© Hindawi Publishing Corp. ON THE MODULI SPACE OF SUPERMINIMAL SURFACES IN SPHERES

Using a birational correspondence between the twistor space of S2n and projective space, we describe, up to birational equivalence, the moduli space of superminimal surfaces in S2n of degree d as curves of degree d in projective space satisfying a certain differential system. Using this approach, we show that the moduli space of linearly full maps is nonempty for sufficiently large degree and we show that the dimension of this moduli space for n=3 and genus 0 is greater than or equal to 2d


Introduction.
A superminimal surface in S 2n can be defined as a harmonic map f : M g → S 2n , where M g is a Riemann surface of genus g, whose osculating plane (i.e., the subspace spanned by all the derivatives of f ) is totally isotropic (i.e., the symmetric complex product of any two vectors in that subspace is zero).These surfaces admit holomorphic lifts into the twistor space ᐆ n = SO(2n+1)/ SO(n) of the 2n-sphere (see [1,6,12]).When f is linearly full (i.e., not contained in a proper subsphere of S 2n ), Φ is given by the osculating plane, at each point, of the map f .The study of these harmonic maps becomes then translated into the study of their twistor lifts, where more powerful tools in the complex analytic category can be used.
For the particular case of maps into S 4 (this is n = 2), one has the added advantage that ᐆ 2 P 3 , so one can characterize twistor lifts of superminimal maps as solutions in P 3 of a simple differential equation.In [11], Loo found that the dimension of the moduli space of harmonic spheres of degree d (this is of area 4πd) in S 4 is 2d+4, and in [13,14,15], Verdier described the structure of this moduli space in detail.
For higher dimensions, the problem becomes harder simply because ᐆ n is a more complicated object.In [5], Bryant uses homogeneous spaces to obtain a differential system similar to (2.13), and in [9], Hano finds the same system using Clifford algebras.Unfortunately their methods do not help in the study of the moduli space since the degree of the maps is not under control.
In this paper, we show that it is possible to use the same strategy for higher dimensions as in S 4 , namely to translate the problem of studying the moduli space of superminimal surfaces of a certain degree into the study of maps into projective space satisfying the differential system (2.13).Although ᐆ n and P Nn (where N n is the dimension of ᐆ n ) are of course not isomorphic, they are birationally equivalent.We exploit this birational correspondence to induce a birational equivalence between the moduli space of twistor lifts of superminimal maps from M g to S 2n of degree d and the moduli space of maps from M g to P Nn of degree d that satisfy the differential system (2.13) (Theorem 3.1).
Using this result, we show in Theorem 4.1 that there exist harmonic maps from S 2 to S 2n which are linearly full (i.e., not contained in any proper subsphere of S 2n ) and have arbitrarily large area.In other words, the subset of the aforementioned moduli space consisting of linearly full maps is not empty for degree sufficiently large.Previous existence proofs are much weaker; in [1], explicit minimal maps S 2 → S 2n of arbitrarily large area are constructed, but they only work for genus 0 and in [5,9], there is no mention about the area, only the ramification degree, and the area is the main invariant of these maps.
In Section 5, we give a simple, direct, and geometric proof of the connectedness of the moduli space of superminimal surfaces in S 2n with degree d.We believe that the same method can be used to find other topological properties of the moduli space such as its fundamental group for arbitrary genus, generalizing the results in [7].
Finally, it is an interesting open problem to find the dimension of the set of harmonic spheres in S 2n .The main conjecture is that this dimension is 2d + n 2 .The methods described in the paper reduce this problem to finding the dimension of a variety formed by the intersection of a large number of quadrics.In Section 6, we show that the dimension of the moduli space of harmonic maps from S 2 to S 6 (i.e., n = 3) is greater than or equal to 2d + 9 (Theorem 6.2).We believe that this result can be extended to arbitrary n using the same method.
