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This is a quite general (and maybe very simple) question about the definition of Gaussian curvature. I have found a lot of "Gauss-Bonnet related questions" (e.g. here) but I didn't find a simple answer to my question.

In John Lee's "Introduction to Riemannian manifolds" (p.250), the Gaussian curvature $K$ of an abstract Riemannian 2-dimensional manifold is defined as $ K = \frac{1}{2} S$, where $S$ is the scalar curvature. This allows to state the Gauss-Bonnet theorem for abstract Riemannian manifolds $(M^2,g)$ without boundary "as usual", that is

$$ \int_M K dM = 2 \pi \chi(M),$$ where $\chi(M)$ is the Euler characteristic of $M$.

The other definition I know for the Gaussian curvature is an extrinsic definition, where $K$ is the determinant of the shape operator (equivalently the product of the principal curvature). This is the definition that appears on wikipedia.

The Theorema Egregium tells us that the two definition coincide when $M$ is embedded in Euclidean space. However, when $M$ is isometrically immersed in another ambiant Riemannian manifold, it seems to me that the two definitions (intrinsic and extrinsic) don't coïncide anymore, since the extrinsic one depends on the geometry of the ambiant space as well (this is exactly what it means to be extrinsic I guess).

So my question is : am I right that these definitions don't coïncide in general, and if yes, is there a reason for which we define the Gaussian curvature of abstract manifolds as half the scalar curvature, other than keeping the "historical" statement of Gauss-Bonnet on abstract manifolds unchanged ?

I find it quite confusing, because I've several times encountered the statement of Gauss-Bonnet on abstract manifolds as above only without defining the Gaussian curvature for abstract manifolds (it is the case on wikipedia for example). Also, I've been trying to prove that $ S = 2K $ for a 2-dimensional manifold immersed in a non flat ambiant space before noticing that the proof I was reading used this as definition for the gaussian curvature instead of the "extrinsic definition".

Maybe I find this confusing because I'm knew to Riemannian geometry, nevertheless it seems to me that it would be less confusing to keep the notion of "Gaussian curvature" as an extrinsically defined notion for immersed manifolds, noticing that when the ambiant space is flat it is in fact intrinsic, and stating Gauss-Bonnet for abstract manifolds instead as $$ \int_M S \, dM = 4 \pi \chi(M). $$

Geometric tortoise
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    What do you mean by the shape operator when $M$ is not a hypersurface in Euclidean space? – Ted Shifrin May 10 '25 at 18:08
  • @TedShifrin sorry, I should have said that I consider only 3-dimensional ambiant manifolds. For a choice of unit normal $\nu$, I mean by shape operator the endomorphism field $\tilde\nabla\nu$, where $\tilde\nabla$ is the connection in the ambiant space. It seems to be a good generalisation of the shape operator as the differential of the Gauss map – Geometric tortoise May 10 '25 at 18:21
  • I realize that I'm not being very precise, so just for the sake of completeness : the definition given in my previous comment of the shape operator is for a 2-dimensional Riemannian submanifold $(M,g)$ in a 3-dimensional Riemannian manifold $(\tilde M,\tilde g)$ with Levi-Civita connection $\tilde\nabla$ – Geometric tortoise May 10 '25 at 18:36
  • OK, so what is the shape operator for $M^2\subset N^3$? Does its determinant make sense? – Ted Shifrin May 10 '25 at 18:37
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    By the way, you definitely want to understand the Gauss equation, relating intrinsic, ambient, and extrinsic (sectional) curvatures. – Ted Shifrin May 10 '25 at 18:45
  • sorry for the confusion. So if $M^2 \subset (N^3,g)$ is a Riemannian submanifold, and $p\in M$, we can choose a unit normal vector field $\nu$ defined near $p $. Then noticing that $ g(\tilde\nabla_X \nu, \nu) = (1/2) X\cdot g(\nu, \nu) = 0$ for all vectors $X$ tangent to $M$, we can see the shape operator $X \mapsto \tilde\nabla_X \nu$ as an endomorphism of $T_p M$ and so I think it makes sense to speak of its determinant at $p$. – Geometric tortoise May 10 '25 at 19:04
  • Dumb question from a passerby: Is there an obstruction to embedding the ambient Riemannian manifold in another Euclidean space and working "up there"? Also, potentially relevant: https://math.stackexchange.com/questions/1255522/the-relationship-between-ricci-and-gaussian-curvatures – Eric Towers May 10 '25 at 19:14
  • OK, good. Of course, you’re assuming the normal bundle is oriented (in particular, trivial), although the determinant will not care. So the Gauss equation tells you that this determinant differs from the intrinsic Gaussian curvature by the sectional curvature of the tangent plane in $N$. (Try various two-spheres in $S^3$.) – Ted Shifrin May 10 '25 at 19:19
  • Right, thank you ! So I guess I will just stop calling this determinant of the shape operator "Gaussian curvature" in the general case, and use instead the intrinsic definition. Also, it's a bit of a detail, but I don't see why I should assume the normal bundle to be trivial in order to define the determinant : can't I just trivialise it locally to find a unit normal vector field in a small enough neighbourhood of each point, that does not need to extend globally on $M$ ? – Geometric tortoise May 10 '25 at 20:37
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    I already said you don’t need it for the determinant, but to get a globally-defined shape operator you do! (Try $\Bbb RP^2\subset\Bbb RP^3$.) – Ted Shifrin May 10 '25 at 23:07

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