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!

  • 2
    $\begingroup$ 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
    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. $\endgroup$ 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. $\endgroup$ Jul 26, 2018 at 19:29

1 Answer 1


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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