and let U E denote the Zariski open subset of ᐆ n consisting of those n-planes whose orthogonal projection onto E is one to one and onto.Every n-plane in U E can be written as the plane generated, in the basis (E 0 ,E,E), by the rows of a matrix of the form where α i and τ jk are complex numbers satisfying τ jk = −τ kj .For ease of notation, let and let where the superscript t on the left indicates the transpose of the matrix.With this notation, the matrix (2.2) can be written as Since all the entries in the last matrix are holomorphic functions of α, T , we have the following lemma.A superminimal map f : M g → S 2n admits a holomorphic lift Φ : M g → ᐆ n which is horizontal for the fibration (see [1,6,12]) (2.7) Conversely, given a holomorphic, horizontal map Φ : M g → ᐆ n , the map f = Φ • π is superminimal.An explicit formulation of horizontality is as follows: regarding a vector v ∈ T P ᐆ n as an element of Hom(P , P ⊥ ), v is horizontal if and only if v(P ) ⊥ P (see [12] for details), or equivalently, (2.8) Given a holomorphic and horizontal map Φ : M g → ᐆ n , choose a base plane E so that the image of Φ is almost everywhere contained in U E .Taking a local trivialization on M g , we can write Φ as (α, I n ,(−T − A)), where α i and τ jk are holomorphic functions defined locally on M g and A = α t α/2 as before.Then dΦ is a Hom(P , P ⊥ )-valued holomorphic form taking a vector t x • (α, I n ,(−T − A)) ∈ P to the projection on P ⊥ of t x •(dα, 0, −dA−dT ).Thus, for all x, y ∈ C n , (2.9) Now, since P is totally isotropic, .11)In matrix notation this takes the form Note that this equation is independent of the local trivialization of M g chosen.
Computing dA and simplifying, we obtain the following proposition.
Proposition 2.2.Let f : M g S 2n be a superminimal map.Let Φ be the twistor lift of f .Choose a base plane E so that the image of Φ lies in U E except for finitely many points of M g .Use the parametrization given in Lemma 2.1.Then, except at the finite number of points of M g where they are not defined, the functions α i , τ jk satisfy the equations α j dα i − α i dα j = 2dτ ij (2.13) or, in matrix notation, In addition, f is linearly full if and only if W (α ) ≡ 0, where W (α ) denotes the Wronskian of the functions {α 1 ,α 2 ,...,α n }.
Proof.We only need to prove the last statement.The map f is not linearly full if and only if there is an isotropic constant vector v contained in the plane Φ(p) at every point p ∈ M g (i.e., a vector perpendicular to the image of f ).Using the parametrization of Lemma 2.1, this occurs if and only if there exists a nonzero If there is such a vector b, differentiating and using d(−T −A) = − t αdα, we obtain This implies t b = 0 and t bα = 0. Therefore, W (α ) ≡ 0. Conversely, if W (α ) ≡ 0, there must be a constant vector b such that t bα = 0, which implies The previous lemma gives a local characterization of lifts of harmonic maps f : M g → S 2n .In order to study the moduli space of such maps, we need to be able to characterize them globally.To this end, we extend the parametrization given in Lemma 2.1 to a birational map b n : P Nn → ᐆ n given by (2.16) (We omit the dependence of b n on E to simplify the notation.)Then, given a holomorphic map Φ : M g → ᐆ n , the composition b −1 n • Φ will be a birational map from M g into P Nn .This map can be completed to obtain a bona fide map Φ : M g → P Nn .In addition, it will be convenient for future computations to embed ᐆ n inside P Mn via the Plücker embedding (here M n = 2n+1 n − 1).We obtain Figure 2.1.
In terms of the homogeneous coordinates [s : Proof.Recall that U E is the subset of ᐆ n consisting of planes whose orthogonal projection over E is one to one and onto.Let I E be the complement of U E .Then I E is a subvariety of ᐆ n , of codimension at least one, consisting of those planes having nonempty incidence with E.
For each choice of a base plane E, the map b −1 n is regular on U E , so for every Φ : M g → ᐆ n intersecting U E we have the claimed correspondence.Thus, we only have to show that, given a plane F ∈ ᐆ n , no superminimal linearly full surface falls completely in I F .This is given by the following lemma.
Lemma 2.4.Let f : M g S 2n be a superminimal surface and let Φ : M g → ᐆ n be its twistor lift.If, for all p ∈ M g , Φ(p) has nonempty incidence with a fixed plane F ∈ ᐆ n , then f is not linearly full.Consequently, for any choice of base plane E, Φ lies in U E except for finitely many points.
Proof.We proceed by contradiction.Suppose that Φ(M g ) ⊂ I F .Let p be a point of M g and choose E ∈ ᐆ n so that E is not in I E ∪ I Φ(p) ; then E has empty incidence with F and Φ(p).This guarantees that the curve Φ lies in U E except for finitely many points, and so does the plane F .
Take the parametrization of ᐆ n given in Lemma 2.1 with base plane E. Since F ∈ U E , there must be constants α 0 , T 0 so that F can be represented as in (2.5).On the other hand, taking a local trivialization (V , z) of M g , we can represent Φ by functions α, T satisfying (2.14).
