### **Main Classification Theorem**

A connected Riemann surface (S, gₛ) admits a holomorphic isometric embedding into (ℂ × ℍ, gₘ) if and only if (S, gₛ) has constant Gaussian curvature Kₛ ∈ {0, -1} and S is simply connected.

Furthermore, the embedding F = (f, g) must be totally geodesic into one of the factors:
1.  If Kₛ = 0, then g(z) = D (constant) and f: (S, gₛ) → ℂ is a holomorphic isometry.
2.  If KS = -1, then f(z) = B (constant) and g: (S, gₛ) → ℍ is a holomorphic isometry.

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### **Proof Derivations**

The proof is established through a series of propositions and lemmas.

#### **1. Preliminaries: The Isometry Condition and Metric Decomposition**

Let S be a connected Riemann surface with a metric gₛ, which in local complex coordinates z can be written as gₛ = ρ(z)²|dz|². The target space is M = ℂ × ℍ, where ℍ = {w ∈ ℂ : Im(w) > 0}. The product metric on M is gₘ = |dw₁|² + |dw₂|² / (Im w₂)². We seek a holomorphic map F: S → M, locally given by F(z) = (f(z), g(z)), such that it is an isometric embedding, i.e., F*gₘ = gₛ.

**Proposition 2.1:** A holomorphic map F = (f, g): (S, ρ²|dz|²) → (ℂ × ℍ, gₘ) is an isometric immersion if and only if
ρ(z)² = |f'(z)|² + |g'(z)|² / (Im g(z))²

* **Proof:**
    We compute the pullback metric F*gₘ. Since f and g are holomorphic, the differentials are dw₁ = f'(z)dz and dw₂ = g'(z)dz. Substituting these into the expression for gₘ:
    F*gₘ = |f'(z)dz|² + |g'(z)dz|² / (Im g(z))²
    F*gₘ = ( |f'(z)|² + |g'(z)|² / (Im g(z))² ) |dz|²

    The isometry condition is F*gₘ = gₛ. Equating our result with gₛ = ρ(z)²|dz|² yields the condition:
    ρ(z)² = |f'(z)|² + |g'(z)|² / (Im g(z))²
    Since ρ(z) > 0, the map is an immersion. ▪

This condition provides a decomposition of the metric gₛ into two parts: gₛ = g₁ + g₂, where:
* g₁ = |f'(z)|²|dz|² is the pullback of the Euclidean metric on ℂ.
* g₂ = (|g'(z)|² / (Im g(z))²)|dz|² is the pullback of the Poincaré metric on ℍ.

Since S is a Riemann surface, gₛ is a Kähler metric. The components g₁ and g₂ are positive semi-definite (1,1)-forms, and in complex dimension one, they are automatically closed, making them Kähler semi-metrics.

#### **2. Parallel Decomposition (Calabi's Lemma)**

The proof relies on a fundamental result from Calabi regarding such decompositions.

**Lemma 2.3:** Let (S, g) be a connected Riemann surface. If g₁ is a Kähler semi-metric on S such that ∇g₁ = 0 (where ∇ is the Levi-Civita connection of g), then g₁ = Cg for some constant C ≥ 0.

* **Proof:**
    1.  Define a (1,1)-tensor field P, which is an endomorphism of the tangent bundle TS, by the relation g₁(X, Y) = g(PX, Y) for all vector fields X, Y.
    2.  The condition that g₁ is parallel with respect to the Levi-Civita connection of g is ∇g₁ = 0. Since ∇g = 0 by definition of the Levi-Civita connection, the product rule for connections implies ∇P = 0. Thus, P is a parallel endomorphism field.
    3.  On a connected manifold, any parallel tensor field must have constant eigenvalues.
    4.  Since (S, g) is a Kähler manifold, the connection ∇ preserves the complex structure J. As P is parallel (∇P = 0), it must commute with J. This means P is complex-linear.
    5.  The complex dimension of S is 1. A complex-linear endomorphism on a 1-dimensional complex vector space must act as multiplication by a complex scalar, C(z).
    6.  Furthermore, the metric g₁ is Hermitian, which implies the endomorphism P is self-adjoint with respect to g. A self-adjoint endomorphism has real eigenvalues, so the scalar C(z) must be a real-valued function.
    7.  The condition ∇P = 0 implies that the function C(z) must be constant on the connected manifold S. Let C(z) = C.
    8.  Since g₁ is a positive semi-definite metric, the constant C must be non-negative, C ≥ 0.
    9.  Therefore, g₁ = Cg. ▪

#### **3. Proof of the Main Theorem**

**Step 1: Rigidity of the Decomposition**

Let F: S → ℂ × ℍ be a holomorphic isometric immersion. The induced decomposition gₛ = g₁ + g₂ consists of Kähler semi-metrics. By Calabi's work (Lemma 2.2 in the source), this decomposition must be parallel, meaning ∇g₁ = 0 and ∇g₂ = 0.

