How to Understand Galois Cohomology Surjectivity in Non-Quasi-Projective Vari...

Hello! This is a fascinating problem that touches on some deep connections between Galois theory and algebraic geometry. Let's work through this together!

What We're Solving:

We need to find an example of a non-quasi-projective algebraic variety X where the natural map θ: E(K/k, X) → H¹(Gal(K/k), Aut_{X_K}(K)) fails to be surjective. This will show that the classical bijection theorem for forms of varieties requires the quasi-projectivity assumption.

The Approach:

The key insight is that quasi-projectivity gives us enough "rigidity" to ensure that every cohomology class can be realized by an actual form of the variety. When we lose this property, we can have "phantom" cohomology classes that don't correspond to actual geometric objects. We'll construct an example using an infinite-dimensional variety.

Step-by-Step Solution:

Step 1: Set up the basic objects Let's work with:

• k = ℝ (real numbers)
• K = ℂ (complex numbers)
• Gal(ℂ/ℝ) = ℤ/2ℤ generated by complex conjugation σ

Step 2: Choose a non-quasi-projective variety Take X = Spec(ℂ[x₁, x₂, x₃, ...]) where we have infinitely many variables. This is:

• An infinite-dimensional affine variety over ℂ
• Definitely not quasi-projective (it's not even finite-dimensional)
• Has a natural ℝ-structure since we can define it over ℝ

Step 3: Understand the automorphism group The automorphism group Aut_{X_ℂ}(ℂ) is enormous - it includes all ℂ-algebra automorphisms of ℂ[x₁, x₂, x₃, ...]. Complex conjugation acts on this group in a complicated way.

Step 4: Construct the problematic cohomology class Consider an automorphism φ of ℂ[x₁, x₂, x₃, ...] that:

• Permutes variables in a way that's not "algebraically realizable" over ℝ
• For instance, φ(xₙ) = xₙ₊₁ for all n, and φ(x₁) = ix₁

This gives us a 1-cocycle: define c(σ) = φ where σ is complex conjugation.

Step 5: Show the cocycle is not in the image The cohomology class [c] ∈ H¹(Gal(ℂ/ℝ), Aut_{X_ℂ}(ℂ)) exists, but there's no actual ℝ-form Y of X such that Y ×_ℝ ℂ ≅ X via an isomorphism that induces this particular cocycle.

Step 6: Explain why quasi-projectivity would fix this If X were quasi-projective:

• We'd have finite-dimensional spaces of sections
• Automorphisms would be more constrained
• Every cohomology class would correspond to an actual geometric twist

The Answer:

The example is X = Spec(ℂ[x₁, x₂, x₃, ...]) over the extension ℂ/ℝ, with the cohomology class defined by the "shift plus scaling" automorphism described above. This demonstrates that without quasi-projectivity, the map θ can fail to be surjective because the infinite-dimensional nature allows for "pathological" automorphisms that can't be realized as coming from actual ℝ-forms of the variety.

Memory Tip:

Think of it this way: quasi-projective varieties are "well-behaved" because they can be embedded in projective space, giving them finite-dimensional constraint. When you remove this constraint, you get "too much freedom" in the automorphism group, leading to cohomology classes that are purely formal rather than geometric!

The intuition is: "Finite dimensions = geometric realizability, infinite dimensions = potential phantoms!"