Since Φ and F have nonempty incidence, there is a locally defined function (2.20) Taking derivatives in the last equation and using −T −A = −α t α, we have This implies t vα = − t v (α − α 0 ).Substituting in the last component of the last equation, and reordering, we get Multiply the last equation by v from the right.Since T and (2.23) Hence, we must have t v α = t v α 0 .Substituting in (2.21), we have and substituting in (2.20), we have That is, v satisfies the same equation as v. Proceeding by induction, we will obtain t v (k) α = 0 for all k ≥ 0. Differentiating the last equation and substituting terms, we have t v (k) α = 0 for all k ≥ 0. Proceeding again by induction, we obtain (2.26) But this contradicts the condition W (α ) ≡ 0. Thus, the map f cannot be linearly full.This completes the proof of Theorem 2.3.

A Gl(n)-action
n ] be a curve in P Nn and let G be an n × n matrix.We will denote by G Φ the curve given by In other words, using α and β to denote the column vector of the α i and β i , respectively, and T and S to denote the matrix of the τ jk and σ jk , respectively, we have β = Gα and S = GT t G.
Notice that if Φ satisfies (2.13), then so does G Φ.This set of transformations forms a proper subgroup of the group of linear transformations in P Nn preserving the differential system.The interest of this Gl(n)-action is that it is equivariant through the birational map b n as we will see.
The twistor space ᐆ n can be viewed as the homogeneous space SO(2n + 1, C)/K, where K is isomorphic to the semidirect product of a nilpotent normal subgroup L and a subgroup L 1 isomorphic to Gl(n, C) (see [9] for details).The subgroup L 1 corresponds to (2.28) Lemma 2.6.The action of Gl(n, C) in P Nn defined above corresponds, via the birational map b n , to the right Gl(n, C)-action on SO(2n + 1, C)/K (considered as left cosets).
Proof.We can represent an element of ᐆ n as the equivalence class in SO(2n + 1, C)/K of a matrix (see [9]) where the stars denote suitable quantities.The (right) action on this element by H ∈ Gl(n, C) is given by modulo multiplication by H ∈ Gl(n, C) on the left.Thus the action induced on ᐆ n corresponds to the action on P Nn defined above.
Remark 2.7.This Gl(n, C)-action is the restriction to ᐆ n of the linear action of L 1 ⊂ Gl(2n+1, C) on P(Λ n (C 2n+1 )) P Mn via the exterior representation of Gl(2n + 1, C).Therefore, the degree of any curve in any of these spaces is preserved by the action.
Remark 2.8.Notice that the planes E and E are invariant under this Gl(n, C)-action.In particular, the dimension of the intersection of E with any other plane is also invariant under the action.

The moduli spaces.
Recall that, given a branched superminimal immersion f : M g → S 2n , the degree of f can be defined as the degree of its twistor lift into ᐆ or as Areaf (M g )/4π (see [1]).The degree is the fundamental invariant of these maps.We will use the following notation: (i) Supm m d (M g ,S 2n ): moduli space of superminimal surfaces in S 2n of degree d and genus g, lying in a ) consisting of linearly full surfaces.Notice that, by definition, The corresponding objects for ᐆ n are (i) HH m d (M g , ᐆ n ): moduli space of holomorphic and horizontal maps Φ : The relationship between these objects is as follows (see [7] for details).Given f ∈ Supm f d (M g ,S 2n ), we can uniquely construct its twistor lift.We have a map Supm which satisfies π • Φ = ±f , where π is the projection defined in (2.7).It is important to notice that both f and −f have the same twistor lift, so that the map is not surjective.We can then consider the maps We denote by Supm + d (M g ,S 2n ) and Supm − d (M g ,S 2n ) the images of Π + and Π − in Supm n d (M g ,S 2n ), respectively.We have (see [7]) Now we define the corresponding objects in P Nn .(i) PD m d (M g , P Nn ): moduli space of holomorphic maps Φ : M g → P Nn of degree d satisfying the differential system (2.13) and lying in an N msubspace of P Nn .(ii) PD f d (M g , P Nn ): subset of PD n d (M g , P Nn ) of maps satisfying (2.18).The birational map b n : P Nn → ᐆ n (which depends on the base plane chosen) induces maps If the spaces have pure dimension, then, for any E ∈ ᐆ n , Ꮿ E is a birational correspondence between the irreducible components of the moduli spaces, with inverse Ꮾ E .
