# What is the difference between “immersion” and “embedding”?

Could someone please explain what "embedding" means (maybe a more intuitive definition)? I read that the Klein bottle and real projective plane cannot be embedded in ${\mathbb R}^3$, but is embedded in ${\mathbb R}^4$. Aren't those two things 3D objects? If so, why aren't they embedded in ${\mathbb R}^3$? Also, I have come across the word "immersion". What is the difference between "immersion" and "embedding"?

Thanks.

• "Embedding" something means to "stick it inside"; intuitively, if $A$ embeds into $B$, that means that there is a way to "stick" $A$ "inside" $B$, or that $B$ 'contains' a copy of $A$. There is no way to put a copy of the Klein bottle inside $\mathbb{R}^3$ because it is not really a 3-dimensional object: you cannot have a Klein bottle in $\mathbb{R}^3$ without self-intersection, even though the Klein bottle doesn't have any (just like the Moebius strip cannot be done in $\mathbb{R}^2$). "Immersion" is usually synonymous with "embedding", but used in certain fields more than in others. – Arturo Magidin Sep 28 '11 at 16:47
• @Arturo: "Immersion" just means that the derivative is injective on the tangent space at each point. The manifold is allowed to self-intersect. – Cheerful Parsnip Sep 28 '11 at 16:54
• So I'm saying the two terms are NOT synonymous. – Cheerful Parsnip Sep 28 '11 at 16:54
• @ArturoMagidin: Thanks! I think I am not sure why the Klein bottle is not a 3D object...? I can see why the Mobius strip is not 2D though. Would you mind explaining a bit please? – surface Sep 28 '11 at 16:55
• @surface: Essentially for the same reason that the Moebius strip is not a 2D object even though it is a 2-dimensional manifold. If you try to construct a Klein bottle in 3-space, you end up necessarily having to self-intersect in order to get the "twist" necessary to make it have only one "side" (i.e., no interior and no exterior); this "twist" can only happen in 4-space if you don't want the bottle to intersect itself, just like the twist of the Moebius strip has to happen in 3-dimensions if you don't want the strip to intersect itself in 2 dimensions. – Arturo Magidin Sep 28 '11 at 16:58

Basically an abstract surface has, at every point two independent directions along the surface. Or even better, there is an entire circle's worth of rays coming out of each point.

An immersion is, roughly, a map of the surface into a bigger manifold (such as $\mathbb R^n$) where there are still two dimensions worth of rays emanating out of each point. So for the usual immersion of a Klein bottle into $\mathbb R^3$, at the circle of self-intersection, each sheet still retains its two dimensional character. So it is an immersion. If you were to instead map the Klein bottle into $\mathbb R^3$ by mapping everything to a point, that would not be an immersion.

• "So for the usual embedding of a Klein bottle into $\mathbb{R}^3$..." There is no such embedding; this statement just further confuses the issue. – user50229 Mar 16 '17 at 16:35
• I.m.h.o. it should indeed be "usual immersion". At the points of self-intersection two sheets of the surface exits, so the mapping is not injective. But when the derivative of the mapping is injective, it is still an immersion. Probably (not sure) when the derivative is not injective, I imagine that in a certain point two sheets of the surface may exist, with equal change in curvature of the sheets. This perhaps means that the two sheets that intersect cannot be distinguished anymore. Without self-intersection it is called an embedding. – Gerard Mar 17 '17 at 15:01
• @user50229; oops. I have changed it. – Cheerful Parsnip Mar 17 '17 at 18:04
• @Gerard: thanks. Yes, "embedding" was an unfortunate choice of word! – Cheerful Parsnip Mar 17 '17 at 18:04

Understandably there are a lot of answers, but if you still have any further questions maybe this will help.

An embedding of a topological space $X$ into a topological space $Y$ is a continuous map $e \colon X \to Y$ such that $e|_X$ is a homeomorphism onto its image.

Both the Klein bottle ($f \colon I^2 / \sim \to \mathbb{R}^3$) is not embedded into $\mathbb{R}^3$, because it has self-intersections; this means that the immersion of the Klein bottle is not a bijection, hence not a homeomorphism, so not an embedding.

As I understand it, an immersion simply means that the tangent spaces are mapped injectively; i.e. that the map $D_p f \colon T_pI^2 \to T_{f(p)}\mathbb{R}^3$ is injective. In the Klein bottle example, at the self-intersection, any point of intersection has two distinct tangent planes, hence this map is injective.

I hope this makes some sense!

An immersion is precisely a local embedding – i.e. for any point x ∈ M there is a neighbourhood [sic], U ⊂ M, of x such that f : U → N is an embedding, and conversely a local embedding is an immersion.

So, an immersion is an embedding, i.e. an isomorphic (homeomorphic) copy, at each point, and vice versa, though the entire image may not be a homeomorphic copy.

But, later, the same article says:

If M is compact, an injective immersion is an embedding, but if M is not compact then injective immersions need not be embeddings; compare to continuous bijections versus homeomorphisms.

I wish I could give an example of a non-compact imbedding/embedding, or continuous bijections versus homeomorphisms, but though I understand the two ideas, roughly, I am not sure what conditions make them conflict...

Reading a paper by Claudio Gorodski (2012), there is a section he called Riemann submanifolds and isometric immersions'', where he considers a Riemann manifold $(M,g)$ and an immersed submanifold $\iota :N \rightarrow M.$ Thus $N$ is a smooth manifold and $\iota$ is an injective immersion. Riemann metric $g$ induces a Riemann metric $g_N$ in $N$. So if $p \in N$, the tangent space $T_pN$ can be viewed as a subspace of $T_pM$ via the injective map $d_{\iota_p}: T_pN \rightarrow T_\iota(p)M.$ We may define $(g_N)_p$ to be the restriction of $g$ to this subspace: $$(g_N)_p(u,v)=g_{\iota_p}(d_{\iota_p}(u),d_{\iota_p}(v)),$$ where $u,v\in T_pN.$ So $g_N$ is a Riemann metric. Indeed $g_N$ is an induced Riemannian metric in $N$, so that $(N,g_N)$ is a Riemannian submanifold of $(M,g.$ It is worthy to note that $g_N$ makes sense even if $\iota$ is an immersion that is not necessarily injective. $g_N$ is said to be the pulled-back metric, $g_N =\iota^*g,$ and $\iota : (N,g_N)\rightarrow(M,g)$ is an isometric immersion. But any immersion must be locally injective. An important particular case is that of Riemannian submanifolds of Euclidean space. A regular curve or surface is an immersion whenever there is a parametrization which is assumed smooth. In such a case the parametrization is locally an embedding.