17
votes
Accepted
When do real analytic functions form a coherent sheaf?
For real-analytic manifolds, coherence of the structure sheaf always holds. The 1-sentence reason is that one can pass to real and imaginary parts on Oka's coherence theorem in several complex ...
15
votes
Are real-analytic functions in $\mathbb{R}^2$ holomorphic after suitable change of coordinates?
The answer is negative. For the non-injective case, the reason is that non-constant complex analytic functions are open, discrete maps, while real analytic functions can be neither open nor discrete (...
- 91.8k
13
votes
Accepted
No analytic surjection $f:M \to N$ when $\dim(M) >\dim(N)$
Any analytic $n$-manifold admits an analytic surjection $f$ from $\mathbb{R}^n$: pick an analytic Riemannian metric (e.g. by embedding to $\mathbb{R}^N$ by Whitney-Grauert-Morrey embedding theorem) ...
- 2,934
13
votes
Accepted
Is the analytic version of the Whitney Approximation Theorem true?
The result is not stated in Grauert's paper. On the other hand, Grauert proves that every real analytic manifold $M$ sits as a real analytic totally real submanifold, and analytic deformation ...
- 26.3k
8
votes
Are conformal maps between Riemannian manifolds real-analytic?
As essentially follows from your argument in the comment to Piotr Hajlasz's answer (now deleted), for $f\colon M\to N$ to be a counterexample, it cannot be a diffeomorphism. Because then you could ...
- 21.5k
7
votes
Accepted
Topology on space of hyperfunctions
This is a comment but will be too long. You start with the following setup: a pair of Fréchet spaces (the holomorphic functions on the complex plane, resp. on the complement of the real line there) ...
- 281
7
votes
Accepted
Symplectic form on a Kähler manifold can be not real analytic?
The answer is positive for any Kähler manifold.
Consider first surfaces. Take a compact Riemann surface $\Sigma$, then any symplectic form on it is associated to a Kähler form, so the answer is yes, ...
- 28.9k
6
votes
Accepted
In the real analytic category, are the fibers of a proper submersion isomorphic?
There exists an analytic Riemann metric on a real-analytic manifold, which follows from embeddability of real analytic manifolds (see The Analytic Embedding of Abstract Real-Analytic Manifolds ...
- 1,842
5
votes
Accepted
Analog of Newlander–Nirenberg theorem for real analytic manifolds
I think this question can be addressed in a few ways. Two great answers have already been given:
Real analytic is a regularity condition, while holomorphic is more algebraic, so you'd need a ...
- 468
5
votes
Accepted
About nuclear-by-exact extensions
Yes, if the quotient $B$ is nuclear, then the extension is locally semisplit by the Choi-Effros lifting theorem. Hence if (in addition) $I$ is exact, then $A$ is exact (see for instance Exercise 3.9.8 ...
- 2,471
5
votes
Symplectic form on a Kähler manifold can be not real analytic?
Not quite an answer, but something related to it.
There is this paper where the authors prove, that for every symplectic manifold $(M,\omega)$ there is an analytical manifold $M^a$ and an analytical ...
- 2,830
5
votes
Are real-analytic functions in $\mathbb{R}^2$ holomorphic after suitable change of coordinates?
To see why the second question cannot have a simple answer, it is sufficient to look at the local context near a fixed-point of a tangent-to-identity mapping, as Alexandre Eremenko suggests. By "...
- 5,428
5
votes
Accepted
Dimension of intersection of real analytic sets
For a counterexample take a sphere in 3 space and a plane tangent to it .
- 4,131
4
votes
Accepted
Contractible real analytic varieties
It is a consequence of Sullivan's work
Sullivan, D., Combinatorial invariants of analytic spaces, Proc. Liverpool Singularities-Sympos. I, Dept. Pure Math. Univ. Liverpool 1969-1970, Lect. Notes Math. ...
- 12.2k
4
votes
Are conformal maps between Riemannian manifolds real-analytic?
This is not an answer. Just a possible vector of attack for this problem.
