It is well known that one can specify a complex structure on a real $C^\infty$ manifold in two equivalent ways: an atlas with holomorphic transition functions between charts and an integrable almost complex structure. One of the directions is straightforward. In the other direction, the celebrated Newlander–Nirenberg theorem states that an integrable almost complex structure induces a holomorphic atlas.

For real analytic manifolds, I know only the atlas method (transition functions between charts are real analytic). My question: does there exist for real analytic manifolds an analog of an almost complex structure, an integrability condition and an analog of the Newlander–Nirenberg theorem?

At the moment, my suspicion is that the right analog should be an almost CR structure and a corresponding integrability condition (whatever those are). But unfortunately I've not really seen this written anywhere.

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    $\begingroup$ This seems unlikely to me. The integrability condition for compkex analytic structures arises from the fact that a holomorphic function must satisfy the Cauchy-Riemann equations. I don't see any analogue for real analytic functions, where the condition appears to me to be just regularity condition. $\endgroup$ – Deane Yang Jun 27 '14 at 13:33
  • $\begingroup$ @DeaneYang, I wouldn't be quite so pessimistic. There is definitely some geometry involved. For instance, having a fixed analytic atlas fixes also a class of compatible embeddings in complex manifolds (and apparently, by results of Grauert, such embeddings always exist). On the other hand, instead of local charts and transition functions, there might be some global geometric structure that picks out the same class of embeddings. $\endgroup$ – Igor Khavkine Jun 27 '14 at 14:56
  • $\begingroup$ The comments below help me understand what you want. If I'm not mistaken, you want some kind of geometric structure that exists on, say, any closed smooth manifold and induces naturally a real analytic structure on the manifold. If so, I can say only that I have no clue how to do this. $\endgroup$ – Deane Yang Jun 28 '14 at 4:23
  • $\begingroup$ After some thought, it seems to me that CR structures do not really address your point. Any smooth real hypersurface of $\mathbb{C}^n$ is a CR manifold, be it real analytic or not. This means that any intrinsic real analytic structure you put on it will fail to match the would-be one induced from the embedding if the embedding is not real analytic. There is an extensive discussion of CR manifolds, both intrinsic and embedded, I've found in the book of S. Dragomir and G. Tomassini, "Differential Geometry and Analysis on CR Manifolds" (Birkhäuser, 2006). $\endgroup$ – Pedro Lauridsen Ribeiro Apr 8 '15 at 1:49

I think this question can be addressed in a few ways. Two great answers have already been given:

  1. Real analytic is a regularity condition, while holomorphic is more algebraic, so you'd need a somewhat different condition than NN.
  2. Every $C^1$ manifold can be given a real analytic structure, so you don't even need a condition.

Let me add a third, more complicated, answer where we ask the real analytic structure to be adapted to the given data. To do this we have to go back and look at what the Newlander-Nirenberg theorem gives us, and then we will be able to adapt it to give a real analytic type result. So first

The Newlander-Nirenberg Theorem: Let $L_1,\ldots, L_m$ be smooth complex vector fields on a smooth manifold $M$. Suppose for each $\zeta\in M$ we have the following:

  • $[L_j,L_k](\zeta)\in \mathrm{span}_{\mathbb{C}}\{L_1(\zeta),\ldots, L_m(\zeta)\}$.
  • $\mathrm{span}_{\mathbb{C}} \{ L_1(\zeta),\ldots, L_m(\zeta)\} \cap \mathrm{span}_{\mathbb{C}} \{ \overline{L_1}(\zeta),\ldots, \overline{L_m}(\zeta)\}=\{0\}$.
  • $\mathrm{span}_{\mathbb{C}}\{L_1(\zeta),\ldots, L_m(\zeta), \overline{L_1}(\zeta),\ldots, \overline{L_m}(\zeta)\} = \mathbb{C} T_{\zeta} M$.

Then, there exists a complex manifold structure on $M$ (compatible with its smooth manifold structure) such that $\forall \zeta\in M$, $\mathrm{span}_{\mathbb{C}} \{L_1(\zeta),\ldots, L_m(\zeta)\} = T^{0,1}M$. Moreover, this complex structure is unique in the sense that if $M$ is given another complex manifold structure (compatible with its smooth manifold structure) such that $\mathrm{span}_{\mathbb{C}} \{L_1(\zeta),\ldots, L_m(\zeta)\} = T^{0,1}M$, then the identity map $M\rightarrow M$ is a biholomorphism between these two complex manifold structures on $M$.