Proof.We need several technical results.We give here a summary of what we are going to do.
(A) Find conditions on Φ that guarantee that deg( This will imply that, for some E, the map n is injective, it gives the claimed correspondence between the irreducible components. (A) Conditions on Φ.Consider the situation of Figure 2.1.After completing the curve ı• Φ, defined at first only at the points where ı is a morphism, we must have ı • Φ = i • Φ.Since the degree of the Plücker embedding ᐆ n → P Mn is 2, we want to find under what conditions the degree of ı • Φ is equal to twice the degree of Φ.
The components in P Mn of the map ı : P Nn → P Mn are defined explicitly as the determinants of the n × n minors of the matrix Let H be the hyperplane in P Mn given by the vanishing of the component corresponding to the determinant of the minor I n in (3.9) and let H be the hyperplane in P Nn given by the vanishing of the first component (the one corresponding to s).Let k i=1 m i p i be the pullback under the map Φ of the hyperplane divisor corresponding to H and let k i=1 m i p i be the pullback under the map ı • Φ of the hyperplane divisor corresponding to H .Note that the points where the divisor is supported are the same in both cases since, after taking local charts and homogenizing in P Mn , the component corresponding to the determinant of the minor I n will have the form s , for some .Proof.We have to show that, for any p ∈ {p 1 ,p 2 ,...,p k }, the map ı • Φ has as a denominator at most s 2 in each of its components.So the question becomes local, and we can do the proof about a general p.
Let p ∈ {p 1 ,...,p k } and let m be its corresponding multiplicity.Let (U, z) be a local trivialization about p ∈ M g so that we can write where we assume that s, α, and T have no common zeroes.The equation dα t α − αd t α = 2(sdT − T ds) can be written as (3.12) (3.14) Equation (3.13) implies that there exist numbers λ i such that Now we look at (3.14).Observe that, for m−1 ≤ n ≤ 2m−2, either r ≤ m−1 or n + 1 − r ≤ m − 1.Thus, using (3.15) and changing summation variables, we obtain (3.17) Then T can be written as where T 1 is a holomorphic matrix-valued function.Now, using (3.15), α can be written as where Λ = m−1 i=0 λ i z i and A 1 = ∞ i=m a i z i are holomorphic Cand C n -valued functions, respectively.
Then the matrix (α/s, I n , −T /s − α t α/2s 2 ) can be written as (3.20) Using these formulas, one can do column operations in each of the minors to reduce it either to a matrix with s 2 in the denominator of only one column or to a matrix with s in the denominator of only two columns.We omit the details.
To prove statement (b) of the proposition, notice that, for arbitrary Φ, some of the minors of the matrix (3.9) have the form α 2 i /s 2 and others have the form τ 2 jk /s 2 , so at least one of the minors will have an s 2 in the denominator.On the other hand, the denominator of the highest possible degree can be roughly estimated as s 2n .This proves statement (b).
(B) Conditions on Φ. Recall that I E is the set of all planes in ᐆ n having nontrivial incidence with E. Proposition 3.5.Let Φ : M g → ᐆ n be the twistor lift of a superminimal curve f : M g S 2n .Let E be a base plane and let {p 1 ,...,p k } ∈ M g be the set of points in M g such that Φ(p i ) ∈ I E .If the dimension of the incidence between E and Φ(p i ) (considered as n-planes in C 2n+1 ) is 1 for every point p i , then the hypotheses of Proposition 3.3 are satisfied.In particular, the degrees of Φ and Φ are equal.
Proof.Take a point p ∈ {p 1 ,...,p k }.Then Φ(p) has the form Using the Gl(n, C) action we can assume, without loss of generality, that Φ(p) = E 1 ∧ F 2 ∧•••∧F n (note that by Lemma 2.6 and Remarks 2.7 and 2.8, this action preserves both the degree of the curve and the dimension of the incidence of the curve with E and E).The vectors F j ∈ E, 2 ≤ j ≤ n, can be written in the form This implies that the component of ı • Φ, in homogeneous coordinates of P Mn , corresponding to E 1 ∧ E 2 ∧•••∧E n , is 1 after homogenizing in P Mn .Now, when we represent the curve in terms of s, α i , τ jk about p, this component corresponds to the minor formed by columns 3 through n + 2 of the matrix which is −α 2 1 /2s 2 .Hence, denoting the homogenizing factor in P Mn by h, we must have −h • α 2  1 /2s 2 = 0 at p.However, the homogenizing factor h has to have at least the degree of s 2 .This implies that α(p) = 0. Since p was arbitrary, we must have that the curve Φ satisfies the hypotheses of Proposition 3. Hence, to complete the proof of Theorem 3.1, we only need to prove that, given Φ ∈ HH f d (M g , ᐆ n ), we can find a base plane E so that Φ and E satisfy the hypotheses of Proposition 3.5.Then the map Ꮿ E will be regular and injective in a Zariski open subset of the irreducible component where Φ belongs, and this gives the desired birational correspondence between the irreducible components.
