# Derivation on real analytic manifolds

Let $M$ be a real analytic manifold. By $C^{\omega}(M)$ we mean the algebra of all analytic functions from $M$ to $\mathbb{R}$. Assume that $D$ is a derivation on $C^{\omega}(M)$ .

Is there a global real analytic vector field $X$ on $M$ such that $D(f)=X.f$ for all $f\in C^{\omega}(M)$?

The motivation:

The smooth version of this statment is true but the proof is based on usage of non analytic functions

• Your question reduces to if $X$ is $C^\infty$ and $X.f$ is analytic for all $f$ analytic does this imply $X$ is analytic? I think the answer is yes. – Ryan Budney Feb 26 '14 at 21:54
• @Ryan: Are you saying that any derivation $\mathcal{C}^{\omega }(M)\rightarrow \mathcal{C}^{\omega }(M)$ extends to a derivation $\mathcal{C}^{\infty }(M)\rightarrow\mathcal{C}^{\infty }(M)$? Why should that be true? Note that in the holomorphic set-up, there may exist derivations which do not come from vector fields. I would not be surprised if this happened also here. – abx Feb 27 '14 at 6:06
• You haven't heard of real-analytic partitions of unity? (Neither have I.) – Gerald Edgar Feb 27 '14 at 14:22
• @GeraldEdgar Prof. Edgar As you pointed out, there is no a real analytic partition of unity, because of "Analytic continuation". – Ali Taghavi Feb 27 '14 at 20:10
• @Ali: Take a family $p:X\rightarrow D$ of curves of genus $g\geq 2$, with $D$ a disk in the moduli space $\mathcal{M}_g$. Then $\mathcal{O}(X)\cong \mathcal{O}(D)$, which has nontrivial derivations (= holomorphic vector fields on $D$). But Kodaira-Spencer deformation theory implies that these do not lift to $X$, in fact $X$ has no nonzero holomorphic vector fields. – abx Feb 27 '14 at 20:31

$$\def\RR{\mathbb{R}}\def\CC{\mathbb{C}}\def\NN{\mathbb{N}}$$I'm going to try once more. As my previous flawed answers should have made obvious, the real analytic category is not my home, so read with caution.

Conveniently, the two big results I need are stated as Corollary 5.43 and 5.44 here:

Cor. 5.43 $$M$$ has a proper real analytic embedding into some $$\RR^n$$. I'll write $$(z_1, \ldots, z_n)$$ for the coordinates on $$\RR^n$$.

Cor. 5.44 (specialized and rephrased) Let $$p$$ be a point of $$M$$ and let $$M$$ be embedded in $$\RR^n$$ as above. Let $$U$$ be a neighborhood of $$p$$ in $$\RR^n$$. Let $$f: M \cup U \to \RR$$ be a function which is real analytic on $$U$$ and on $$M$$. Let $$d$$ be a nonnegative integer. Then there is a real analytic function $$g: \RR^n \to \RR$$ such that $$g|_M=f$$ and $$(g-f)|_U$$ vanishes to order $$d$$ at $$p$$.

We also need the following:

Easy Lemma: Let $$f: \RR^n \to \RR$$ be a real analytic function for which $$f(z_1, \ldots, z_{n-1},0)$$ is identically zero. Then $$f(z_1, z_2, \ldots, z_{n-1}, z_n)/z_n$$ extends to a real analytic function $$\RR^n \to \RR$$.

Proof This is a local statement, and can be easily checked locally using convergent power series. $$\square$$

Key Lemma: Let $$g: \RR^n \to \RR$$ be a real analytic function vanishing to order $$d$$ at $$0$$. Then we can write $$g(z)$$ as a finite sum $$\sum z^a p_a(z)$$ where $$z^a$$ are monomials of degree $$d$$ and $$p_a(z)$$ are real analytic functions.

Proof By induction on $$d$$ and $$n$$. The base case $$d=0$$ is trivial: Write $$g(z) = 1 \cdot g(z)$$; the base case $$n=0$$ is vacuous. We assume $$d$$ and $$n>0$$. Define $$g(z_1, \ldots, z_n)$$ by $$f(z_1, \ldots, z_n) = f(z_1, \ldots, z_{n-1}, 0) + z_n g(z_1, \ldots, z_{n-1}).$$ By the Easy Lemma, $$g$$ is analytic and, by a basic computation, it vanishes to order $$d-1$$. By induction on $$d$$, we can write $$g = \sum z^c r_c(z)$$ where the monomials $$z^c$$ have degree $$d-1$$. By induction on the number of variables, we can write $$f(z_1, \ldots, z_{n-1},0)$$ as $$\sum z^b p_b(z_1, \ldots, z_{n-1})$$ where the monomial $$z^b$$ have degree $$d$$. So $$f(z_1, \ldots, z_n) = \sum z^b p_b(z_1, \ldots, z_{n-1}) + \sum (z^c z_n) q_c(z_1,\ldots, z_n),$$ an expression of the desired form. $$\square$$

We now prove the result.