Now let's turn to the real setting, and address the original question. Let $X_1,\ldots, X_q$ be $C^1$ vector fields on a $C^2$ manifold $M$ such that $\forall x\in M$, $\mathrm{span}_{\mathbb{R}} \{X_1(x),\ldots, X_q(x)\} = T_x M$.

Question: When is there a real analytic structure on $M$, compatible with its $C^2$ structure, such that $X_1,\ldots, X_q$ are real analytic with respect to this structure? When such a structure exists, we will see it is unique (in the strongest possible sense).

We will answer this question by giving a condition which looks very similar to the Newlander-Nirenberg theorem (but is actually orthogonal to that condition, as we will see).

Even though $M$ is only a $C^2$ manifold, we can define what it means for a function to be real analytic with respect to the vector fields $X_1,\ldots, X_q$. Let $V\subseteq M$ be open. For $r>0$ we define $C_{X}^{\omega,r}(V)$ to be the space of those $f:V\rightarrow \mathbb{R}$ such that the following norm is finite: $$ \| f\|_{C_X^{\omega,r}(V)} = \sum_{m=0}^{\infty} \frac{r^m}{m!} \sum_{|\alpha|=m} \| X^{\alpha} f\|_{C(V)}.$$ In defining $X^{\alpha}$ we have used ordered multi-index notation; i.e., $\alpha$ is a list of elements of $\{1,\ldots, q\}$ and $|\alpha|$ is the length of the list. For example $X^{(1,2,1,2)}= X_1X_2X_1X_2$ and $|(1,2,1,2)|=4$. We set $C_X^{\omega}(V):= \bigcup_{r>0} C_X^{\omega,r}(V)$. The space $C_X^{\omega}(V)$ is "coordinate-free": it does not depend on any choice of coordinate system or atlas. This space was originally defined in greater generality by Nelson (Ann. of Math. (2) 70 (1959), 572-615).

Theorem: There is a real analytic structure on $M$, compatible with its $C^2$ structure, such that $X_1,\ldots, X_q$ are real analytic with respect to this structure if and only if:

  • For every $x\in M$, there is a neighborhood $V_x$ of $x$, and functions $c_{j,k}^{l,x}\in C_{X}^{\omega}(V_x)$ such that $[X_j,X_k]=\sum_{l=1}^q c_{j,k}^{l,x} X_l$ on $V_x$.

When this real analytic structure exists, it is unique in the sense that if $M$ is given another real analytic structure, compatible with its $C^2$ structure, such that $X_1,\ldots, X_q$ are real analytic with respect to this second structure, then the identity map $M\rightarrow M$ is a real analytic diffeomorphism between these two real analytic structures.

So there's the answer to our question--and it has the same "feel" as the NN theorem: it uses an understanding of commutators of given vector fields to give a unique structure on the manifold.

There's nothing special about real analytic in the above. You can replace real analytic with an appropriate space of functions which are $C^\infty$ with respect to $X_1,\ldots, X_q$ and get a corresponding result about $C^\infty$ structures. You can do it for a finite level of smoothness, too, though to obtain a sharp if and only if Zygmund spaces are used (instead of the more familiar $C^m$ spaces).

This is all done in the series of papers (joint with Stovall): 1, 2, 3

No back to my comment that this was actually orthogonal to the NN theroem. One could ask the following:

Question: Suppose $L_1,\ldots, L_m$ satisfy the conditions of the NN theorem (above). When are $L_1,\ldots, L_m$ real analytic with respect to the complex structure gauranteed by the NN theorem?

One can answer this by giving a condition very similar to the real theory given above. You can even get a theory which unifies both the real and complex into one theorem. This is all contained in this paper.

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    $\begingroup$ Wow, this is essentially the kind of answer that I was hoping for. Thank you! And these results seem to be coming from your own papers, so I congratulate you on doing very interesting work! A small question. The number of vector fields, $q$, does it have to equal the dimension of the manifold $M$, or could it be larger? For topological reasons, not all manifolds are framed. In these cases, with $q=\dim M$, one could not keep the vector fields globally defined and spanning $TM$. But with a larger number of vector fields, this is of course possible. $\endgroup$ – Igor Khavkine Dec 22 '18 at 21:38
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    $\begingroup$ $q$ can be greater than the dimension. It only has to be a spanning set, not a basis. That actually takes a little bit of extra work. Also, this is mostly a local thing, so global topological issues don't come up too much. Because of this, I think it should be possible to turn this into a theorem about some kind of compatible families of vector fields defined locally. But my interests were towards a different sort of question, so I haven't pursued that yet. $\endgroup$ – Brian Street Dec 23 '18 at 23:49

It is a theorem of Whitney that every closed $C^{\infty}$-manifold admits a real analytic structure. Furthermore, by a theorem of Morrey and Grauert, this real analytic structure is unique. In any case, as there is no integrability condition to satisfy, I think this probably answers your question in the negative. See this thread Can every manifold be given an analytic structure? for a more detailed discussion.