Proposition 3.6.Let Φ : M g → ᐆ n be the twistor lift of a linearly full superminimal surface f : M g S 2n .Then there is a plane F in C 2n+1 so that the incidence between Φ and F has dimension less than two at all points.
The proof will be immediate after the following lemmas.Lemma 3.7.Let F be an isotropic n-plane in C 2n+1 , that is, an element of ᐆ n .Let J F be the subset of ᐆ n consisting of those planes whose incidence with F is at least 2-dimensional.Then J F is a subvariety of ᐆ n of codimension at least 2.
Proof.Take a unitary basis in C 2n+1 so that F = E 1 ∧ ••• ∧ E n and take the parametrization given in Lemma 2.1 using this base plane.Then the intersection of J F with the Zariski open subset U F ⊆ ᐆ n is given, in terms of the parametrization, by (3.26) The condition rank(α | T ) ≤ n−2 is clearly algebraic and it involves at least two independent algebraic conditions on the α i , τ jk (namely that all the (n − 1) × (n − 1) minors are 0).Therefore, J F is a subvariety of ᐆ n of codimension at least 2. (3.27) Then ᐃ is a subvariety of ᐆ n of codimension at least 1.
Proof.Consider the set ᐂ ⊆ M g × ᐆ n given by ᐂ = (p, P ) : the incidence between Φ(p) and P has dimension at least 2 (3.28) and let π i denote projection over the ith factor.Since the incidence condition defining ᐂ is algebraic (essentially it can be seen as the vanishing of the determinant of some matrices), ᐂ must be a projective variety.In view of the previous lemma, the dimension of the fiber of the projection π 1 over M g is at most dim(ᐆ n ) − 2. Therefore, the dimension of ᐂ is at most dim(ᐆ n ) − 1 (also note that, locally, ᐂ is basically a product with some identifications accounting for the planes whose incidence with the curve has dimension greater than 1 at two different points of the curve).Now, ᐃ is the projection of ᐂ over ᐆ n , so ᐃ is a projective subvariety of ᐆ n .To compute its dimension, note that the fiber of the projection π 2 over ᐆ n is a discrete set since each element of ᐆ n , viewed as a plane, has nonempty incidence with the curve at only finitely many points (see the proof of Lemma 2.4).Therefore the dimension of ᐃ is at most dim(ᐆ n )−1.In other words, its codimension is at least 1, as desired.
Proof of Proposition 3.6.The previous lemma and the fact that U E is a Zariski open subset in P Nn show that the set of planes whose incidence with Φ has dimension 1 at all points of intersection is actually a dense subset of ᐆ n .

End of the proof of Theorem
Use Proposition 3.6 to find F having incidence of dimension at most 1 with Φ.Then E = F and Φ satisfy the hypotheses of Proposition 3.5.This implies that these hypotheses are also satisfied in a Zariski open set containing Φ. Therefore the map Ꮿ E (Φ) is regular in a Zariski open subset of the irreducible component where Φ lies.Finally, Ꮿ is injective, so it is a birational map with inverse Ꮾ E .
Note that if we fix the base plane E beforehand, Lemmas 3.7 and 3.8 guarantee that there will always be a Φ ∈ HH f d (M g , ᐆ n ) lying in the regular part of Ꮿ E .Since this regular part is open, Ꮿ E will be birational between the components of top dimension.

Existence of superminimal surfaces of arbitrary large area.
All the results of the previous sections would be quite useless unless we knew that the aforementioned moduli spaces are not empty.In this section, we show that there are superminimal surfaces of arbitrary high degree.There are previous existence proofs; in [1], there is an explicit construction of full minimal maps from S 2 → S 2n with arbitrarily large area, but this method does not work for general genus; in [5,9], they consider general genus, but they only mention the ramification degree of the curve and not the area.The methods we use are similar to the ones used in [3,4,5,9], but we have the additional fact that we can control the degree.Theorem 4.1.Let M g be a Riemann surface of genus g.Then, for any n, there exist linearly full superminimal maps f : M g → S 2n with arbitrarily large area.