Applying Lemma 2.3 to our connected Riemann surface S, we conclude there exist constants C₁, C₂ ≥ 0 such that:
g₁ = C₁gₛ
g₂ = C₂gₛ

Summing these gives g₁ + g₂ = (C₁ + C₂)gₛ. Since gₛ = g₁ + g₂, we have gₛ = (C₁ + C₂)gₛ. As gₛ is a non-degenerate metric, it follows that C₁ + C₂ = 1.

**Proposition 3.1:** The decomposition must be trivial. That is, either (C₁=1, C₂=0) or (C₁=0, C₂=1).

* **Proof:**
    Assume for the sake of contradiction that the decomposition is non-trivial, meaning C₁ > 0 and C₂ > 0.
    1.  If C₁ > 0, then g₁ = C₁gₛ is a genuine metric (not just semi-metric). The map f: S → ℂ induces this metric. As a map between Riemann surfaces, f is a local isometry between (S, g₁) and (ℂ, g_ℂ). The Gaussian curvature K₁ of g₁ is related to the curvature Kₛ of gₛ by K₁ = Kₛ / C₁. Since a local isometry preserves curvature, K₁ must equal the curvature of ℂ, which is 0. Thus, Kₛ / C₁ = 0, which implies Kₛ = 0.
    2.  If C₂ > 0, then g₂ = C₂gₛ is also a genuine metric. The map g: S → ℍ is a local isometry between (S, g₂) and (ℍ, g_ℍ). The Gaussian curvature K₂ of g₂ is related to Kₛ by K₂ = Kₛ / C₂. A local isometry requires K₂ to equal the curvature of ℍ, which is -1. Thus, Kₛ / C₂ = -1, which implies Kₛ = -C₂.
    3.  Combining these two results, we have Kₛ = 0 and Kₛ = -C₂. This implies C₂ = 0, which contradicts our initial assumption that C₂ > 0.
    4.  The contradiction forces us to conclude that the assumption was false. The decomposition cannot be non-trivial. Therefore, one of the constants must be zero. ▪

**Step 2: Geometric Consequences**

We analyze the two cases allowed by Proposition 3.1.

* **Case 1: C₁ = 1 and C₂ = 0.**
    The condition C₂ = 0 implies g₂ = 0. From the definition of g₂, this means |g'(z)|² / (Im g(z))² = 0, which requires g'(z) = 0. Since S is connected, g(z) must be a constant, g(z) = D ∈ ℍ.
    The condition C₁ = 1 implies g₁ = gₛ, so |f'(z)|²|dz|² = ρ(z)²|dz|². This means f: (S, gₛ) → ℂ is a holomorphic local isometry. For such an isometry to exist, the Gaussian curvatures must match: Kₛ = K_ℂ = 0.

* **Case 2: C₁ = 0 and C₂ = 1.**
    The condition C₁ = 0 implies g₁ = 0. This means |f'(z)|² = 0, which requires f'(z) = 0. Since S is connected, f(z) must be a constant, f(z) = B ∈ ℂ.
    The condition C₂ = 1 implies g₂ = gₛ, so (|g'(z)|² / (Im g(z))²)|dz|² = ρ(z)²|dz|². This means g: (S, gₛ) → ℍ is a holomorphic local isometry. This requires Kₛ = K_ℍ = -1.

In both cases, the embedding is totally geodesic, as it maps entirely into one factor while being constant in the other.

**Step 3: Topological Constraints**

**Proposition 3.2:** If a connected Riemann surface (S, gₛ) admits a holomorphic isometric immersion into ℂ × ℍ, then S must be simply connected.

