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If $V$ is a real vector space, then the complexification of $V$ is formally defined as $V^{\mathbb{C}}=V\otimes_{\mathbb{R}}\mathbb{C}$. Is there an analogous complexification operation for a real $n$-dimensional Riemannian manifold $(M,g)$?

Idea: The notion of complexification exists for Lie groups, so perhaps one can "complexify" a real Riemannian manifold by realizing it as a Lie group (or the quotient of one). It seems that under complexification of a real manifold some additional information must be added to determine a complex structure.

The reason I ask this is because I am looking through the Riemannian holonomy section of this article and it states that "the complexified holonomies $SO(n,\mathbb{C})$, $G_2(\mathbb{C})$, and $Spin(7,\mathbb{C})$ may be realized from complexifying real analytic Riemannian manifolds." What precisely does complexifying a real analytic Riemannian manifold mean in this context?

Any help would be much appreciated!

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    $\begingroup$ For holonomy groups, keep in mind that these are reduced holonomy groups, i.e. dependent only on holonomy around contractible loops, and that by the Ambrose-Singer theorem, the reduced holonomy is determined by the germ of the connection at a point. So you can complexify in local coordinates near a point, in any coordinate system in which the connection is real analytic. You don't need to complexify the entire manifold, just a neighborhood of a point. Note that $G_2$ and $Spin(7)$ holonomy metrics are Ricci flat, so real analytic in any harmonic local coordinates. $\endgroup$
    – Ben McKay
    Commented Jul 25, 2018 at 10:46
  • $\begingroup$ This has duplicates at mse and also here. I think it ought to be moved to mse so that the copy can be closed there. $\endgroup$
    – Francois Ziegler
    Commented Jul 26, 2018 at 19:18
  • $\begingroup$ @Francois Ziegler The former is is a cross-post of mine on MSE. I was not aware of the latter. I can always remove the MSE one if you like. $\endgroup$
    – Sergio Charles
    Commented Jul 26, 2018 at 19:29

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I believe the following is meant:

Every smooth (real) manifold $M$ has a (unique) real-analytic structure compatible with the smooth structure. So, cover $M$ with real-analytic charts, i.e. whose transition functions are real-analytic diffeomorphisms $$ \phi_{ij}:=\phi_j^{-1}\circ\phi_i: U_{ij}:=\phi_i^{-1}(\phi_i(U_i)\cap\phi_j(U_j))\to U_{ji} $$ One can find open subsets $U_i^{\mathbb{C}}\subseteq\mathbb{C}^n$ with $U_i^{\mathbb{C}}\cap\mathbb{R}^n=U_i$ and $U_{ij}^{\mathbb{C}}\cap\mathbb{R}^n=U_{ij}$ such that the (real-analytic) $\phi_{ij}$ extend to biholomorphisms $\phi_{ij}^{\mathbb{C}}:U_{ij}^{\mathbb{C}}\to U_{ji}^{\mathbb{C}}$ satisfying the usual cocycle conditions. Then the complexification $M^{\mathbb{C}}$ is defined as a quotient space of the disjoint union, $\left(\coprod_i U_i^{\mathbb{C}}\right)/\sim$, where $z_i\sim z_j$ iff $z_i\in U_{ij}^{\mathbb{C}}$ and $z_j = \phi_{ij}^{\mathbb{C}}(z_i)$ (this works because of the cocycle conditions). The maps $U_i^{\mathbb{C}}\hookrightarrow\coprod U_i^{\mathbb{C}}$ induce coordinate charts $U_i^{\mathbb{C}}\to M^{\mathbb{C}}$ with biholomorphic transition functions.

This and the details around it are part (of the proof of) Bruhat-Whitney's theorem* on the existence of $M^{\mathbb{C}}$. Moreover, complexification is functorial in the obvious way. By Grauert, $M^{\mathbb{C}}$ is in fact a Stein manifold.

*F. Bruhat and H. Whitney, Quelques propriétés fondamentales des ensembles analytiques-réels, Comment. Math. Helv. 33, 132-160 (1959).

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  • $\begingroup$ Very interesting that someone knows formal and exact answers with reference to such question. I must ask, if you do not mind if the complex manifold $M^{\mathbb{C}}$ is unique in terms of the topology, differential structure etc? $\endgroup$
    – Bastam Tajik
    Commented Jul 5 at 9:26
  • $\begingroup$ Specially that I am curious about the case where a Lorentzian metirc assigned to the real manifold $M$. Can one analytically uniquely continue the metric from $M$ all over the complexified manifold $M^{\mathbb{C}}$ (for a specific choice of $M^{\mathbb{C}}$ in case it is not unique anyhow) $\endgroup$
    – Bastam Tajik
    Commented Jul 5 at 11:20
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    $\begingroup$ @BastamTajik: Apologies for the delayed reply! My inbox was overflowing with notifications. Since these are separate questions, I believe the best MO practice would be for you to ask them as two separate posts and refer to the original question above. That way they will be seen by many more MO users, which raises the chances for more answers and also for more users to benefit from that information. I'm pretty busy at the moment, but I will try and think about your questions in the mean time. $\endgroup$
    – M.G.
    Commented Jul 12 at 10:17
  • $\begingroup$ Right, but the question is not answered completely yet, since the original manifold $M$ is assigned a metric $g$ and it's not outlined how this metric is continued consistently all over $M^{\mathbb{C}}$. In other words the question is rather geometrical and not only topological. Thanks @M.G. $\endgroup$
    – Bastam Tajik
    Commented Jul 13 at 14:18

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