All of my geometric intuition for "immersed" versus "embedded" surface is contained in my geometric intuition for "immersions" versus "embeddings". In particular, as many people have pointed out, immersions need not be injective. But, of course, even injective immersions need not be embeddings. As a very simple example, consider the map $f : (-\pi, \pi) \to \mathbb{R}^2$ given by
$$
f(t) = (\sin t, \sin 2t).
$$
The image of this map is a sort of "figure 8" in the plane, traced out starting at the origin, moving through quadrants II, III, I, and IV, in that order, as $t$ moves from $-\pi$ to $\pi$. It's easy to see that $f$ is an injective immersion, but $f$ is not an embedding, since every neighborhood of $\mathbb{R}^2$ containing the origin also contains points of the form $f(-\pi + \epsilon)$ and $f(\pi - \epsilon)$ for all sufficiently small $\epsilon$.
As another one-dimensional example of this type, you could consider the closed topologist's sine curve with a loop, which is the graph of $g(x) = \sin(1/x)$ for $x \in (0, 1]$ together with the $y$-axis between $y = -1$ and $y = 1$ together with a "loop" smoothly connecting the point $(0, -1)$ to the point $(1, \sin(1))$. It's clear that there is some injective immersion $f : [0, \infty) \to \mathbb{R}^2$ whose image is this curve, and this immersion is not an embedding.
You can, of course, easily make either of these example into a surface by considering $h: (-\pi, \pi) \times (0,1) \to \mathbb{R}^3$ given by $h(t, s) = (f(t), s)$.
On the other hand, I don't really know any examples of injective immersions of surfaces which aren't embeddings that are "interesting" in a way that's fundamentally different from the examples above. The idea I have is that immersions are allowed to "approach themselves" or "limit onto themselves" in crazy ways that embeddings are not. In particular, if $f : X \to Y$ is an injective immersion, the topology on $f(X)$ as a subspace of $Y$ might be very different than the topology on $X$.
Let's first clarify the terminology and setting. Let's assume that we have an embedded submanifold $S$ (e.g. a surface) in a Riemannian manifold $M$ (e.g. Euclidean 3-space).
I assume that by intrinsic curvature of $S$, you mean the Gaussian curvature of a surface. For higher dimensional manifolds, this generalizes to the sectional curvature, but this is a little more complicated: it assigns a number to each 2-dimensional subspace of the tangent space, namely the Gaussian curvature of the submanifold (surface) tangent to that plane.
"Extrinsic curvature" could mean several things, but for a hypersurface (e.g. an embedded surface $S$ in Euclidean 3-space), we could summarize by saying that an extrinsic curvature is a quantity defined by the second fundamental form, or equivalently its associated shape operator $B$. If you don't know what these things are, it's ok, you can keep reading. Think of the shape operator as a symmetric matrix depending on $p \in S$. The main "extrinsic curvatures" that are worth considering are:
- The eigenvalues of $B$, called principal curvatures. Note that they are equal to the curvature of curves lying in $S$, seen as curves in $M$.
- The trace of $B$ (maybe divided by the dimension), called the mean curvature, equal to the sum (or average) of the principal curvatures.
For a surface in a 3-dimensional manifold, the Gauss equation says that:
$$ \det B = K_S - K_M$$
where $K_S$ is the Gauss curvature of $S$ and $K_M$ is the sectional curvature in $M$ of the plane tangent to $S$. This equation is probably one of things you're looking for answering your question: it tells you the relation between the second fundamental form (defining the "extrinsic curvatures"), the intrinsic curvature of $S$ and the intrinsic curvature of $M$.
Now let's answer your question more precisely. As you can tell from Gauss equation, if all the extrinsic curvatures are zero, i.e. $B$ vanishes (FYI, in this case one says that $S$ is a totally geodesic submanifold), then the intrinsic curvature of $S$ is equal to the intrinsic curvature of $M$. In particular, if $M$ has zero curvature (i.e. Euclidean 3-space), then a submanifold for which all the extrinsic curvatures are zero also has zero intrinsic curvature.