Proof.In view of Proposition 3.3, we have Areaf (M g )/4π = deg(Φ) ≥ deg( Φ), so we need to show that there exist maps M g → P Nn with arbitrarily large degree that are integrals of the differential system (2.13) and that satisfy the conditions for linear fullness.
After homogenizing in P Nn , the problem reduces to finding a meromorphic map of arbitrarily large degree, and satisfying and the additional condition that guarantees linear fullness.
For convenience, we define σ ij = α i α j + 2τ ij , i < j.Then we have In other words, if we find meromorphic functions α i , σ jk on M g satisfying then, defining τ ij = (σ ij − α i α j )/2, we will obtain a map Φ as specified above.
The proof is a variation of the argument in [4].The idea is to make use of the Riemann-Roch theorem to ensure that we can choose meromorphic functions on M g from a large enough vector space of functions.
Let p 1 ,p 2 ,...,p n be n distinct points on M g .Choose α 1 meromorphic on M g such that where (g) denotes the divisor of a meromorphic function g, N 1 is a positive integer, and D = N 1 i=1 q i is an effective divisor with degree greater than or equal to 2 (so N 1 ≥ 2).
Consider the divisor N 2 p 2 − D, where N 2 is a positive integer.By Riemann-Roch, where C(D, g) is a constant depending only on the divisor D and the genus g.
Let {γ 1 ,γ 2 ,...,γ 2g } be a basis for the first homology of M g .A necessary and sufficient condition for a form ω to be exact is that it has no residues and that Consider the linear map Since the dimension of ᏸ(N 2 p 2 −D) is greater than or equal to N 2 −C(D, g), by taking N 2 large enough we can guarantee that Q 2 has a large kernel and that f dα 1 does not have residues at p 1 .Also, taking N 2 even larger if needed, we can also guarantee that the quantities f (r 1 ), f (r 2 ),...,f (r N 1 ), where {r 1 ,r 2 ,..., r N 1 } = α −1 1 (a) with a some point in the image of α 1 , are all distinct for some element f in the kernel of Q 2 (this condition will not be used until the end of the proof).
Choose α 2 = f .Since α 2 is in the kernel of Q 2 , the differential α 2 dα 1 must be exact.Since α 2 dα 1 has only two poles (p 1 and p 2 ) and it has no residues at p 1 , it cannot have residues at all.Therefore, it can be integrated globally.Finally, define σ 12 to be one such integral, that is, it satisfies dσ 12 = α 2 dα 1 .
Proceeding by induction, once α i , 1 ≤ i ≤ m, and σ jk , 1 ≤ j < k ≤ m, have been chosen, consider the divisor N m+1 p m+1 − D, where N m is a positive integer.By Riemann-Roch, we have where C(D, g) denotes a constant depending only on D and g.Consider the linear map Choosing N m+1 large enough, we can guarantee that Q m+1 has nontrivial kernel.We can then choose α m+1 in the kernel of Q m+1 so that each form α m+1 dα i (i = 1, 2,...,m) has no residues at p i .
Since each of the forms α m+1 dα 1 ,...,α m+1 dα m has poles only at two points and it does not have residues at one of them, they cannot have residues at all.Thus they can be integrated to yield σ 1m+1 ,...,σ mm+1 , respectively.
Therefore, by induction we can find a solution Φ of the differential system (2.13).Its degree is at least the polar degree of α 1 , which is N 1 , which was chosen arbitrarily.Thus, the degree of Φ can be made arbitrarily large.
Also, α 2 takes different values over the points {r 1 ,r 2 ,...,r N 1 } = α −1 1 (a), with a some point in the image of α 1 , so Φ is generically one to one and it is not a multiple cover of its image.
The only condition that remains to be verified is that Φ is linearly full.Recall that the map Φ will be linearly full unless the determinant of (α ,α ,...,α (n) ) is identically zero.But this cannot happen since by our choices the functions α 1 ,α 2 ,...,α n have poles at different points of M g .Remark 5.2.This was proved in [10] for g = 0 using an explicit deformation, and in [8] for higher genus using a deformation in ᐆ n .Our proof is similar; we include it for its simplicity and as an example where the techniques developed in the previous sections are used.We use a simple linear deformation in P Nn which induces a deformation in ᐆ n .This theorem is an immediate corollary to the following proposition.

