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The title of this post is both misleading and not.

Let say we have smooth complex vector bundles $E\to X$ and $F\to X$ over a smooth $n$-manifold $X$ of the same rank, say $k$. Assume $X$ is compact and connected. Then both $E$ and $F$ are connected as smooth $(n+k)$-manifolds. Thus, as smooth manifolds, we can form the connected sum $E\# F$.

My questions are

  1. Does $E\# F$ form a complex vector bundle over $X$?
  2. If $E\#F\to X$ is a complex vector bundle, does $E\to X$ embed into $E\# F\to X$?

(If the answer to 1 is yes, I imagine we would have to choose the $n$-discs in $E$ and $F$ to be deleted and the corresponding diffeomorphism very carefully)

Or, better yet, does there exist a "vector bundle version of connected sum of manifolds" (in the above sense) such that $E\to X$ embeds into the resulting vector bundle?

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  • $\begingroup$ Regarding your second question, I'm not sure why you would expect this. In the manifold case, $M$ does not embed into $M\# N$. Maybe you have some other analogy in mind. $\endgroup$ Commented Jan 2 at 9:50
  • $\begingroup$ You are right that I indeed mixed it up with something else. $\endgroup$ Commented Jan 2 at 16:41

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Consider $E = F = \mathbb{C}^k$ as vector bundles over $X = \{\text{pt}\}$. The connected sum $E\# F$ is diffeomorphic to $S^{2k-1}\times\mathbb{R}$. The total space of any vector bundle over $X$ is contractible, so $E\# F$ is not a vector bundle over $X$.

As I mentioned in the comments below, given two complex vector bundles $E \to X$ and $F \to Y$ of the same rank, there is a vector bundle over $X\# Y$ which you may want to denote by $E\# F$ (although, it could be confusing to do so as the total space is not the connected sum of the total spaces of $E$ and $F$).

Choose basepoints for $X$ and $Y$ and choose basepoint-preserving classifying maps $f : X \to BU(k)$ and $g : Y \to BU(k)$ for $E$ and $F$ respectively. Then $f\vee g : X\vee Y \to BU(k)$ classifies a vector bundle. Now note there is a map $c : X\# Y \to X\vee Y$ which crushes the neck created during the connected sum procedure. Pulling back the bundle on $X\vee Y$ to $X\# Y$ by the map $c$ yields the bundle I was alluding to. When $X = Y = S^n$, this construction is the binary operation on $\pi_n(BU(k))$.

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  • $\begingroup$ thanks for the answer. I still hope there exists a vector bundle version of connected sum of manifolds (probably through a different construction) that kind of preserves some properties of connected sum of manifolds. $\endgroup$ Commented Jan 2 at 8:37
  • $\begingroup$ Given complex vector bundles $E \to X$ and $F \to Y$ of the same rank, there is a bundle which one might denote $E\# F \to X\# Y$. This also works for orientable real bundles too. $\endgroup$ Commented Jan 2 at 9:58
  • $\begingroup$ May I know what is the bundle $E\# F\to X\# Y$? For me, $Y=X$ is good enough. $\endgroup$ Commented Jan 2 at 16:38

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