# ag.algebraic geometry – Is there an intrinsic Gauss map?

$$DeclareMathOperatorGL{GL}DeclareMathOperatorGr{Gr}$$This may all be well known, too vague, or stupid; my apologies.

The Gauss map is defined by embedding your $$k$$-dimensional smooth scheme/manifold $$X$$ into an $$n$$-dimensional affine space $$M = mathbb{A}^n$$ (or some other space with trivialized tangent bundle like an abelian variety) and looking at how tangent planes of $$X$$ sit inside the tangent space of $$M$$. It heavily depends on the embedding. However, so does the normal cone, and there’s an intrinsic normal cone obtained by locally choosing such an embedding, looking at the normal cone, and modding out by “choices” in the form of the action of the tangent bundle of $$M$$.

Q1 : Precisely how does the Gauss map change as you vary the embedding? What if two embeddings are related by deformation?

The Grassmannian target of the Gauss map has an obvious action of the tangent space of $$M$$, but I feel that quotienting out by that simply kills everything. Maybe that’s the right thing to do for taking the stack quotient. One still has to check the result is independent of embedding.

The way one checks that the result is independent of choices in the normal cone case is via seeing how it behaves in triangles $$X to Y to Z$$. Presumably there’s no harm in defining a relative Gauss map after choosing an embedding $$X subseteq M to Y$$ into affine space over the target, and I was wondering if there could be a relationship between the maps coming from the triangle. Initially, assume $$Z$$ is a point:

Q2 : Given $$X to Y$$, is there a relationship between their two Gauss maps?

Lastly, do you need $$M$$ to have trivialized tangent bundle? Naively yes, but if you pass to the frame bundle of $$M$$‘s tangent bundle, you get a $$GL_n$$-torsor $$F to M$$ on which the tangent bundle canonically trivializes, and one pulls back $$X$$ to $$F$$ to get a well-defined Gauss map. I.e., $$X subseteq M$$ induces a $$GL_n$$-torsor $$widetilde{X} to X$$ and a Gauss map $$widetilde{X} to Gr(k, n)$$. Isn’t this pulled back from a map $$X to (Gr(k, n)/GL_n)$$ that avoids making such a choice?

I don’t claim to be familiar with the Gauss map, just curious and naive. I was hoping there would be a geometric way to see intrinsic vs. extrinsic notions of curvature from this picture. Thanks for your patience.