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    $\begingroup$ Perhaps it is important to note that analytic structures on a smooth manifold are unique only up to analytic diffeomorphisms. In fact, there are analytic structures on the real line which refine the usual smooth structure, yet do not give the same maximal analytic atlas. By way of contrast, any two complex structures on a given almost complex manifold are necessarily the same, i.e. induce the same maximal complex-holomorphic atlas. $\endgroup$ – Ricardo Andrade Jun 26 '14 at 21:32
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    $\begingroup$ I understand Igor, but seeing as there is not a differential equation which an analytic function on $\mathbb{R}^n$ satisfies which guarantees it's analytic (I don't think), I would be somewhat surprised that you could capture the real analytic condition via the vanishing of some tensor field. Not that this is all that helpful. $\endgroup$ – Andy Sanders Jun 27 '14 at 1:38
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    $\begingroup$ So that I can understand the question better, let me ask whether the following examples would count: In dimension $n=1$, let the geometric structure simply be a metric $g$ on $M$ (subject to no conditions). Then the local coordinates $t$ on $M$ for which $g = \mathrm{d}t^2$ generate a real-analytic atlas on $M$. In dimension $2$, let the geometric structure simply be a metric $g$ on $M$ subject to the (third-order) differential equation that the Gauss curvature of $g$ be (locally) constant. Again, the (local) Gauss normal exponential coordinates for $g$ generate an analytic atlas on $M$. $\endgroup$ – Robert Bryant Jun 27 '14 at 13:51
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    $\begingroup$ @IgorKhavkine: OK. However, as far as I know, it's an open question as to whether there is any analogous geometric structure in dimension $3$ that exists on all $3$-manifolds. For example, not every $3$-manifold carries a flat projective structure (though several of Thurston's 8 geometries do), which would, of course, define an underlying analytic structure. Whether there is some higher order differential equation on metrics in dimension $3$ whose solutions are all locally analytic and, moreover some metric satisfying these equations exists on every $3$-manifold is an interesting question. $\endgroup$ – Robert Bryant Jun 27 '14 at 15:40
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    $\begingroup$ @RobertBryant: I have two remarks on this: 1. In dimensions $\ge 4$ the associated real-analytic pseudogroup would have to be infinite-dimensional. 2. In 3d every closed 3-manifold admits a "birational structure", an atlas one where transition maps are birational transformations of $R^3$. (Maybe this also works in higher dimensions, I do not know.) Sadly, I do not know how to capture it via any finite order tensor since the degree is unbounded. $\endgroup$ – Misha Jun 27 '14 at 20:05

Here's an interesting and relevant result from DeTurck & Kazdan's Some regularity theorems in Riemannian geometry Ann. sci. de l'ENS 14 249-260 (1981):

Theorem 5.2: Let $(\mathscr{M},g)$ be a connected Einstein manifold [$g$ is Riemannian and $\mathrm{Ricc}[g] = c g$ for any $c\in \mathbb{R}$] of class $C^2$ with $\dim \mathscr{M} \ge 3$. Then $g$ is real analytic in harmonic and geodesic normal coordinates.

From the discussion in Section 1 of the same paper, this means that every point of $p\in \mathscr{M}$ possesses a chart with harmonic coordinates [$(x^i)$ is a harmonic chart if $\Delta_g x^i = 0$] in which, by the above theorem, $g$ is real analytic. I think then that an Einstein (Riemannian) metric on a manifold defines a special analytic atlas (the fact that the set of all harmonic charts constitutes an answer is not stated in the paper, but I think one can make an argument for it using the same methods).

I'd be curious to see whether this atlas can be interpreted as giving a CR structure to the manifold (e.g., by restricting the complex structure from a Grauert embedding of $\mathscr{M}$ into a complex manifold).

  • $\begingroup$ I'm not sure I understand what you're asking for. I'm under the impression that your original question asks whether there is an integrability condition (this is presumably a system of PDE's) satisfied by any real analytic manifold. The DeTurck-Kazdan theorem gives a sufficient (and not necessary) condition of a manifold to be real analytic. It relies heavily on the fact that there exists a Riemannian metric that satisfies a nonlinear elliptic PDE, which is a very strong assumption. $\endgroup$ – Deane Yang Apr 7 '15 at 15:52

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