This might be standard if true (if so, I shall be grateful if provided with a reference). Given a smooth map from a Stein manifold $X$ to $\operatorname{Gr}(k,n)$ (the Grassmannian of $k$ planes in $\mathbb{C}^n$), is there a holomorphic map in its homotopy class?

  • 5
    $\begingroup$ Yes. It follows from a theorem of Grauert, see his paper "Analytische Faserungen über holomorph-vollständigen Räumen" Math. Ann. 135 1958 263–273. $\endgroup$ – ulrich May 17 '13 at 4:00
  • 1
    $\begingroup$ These types of results are sometimes also called "Grauert's Oka principle". So perhaps there are earlier partial results of Oka (e.g., when $k$ equals $1$). $\endgroup$ – Jason Starr May 17 '13 at 12:30

As has been established in the comments, the answer to your question is yes. It is a special case of a general result known as the Oka principle which has been strengthened over time. The key is that $\operatorname{Gr}(k, n)$ is an Oka manifold as are all homogeneous spaces for complex Lie groups.

Theorem: Let $X$ be a Stein manifold and $Y$ an Oka manifold. The inclusion

$$\mathcal{O}(X, Y) \hookrightarrow \mathcal{C}(X, Y)$$

is a weak homotopy equivalence. In particular, every continuous map is homotopic to a holomorphic map.

Here $\mathcal{O}(X, Y)$ denotes the collection of holomorphic maps $X \to Y$, $\mathcal{C}(X, Y)$ denotes the collection of continuous maps $X \to Y$, and both spaces are equipped with the compact-open topology.

There are more general versions of the result stated above. In particular, there is an analagous result for holomorphic and continuous sections of stratified holomorphic fiber bundles over Stein manifolds with Oka fibers - the above statement corresponds to the trivial bundle $X\times Y \to X$.

If you are interested in learning more about this story and its history, have a look at Stein Manifolds and Holomorphic Mappings (The Homotopy Principle in Complex Analysis) by Forstnerič, in particular Chapter $5$.

Allow me to explain somewhat carefully (mostly for my own benefit) why the claim "every continuous map is homotopic to a holomorphic map" follows from the first part of the theorem.

A topological space $X$ is compactly generated if it is Hausdorff and $A \subseteq X$ is closed if and only if $A\cap K$ is closed in $K$ for every compact $K \subseteq X$.

Such spaces are ubiquitous. Every locally compact Hausdorff space is compactly generated, in particular topological manifolds. In addition, metric spaces and CW complexes are also compactly generated.

In his paper A Convenient Category of Topological Spaces, Steenrod showed that the category of compactly generated spaces has the properties one would desire. In particular, he proved the following (Theorem $5.4$).

Theorem: Let $W$, $X$, and $Y$ be compactly generated spaces. The map

\begin{align*} \mu : \mathcal{C}(X\times W, Y) &\to \mathcal{C}(W, \mathcal{C}(X, Y))\\ f &\mapsto \mu f \end{align*}

where $(\mu f(w))(x) = f(x, w)$ is a homeomorphism.

Applying the above to $W = [0, 1]$, we see that

$$\mathcal{C}([0, 1], \mathcal{C}(X, Y)) = \mathcal{C}(X\times [0,1], Y).$$

So a path between $f$ and $g$ in $\mathcal{C}(X, Y)$ gives rise to a homotopy between $f$ and $g$, and vice versa. Therefore, we see that

$$\pi_0(\mathcal{C}(X, Y)) = [X, Y]$$

where the right hand side denotes the collection of homotopy classes of continuous maps $X \to Y$.

The inclusion $\mathcal{O}(X, Y) \hookrightarrow \mathcal{C}(X, Y)$ induces an isomorphism $\pi_0(\mathcal{O}(X, Y)) \cong \pi_0(\mathcal{C}(X, Y))$. That is, each path-connected component of $\mathcal{C}(X, Y)$ contains a unique path-connected component of $\mathcal{O}(X, Y)$. So for every $f \in \mathcal{C}(X, Y)$, there is $g \in \mathcal{O}(X, Y)$ in the same path-connected component as $f$, and hence homotopic to $f$. Therefore, every continuous map $X \to Y$ is homotopic to a holomorphic one.

Moreover, if $f$ and $g$ are holomorphic maps which are homotopic, $f$ and $g$ belong to the same path-connected component of $\mathcal{C}(X, Y)$, and hence the same path-connected component of $\mathcal{O}(X, Y)$. Therefore $f$ and $g$ are homotopic through holomorphic maps. That is, there is a homotopy $H : X\times [0, 1] \to Y$ between $f$ and $g$ such that $H(\cdot, t) : X \to Y$ is holomorphic for all $t\in [0, 1]$.

Given the title of your question, it seems you had the Oka-Grauert Theorem in mind. Namely, the fact that on a Stein manifold, the classification of topological complex vector bundles coincides with the classification of holomorphic vector bundles. A proof of this theorem using classifying maps to Grassmannians and the Oka Principle is also discussed in Chapter $5$, see pages $190 - 191$.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.