* **Proof:**
    Let π: S̃ → S be the universal covering map, and let Γ be the group of deck transformations. The metric gₛ lifts to a metric g̃ₛ = π*gₛ on S̃. The immersion F: S → M lifts to an immersion F̃ = F ∘ π: S̃ → M.
    For F to be well-defined on S (i.e., descending from the cover S̃), the lifted map F̃ must be invariant under the action of Γ. That is, for any deck transformation γ ∈ Γ and any point z ∈ S̃:
    F̃(γz) = F̃(z)

    We analyze this condition for the two cases:
    * **Case 1: Kₛ = 0.**
        The universal cover (S̃, g̃ₛ) is a flat, simply connected Riemann surface, so it is globally isometric to the Euclidean plane (ℂ, c²|dz|²) for some c > 0. We can identify S̃ with ℂ. The deck transformations γ ∈ Γ are holomorphic isometries of ℂ, so they have the form γ(z) = az + b with |a| = 1.
        The lifted immersion is F̃(z) = (f̃(z), D), where f̃: ℂ → ℂ is a local isometry. This implies |f̃'(z)| = c, so f̃ must be of the form f̃(z) = Az + B with |A| = c.
        The invariance condition becomes:
        (A(az + b) + B, D) = (Az + B, D)
        A(az + b) + B = Az + B
        Aaz + Ab = Az
        This must hold for all z ∈ ℂ. This implies A(a-1) = 0 and Ab = 0. Since A ≠ 0 (because c > 0), we must have a = 1 and b = 0. This means γ is the identity transformation. As this holds for all γ ∈ Γ, the group Γ must be trivial. If the deck transformation group is trivial, S is its own universal cover, so S must be simply connected.

    * **Case 2: Kₛ = -1.**
        The universal cover (S̃, g̃S) is a simply connected Riemann surface with constant curvature -1, so it is globally isometric to the hyperbolic plane (ℍ, c²ds\_ℍ²). We identify S̃ with ℍ. The deck transformations γ ∈ Γ are elements of Aut(ℍ).
        The lifted immersion is F̃(z) = (B, g̃(z)), where g̃: ℍ → ℍ is a local isometry. Since ℍ is simply connected, any local isometry is a global isometry, so g̃ ∈ Aut(ℍ) (up to scaling).
        The invariance condition becomes:
        (B, g̃(γz)) = (B, g̃(z))
        g̃(γz) = g̃(z)
        Since g̃ is an isometry, it is injective. Therefore, the above equality implies γz = z for all z ∈ ℍ. This means γ must be the identity transformation. The group Γ is trivial, and thus S is simply connected.

This completes the proof of the main theorem. The necessity of the conditions has been shown. The sufficiency is clear: if S is simply connected with Kₛ = 0, it is isometric to ℂ, which can be isometrically embedded as F(z) = (z, D). If S is simply connected with Kₛ = -1, it is isometric to ℍ, which can be isometrically embedded as F(z) = (B, z). ▪

#### **4. Application to the Euclidean Plane (Corollary 1.2)**

The holomorphic isometric embeddings F: (ℂ, |dz|²) → ℂ × ℍ are precisely the maps of the form F(z) = (Az + B, D), where A, B ∈ ℂ, D ∈ ℍ, and |A| = 1.

* **Proof:**
    We apply the main theorem to the case (S, gₛ) = (ℂ, |dz|²).
    1.  The source manifold S = ℂ is simply connected.
    2.  The metric gₛ = |dz|² is flat, so its Gaussian curvature is Kₛ = 0.
    3.  According to the main theorem, for Kₛ = 0, the embedding must be of the form F(z) = (f(z), D), where D ∈ ℍ is a constant.
    4.  Furthermore, the map f: (ℂ, |dz|²) → (ℂ, |dw₁|²) must be a holomorphic isometry.
    5.  The holomorphic isometries of the complex plane ℂ are precisely the affine maps f(z) = Az + B, where the isometry condition requires |A| = 1.
    6.  Combining these facts, the embedding must be of the form F(z) = (Az + B, D). ▪

References
[1] E. Calabi. "Isometric imbedding of complex manifolds". Annals of Mathematics, 58 (1): 123,
1953.
[2] P. Griffiths and J. Harris. Principles of Algebraic Geometry. Wiley-Interscience, 1978.
[3] S. Kobayashi and K. Nomizu. Foundations of Differential Geometry, Vol. I & II. Wiley-
Interscience, 1963, 1969.