then $(\alpha\varphi,\psi)$ realizes $M\#_+M^{\prime}$ and you can check that the diffeomorphism $M\setminus\varphi(\mathrm{int} B^n)\rightarrow M\setminus\alpha\varphi(\mathrm{int} B^n)$ induced by $\alpha$ glues together with the identity on $M^{\prime}\setminus\varphi^{\prime}(\mathrm{int} B^n)$ to yield a diffeomorphism $M\#_-M^{\prime}\cong M\#_+M^{\prime}$. The same argument works with the factors interchanged, of course. In total, we obtain that if one of $M$ or $M^{\prime}$ possesses an orientation-reversing diffeomorphism, there is only one diffeomorphism type of manifolds that can be obtained from connected sum constructions, but if neither of them has an orientation-reversing diffeomorphism, you can get two distinct diffeomorphism types (I do not know if they always are).