Devoir 1. Valerio Proietti. November 5, 2014 EXERCISE 1. z f = g. supp g EXERCISE 2

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1 GÉOMÉTRIE ANALITIQUE COMPLEXE UNIVERSITÉ PARIS-SUD Devoir Valerio Proietti November 5, 04 EXERCISE Thanks to Corollary 3.4 of [, p. ] we know that /πz is a fundamental solution of the operator z on C. As a consequence, if g is a smooth function (or even a distribution) with compact support in C, then the convolution f = /πz g is a solution of the equation To see the necessary condition, observe that z f = g f, zn z = g, zn = 0 since z z n is holomorphic. It is also clear that our solution has the required regularity, since we can write f = πz g = g πz = g (w z) dλ(z) C πz and take advantage of the smoothness of g. Conversely, if the condition on the powers of z is satisfied, by means of the series expansion w z = w z w = n 0 z n w n z : z < w we can write, for all w > R, 0 = f (w) = π(w z) dλ(z) = C supp g π(w z) dλ(z) = n 0 supp g π dλ(z) where R is such that supp g D(0,R) and the switch between integral and series is due to uniform convergence of the geometric series. EXERCISE Since T is normal, we know that the coefficients of T and dt (when written in local coordinates) are bounded linear functionals on the space of continuous functions with compact support. As a consequence of Riesz-Markov representation theorem, we conclude that the

2 generic coefficient of T (which will be denoted t in the sequel) is a (complex) Radon measure. Recall that such measures are inner regular, id est µ t (E) = supµ(f ) F E, F is compact} Since our manifolds have countable basis, we know that every open set admits an exhaustion by compact sets. We can safely assume that Y is closed. These preliminaries are useful for the following construction: given a compact set F M Y, we build a bump function φ which equals on F and has compact support in M Y. Then, thanks to the hypothesis on the support of T, we have t(φ) = 0 = µ t (F ) = 0 = suppµ t Y We now choose a local chart for which Y = x = = x q = 0}. Thus for every compactly supported function f on M and j q, we can write (x j t)(f ) = x j f dµ t = x j f dµ t + x j f dµ t = 0 M x j =0} M x j =0} and conclude x j T = 0. An analogous argoment shows x j dt = 0. For a (p )-form ω, we have 0 = (x j dt )(ω) = dt (x j ω) = ( ) m p+ [T (dx j ω) + T (x j dω)] = ( ) m p+ [T (dx j ω) + (x j T )(dω)] = ( ) m p+ T (dx j ω) and for ω was arbitrarily chosen, we can infer that T is zero on all forms which have dx j in their local expression. This is equivalent to say that we can factor out dx dx q in the local expression for T. By means of partition of unity, we know that C (X ) is a fine sheaf, and since D p (X ) is a C (X )-module, it is soft. This basically means that D p (X ) D p (Y ): η η is surjective. We are now in a position to define Θ. Taken η D p (Y ), we pose Θ η = T η. This is well defined because we proved suppµ t Y, and on this set any two extensions coincide. 3 We have ω = ω I,J dx I,J I + J = p, I,..., q}, J q +,...,m} T ω = (i Θ)ω = Θ(i ω) = EXERCISE 3 0 if I T ω = Θω We have v = I =q v I d z I and we choose k minimum such that no d z i occurs in this sum for i > k. We proceed by induction on k, the case k = being solved by application of Exercise. We write v = f d z k + g so that only d z i for i,...,k } is found in the expression of f, g. By assumption, we have 0 = v = (f ) d z k + g, this clearly implies else v I z i = 0 i > k Strictly speaking, this is not needed, but it s psychologically helpful. Note that the support of a measure is the complement of the union of open sets of measure zero. This sheaf theoretical language breaks down to the fact that we can extend a smooth function on Y to a smooth function on X. 3 Another possibility would be a local definition of Θ. In some sense, this would be even easier, since the coefficients of T are measures, and we have no problems in taking the restriction (to Y ) of a measure.

3 Therefore, the functions v I are holomorphic in the variables z i, i > k. We now take a bump function φ which equals on D(0,r ) k and define the functions F I = (φ(z k )v I ) πz k where the convolution is considered only with respect to z k. We know that F I z k = v I on D(0,r ) k C, in addition F I is holomorphic in z k+,..., z m and differentiable in the other variables. Having defined F = ( ) q k I F I d z I k} we can calculate i F = 0 i > k f d z k i = k i = z i d z i Hence, v +F is -closed and does not involve any d z i for i k. This terminates the inductive step and the proof. EXERCISE 4 We begin remarking that a simple computation yields ( log z ) = 4 ( log z ) = 0 j z z j } on the set C j. The point left is where the function attains, hence u j = log z j is subharmonic. We claim u is harmonic on the set S = C 0 j j N+}, where C 0 denotes the punctured plane. To see this, choose z 0 S and set z0 m = min > 0 j N + j Note that m exist otherwise there would be a subsequence j (n) such that 0 < z 0 j (n) < n. For z D(z 0,m/), we have m m z0 z z 0 z z z 0 + z 0 m j j j + z 0 + () Due to continuity of log ( ), we conclude u j M for some M > 0 and for all j positive integers. As a consequence, u = j u j and j j j z u j = j j z j converge uniformly, and this is sufficient to interchange derivative and series. We still need to check subharmonicity in 0, but this is easy, as the partial sums of u(0) form an increasing sequence on which we can apply Beppo-Levi theorem on monotone convergence, to obtain u(0) µ S (u;0,r ) r R 3

4 because the inequality is valid for each term of the series (notation is as in [, p. 33]). Being the exponential map increasing and convex, we see that e u is also subharmonic as well as locally bounded, since () proves a local upper bound for u (valid for all complex numbers z 0 ). Lastly, using the sequence z n = n, we see eu (z n ) = 0 whereas e u (0) > 0. EXERCISE 5 If we denote the ellipsoid by Ω, then the function ρ(z) = z + z 4 + z 3 6, is the defining function of Ω, i.e., ρ < 0 on Ω and ρ = 0,dρ 0 on Ω. On the holomorphic tangent space h T Ω, the Levi form is defined by L Ω,z = ρ(z) z i,j n ρ z i z j dz i dz j Therefore, carrying out simple calculations, we find 0 0 L Ω,z = ( z + 4 z z 3 0 ) 0 4 z z 3 4 Hence, 3 if z z 0} z 3 0} rankl Ω,z = if z z = 0} z 3 = 0} if z D(0,) 0} 0} (Here, is the simmetric difference of sets). As a sidenote, applying Theorem 7. in [, p. 59], we see that Ω is pseudoconvex. Let K be the holomorphic hull of K and set EXERCISE 6 K = (z, z ) C z r, z r, z /logr z logr e} By definition, f O (C) and z K, we have f (z) sup f () K We choose three functions: the projections, and (z, z ) z /logr z /logr. Applying definition () to the former yields conditions and, whilst condition is obtained with the latter. Hence K K. For the reverse inclusion, consider U α = S α D(0,(r,r )), where S α = (z, z ) C z /logr z /logr = α} and α (0,e]. For the implicit function theorem each S α is a Riemann surface. Now observe that U α = S α D(0,(r,r )) K, simply taking the modulus on the condition defining the surface. Then the inclusion map satisfies the hypothesis of Proposition 6.6d in [, p. 47], thus we get U α K. To conclude, note that every point of K is in U α for some α: K e α =0 U α K 4

5 REFERENCES [] J.-P. D ly. Complex analytic and differential geometry. 0. URL: www- fourier. ujf-grenoble.fr/~d ly/manuscripts/agbook.pdf (cit. on pp., 4). 5