That being said, if by "the extrinsic curvature" we just signify the mean curvature $\mathrm{tr}(B)$, we are just looking for surfaces with zero mean curvature, these things are called minimal surfaces. Now the question is: are there minimal surfaces that are not totally geodesic (basically, minimal surfaces that are not planes in Euclidean 3-space)? You can guess that the answer is probably "yes, there are plenty", because that's pretty much asking if there are some symmetric matrices $B$ whose trace is zero, but are not the zero matrix. In fact, the "fundamental theorem of surface theory" basically guarantees that there are many examples. You'll see examples by looking for images of "minimal surface" on the web.
I tried to give a complete and detailed answer, I hope it was useful, but if you're looking for the short answer: yes, any minimal surface in Euclidean 3-space that is not a plane has zero mean curvature but nonzero Gaussian curvature. For example, the catenoid:
Best Answer
The terms "intrinsic" and "extrinsic" are confusing when trying to be defined in introductions to differential geometry, and for good reason: the standard definition of surfaces at this level float between things like "a subset of $\mathbb{R}^3$ such that etc etc".
A surface is a topological space. A topological space with a particular set of properties, and its definition generalizes to manifolds, which are "surfaces" of higher dimensions. A Riemannian manifold is a manifold together with a Riemannian metric, which is a kind of object defined on a space associated to the manifold. It is common to denote a Riemannian manifold by $(M,g)$, where $M$ is the manifold (and its underlying topology, which we usually omit from notation) and $g$ is the metric. I did not expose precisely what means to be a Riemannian manifold, but nowhere in the definition I'm alluding to is supposed that a Riemannian manifold is inside some $\mathbb{R}^n$.
Before proceeding, an analogy may go well (although, as all analogies, it is not perfect). Consider your nickname: "bubba", as defined by a concatenation of characters. Do you need a paper in order to conceive your name? Or a blackboard? Your nickname has an abstract existence on itself. If I were to ask, say: "How big is the 'u' on your name?", this question would make little sense. It depends on how you write it on paper. The length of the letters is an extrinsic property. However, having five letters is an intrinsic property: it doesn't matter how/where it is written, it is a result of how your name is defined.
Now, moving on. We then usually say that a property of a Riemannian manifold $(M,g)$ is intrinsic if it is a byproduct only of the topology on $M$, and $g$. One example of an intrisic property is the fact that any smooth function on the torus $T^2$ has at least two critical points (in fact, the lower bound is a little bigger). This is a consequence of the fact that $T^2$ is compact. You may say that we know that $T^2$ is compact since it is a subset of $\mathbb{R}^{3}$, but what if I told you that my $T^2$ is $S^1 \times S^1$? This lives inside $\mathbb{R}^4$ instead, and is completely different setwise than what you imagine as a standard "doughnut torus". If I said that my $T^2$ is the square with convenient identifications, then this $T^2$ isn't in any $\mathbb{R}^n$ setwise-ly speaking. Intrinsic properties receive this name because they do not depend on how you envision them, only on the structure the spaces have.
This has a lot of theoretical and practical applications. But I think there is a reason why this terminology is not so abundant in all mathematics, and it is due to the practical applications of geometry. For example, Gauss's result that the curvature is an intrinsic information is marvelous: it says that something that you can define using the way that a normal vector field varies (and a normal vector field clearly depends on how you put your surface in space) can be computed directly through measures which are related to the tangent space and the Riemannian metric (namely, the first fundamental form - it may not be clear how the tangent space is something intrinsic if you think about it geometrically, so I suggest you look up for one of the abstract definitions of tangent space), and therefore are intrinsic - it doesn't matter how you are "inside" space. In fact, it doesn't matter that you are "inside" space.
For instance, this has a lot of importance in general relativity (although the setup is not exactly Riemannian manifolds): you may have heard that spacetime is curved. This terminology can be quite confusing, and sometimes people try to explain the concept by analogy with how balls curve a rubber sheet etc. However, a big part of the success of the theory is precisely that we don't need that our space is curved inside anything: we don't need to ask "what is outside", and it doesn't make sense a priori (and it should not). It is "curved" in a way that we can define only by means of itself, making it measurable and not a pseudo-science concept.
Now, back to the beginning, it is perfectly understandable that the "intrinsic/extrinsic" duality (and its usefulness) is a little cloudy if you do not know the abstract definitions. If the above discussion does not clear some things up, I think it may be wise to wait for (or go for) the abstract definitions.