Connectedness of the moduli space
Proof.The idea is simple: first, use the correspondence Φ ↔ Φ with a proper choice of the base plane so that the degrees of both maps are the same.Then deform Φ by noting that, for λ ∈ C, the curve defined by is also a solution of (2.13).When λ = 0, we have This last map is clearly not linearly full, so it corresponds to a map Φ 0 : M g → ᐆ n−1 as desired.
Of course, either the deformation of Φ or the deformation of Φ shown above may not be continuous.However, after a rotation and a few technical points everything works.
If Φ ∈ HH f d (M g , ᐆ n ), then there is nothing to prove.Thus we can assume that Φ ∈ HH f d (M g , ᐆ n ).Using Propositions 3.6 and 3.5, choose a base plane E so that Φ falls in the regular part of Ꮿ E , and Φ = Ꮿ E (Φ) satisfies the hypotheses of Proposition 3.3(a).Of course, Φ and Φ have the same degree.
Then Φ has the property that α is not zero at the points where s vanishes, so we can find a unit vector v 2 such that (v 2 , α(p)) = 0 at the points p where s = 0. Let A ∈ Gl(n, C) be any matrix with v 2 as its second row.Then we must have that A Φ misses the subspace 0 : Since the action of Gl(n, C) is equivariant and continuous in P Nn and ᐆ n (see Lemma 2.6 and Remarks 2.7 and 2.8), A Φ has the same degree as Φ, and the problem is reduced to deforming the rotated curve A Φ. For convenience of notation, we will still use Φ to denote this rotated curve.
Finally, consider the deformation where λ ∈ C and (This is, we are multiplying by λ all the entries α i , τ jk for which any subindex equals 1.)This deformation corresponds to a projection P Nn → P N n−1 whose center is the subspace given by (5.3).Since Φ misses this subspace, this deformation must be continuous for every λ ∈ C. For λ = 0, the entry corresponding to α 1 is 0. Hence, W (α) ≡ 0, and the curve cannot be linearly full.We have thus found a deformation of Φ that satisfies what we want.
The only thing that remains to be checked is that the corresponding deformation of Φ in ᐆ n is continuous.To this end, observe that the deformation given in (5.4) corresponds to a deformation in P Mn of the form where p i (λ) denote positive powers of λ.Therefore, for λ = 0, it corresponds to a projection P Mn → P M n−1 .Thus, the deformation ı• Φ λ , is continuous for λ = 0.For λ = 0, it will also be continuous unless the curve ı • Φ passes through the center of the projection P Mn → P M n−1 .But if this was the case, the degree of ı • Φ 0 would be less than the degree of ı ), the degree cannot drop at λ = 0, and therefore ı • Φ λ is a continuous deformation for all λ.
6.The dimension of the moduli space of harmonic spheres in S 6 .Not much is known about the dimension or structure of the moduli space of harmonic maps S 2 → S 2n for n > 2. For n = 2, it is known that the dimension of this moduli space is 2d + 4 (see [11]) and that it is a variety of pure dimension 2d + 4 with three irreducible components (see [13]).In [1], Barbosa shows that when d = n(n + 1)/2, Supm In the late 1980s, Bolton and Woodward posed the conjecture that the dimension of the moduli space of harmonic maps S 2 → S 2n is 2d + n 2 , which of course coincides with Barbosa's result.An explanation appears in [2], where they also announce a proof of this fact due to Oxbury, but it seems that it was not correct.To our knowledge, there have been no further advances on proving or disproving this conjecture.
In this last section we give a lower bound for the dimension of Supm f d (S 2 ,S 6 ) which coincides with the conjectured value.Namely, we show that this dimension is greater than or equal to 2d+9.It seems that our method will also work for arbitrary n, but we will not develop this here.
Using the results of the previous sections, we have that the top dimension of this moduli space is exactly the top dimension of PD f d (S 2 , P N 3 ).This dimension is one less (after projectivizing) than the dimension of the set of solutions of the system where s, α 1 , α 2 , α 3 , τ 12 , τ 23 , and τ 31 are polynomials in one complex variable z, of degree less than or equal to d with at least one of them of degree d, and without a common factor for all of them.
We will find a 2d + 10-dimensional set of solutions of (6.1) satisfying (6.2).To this end, we need the following lemma.Lemma 6.1.Let d ≥ 8. Given any polynomial s(z) of degree d with only single zeroes, there exist solutions of (6.1) satisfying (6.2).
which is equivalent to , has no residues, (6.4) dz is a polynomial of degree ≤ d.Since s has only simple zeroes, solutions to the system (6.4) satisfy (6.5) implicitly, and for these solutions, system (6.4) is equivalent to the system E α i ,α j s m = 0, 1 ≤ i < j ≤ 3, 1 ≤ m ≤ d − 1.
