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Hodge decomposition and polarization of a point

Example

Assume AC. The one-point space is a compact Kähler manifold of complex dimension zero in the sense of Compact Kähler and rational Hodge polarization conventions. Its rational cohomology is $H^{0} (p t; Q) = Q$ with the single Hodge type $(0, 0)$ , all higher cohomology vanishes, and $Q (x, y) = x y$ is a rational polarization. The hard-Lefschetz map $L^{0}$ in degree zero is the identity.

Facts & Assumptions

Given: The Axiom of Choice and the one-point space $p t$ with the dimension-zero conventions of Compact Kähler and rational Hodge polarization conventions.

[F1]

For $n = 0$ a complex manifold is a discrete space, compactness makes it a finite set of points, there are no tangent vectors and no forms of positive degree, $ω = 0$ , and $L, Λ, \partial, \overset{ˉ}{\partial}, d, \partial^{*}, {\overset{ˉ}{\partial}}^{*}, d^{*}$ and all Laplacians are zero, with the Hodge star the identity on degree zero and $d V$ the counting measure; the same item defines rational pure Hodge structures, the Weil operator, polarizations (nondegenerate $Q$ -bilinear $Q$ with $Q (y, x) = (- 1)^{k} Q (x, y)$ , $Q (V^{p, q}, V^{a, b}) = 0$ unless $(a, b) = (q, p)$ , and $h (x, y) = Q (C x, \overset{ˉ}{y})$ positive definite Hermitian) and the Tate shift (Compact Kähler and rational Hodge polarization conventions).

[F2]

For a compact Kähler manifold $X$ , every de Rham class has a unique smooth harmonic representative and $H^{k} (X; C) = ⨁_{p + q = k} H^{p, q} (X)$ with $\overline{H^{p, q} (X)} = H^{q, p} (X)$ , where $H^{p, q} (X)$ is the set of classes with a $d$ -closed smooth representative of type $(p, q)$ (Compact Kähler Hodge decomposition from elliptic harmonic forms).

[F3]

Assume AC. For a compact smooth manifold the natural coefficient maps give $H^{k} (X; Q) \otimes_{Q} C ≅ H^{k} (X; C)$ for every degree, with all groups finite-dimensional, and the isomorphisms are natural and unital (Rational cohomology scalar comparison on compact smooth manifolds); AC is The Axiom of Choice.

Verification

1.1F1F2givenalgebra

The one-point space is a compact complex manifold of complex dimension zero, so by [F1] it has no forms of positive degree and its complex of complex-valued forms is $C$ in degree zero with zero differential; hence $H^{0} (p t; C) = C$ and $H^{k} (p t; C) = 0$ for $k > 0$ , and $ω = 0$ with all operators zero. By [F2] each degree-zero class $[λ]$ has the unique smooth harmonic representative equal to the function with value $λ$ , and $H^{0} (p t; C) = H^{0, 0} (p t)$ is purely of type $(0, 0)$ while both sides of the conjugation symmetry are this same space and all other Hodge pieces vanish.

2.1step 1.1F3givenalgebra

By [F3], whose AC hypothesis is the Given, the natural coefficient map $H^{0} (p t; Q) \otimes_{Q} C \to H^{0} (p t; C) = C$ is an isomorphism, and its unitality sends the rational unit to the class of the function $1$ ; hence $H^{0} (p t; Q) = Q$ with rational structure spanning the type- $(0, 0)$ piece, and $H^{k} (p t; Q) = 0$ for $k > 0$ because the comparison is an isomorphism in every degree and the complex cohomology vanishes there.

3.1step 1.1step 2.1F1givenalgebra∎

By [F1] $L = ω \land = 0$ and $Λ = 0$ , so $L^{0}$ is the identity operator on the one-dimensional space $H^{0} (p t; C)$ , which is the asserted degree-zero hard-Lefschetz map. For the polarization, $Q (x, y) = x y$ on $V = Q$ is $Q$ -bilinear and nondegenerate ( $Q (1, 1) = 1$ ); it satisfies $Q (y, x) = Q (x, y) = (- 1)^{0} Q (x, y)$ ; the type condition $Q (V^{p, q}, V^{a, b}) = 0$ unless $(a, b) = (q, p)$ only involves the pair $(p, q) = (a, b) = (0, 0)$ , where $(0, 0) = (q, p)$ holds, so it is satisfied; and the Weil operator of the weight-zero structure is $C = i^{0 - 0} = i d$ , so $h (x, y) = Q (C x, \overset{ˉ}{y}) = x \overset{ˉ}{y}$ is positive definite Hermitian on $V_{C} = C$ . Therefore $Q$ is a rational polarization of the rational pure Hodge structure $H^{0} (p t; Q)$ , of weight zero, as claimed.

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