Can two different types of manifolds be coupled to eachother? or stitched together? If possible, how would this be done?

For example, can I have a pseudo-riemannian manifold be coupled/attached to a Kahler manifold/symplectic form at their boundaries?

  • $\begingroup$ Welcome to Math.SE! <> This question is Very General and Open-Ended. Although it's not stated, presumably you have in mind that some structure on the "summands" extends to the "sum." The details are important because the answer depends on local symmetries of, and global obstructions to, the structure(s). Could you focus the question to ask about attaching specific types of manifold, including the geometric constraints you want on the "sum"? (I'm voting to close in the meantime since as written this is not really answerable.) $\endgroup$ Commented Sep 25, 2022 at 12:21
  • $\begingroup$ I guess another way to word it is, what objects am I looking to match at the boundaries to attach them smoothly? ie: the opens sets of each manifold, the charts, connections? An analog would be for functions we would match the derivatives to k order. The specific manifolds I have in mind are the R4 pseudo-riemannian manifold of General Relativity and the Kahler manifolds which cover the same riemannian manifold but also admit a complex/symplectic structure. $\endgroup$
    – B K
    Commented Sep 25, 2022 at 20:23
  • $\begingroup$ The thing is, if we have two different structures, say a Lorentzian metric and a Kähler metric, they are never equal at a point. In the function-patching analogy it's as if we're trying to glue a vector field on one set to a one-form on a set sharing a common boundary. There has to be some type of global object we intend to construct. <> The connected sum operation on (topological or smooth) manifolds is analogous to patching functions, but I still don't see how to make sense of the type of example in your comment. $\endgroup$ Commented Sep 26, 2022 at 11:42
  • $\begingroup$ I see, so instead of patching a Lorentzian and Kahler metric together, it would be more sensible to find the Lorentzian metric that is a subset of the Kahler metric such that the metrics are the same when only considering real valued terms? $\endgroup$
    – B K
    Commented Sep 26, 2022 at 23:41


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