Lemma Let $$f$$ be a real analytic function on $$M$$ that vanishes to second order at $$p$$. Then $$(D f)(p)=0$$.

Proof Embed $$M$$ in $$\RR^n$$ by Cor 5.43. Extend $$f$$ to a function on $$\RR^n$$ also vanishing to order $$2$$ at $$p$$ and, without loss of generality, translate $$p$$ to $$0$$. By the Key Lemma, we can write $$f(z) = \sum z_i z_j c_{ij}(z)$$. Then $$(D f)(0) = \sum (D z_i)(0) \cdot 0 \cdot c_{ij}(0) + \sum 0 \cdot (D z_j)(0) \cdot c_{ij}(0) + \sum 0 \cdot 0 \cdot (D c_{ij})(0) =0. \quad \square$$

Lemma $$(D f)(p)$$ depends only on $$(df)(p)$$.

Proof Suppose that $$(d f_1)(p) = (d f_2)(p)$$. Define $$q(z)$$ by the equation $$f_2(z) = f_1(z) + (f_2(p)-f_1(p)) + q(z)$$, then $$q$$ vanishes to order $$2$$ at $$p$$ so $$(D q)(p) =0$$. Also, $$D$$ of a constant function is $$0$$. So $$(D f_1)(p) = (D f_2)(p)$$. $$\square$$

The previous lemma shows that there is a well defined map $$v(p) : T^{\ast}_p M \to \RR$$ so that $$(D f)(p) = v(p)(d f)$$. (Well, we also need to see that $$C^{\omega}(M) \to T^{\ast}_p M$$ is surjective, but that is obvious because the functions $$z_i$$ span the cotangent space.) It is easy to see that $$v(p)$$ is linear. So $$v(p)$$ is an element of $$(T_{\ast})_p M$$. We have now created a section $$v: M \to T_{\ast} M$$ so that $$(D f)(p) = \langle v(p), df \rangle$$. It remains to see that $$v$$ is real analytic.

This question is local. Let $$(z_1, \ldots, z_d)$$ give local coordinates on $$M$$ near some point $$p$$. Then we can write $$v(p) = \sum_{i=1}^d v_i(p) \frac{\partial}{\partial z_i}.$$ We need to show that the functions $$v_i$$ are real analytic. Since $$v_i = D z_i$$, this is clear. QED.

• Thank you very much for the second answer. Perhaps there is a misunderstanding for me about the concept of "real analytic function". a real analytic function is a map which has a LOCAL Taylor series expansion for example $1/(x^{2}+1)$ is a real analytic function from $\mathbb{R} \to \mathbb{R}$. this function has no a global Taylor series. But in your key lemma you used the global expansion. In your proof of the lemma(after the Key lemma) is it essential to have a global taylor expansion? – Ali Taghavi Mar 3 '14 at 19:43
• One more try here. I think this works! – David E Speyer Mar 4 '14 at 13:12
• Honestly, I enjoyed your very interesting answer.thank you very much for your help. – Ali Taghavi Mar 5 '14 at 14:13
• More generally, real-analytic geometry is much closer to $C^{\infty}$ geometry than to complex-analytic geometry due to Grauert's theorem that every real-analytic manifold is a real-analytic submanifold of a Stein space (and the Morrey-Grauert theorem that gives a unique real-analytic structure to any smooth manifold). The question above is one of many consequences of Grauert's results (which should be more widely known); see the last Theorem in the Introduction of this 1981 paper: archive.numdam.org/ARCHIVE/CM/CM_1981__43_2/CM_1981__43_2_239_0/… – user76758 Mar 9 '14 at 13:57
• What can be said about the vector field analogy of the result you proved?: The classification of all derivation of $C^{\omega}(M)$ when $M$ is an analytic manifold? Please see this question mathoverflow.net/questions/330065/… – Ali Taghavi Apr 27 '19 at 18:28