According to the Theorem 11.4.6 of Harmonic Morphisms Between Riemannian Manifolds by Paul Baird and John C. Wood, every non-...
- 8,597
4
votes
Accepted
Identity Theorem for Real-Analytic Hypersurfaces
Yes: two closed irreducible analytic hypersurfaces are identical just when they are identical near some point. Careful that they are closed. But your hypothesis ensures that they are irreducible.
- 26.3k
3
votes
Accepted
Decomposition of a real analytic variety
You may want to look at Lojasiewicz's structure theorem; for a statement and proof (I won't reproduce it here since the complete theorem statement is over a page long) see Chapter 6, section 3 of ...
- 39k
3
votes
Is the analytic version of the Whitney Approximation Theorem true?
The book
F. Guaraldo, P. Macrì, A. Tancredi
"Topics on Real Analytic Spaces", Advanced Lectures in Mathematics, Braunschweig/Wiesbaden: Friedrich Vieweg & Sohn, pp. X+163 (1986), ISBN:3-...
- 2,934
3
votes
Polynomial vector field tangent to a given analytic simple closed curve
The answer is "no". In fact, it is still "no" for germs of curves : generically, a germ of an analytic curve $\gamma : (\mathbb R,0)\rightarrow (\mathbb R^2,0)$ is not tangent to any polynomial vector ...
- 5,428
3
votes
Accepted
Polynomial vector field tangent to a given analytic simple closed curve
I see this as very unlikely. A polynomial vector field would have a slope function that is a rational function of two variables with a finite number of coefficients.
As a consequence, if you take ...
- 5,186
3
votes
Accepted
Does Noetherianity imply division theorem?
This is a consequence of the noetherianity of $\mathcal{O}_n$, in some way.
Let me write $A_n$ for the ring of real formal formal series around the origin, i.e., $A_n=\mathbb{R}[[t_1,\dotsc,t_n]]$. ...
- 2,928
2
votes
Contractible real analytic varieties
I think that the answer is negative for non-compact real analytic manifolds.
Consider the curve $$\gamma \colon (0, \, + \infty) \longrightarrow \mathbb{R}^2, \quad \gamma(t)=(e^{-t} \cos t, \, e^{-t} ...
- 66.3k
2
votes
Accepted
Analyticity of central stable manifolds
Quick answer to the first question: no, there is no reason why it should be analytic. Take e.g. the parametric vector field (written as a Lie derivative)$$X(x,y):=-x^3\partial_x+(y+\alpha x)\partial_y~...
- 5,428
2
votes
When do volumes depend real-analytically on the parameters defining the regions?
In your setting, it seems that your function $V(r)$ depends too much on the shape of the domain $B$. Take for instance $B= \{0\leq y \leq h(x)\} \subset \mathbb R^2$ for a continuous but very wild ...
- 1,908
2
votes
Are continuous rational functions arc-analytic?
If $X$ is all of $\mathbf R^n$ then the answer to the first question is "yes", I think. Indeed, for any continuous rational function $f$ on $\mathbf R^n$ there is a stratification of $\mathbf R^n$ in ...
- 1,424
2
votes
A special oscillatory orbit in space
Of course not. The image of a $C^1$ function on $\mathbb R^+$ has $\sigma$-finite Hausdorff $1$-dimensional measure, and therefore has $2$-dimensional measure $0$.
In fact, $C^1$ can even be weakened ...
- 54.2k
2
votes
On a case of real-analytic interpolation
If you are happy with a sufficient condition, you can get one from the necessary and sufficient condition for the interpolation by bounded holomorphic functions on the complex unit disc $\mathbb{D}$, ...
- 21.5k
1
vote
Real analytic function of one variable with given set of values
It's not always possible to find such a function, for example take $x_n\to 0-$, $y_n=-e^{-1/x_n^2}$. A holomorphic function $f: D_r(0)\to\mathbb C$, $f\not\equiv 0$, would satisfy $|f(x)|\gtrsim |x|^k$...
- 24.2k
Only top scored, non community-wiki answers of a minimum length are eligible