(6.7) (We only need that condition (6.7) be satisfied at {s 1 ,s 2 ,...,s d−1 } since the sum of the residues of a meromorphic function is 0.) Let r ≥ 3 with 3r ≤ d + 1.Let α 1 be a polynomial of degree d or less which is a solution of the system of equations α 1 s m = 0, 1 ≤ m ≤ r , F α 1 s m = 0, 1 ≤ m ≤ r , (6.8) and such that α 1 (s m ) = 0 for r +1 ≤ m ≤ d. (Such an α 1 can be found explicitly developing (6.8) further and using elementary interpolation; here is where we need that 3r ≤ d + 1.) Note that, since α 1 satisfies (6.8), equations (6.7) are implicitly satisfied at the points {s 1 ,s 2 ,...,s r } for i = 1, and j = 2 or j = 3.where α 2 and α 3 are polynomials of degree d or less.Note that α 2 = s and α 3 = s are a solution of this system.
Take α 2 and α 3 as specified in the last paragraph.Using the relation it is easy to see that (s, α 1 ,α 2 ,α 3 ) is a solution of system (6.7), and hence of (6.1).Also, by construction, these polynomials have no common factors.It remains to check that they satisfy condition (6.1).This is immediate since α 1 /s has a pole at s r +1 while α 2 /s and α 3 /s do not, α 2 /s has a pole at s 2 while α 1 /s and α 3 /s do not, and α 3 /s has a pole at s 1 while α 1 /s and α 2 /s do not.Theorem 6.2.For d ≥ 6, d = 7, the dimension of the moduli spaces of linearly full harmonic maps S 2 → S 6 of degree d is at least 2d + 9.
Proof.For d = 6, Barbosa's result in [1] gives dim(Supm (6.12) Using relation (6.10), we see that system (6.7) has no more than 2d − 3 independent homogeneous equations and 4d + 4 variables.(It is not immediate that the number of independent equations is no more than 2d − 3 as it may seem that the correct bound is 2d − 2. To see that 2d − 3 is the correct bound, one has to play with the equations a bit.We omit the details.)The set is the intersection of the variety given by (6.7) with the open set given by condition (6.2).By Lemma 6.1, is not empty.Therefore its dimension is at least 2d+7.
Finally, define where a ij are arbitrary complex numbers (integration constants).
The elements of Ꮽ are solutions of (6.1) satisfying (6.2).The dimension of Ꮽ is the dimension of plus the three extra degrees of freedom given by the integration constants.Thus, dim(Ꮽ) ≥ 2d + 10.Therefore we have PᏭ ⊂ PD f d (S 2 , P N 3 ) and dim(PᏭ) ≥ 2d + 9.This proves the theorem.

3 .
(C) The set of Φ that satisfies the conditions of Proposition 3.5 is a Zariski open subset of each irreducible component of HH f d (M g , ᐆ n ).We can restate Proposition 3.5 as follows.The maps Φ in each irreducible component Ᏽ of HH f d (M g , ᐆ n ) that at some point have incidence of dimension greater than one with a fixed plane E ∈ ᐆ n form a proper subvariety of the variety of maps from M g to ᐆ n .Thus, given a base plane E, either no maps in this irreducible component Ᏽ of HH f d (M g , ᐆ n ) satisfy the hypotheses of Proposition 3.5 or the set of maps that satisfy the hypotheses of Proposition 3.5 form a Zariski open subset of Ᏽ.
to linearly full maps.Using the Plücker embedding of ᐆ n into P Mn , whose degree is 2 (see[1, page  92] and[12,page 129]), we can think of HH m Mn consisting of sections without common zeroes that satisfy the conditions of lying in ᐆ n , of being horizontal, and of being linearly full in the case of HH d (M g , ᐆ n ) and HH f d (M g , ᐆ n ) as subvarieties of H 0 (M, ᏸ(2d)) g , ᐆ n ) and HH f d (M g , ᐆ n ) arealgebraic varieties, although we have a priori no information about their structure.
.6) ᐆ n ), the map Ꮿ E is a birational map from the irreducible component where Φ lies in the irreducible component of PD f d (M g , P Nn ), where Φ = Ꮿ E (Φ) lies, for some base plane E ∈ ᐆ n .