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An embedding is an injective map into a universal, simpler model object. Many embedding theorems are without obstruction, in the sense that every object which you wish to embed can be embedded. Examples of such theorems are Yoneda lemma, algebraic closure of fields, Nash embedding theorem for Riemannian manifolds. unconditional.

I'm interested in embedding theorems with obstruction. Do you have examples of theorems that give an obstruction to embedding? In the case where there is an obstruction, would you consider the obstruction to be local or instrinsic? Most embedding problems are possible locally, but there is often a local-global obstruction.

The example that led me to this question is Kodiara's embedding theorem that gives an obstruction for a complex manifold to be a submanifold of complex projective space. Here the obstruction is that the manifold must carry a positive line bundle. Positivity of curvature is a local criterion.

PS. Sorry, but I really don't know how to tag this question.

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    $\begingroup$ I would say that although the word "embedding" has a fairly universal meaning of being an injective map, the specific reasons why you want an embedding can vary according to the subject. There is clearly a (not necessarily correct) philosophy that if you take an object you want to understand it better and embed it somehow into a bigger but simpler (even canonical) object, then this will make it easier to analyze the original object. But I believe the specific criteria are for the bigger object or conditions you want for the embedding depends quite strongly on the specific circumstances. $\endgroup$
    – Deane Yang
    Nov 13, 2010 at 16:29
  • $\begingroup$ Hi Deane, in instances where we cannot embed into the canonical object, what does the obstruction to embedding teach us? $\endgroup$
    – user2529
    Nov 14, 2010 at 2:19
  • $\begingroup$ @Colin, at that level of generality, nothing. $\endgroup$
    – Mariano Suárez-Álvarez
    Nov 15, 2010 at 6:21
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    $\begingroup$ I don't view the target of embedding and non-embedding theorems as being a canonical object (which usually has a universal property that implies an embedding theorem) but merely a "simplest" object. I also think embedding theorems were more popular in the past, when people were less comfortable working with things like manifolds and algebraic varieties unless they were embedded in a more familiar space. Today, we know that the embedded object is often a lot messier than the intrinsic one. Embeddings of functional spaces (see Serre below) are however, crucial to nonlinear PDE's. $\endgroup$
    – Deane Yang
    Nov 15, 2010 at 13:59

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I'm not sure whether you look for such an answer, because it comes from analysis. Analysts use various functional spaces, especially the Sobolev spaces. $W^{s,p}(\Omega;\mathbb R)$ is, roughly speaking, the set of functions with $s$ derivatives in $L^p(\Omega)$ (but $s\ge0$ needs not be an integer).

Sobolev embedding. If $\Omega$ is an open subset with a smooth boundary, and if $\frac1q=\frac1p-\frac{s}{n}$ with $1\le p< q<\infty$, then $W^{s,p}(\Omega;\mathbb R)$ embeds into $L^q(\Omega)$. If instead $sp>n$, then $W^{s,p}(\Omega;\mathbb R)$ embeds into ${\mathcal C}^\alpha(\bar\Omega)$ where $\alpha:=s-\frac{n}{p}$, unless this exponent is an integer.

When the target $\mathbb R$ is replaced by a manifold, the situation may not be so nice. Embedding theorems are related to norm inequalities, which are usually proved first for ${\mathcal C}^\infty$-fields, then extended by means of density of ${\mathcal C}^\infty$ in $W^{s,p}$.

Obstruction (Bethuel 1991). Assume that $p< n$, and let $N$ be a compact manifold of dimension $k$. Then ${\mathcal C}^\infty(\Omega,N)$ is dense in $W^{1,p}(\Omega;N)$ if and only if $\pi_{[p]}(N)=0$, where $[p]$ is the largest integer $\le p$.

The consequence of this is that in some situations, there is a discrepency between $W^{s,p}$ and the closure of ${\mathcal C}^\infty$ under the $W^{s,p}$-norm.

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  • $\begingroup$ Does $\Omega$ refer so some subset of $N$? May I know heuristically why a homotopy group features in the obstruction, and why the obstruction involves only $N$ and not $\Omega$? $\endgroup$
    – user2529
    Nov 13, 2010 at 13:59
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    $\begingroup$ @Colin. The obstruction is of local nature. Therefore, as long as $\Omega$ is a smooth manifold, only the topology of $N$ matters. A typical example, related to the mathematics of liquid cristals, is that of $N=S^2$, the unit sphere in $\mathbb R^3$, whereas $\Omega$ is an open subset of $\mathbb R^3$. Then a degree can be defined over $W^{1,p}$ and is non-trivial. It would be trivial if ${\mathcal C}^\infty$ had been dense. $\endgroup$
    – Denis Serre
    Nov 13, 2010 at 14:41

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